This board is a wealth of knowledge, and it seems that no question goes unanswered. From noobs like me to seasoned professionals with backgrounds in electrical engineering to the folks in the middle, it is an awesome clearininghouse and discussion place to learn and show.
But.....
I think one thing that is a bit lacking (and has certainly been a steep learning curve for me) is the basic understanding of what each component does in a circuit. I can read schematics. I can easily identify certain things and know what they do. But that's after months of reading and studying, and my knowledge is still incomplete.
What's that saying about teaching a man to fish?
In that vein, I'd like to start a thread where anyone who is interested can join in and ask or tell--what does each bit do?
The goal is to keep this discussion in beginner's terms. No detailed math or theory. The idea is that if the key concept, word or phrase is explained here, a deeper dive into the DIYStompboxes Lake of Knowledge would be a natural next step.
I propose we go through one circuit at at time.
Fill in the blanks below. Keep the explanation to 2 sentences or less. Assume us noobs will take it upon ourselves to search and learn based on the starting point provide here.
I'll edit this original post as contributors add more info.
So, here's the Electro Harmonix LPB-1 Booster: This circuit provides a simple boost with a variable output control.
-LPB Booster (http://www.diystompboxes.com/smfforum/index.php?action=search2;search=lpb%20booster)
(http://www.beavisaudio.com/techpages/HIW/EXHLPB1.gif)
9vDC - This is the source of positive voltage and current. Most effects run on a 9 volt battery or AC adaptor. This is where the 9V positive connection is made. The combination of the + voltage source and ground make up the power source. All current coming out of the +9V source must eventually get back to the battery through the ground connection.
SW1 - Switch to turn the power on and off. This interrupts the current from the primary power source, which is a battery or other DC voltage supply. Interrupting either the +9V path or the ground path will deactivate the circuit by by breaking the path from the positive voltage to ground.
R1 and R2 - These components together establish the bias point for the base of transistor Q1. They form a voltage divider which sets the bias voltage, and an equivalent series resistance which limits the base bias current, as well as forming a part of the input impedance.
- Voltage Divider (http://www.diystompboxes.com/smfforum/index.php?action=search2;search=voltage%20divder)
- Impedance (http://www.diystompboxes.com/smfforum/index.php?action=search2;search=impedance)
- Transistor Bias (http://www.diystompboxes.com/smfforum/index.php?action=search2;search=transistor%20bias)
C1 and R2 -Put simply, these two components, taken together, provide a High Pass Filter. A high pass filter allows higher frequencies to pass through, while simultaneously attenuating lower frequencies. A more detailed explanation: C1 and the parallel combination of R1 and R2, along with the base of Q1 provide a high pass filter at the input. The primary purpose of this high pass filter is to block the DC bias voltage from being disturbed by whatever is connected to the input. Exactly what frequency the C1 and R1/R2 filter cuts off at may or may not be important, depending on the application. Often it is set below all audio frequencies so everything passes. Occasionally it is twiddled to make a treble filter, as in some treble boosters. The finessing of the exact rolloff point of this filter is tricky for beginners, as there are a lot of interactions. But making C1 big enough to pass all audio isn't - just make it big.
- High Pass Filter (http://www.diystompboxes.com/smfforum/index.php?action=search2;search=high%20pass%20filter)
- High Pass Filter (http://en.wikipedia.org/wiki/High-pass_filter)
- Tone Control (http://www.diystompboxes.com/smfforum/index.php?action=search2;search=tone%20control)
R3 - This is the component which converts the amplified current in Q1's collector into an output voltage. Any base current is multiplied by the transistor hfe and is converted into an output voltage by ohm's law in this resistor. The value of R3 sets the output impedance to a large degree.
- Transistor HFE (http://www.diystompboxes.com/smfforum/index.php?action=search2;search=transistor%20HFE)
- Output Impedance (http://www.diystompboxes.com/smfforum/index.php?action=search2;search=output%20impedance)
R4 -This resistor is a feedback stabilizing resistor. Any current which goes through the base OR the collector goes through here. As such, the collector current going through this resistor raises the base voltage, and reduces the base current - negative feedback of the amount of collector current into the base. R4 also sets the voltage gain of the circuit, to a large degree. Since the base current is small (usually 1/100th or smaller part of the collector current), the R4 current is almost identical to the collector current. Therefore, the voltages across the resistors must be in proportion to their resistances. Since the base voltage must be almost identical to the emitter voltage (that is, only one diode drop, or 0.5 to 0.7V higher), then the voltage at the emitter reflects the voltage at the base. And the collector voltage is R3/R4 times as much.
- Feedback Stabilizing Resistor (http://www.diystompboxes.com/smfforum/index.php?action=search2;search=negative%20feedback)
- Negative Feedback (http://www.diystompboxes.com/smfforum/index.php?action=search2;search=negative%20feedback)
R5 - 100 K potentiometer which acts as a Volume or Output control. This pot adjusts the amount of output signal by bleeding the boosted output to ground. As you turn the pot up, you are increasing resistance so less signal is bled to ground, hence more comes out of the output jack.
- Volume Control (http://www.diystompboxes.com/smfforum/index.php?action=search2;search=volume%20control)
C2 - This capacitor acts as Decoupling or Ouptut cap. It prevents the direct current supplied by the 9V source from getting into the output signal. Typically, you can increase the value of the output capacitor to let more bass through. Conversely, you can decrease the value to decrease the lower frequencies and end up with a more trebly/thin sound.
- Decoupling Cap (http://www.diystompboxes.com/smfforum/index.php?action=search2;search=decoupling%20cap)
- Output Capacitor (http://www.diystompboxes.com/smfforum/index.php?action=search2;search=output%20capacitor)
Q1 - This is an NPN transistor that provides the core of the booster. Original versions specify the 2N5133 part, but alternates such as 2N3904 / 2N5088 / BC549 can also be used.
- NPN Transistor Boost (http://www.diystompboxes.com/smfforum/index.php?action=search2;search=NPN%20transistor%20)
- How Transistors Work (http://www.williamson-labs.com/480_xtor.htm)
C1 and R2 make a high pass filter which lets more treble into the sound. The resistor also cuts out all clicks and other unpleasent sounds from the signal, and the capacitor also filters DC voltage from getting into your guitar. I think C2 is used as the output buffer.
Quote from: call1800ksmyazz on July 26, 2006, 10:38:20 AM
C1 and R2 make a high pass filter which lets more treble into the sound. The resistor also cuts out all clicks and other unpleasent sounds from the signal, and the capacitor also filters DC voltage from getting into your guitar. I think C2 is used as the output buffer.
Thanks! I'll keep the main post edited as more folks contribute.
K thanks. Any where the C2 should be, it says C1 again. And I think thats a buffer (C2)
Q1 - this is an NPN transistor ! ( circuit is negative ground "as normal" for NPN circuits )
Q1 - can be 2N3904 / 2N5088 / BC549 etc etc
C2 - decoupling/output cap, stops DC getting from the 9v line, increase the size to let through more bass in the signal
decrease size to make the signal more "trebly" and thin .
R3 - regulates the voltage applied to the transistors collector ( approx 6 to 7 volts here )
MM
Quote9vDC - This is the power source. Most effects run on a 9 volt battery or AC adaptor. This is where the 9V positive connection is made.
A slight difference. This is the source of positve voltage and current. The combination of the + voltage source and ground make up the power source. All current coming out of the +9V source must eventually get back to the battery through the ground connection.
QuoteSW1 - Switch to turn the power on and off. Pretty simple--it either connects or disconnects the 9v posistive connetion
Correct, with the above caveat. This interrupts the current from the primary power source, which is a battery or other DC voltage supply. Interrupting either the +9V path or the ground path will deactivate the circuit by denying power to it.
R1 and R2: These components together establish the bias point for the base of transistor Q1. They form a voltage divider which sets the bias voltage, and an equivalent series resistance which limits the base bias current, as well as forming a part of the input impedance.
Quote
C1 and R2: Taken together, these components provide a High Pass Filter .
Almost. C1 and the parallel combination of R1 and R2, along with the base of Q1 provide a high pass filter at the input. The primary purpose of this high pass filter is to block the DC bias voltage from being disturbed by whatever is connected to the input. Exactly what frequency the C1 and R1/R2 filter cuts off at may or may not be important, depending on the application. Often it is set below all audio frequencies so everything passes. Occasionally it is twiddled to make a treble filter, as in some treble boosters. The finessing of the exact rolloff point of this filter is tricky for beginners, as there are a lot of interactions. But making C1 big enough to pass all audio isn't - just make it big.
R3: This is the component which converts the amplified current in Q1's collector into an output voltage. Any base current is multiplied by the transistor hfe and is converted into an output voltage by ohm's law in this resistor. The value of R3 sets the output impedance to a large degree.
R4: This resistor is a feedback stabilizing resistor. Any current which goes through the base OR the collector goes through here. As such, the collector current going through this resistor raises the base voltage, and reduces the base current - negative feedback of the amount of collector current into the base. R4 also sets the voltage gain of the circuit, to a large degree. Since the base current is small (usually 1/100th or smaller part of the collector current), the R4 current is almost identical to the collector current. Therefore, the voltages across the resistors must be in proportion to their resistances. Since the base voltage must be almost identical to the emitter voltage (that is, only one diode drop, or 0.5 to 0.7V higher), then the voltage at the emitter reflects the voltage at the base. And the collector voltage is R3/R4 times as much.
QuoteR5: 100 K potentiometer which acts as a Volume or Output control. This pot adjusts the amount of output signal by bleeding the boosted output to ground. As you turn the pot up, you are increasing resistance so less signal is bled to ground, hence more comes out of the output jack.
http://www.diystompboxes.com/smfforum/index.php?action=search2;search=volume%20control
Yep.
C1: See above.
QuoteQ1 - This is a PNP transistor. It provides the core of the booster.
Actually, it's an NPN transistor.
Awesome! Thanks guys--I've updated the original post to merge in the comments/explanations and also provide some more links.
You should start out with some basic electronics training. Places like this are not a bad place to start:
http://www.tpub.com/content/neets/index.htm (http://www.tpub.com/content/neets/index.htm)
Doug
Hi Doug, I've actually read through quite a few books at this point and have a very basic understanding of some of the fundamentals.
The idea for this thread is to get help in translating how that basic electronics knowledge fits into stompbox circuits specifically.
And to share the group's knowledge in a specific context.
Thanks!
-dano
Quote from: Doug_H on July 26, 2006, 12:16:23 PM
You should start out with some basic electronics training. Places like this are not a bad place to start:
http://www.tpub.com/content/neets/index.htm (http://www.tpub.com/content/neets/index.htm)
Doug
Three cheers for DIYstompboxes.com and Aron Nelson for creating this forum!!!
Two points:
1) Switching a pedal in this method will work, but these days it seems a lot better to use a bypass rather than a power switch. Instead of putting a switch between +9v and the current, it is better, as we all know, to put a DPDT switch between Input and output, where in one position in & out are directly connected (bypassing the effect) and in the other the effect is in between in & out (activating it). Plenty has been written on this.
2) If we do make a bypass switch, we would probably want to add a large resistor (2M2) before C1 going to ground to prevent popping. what it would do is allow capacitor to discharge its voltage (current?) to ground when the pedal is bypassed.
If any of this is incorrect, please feel free (R.G.........) to correct me :)
It's exciting to read that other people seem to be about where I am with all of this information. Hehe, the experience of 'reading quite a few books' and 'having a very basic understanding of some of the fundamentals' is humbling at least... ;D I myself have a small background with hobby electronics. I fiddled a bit with certain applications in high school but never went as far as to design my own circuits or really understand the workings of individual components. Now I've found a way to bridge my forever urge to create music with my intrigue for electronics. Pretty f*in' bitchin! Right now I'm reading textbooks for fun and as I go to bed (They never get me to do that back in school!) and over the last weekend I made some substantial jumps in understanding capacitors, resistors, and transistors (coinciding with my first active pedal build). I can now honestly say "I kindof understand some of the basics of whats going on in the simplest of circuits" and it feels great!! I big *CHEERS* to all of you out there like me that are learning, enjoying, and creating all at the same time!!
CHEERS
zpyder
Just wanted to say: Great thread Dano!
I feel I am in the same position as you, having done a lot of reading, but still not understanding on a practical level why each component is there in a circuit. I'm quite sure there are more of us out there. Keep up the good work and thank you.
Quote from: WildMountain on July 26, 2006, 02:49:59 PM
Just wanted to say: Great thread Dano!
I feel I am in the same position as you, having done a lot of reading, but still not understanding on a practical level why each component is there in a circuit. I'm quite sure there are more of us out there. Keep up the good work and thank you.
Thanks go to the folks who filled in the blanks. I learned a lot just in this one post. We should do another one.
I think it is worthwhile, too. Dano's motivations are much the same as mine when I wrote up the "Technology of... " series. It helps to get under the covers, dissect the circuit into pieces and think about "what does that do?"
After a while of doing this, you come to understand that electronics has two modes of understanding to be done: the innards of how a block works, and how to string blocks together. There are very few gigantic clots of all-interconnected circuits where you cannot subdivide it and start understanding. Most effects are simple strings, daisy chains of a few kinds of functional blocks. That was the point behind my "FX Buss" stuff - sub-bits of effects that could be strung together, necklace style, into new effects.
So learn your blocks. Then start stringing.
Great thread! I'm a newbie so I still don't have anything to add but I learn A LOT from reading this- thanks a lot guys!
This truly is a GREAT thread.
Very enlightening!
This would be great to add to a web site or a section of a site for future reference.
:oWHERE HAVE YOU BEEN ALL MY LIFE??????... erm... i mean this thread ::)
Great idea Dano. Thanks for taking the time. In some ways electronics is like making bread: you can do a lot and even be successful without having much of a clue about how it all works. But speaking for myself, I always feel like I could use more theory and explanation, so it's really useful to read a good discussion just the same.
On a maybe trivial point, I was wondering if the diagram could make more clear to some viewers that the input and outputs also are connected to ground. As it is, the diagram suggests that you just plug your guitar cord into a resistor or a capacitor, whereas there's actually a completion of the circuit by way of the "ground" portion of the cord/jack. But like I said, that's maybe obvious enough to others.
Great writing so far! Can't wait for the next lesson. :D
Quote from: idlechatterbox on July 27, 2006, 09:57:52 AM
On a maybe trivial point, I was wondering if the diagram could make more clear to some viewers that the input and outputs also are connected to ground. As it is, the diagram suggests that you just plug your guitar cord into a resistor or a capacitor, whereas there's actually a completion of the circuit by way of the "ground" portion of the cord/jack. But like I said, that's maybe obvious enough to others.
Great writing so far! Can't wait for the next lesson. :D
Good feedback. Schematic updated.
Ok, well the first one was fun. I thought for a follow-on, it would be cool to take that basic single transistor boost as in the LPB1, and add another transistor to make it a fuzz. I searched through a lot of schematics to find something that was both similiar to the LPB1 (to keep in context) and simple. I happened across Hemmo's webpage and found his "Standard Fuzz".
Here is a redrawn schematic. I'll fill in some of the bits I know based on the feedback from the LPB1.
As before, I'll need help in explaining what some of parts do, or corrections to my errors.
(http://www.beavisaudio.com/techpages/HIW/StdHemmo.gif)
What it does This is a simple 9-volt powered transistor device that can provide either a boost or a fuzz sound. Using only the first transistor provides the boost. Cascading that signal into the second transistor provides the clipping to generate fuzz.
9vDC - This is the source of positive voltage and current. Most effects run on a 9 volt battery or AC adaptor. This is where the 9V positive connection is made. The combination of the + voltage source and ground make up the power source. All current coming out of the +9V source must eventually get back to the battery through the ground connection.
SW1 - Switch to turn the power on and off. This interrupts the current from the primary power source, which is a battery or other DC voltage supply. Interrupting either the +9V path or the ground path will deactivate the circuit by by breaking the path from the positive voltage to ground.
R1 and R3 - These components together establish the bias point for the base of transistor Q1. They form a voltage divider which sets the bias voltage, and an equivalent series resistance which limits the base bias current, as well as forming a part of the input impedance.
R2 - This is the component which converts the amplified current in Q1's collector into an output voltage. Any base current is multiplied by the transistor hfe and is converted into an output voltage by ohm's law in this resistor. Remember that transistors are Current devices, so we need to convert the current at Q1's collector to a voltage so it can continue on to the next part of the circuit. The value of R2 sets the output impedance to a large degree.
C1 and R3 - C1 and the parallel combination of R1 and R2, along with the base of Q1 provide a high pass filter at the input. The primary purpose of this high pass filter is to block the DC bias voltage from being disturbed by whatever is connected to the input. Exactly what frequency the C1 and R1/R2 filter cuts off at may or may not be important, depending on the application. Often it is set below all audio frequencies so everything passes. Occasionally it is twiddled to make a treble filter, as in some treble boosters. The finessing of the exact rolloff point of this filter is tricky for beginners, as there are a lot of interactions. But making C1 big enough to pass all audio isn't - just make it big.
R4 -This resistor is a feedback stabilizing resistor. Any current which goes through the base OR the collector goes through here. As such, the collector current going through this resistor raises the base voltage, and reduces the base current - negative feedback of the amount of collector current into the base. R4 also sets the voltage gain of the circuit, to a large degree. Since the base current is small (usually 1/100th or smaller part of the collector current), the R4 current is almost identical to the collector current. Therefore, the voltages across the resistors must be in proportion to their resistances. Since the base voltage must be almost identical to the emitter voltage (that is, only one diode drop, or 0.5 to 0.7V higher), then the voltage at the emitter reflects the voltage at the base. And the collector voltage is R2/R4 times as much.
C3 Short Description: C3 is there to make the gain of Q1 be as large as it can be while still keeping the DC stabilization of R4. Long Description: C3 has an AC impedance that tends toward zero as the AC frequency goes up. In this case, it "shorts" R4's feedback effect at AC frequencies where the impedance of C3 is less than R4's 10K ohms. That is, at frequencies over F = 1/(2*pi*R4*C3). Since the gain is R2 divided by the impedance in the emitter, the gain goes to infinity, right? Wrong. There is a resistor internal to the transistor's base-emitter junction, called the Shockley resistance that keeps the gain in the same Rc/Re relationship. It's just that the Shockley resistance is not only an internal part of the transistor where you can't see it, it is also a variable. The Shockley resistor is approximately Rs= 0.026V/Ie. It's small, in the few ohms to hundreds of ohms range, so the gain gets big. Here's another description: C3 sets a shelf frequency. The C3 bypasses the R4 resistor setting an EQ shelf. In amps it called "cathode bypass". Higher values let more frequencies through, more gain and bass. Lower values form sort of a high pass shelf or more gain in theupper mids and high end.
Q1 - This is the first gain transistor. It boosts the input signal but should not normally clip/distort the signal. This can be any NPN such as a BC549, 2N3904, MPSA18, etc. Note that with high gain transistors, it is possible that the transistor's gain will let it distort some. This depends on the exact bias point. It is easy to bang into the + power supply if the bias point of the collector is near the + supply, and vice versa with ground.
C2 - This is a coupling capacitor that blocks Q1's collector voltage from reaching the base of Q2. See the C4 description for a C2 suggestion on de-popping the fuzz switch.
R5 - As with R2, this component converts the amplified current in Q2's collector into an output voltage.
Q2 - The second NPN transistor takes the signal from the output of the first transistor (which boosted the signal). This boost causes Q2 to clip the signal as it reaches its operating threshold. Same part type as Q1.
R6 - Same as R4, but for Q2
C4 – As with C2, this capacitor blocks the DC voltage on Q2's collector from appearing at the output when the switch is in the "fuzz" position. As suggested by R.G., there ought to be a high value pulldown resistor of maybe 1 to 4.7M to ground on the outboard side of this to prevent switch pop.
C5 – This capacitor does the same thing as C3, but for the Q2 transistor. I.e. it makes the gain of Q2 be as large as it can be while still keeping the DC stabilization of R6.
SW2 - This switch controls which signal path goes to the output jack. When the switch is in the down position (as shown in the schematic), the output jack receives only the output of Q1, i.e. the boosted signal. Q2 continues to operate and clip the signal; we are just bypassing it. With the switch in the up position, the signal coming from Q2 is sent to the output jack, giving us the fuzz sound.
QuoteC3 ?
C3 has an AC impedance that tends toward zero as the AC frequency goes up. In this case, it "shorts" R4's feedback effect at AC frequencies where the impedance of C3 is less than R4's 10K ohms. That is, at frequencies over F = 1/(2*pi*R4*C3). Since the gain is R2 divided by the impedance in the emitter, the gain goes to infinity, right?
Wrong. There is a resistor internal to the transistor's base-emitter junction, called the Shockley resistance that keeps the gain in the same Rc/Re relationship. It's just that the Shockley resistance is not only an internal part of the transistor where you can't see it, it is also a variable. The Shockley resistor is approximately Rs= 0.026V/Ie. It's small, in the few ohms to hundreds of ohms range, so the gain gets big.
C3 is there to make the gain of Q1 be as large as it can be while still keeping the DC stabilization of R4.
QuoteQ1 - This is the first gain transistor. It boosts the input signal but does not clip/distort the signal. This can be any NPN such as a BC549, 2N3904, MPSA18, etc.
Note that with high gain transistors, it is possible that the transistor's gain will let it distort some. This depends on the exact bias point. It is easy to bang into the + power supply if the bias point of the collector is near the + supply, and vice versa with ground.
QuoteC2 - This is a coupling capacitor that blocks DC voltage from reaching the base of Q2 ?
Correct - and this is problematic. I took one look at the C2 part of the schemo and decided that the schemo is drawn incorrectly. That may have been premature, but it's certainly not standard design practice.
C3 does block the Q1 collector voltage from Q2 base. But there is no other source of bias for Q2's base, so Q2's base will not conduct at all until the signal at its base exceeds +0.6V. Then it operates Q2 at high gain (the R6/C5 pair does the same thing as R4 and C3 do for Q1). This would clip off everything below +0.6V from Q1's collector. It will certainly be distorted, conducting only on big positive peaks.
It is possible that this was intentional since the idea was to make distortion with this section. It is also possible that it's a schematic copying error.
The other bug that's related to this is that there is no DC blocking capacitor between Q1 collector and the boost/fuzz switch. Q1's DC level will appear on the output when the switch is in the boost postion, and that will make a major pop. Was there not a capacitor there to block the Q1 collector voltage?
C4 - Blocks the DC voltage on Q2's collector from appearing at the output when the switch is in the "fuzz" position. There ought to be a high value pulldown resistor of maybe 1 to 4.7M to ground on the outboard side of this to prevent switch pop.
C5 - Does the high-gain AC thing for Q2.
QuoteSW2 - This switch controls which signal path goes to the output jack. When the switch is in the down position (as shown in the schematic), the output jack receives only the output of Q1, i.e. the boosted signal. Q2 continues to operate and clip the signal; we are just bypassing it. With the switch in the up position, the signal coming from Q2 is sent to the output jack, giving us the fuzz sound.
See the comments on Q1 collector DC level. There should be a capacitor in series between Q1 collector and the switch and another high value resistor to ground on the outboard side here too, IMHO.
OooH oooH! I know one.... C3 sets a shelf frequency. The C3 bypasses the resistor setting an EQ shelf. In amps it called "cathode bypass".
Higher values let more frequencies through, more gain and bass. Lower values form sort of a high pass shelf or more gain in theupper mids and high end.
John
Quote from: R.G. on July 27, 2006, 04:02:19 PM
It is possible that this was intentional since the idea was to make distortion with this section. It is also possible that it's a schematic copying error.
Yep, transcription error on my part. I'll fix this in tomorrow morning--thanks for the Keen eye.
Regarding the rest of the information, truly thanks much. I'll start digesting it now and edit the post.
At the risk of sounding like a total dweeb, this is the post I've been waiting for all my life. Or at least all of my circuit building life. A component by component walkthrough of a circuit that simply answers, "what's that there for". THANKS
Anyone wanna know how class II biological safety cabinetry works? I know that.
Thanks for the feedback everybody. I've corrected the schematic--it had a transcription error where C2 was between Q1 and Q2. It should now be correct.
I've updated the text.
Let's do a little investigation into the secrets of biasing.
With the picture of circuit 2 in mind - what are the voltages that appear on the transistor pins? Believe it or not, you only need ohm's law and two facts about transistors to figure that out.
Since Q2's base is DC-tied to Q1's collector, we need to figure out Q1 first.
For bipolar transistors, the first fact - the secret to the whole mess, actually - is that the base and emitter are connected by a diode. The base emitter junction will pass any amount of current up to burning out the lead wires in order to keep the base no more than one diode drop above the emitter. So the base goes at whatever voltage it's told to, and the emitter follows, unless something burns out and it can't. We'll check that to be sure we did it right.
So what voltage is the base at? Easy - R1 and R3 set it. It's a simple voltage divider. With 9V power in, and the values shown, the base voltage tries to be Vb= 9V*(R3/(R1+R3)) = 2.8125V. Of course that's only true if the base pulls no current, which we know is not true, but may be almost true. Here's that second fact about BJTs - the base current is often so small you can ignore it. Just like our guess about the base emitter not burning out, we'll ignore this and then come back and check it.
So if the base of Q1 sits at 2.8125V, the emitter must be at one diode drop down (fact #1). A diode drop is ill defined - it may be as high as 0.7V for silicon, and may be as low as 0.4V for tiny currents in the same junction. So like good designers everywhere, we guess, then come back and check. Let's guess that the current will be small, along with the guess we made about the base current anyway, so the voltage will be on the low side. We'll call it 0.5V, which I know from experience is common in low signal circuits, and you will someday too.
That puts the emitter at 2.3125V. If the emitter is at 2.3125V, then the current in the emitter resistor R4 must be I = 2.3125/10K = 231uA.
Now is the time to check one of our previous guesses. If the emitter current is 231uA and the transistor has a gain of even 100, then the base current is 2.31uA. R1 and R3 are actually passing 9V /(R1 + R3) = 18.75uA if the base current were zero. It's really 2uA, or about 1/9th of the total current. That's not really small enough to ignore. So we compensate. At this point I wish I'd picked my own circuit to illustrate because while the compensation is easy, it introduces a new concept I hadn't planned to saddle you with yet.
UGH. Extra concept: Thevenin equivalent circuit. For a resistor divider (the math works out, just trust me on this for now) the circuit works with a load on the center of the divider so that you can replace the voltage divider with a single voltage source in series with a single resistor and have the results be equivalent to what happens with the voltage divider and two resistors. The Thevenin Equivalent Voltage Source is equal to the unloaded voltage at the center of the divider, or 2.8125V in this case as we calculated. The series resistor is equal to the parallel combination of the two resistors in the divider, or Req = 330K||150K = 103.125K
Now we can correct our assumption on the base voltage. It's really 2.8125V - (2.3uA*103K) = 2.8125- 0.237 = 2.575V. That makes the emitter voltage change!! ARG! How will we ever figure this out?
Easy - it converges. We make a new estimate of the emitter voltage, now 2.075V instead of 2.375V. The emitter current is now 207uA instead of 231, and the base current is now 2.07uA instead of 2.31uA. That gives us a new estimate of base voltage of 2.8125 - (2.07uA*103K) = 2.598V, and a third estimate of emitter voltage of 2.098V. As you can see, each repeat of the calculation leads to a closer and closer estimate of the true value. We're at two iterations, and we already have the emitter changing only 1% with the interation, so let's call that good. The error just gets smaller. Besides, we have no earthly idea what the actual transistor gain is. We said 100, but that's on the low side for modern transistors. It could easily be 200 or 500. The bigger the gain, the smaller the error that the base current causes and the more the first estimate is perfect. And we also had to guess at the base-emitter voltage being 0.5V. It could have been 0.45 V or 0.55V. In either case, it does not make much difference in the transistor's biasing.
So back to the main track.
That puts the emitter at 2.098V. If the emitter is at 2.098V, then the current in the emitter resistor R4 must be I = 2.098/10K = 209.8uA. (and the 10K resistor has a tolerance of +/-5%, just like R1 and R3 do... 8-) ).
We know that the base current is less than 1% of this, so we'll ignore it and assume that the collector current is also 209.8uA, and so the collector voltage must be the voltage drop through R3 lower than 9V. Q3's collector sits at 9V - R3*209.8uA = 9V - 3.776 = 5.22V
This is an estimate. As I've noted, we have guessed at the most likely values for transistor gain ( >100) and Vbe (0.4 to 0.7V) and resistor values ( actually +/-5%). But if you built ten thousand of these, the average of the actual values would lie very close to the values we've calculated.
Q2 turns out to be easy. Its base is nailed to the 5.22V of Q1's collector by that copper wire holding them together. So its emitter is at 5.22V-0.5V = 4.7236V, the emitter current is 4.7236V/ 10K= 472uA, and the collector is down from 9V by the same amount.
UGH. I gotta keep my mouth shut and pick my own example circuits. Notice that the voltage on the emitter of Q2 is estimated to be 4.72V? That's over half of 9V. And the collector with a 10K resistor is over half of 9V too. That does not compute. They can't BOTH be more than half of 9V away from their power connections, because 9V is all you have.
What's really happening here is this.
(1) This may be another circuit transcription error. Maybe not.
(2) If not, Q2 is sitting there saturated all the time. It comes OUT of saturation when the swing on Q1's collector turns it more off. That's consistent with it generating lots of distortion.
So that's a blow-by-blow of how this thing is biased.
Dano - are the values correct for Q2 emitter and collector resistors?
Quote from: R.G. on July 28, 2006, 10:12:18 AM
Dano - are the values correct for Q2 emitter and collector resistors?
Yep, I double-checked them against the original schematic. The author talks about the Q1 stage as heavily boosted and the Q2 stage as "total fuzz mayhem." My guess is that he took the empirical approach in designing this circuit. I.e. stick bits on the breadboard without too much concern as to well calculated bias levels.
Which brings up the interesting yin and yang of DIY design:
- As beginners, we create things based on schematics that others have done. While we work to learn the actual theory and math behind the circuit, we learn empirically by listening to what it sounds like. And we make lots of mistakes along the way.
- As professionals, we approach circuit design from the engineering view: design it in theory and do the calculations to avoid any mistakes. Then build it and empirically tweak it to taste.
Of course, that is an oversimplification. But I think it does explain part of the DIY impedance mismatch: those wanting to learn the building blocks first, being enlightened by those wishing to explain the underlying details of how it actually works. A fascinating process to watch and be part of. My goal for this thread is to attempt to build a buffer that smoothes out this impedance mismatch.
Thought it would be nice to finish out the week with some discussion of another thing that I don't know enough about:impedance mismatches and buffers. I did a fair amount of reading to come up with the following description of the issues. And a couple of schematics to trace through.
Why Buffers?
A guitar pickup is a high impedance device. Impedance is the interplay between resistance, inductance, and capacitance. It exists not just in your guitar pickups, but in the overall mix of pickups/cables/effects/loads (amp, etc.). Impedance mismatches will cause a loss in frequency response unless properly managed. This is typically manifested as a loss of treble response and general tone "muddiness".
Additionally, your pickups have very little current drive ability. So what we can do is introduce a buffer into the signal chain. A buffer is typically a unity gain device (i.e. the output level is the same as the input level; no gain is introduced.) The buffer converts the high impedance of the pickups to a lower value, and introduces the ability to drive your signal with a higher current along the overall chain described earlier. This ability to increase current capability comes about because the buffer is an active device--in other words, it has a power source that can drive active components.
Buffers can be built with a variety of circuits and components. Transistors, tubes, and op-amps have all been used in this application. It should be noted that no buffer is sonically "transparent". The goal of a well designed buffer is to solve the impedance mismatch problem while adding as little coloration to the sound as possible. Buffers can appear in many places: any active circuitry in your guitar itself, as a standalone buffer box, or within the circuitry of a stompbox.
Many stompboxes will have an input buffer to ensure that the input impedance that arrives at the "front" of the stompbox's circuit are at levels conducive to lower noise and better frequency response. (An exception to this design practice is the class of circuits that are actually highly dependent the guitar pickup's high impedance. For example, the classic Fuzz Face circuit relies to some extent on being directly connected to your guitar for part of its tone.)
In true-bypass designs, this "built-in" buffer is bypassed when you switch your effect off. So if you have a long string of true-bypass effects, and all are turned off, there is no buffer in the chain. On the other hand, most mass-produced commercial pedals have buffers that are on all the time. This is typically not a problem when you have just one of these at the beginning of your effects chain. But stack enough of these "always on" buffers in a chain and your tone will be changed, most likely in an adverse manner. So the correct approach would be a balance between no buffers and too many buffers. Arguments abound as to where you would place a buffer. It seems logical to have the buffer at the beginning of the effects chain in order to convert and massage the guitar's signal as early as possible in the chain. Others claim that a buffer at the end of the chain is the way to go. Regardless, the buffer is a useful device.
So we'll look at two examples.
A JFET Buffer – This circuit uses a single JFET as the buffer. The schematic was originally proposed by Jack Orman and had been modified into a stompbox form by General Guitar Gadgets. My version keeps the stompbox topology, but omits the on/off switch and LED.
(http://www.beavisaudio.com/techpages/HIW/buffer_jfet.gif)
9vDC – Like everything so far, we'll use a 9 volt battery for the power supply. It is interesting to note that some commercial buffer designs use 18 volts for the supply to ensure maximum headroom and to eliminate the possibility of clipping in the buffer.
R1 and R2 – These two resistors perform two very important functions. First, they form a voltage divider. At the top of R1, the circuit sees 9 volts. Where R1 and R2 connect, the circuit sees 4.5 volts. This voltage divider is used to set the reference or bias voltage for the transistor. Second, the values we choose for R1 and R2 determine the input impedance of the circuit. Remember that our buffer wants to present a high input impedance to the guitar pickups, but a lower impedance to the following chain (whether that be the remainder of a stompbox, or other effects and cables in our chain.) We determine the output impedance of this circuit by calculating the value of R1 and R2 in parallel which gives us an input impedance of 500k. If we wanted our buffer to maintain the typical pickup level impedance of 1 mega ohm, we would simply increase R1 and R2 to around 2 mega ohm values.
C1 and R2 -Put simply, these two components, taken together, provide a High Pass Filter. A high pass filter allows higher frequencies to pass through, while simultaneously attenuating lower frequencies. A more detailed explanation: C1 and the parallel combination of R1 and R2, along with the base of Q1 provide a high pass filter at the input. The primary purpose of this high pass filter is to block the DC bias voltage from being disturbed by whatever is connected to the input. Exactly what frequency the C1 and R1/R2 filter cuts off at may or may not be important, depending on the application. Often it is set below all audio frequencies so everything passes. Occasionally it is twiddled to make a treble filter, as in some treble boosters. The finessing of the exact rolloff point of this filter is tricky for beginners, as there are a lot of interactions. But making C1 big enough to pass all audio isn't - just make it big.
Q1 – This is the JFET transistor that forms the active part of the circuit. The part shown is a J201, but the following parts could be substituted: 2N5457, 2N5089 or 2N5088. JFETs are good parts for discrete buffers because the present a higher input impedance than bipolar devices.
R3 - This resistor is a feedback stabilizing resistor. ? I had text here from RG on the original LBP schematic, but that was for a bipolar NPN. Wonder if the same text applies here?
C2 - This capacitor acts as Decoupling or Output cap. It prevents the direct current supplied by the 9V source from getting into the output signal.
An Opamp Buffer
Ok, that was fun. Now let's look at accomplishing the same goal with an opamp instead of a transistor. Operational Amplifiers make great buffers. And they also reduce the circuit's part count. This design is based on General Guitar Gadget's Opamp Buffer project. The switching and LED have been omitted for clarity.
In the following schematic, you'll see a TL071 opamp plus the three building blocks we've covered so far:
- The DC power source and voltage divider
- The input coupling stage
- The output coupling stage.
(http://www.beavisaudio.com/techpages/HIW/buffer_opamp.gif)
C1 and R2 – This forms the input stage, just as described in the circuits previously. The components, taken together, form a high-pass filter (depending on C1's value) and stabilize the bias point.
R1 and R2 – The trusty voltage divider.
U1 The opamp. The part specified for this is the TL071. Any similar single opamp could be used. The input of the circuit goes to pin 3 which is the non-inverting input, which is also connected to the reference voltage point. The connection between pins 2 and 6 forms a negative feedback loop. The configuration in this schematic has U1 acting as a unity gain amplifier. I.e. the output signal will be at the same level as the input signal. Pin 7 is connected to 9 volts.
C2 – The output stage. Prevents DC from being sent on to the signal path.
Resources
Basic Buffers: http://www.muzique.com/lab/buffers.htm (http://www.muzique.com/lab/buffers.htm)
Buffers: http://www.customaudioelectronics.com/frequently_asked_questions.htm (http://www.customaudioelectronics.com/frequently_asked_questions.htm)
Loading and Cables: http://howard.davis2.home.att.net/LoadingandCables.htm (http://howard.davis2.home.att.net/LoadingandCables.htm)
Discrete Buffer: http://www.generalguitargadgets.com/index.php?option=content&task=view&id=69&Itemid=100 (http://www.generalguitargadgets.com/index.php?option=content&task=view&id=69&Itemid=100)
Op-Amp Buffer: http://www.generalguitargadgets.com/index.php?option=content&task=view&id=156&Itemid=190 (http://www.generalguitargadgets.com/index.php?option=content&task=view&id=156&Itemid=190)
That's it for buffers for now. I'm sure folks will have a field day with this one, finding errors and making suggestions :)
First, sure, I wouldn't mind knowing a bit about "class II biological safety cabinetry" (or Class I for that matter!) :icon_wink: My day job is about 1/10th as exciting as "biological" and "safety" imply.
Second, Dano12, I'm really impressed with what you've written here, and I was impressed already. Nice work and good idea overall to involve other brains in the process. Look out wikipedia :P
Good thread, I've been looking for something like this. I'll be watching!
QuoteImpedance mismatches will cause a loss in frequency response unless properly managed. This is typically manifested as a loss of treble response and general tone "muddiness".
This is true for inductive-pickup guitars particularly. In a more general view,
(1) Primarily resistive impedance mismatches cause LEVEL changes, depending on the direction of mismatch. If you have a voltage source and you want to preserve the voltage signal level, you want the receiver to be a much higher impedance than the source. This is the case for guitars. If you have a current source and you want to preserve the current signal level, you want the receiver to be a much lower impedance than the source. This is the case for photovoltaic equipment.
(2) Reactive mismatches (i.e. inductors and capacitors) cause frequency response imbalances. The one that gets us as guitarists most of the time is the loss of treble to a low impedance load. That happens because a guitar pickup is a resistive/inductive source. It's impedance changes from the DC value of the wire resistance to the inductive value of the coil as frequency rises. So if you load that down with a pure resistor, the rise of impedance in the pickup with higher frequencies means that the load resistor is effectively a bigger load at high frequecies, even though the resistor load did not change. So the trebles are loaded down more, and get lost. Capacitor loads, like long cables, give you a double whammy effect here because the cable capacitance is dropping with frequency.
The opposite happens with primarily capcitive pickups, like piezos. A piezo pickup is primarily capacitive, so if you load it with a resistive load, you lose the bass frequencies. Piezos need high impedance buffers to keep bass.
QuoteA buffer is typically a unity gain device
A buffer is typically
but not always a unity gain device. (added just for emphasis for beginner readers)
QuoteThis is typically not a problem when you have just one of these at the beginning of your effects chain. But stack enough of these "always on" buffers in a chain and your tone will be changed, most likely in an adverse manner.
To the EE purist, there is only one disadvantage to many buffers if they're well designed - noise. Each active component you put in the path adds noise. Some loss of highs could be benign, as hiss is primarily a high frequency problem, and a little treble loss keeps this in check. Each buffer preserves the signal - and the noise - that it gets. Each buffer adds a touch of its own noise.
QuoteIt should be noted that no buffer is sonically "transparent".
A
well designed buffer is not going to change your tone at all. A well designed buffer is frequency flat from well below to well above your sound frequencies. The catch here is that what passes for a buffer in musical instrument design is sadly lacking in flatness, so the tone effect of many of them is a crap shoot. The idea that a buffer cannot be transparent - whatever that means - is a construction of the hifi tweakos. There are and can be buffers that are frequency-accurate over the entire audio range. But simple ones like in musical effects are not necessarily that way.
JFET buffer
R1 and R2 – In addition to the description: A lower noise way to do this is to make R1 and R2 be much lower valued, perhaps 10K, and couple their junction to ground with a capacitor. Take a high value resistor of 1M or over to the junction of the JFET gate and C1. This eliminates all of the current noise that the bias resistors would produce, and 1/2 of the thermal noise. The technique is know as "noiseless biasing", and is a neat trick that most beginners never get presented.
Q1 – The circuit as shown is best suited to the J201 JFET. This is because the J201 is an atypical JFET. The J201 has a very small Vgsoff, of about 0.2 to 0.5V. That means that its source will be pulled to within that voltage of its gate if the device can pass enough current. Other JFETs like the 2N5457 may have a Vgsoff of up to 5V. That means that the source will self-bias up to 5V higher than the gate. In this case, R1 is lower than infinity (i.e. an open circuit) because the J201 is being used. A 2N5457 might be better served with an open circuit for R1. JFETs have a number of gotchas associated with them in biasing. While BJTs like the 2N5088 are a breeze to bias like I described earlier, JFETs in general are NOT.
QuoteR3 - This resistor is a feedback stabilizing resistor. ? I had text here from RG on the original LBP schematic, but that was for a bipolar NPN. Wonder if the same text applies here?
It does, and even more. R3 does the feedback stabilizing job for a JFET, but it also plays an important role in the biasing beyond that. JFETs are typically depletion mode devices which means that they are normally fully on and you have to do something to the gate to turn them off. In most circuits that something is using an R3 to hold the source at a voltage higher than the gate. How much higher is determined by the Vgsoff of the JFET and its transconductance. In this circuit, the voltage at the top of R3 will be higher than the voltage at R1/R2 by the percentage of the Vgsoff needed to back the JFET down to just the right current to make that voltage work out for the gain of this particular JFET.
Confusing? Yes. Easy to handly by guess and approximate, like with the bipolar? No. This is one reason almost all beginner designs with JFETs have trimmer pots. They're to compensate for that oddity of complex design and variation between devices.
Quote from: R.G. on July 29, 2006, 10:37:43 AM
JFET buffer
R1 and R2 – In addition to the description: A lower noise way to do this is to make R1 and R2 be much lower valued, perhaps 10K, and couple their junction to ground with a capacitor. Take a high value resistor of 1M or over to the junction of the JFET gate and C1. This eliminates all of the current noise that the bias resistors would produce, and 1/2 of the thermal noise. The technique is know as "noiseless biasing", and is a neat trick that most beginners never get presented.
First off, I'm so glad I found this forum, especially so since it's incredibly active unlike others which are good for research but difficult when trying to get a question answered!
My question has to do with the quoted passage. First off, when you say couple the junction to ground with a capacitor, you mean there is a capacitor connected between the junction of the voltage divider and the gate of the JFET and then between the cap and the JFET there is a large value resistor to ground, right? Now, assuming my visualization of this circuit is correct, could you perhaps go into a little more detail about why this technique eliminates current noise, or is that beyond the scope of this discussion?
And finally, to the question that originally got me to post. Why do you sometimes use resistors in high/low pass filters and other times not. I've seen plenty of coupling caps used between stages without resistors to ground, and then I've read in Art of Electronics that it is a must to have a resistor to ground.
And my other problem(s) with capacitors stem from the fact that I have trouble visualizing what is happening on a particle level in, for example, a high pass filter.
Now its of course understood that they do not pass DC and this is becuase if just a DC voltage is placed across one, it will charge until its plates equal that voltage and then nothing more can really happen. Of course I'm guessing there must be a brief period, while its charging, where the cap is "converting" that voltage into a rising current before it stops, correct?
To explain my other confusion, let me try to set up an example. Say we have a cap already blocking a 5VDC voltage, so the cap's plates are at 5VDC. Now we inject an AC signal, say a sine wave, into this mix. On the positive part of the sinewave, the voltage is rising above 5V on the cap and then falling again. When it goes negative, the voltage, according to my understanding, will go below 5VDC by the same amount that it rose above it. Now it seems to me that when its on the negative side, becuase the cap is now below the DC voltage, the present DC voltage would expedite the recharging of the cap back to 5V effectively distorting the lower part of the signal. However I'm sure that isn't true, but I'm not sure what I'm missing in my understanding.. Sorry if I'm completely off base with this. I've read so much about this stuff but I still feel like I know absolutely nothing about it because I can't visualize/explain what is happening.
QuoteMy question has to do with the quoted passage. First off, when you say couple the junction to ground with a capacitor, you mean there is a capacitor connected between the junction of the voltage divider and the gate of the JFET and then between the cap and the JFET there is a large value resistor to ground, right? Now, assuming my visualization of this circuit is correct, could you perhaps go into a little more detail about why this technique eliminates current noise, or is that beyond the scope of this discussion?
I did a really rotten job of describing what I meant. Lemme try again.
Imagine that for the circuit of the JFET buffer, we remove R1 and R2 entirely. Then we make a new voltage divider with Ra = 10K and Rb= 10K, but this one is not connected to the gate yet, just two resistors between +9V and ground. We place a capacitor Ca from the middle of this new divider to ground. This gives us a bias voltage of half the 9V supply, and it is tightly coupled to ground by Ca. The Ca connection to ground shunts any noise from Ra and Rb to ground. We still have signal coming in to the JFET through C1, but the gate now floats, no bias on it. To bias the JFET, we take a new resistor Rc = 1M from the junction of Ra and Rb to the junction of C1 and the gate of the JFET.
The new resistor does a couple of things for us. It biases up the JFET to the same place as the old R1 and R2 did, but it has almost zero current going through it, only the leakage of the JFET gate junction. So any noise which would have been generated by the current going through the new Rc is almost zero, because it has almost no current flowing. It has only its thermal noise, which we can't escape. The old R1/R2 divider not only generated the thermal noise of two resistors, but also had a healthy does of current noise from the currents running through them. That is missing from the new setup.
QuoteAnd finally, to the question that originally got me to post. Why do you sometimes use resistors in high/low pass filters and other times not. I've seen plenty of coupling caps used between stages without resistors to ground, and then I've read in Art of Electronics that it is a must to have a resistor to ground.
The term "resistor to ground" is tricky. A resistor to the power supply is equal to a resistor to ground because the power supply is effectively a short to ground for AC signals. A resistor divider, one up and one down, is equal to a resistor to ground. If you are doing a low pass filter, there may be a series resistor and a shunt cap to ground. It can be difficult to point out exactly what part or combination of parts defines the "resistor", as it can also be the internal resistance of a part, like the Rds of an FET.
QuoteAnd my other problem(s) with capacitors stem from the fact that I have trouble visualizing what is happening on a particle level in, for example, a high pass filter.
Now its of course understood that they do not pass DC and this is becuase if just a DC voltage is placed across one, it will charge until its plates equal that voltage and then nothing more can really happen. Of course I'm guessing there must be a brief period, while its charging, where the cap is "converting" that voltage into a rising current before it stops, correct?
Good point. Let's talk about it for a minute. A voltage is a pressure of electrons. If you apply a voltage across some material, the voltage supplies pressure that tries to push the electrons away from the negative voltage and toward the positive voltage. How fast or slow the electrons move depends on the pressure and the resistance to electron motion that the material has. If we push one electron into a capacitor from one side, that electron pushes on the electrons on the other side, as well as the unspecified positive side of our electron pressure pump pulling on the electrons on the positive side, and eventually one electron leaves the positive side. This happens electron by electron until the voltage across the capacitor equals the voltage being applied to it by the pump, and when that happens, there is no more pressure left to move electrons. The negative side is as full of negative-charge electrons as there is pressure available to push them in (and pull them off the positive side of course).
And your intuition about the rising current is correct. Let's assume we have a resistor in series with the negative side of the cap, and our voltage pressure pump connected negative side to the resistor, positive side to the cap. When we turn the pressure on, the cap has no voltage across it, because we have to move electrons in and out to make a voltage appear. If there is no voltage across the cap, all the voltage appears across the resistance, and electrons start moving out of the resistor and into the cap at a rate determined by the pressure (voltage) and resistance. Ohm's law tells us how many electrons move (i.e. the current). The electrons move onto the negative side of the cap and off the positive side, pulled/pushed by the pump. But the electrons pile up in the cap, and a voltage builds up across the cap. This lessens the voltage across the resistor, lessening the rate at which electrons move through the resistor. So at first, lots of electrons move because there is little voltage on the cap, lots of voltage across the resistor. As the cap fills up, the voltage on the cap rises and the voltage across the resistor falls. So every time we charge a cap, there is an instant surge of current that tails off as the capacitor charges through the resistance - however small that resistance is.
QuoteTo explain my other confusion, let me try to set up an example. Say we have a cap already blocking a 5VDC voltage, so the cap's plates are at 5VDC. Now we inject an AC signal, say a sine wave, into this mix. On the positive part of the sinewave, the voltage is rising above 5V on the cap and then falling again. When it goes negative, the voltage, according to my understanding, will go below 5VDC by the same amount that it rose above it. Now it seems to me that when its on the negative side, becuase the cap is now below the DC voltage, the present DC voltage would expedite the recharging of the cap back to 5V effectively distorting the lower part of the signal. However I'm sure that isn't true, but I'm not sure what I'm missing in my understanding.
That's a question that puzzles you because it's not possible to determine what happens without knowing what the other parts are. If you have a cap charged to 5Vdc directly from a 100lb 5V battery with half inch thick copper cables, it's quite different from a cap charged to 5Vdc from a 5V battery through a 1M resistor. And it depends even more on what drives the AC signal. Is that from a 10MegaWatt AC power station generator or a toy oscillator through a 100K resistor?
In the first case, a cap held to 5V by a perfect voltage source, you can't inject an AC signal. The DC source fights to keep the voltage constant, and the generator with the biggest power rating wins. In the second case, the cap held to 5V through a 1M resistor can be "pushed around" by almost anything because the forces holding the DC in the cap are so weak.
So the situation you suggest is incomplete. A complete description involves the capacitor, two voltage sources, and two source impedances before you can calculate anything about what happens. What does not happen is distortion, as all the parts are linear.
wow... R.G. for the win!!! a very enlightening read and much appreciated.
Quote from: R.G. on July 29, 2006, 01:08:54 PM
What does not happen is distortion, as all the parts are linear.
I was thinking that the power supply might be "thinking" to itself, "ok the voltage is going above my 5V so I'll just let it go with that cause I'm doing my job", but when it goes below 5V then the power supply then thinks it has to fight to maintain at least the 5V it's "responsible" for; thus causing a distortion on only part of the waveform. But I can see that the situation is much more complex than that, and that the powersupply would "fight" equally above and below; it's success being dependant on the synergy of all the actors at play in this circuit and their individual specifications.
Also, that noise reduction technique is really clever, I'll remember it when I figure out enough of this stuff to start making my own designs. Thanks again!!
Ack! Sorry if I come across as though I'm abusing your generosity in answering all these noob questions! It's just that I find that your explanations are working very well for me! With that said, I read your tube screamer tech article and came up with another question related to the biasing post you made above.
In the input buffer section of the article you show an BJT biased with 4.5V through a resistor. Now my question is how is that different from applying the bias voltage directly to the base as in earlier examples, and what are the reasonings for doing that? I spent some time thinking about how this works and I came up with the following explanation:
With no input signal, the base is receiving a 4.5/510k bias current. As the input voltage increases, the input current is now decreasing becuase the voltage drop is decreasing, and vice versa for the negative portion of the input signal. Now this would have the effect of inverting the output with respect to the input, which seems like a good way to counter the natural inversion of using a transistor with a collector voltage output, but that isn't the case here. So why do that?
Also, assuming my understanding is correct, what is the significance of the value 510k for that resistor? Is it just a nice compromise between input impedance and bias/input current?
hi, what does NC mean on a shem? if anyone can tell me id be greatful
QuoteIn the input buffer section of the article you show an BJT biased with 4.5V through a resistor. Now my question is how is that different from applying the bias voltage directly to the base as in earlier examples, and what are the reasonings for doing that? I spent some time thinking about how this works and I came up with the following explanation:
With no input signal, the base is receiving a 4.5/510k bias current. As the input voltage increases, the input current is now decreasing becuase the voltage drop is decreasing, and vice versa for the negative portion of the input signal. Now this would have the effect of inverting the output with respect to the input, which seems like a good way to counter the natural inversion of using a transistor with a collector voltage output, but that isn't the case here. So why do that?
It's a little early in your learning curve, but you need to learn it sometime. We're seeing the principle of superposition in action. When trying to figure out what two different things do to a circuit, you can usually figure out what each one does separately, then add them. That's what's happening here. The 510K resistor from +4.5V to the base of the transistor is the noiseless biasing I mentioned about the JFET buffer, but applied to a BJT in this case. The bias voltage is made by two resistors and a cap to ground, just as in the JFET example in this thread. The 4.5V supply through the 510K resistor supplies a trickle of current to the base, enough to pull the emitter up to about one diode drop less than the base voltage. The base is not sitting exactly at 4.5V, because there is some base current, as I described in the section on biasing the two-transistor setup. So the 4.5V/510K set the DC condition. On top of that, the input signal through the input capacitor adds to the DC bias voltage. The signal's wiggling around is transmitted to the base, and just like with the DC bias voltage, the DC plus wiggling is transmitted to the emitter almost unchanged because of the very high gain of the transistor. The emitter follows the base - it's an emitter follower, which is where the name came from.
QuoteAlso, assuming my understanding is correct, what is the significance of the value 510k for that resistor? Is it just a nice compromise between input impedance and bias/input current?
You're dead right on this one. It's a compromise of wanting it to be as high as possible because you don't want to load the signal, and as low as possible to get the base to be up at 4.5V. The base sags by Ib times the resistor. So you want it big for input signal unloading, small for bias stability.
Hubble:
In my experience, "NC" means no connection. That is likely the only thing I will be able to contribute to this thread.
ok. cool thanks!
I so wish I could wrap my head around this stuff. ??? :icon_confused:
Quote from: captntasty on August 02, 2006, 09:05:28 PM
I so wish I could wrap my head around this stuff. ??? :icon_confused:
Don't worry. You don't need to understand all this stuff.
At least not now.
Learn how to identify some basic stuff and don't be afraid to experiment. Half of what R.G. has said still hasn't sunk in for me. But I learned enough about biasing to actually find the right bits on the breadboard and play around.
Quote from: R.G. on July 29, 2006, 10:37:43 AM
Quote
JFET buffer
R1 and R2 – In addition to the description: A lower noise way to do this is to make R1 and R2 be much lower valued, perhaps 10K, and couple their junction to ground with a capacitor. Take a high value resistor of 1M or over to the junction of the JFET gate and C1. This eliminates all of the current noise that the bias resistors would produce, and 1/2 of the thermal noise. The technique is know as "noiseless biasing", and is a neat trick that most beginners never get presented.
MXR Micro Amp and Distortion+ biasing, I think. :)
A little example to go along with all this theory.
This thread rules, thanks RG and everyone!
So what's next. How about diode clipping? I know a little about it but it is such a rich vein, we might as well mine it....
Maybe cover how diodes actually clip a signal, different types, in the opamp feedback loop vs. the output, germanium, silicon and MOSFET devices, symmetrical vs. asymmetrical, etc.
And how about we add in using tubes as clipping diodes (something I know nothing about, let alone whether its possible but it sounds wicked cool).
Your humble servant,
-dano
Quote from: dano12 on August 02, 2006, 09:47:04 PM
So what's next. How about diode clipping? I know a little about it but it is such a rich vein, we might as well mine it....
Maybe cover how diodes actually clip a signal, different types, in the opamp feedback loop vs. the output, germanium, silicon and MOSFET devices, symmetrical vs. asymmetrical, etc.
And how about we add in using tubes as clipping diodes (something I know nothing about, let alone whether its possible but it sounds wicked cool).
Your humble servant,
-dano
DISCLAIMER: LISTEN TO ME WITH CAUTION UNTIL VERIFIED BY R.G. AND FRIENDS :)
Being more a student of this thread, there isn't much I can offer, but I can at least share a revelation I had about clipping diodes while reading one of R.G.'s articles. I think I was initially thrown off by the phrase "clipping diode". I imaged that the diode itself was doing the clipping, and in a way it is. However I found it easier to think of it as a "clipping-enabling diode". What I mean is that, at least when placed in the feedback loop, the diode prevents the gain device (transistor, opamp) from continuing it's linear increase once the diodes drop voltage is broached; the diode on it's own won't do this. This is probably best explained with an example.
Using an ideal opamp and a diode drop of .7V as an example, imagine x volts at one input of the opamp. Now because the opamp's job is to maintain equal voltages at it's inputs, the opamp has to swing its output terminal such that this disparity between the inputs (caused by the presence of a signal on one, and not the other) is alleviated. The ouput needs to be connected to the other input for it's output to cause a change in the difference between the two inputs; this is the feedback loop. If the output was connected to the non-signal input with a short circuit, there would be no gain because the output would only have to match the input for things to be equal. However with a resistor in the way, the opamp has to work a little harder to normalize its inputs, so it has to put out more voltage to achieve the same effect, hence gain. The diodes come in when the signal approaches the limit (set by the diode drop). The diode turns on and short circuits the gain resistors limiting the ouput from rising any further, hence a flat, "clipped" waveform.
Now because the diode drop is fixed, one way to vary the gain is to vary the level of the input signal into opamp with another gainstage. So it's really approaching the problem from the other end; rather than adjusting the clipping point (we can't, the .7V is fixed), we adjust how hard the signal is driven into this clipping stage, effectively changing the clipping point. Sorta like how some compressors don't have a threshold setting (it's fixed and internal), just an input gain, which drives more or less of the input above the threshold.
I hope my version of the story is correct, I'll feel like an ass if it isn't!! Not the full picture by any means, but it got me thinking on the (hopefully)right track. Hope it helps.
QuoteMaybe cover how diodes actually clip a signal, different types, in the opamp feedback loop vs. the output, germanium, silicon and MOSFET devices, symmetrical vs. asymmetrical, etc.
That's a good example of a simple question with no simple answer. Each case is a whole study in itself.
A diode is a good example of a simple non-linear resistor. For a normal resistor, the current through it depends linearly on the voltage across it. So ohm's law ( I=V/R) holds and a graph of I versus V is a sloped straight line. For a diode, the current is best described by I = K e^mV where K is some constant, m is some other constant, and "^" means "to the power of". "e" is the base of natural logarithms, 2.718.... What this does is that as long as mV is below 1, the resulting current is below K. When the term mV gets over 1, then you get current which turns up and heads for the moon. For silicon, the K and m terms make the turnover voltage about 0.7V.
What that means is that a diode hardly conducts at all (i.e., effective resistance is big) until the turnover point, and then its resistance drops to quite small.
So let's clip a signal. We have this here signal generator that puts out an AC signal voltage which we can adjust from 1uV peak to 10V peak. Since our signal generator is perfect ('cause it's imaginary 8-) ) we put some resistance in series with it to fake a real signal generator. So let's say we have a 1K resistor in series to a diode to ground. We put an oscilloscope across the diode.
What we see is that for small signal the signal is unchanged. That's because the diode resistance is much, much higher than 1K for all signal voltages. As we turn the signal voltage up, when the signal gets somewhere between +0.4 and +0.6V from anode to cathode, the resistance of the diode changes from very high to below 1K. When that happens, the signal sees a voltage divider of the 1K signal generator and the resistance of the diode. Within a couple of tenths of a volt, the diode resistance goes from over 1K to much less than 1K, so above 0.6 to 0.7V, the signal flattens off as the 1K prevents the signal generator from providing enough current to make the diode voltage go up anymore. So the signal appears to be clipped off at 0.6 to 0.7V for as long as the signal is greater than that.
For signals reversing the diode voltage, the diode never conducts, it always looks like a big resistor and so the undisturbed signal appears for those voltages.
If we put another diode in anti-parallel to the first, anode to cathode both ends, then the clipping effect happens for both polarities of signal, and we get a signal clipped on both top and bottom.
Questions on this one?
Wow I haven't even posted the next set of schematic/questions and you guys are already posting the answers.
Not only are you guys great resources, you are also apparently pyschic. :)
Someone point me to a starting point for tube clipping (if there is such a thing) and I'll get busy.
OK, no questions on Diode Clipping 1. I must have had one of my rare lucid moments. :icon_lol:
So with that in hand, what's the difference in diodes?
Diodes differ primarily in conduction voltage and knee sharpness.
Conduction voltage is the voltage at which the current per added volt across the diode starts to increase rapidly.
Knee sharpness is the degree to which the sudden change in conduction makes a sharp corner versus a rounded turn as current increases.
Conduction voltage determines how big a signal the diode will pass before clipping. LEDs are made out of non-mainstream semiconductors like Gallium Phosphide, Gallium Aluminum Phosphide, etc. Weird stuff. The different materials change the values of K and m in their diode equation, I = K e^m*V. What that does is make them have conduction voltages where the current increases rapidly of 1.5 to 3.0V for normal LEDs, about 2-4 times what it takes for a normal silicon junction to turn on. And you have to feed them a signal with peaks that big or bigger before they will turn on at all. Germanium's values of K and m make it start conducting somewhere between 0.2 and 0.3V.
Knee sharpness is what determines the abruptness of conduction, and has an effect on the harshness of distortion. The rounder the diode knee in comparison to the signal height, the smoother the distortion. This is true in general for distortion - the more sudden the change in clipping, the harsher and buzzier the sound. For diodes, there is not a huge range of knee softnesses. They are all about the same roundness. Not exact, but similar. Germanium often sounds smoother as a clipping diode if you feed it a small signal because the roundess is similar and the cutin voltage is small, so for small signals, more of the signal height is compressed into the conduction knee. If you feed it a big signal, it sounds buzzier because the signal is too big to compress into the knee. LEDs have softer clipping than silicon by another path - they have high junction capacitances, so they tend to smooth the corners a little bit capacitively, not in their diode equation.
Questions?
Quote from: R.G. on August 05, 2006, 07:53:27 AM
Questions?
Questions? But of course. I understand how diodes to ground produce clipping (as in the Rat).
But what about diodes in the feedback loop (if that's what it is...) as in the TubeScreamer? How is it clipping if the diodes aren't going to ground? I think I'm missing something fundamental here...
Let's think for a moment about how opamps work.
First, inverting opamps. The non-inverting input is tied to some reference voltage between the most positive and most negative power supplies, so we'll call it ground, or 0V. The opamp amplifies the difference between the + and - inputs by about a zillion, and that voltage appears on the output. If we put a resistor from output to the - input, then the more positive the + input gets, the harder it drives the - input, and that forces the - input toward the + input's voltage, so the whole thing balances out with the + and - inputs at the same voltage, plus or minus a gnat's eyelash of error voltage.
What happens if we put some current into the - input? It drives the - input higher, which causes the amplifier to amplify that voltage by negative one zillion, and that voltage appears on the output, which sucks current out of the - input through the feedback resistor until the + and - inputs are at the same voltage again. In fact, if we force a constant current into the -input, the output drops low enough to make the feedback resistor just suck that same amount of current OUT of the - input to rebalance the two inputs. And since the current coming out of the - input is equal to the current coming in, the - input neither eats nor sources any of that current. The output of the opamp eats the current through the feedback resistor. Thanks to Georg Ohm, we know that the voltage across that feedback resistor will be the current times the resistor, or Iin*Rf. And since the two inputs are both sitting at 0V, because they have to, then the output votlage at the opamp is -Iin*Rf.
Now we make our current source be a voltage through an input resistor. That is, we put a voltage source thorugh Rin. Since the - input must be at 0V matching the + input, then the current into the input must be Iin = Vin/Rin. And the output voltage will be Vout = -Iin*Rf, so the output voltage must be Vout = -(Vin/Rin)*Rf and a little bit of algebra gives us the gain Vout/Vin = -Rf/Rin.
Yes, yes, I know. I'm getting to the point now.
What happens if RF is paralleled by two diodes in anti-parallel?
We still have exactly the same input current, or Iin = Vin/Rin. As Vin increases away from 0 in either direction, the output voltage changes by Vout = -Vin* Rf/Rin. And it does so right up to the point where one of those diodes sees a voltage across it where it starts to conduct. When the conduction starts, the diode resistance changes from essentially an open circuit to a much lower resistance, as low as a few ohms if you drive it really hard (i.e., with lots of current). And when that happens, the gain of the whole setup changes. We still have the current drawn from the - input by the output being equal to Iin. But it's now being drawn through a diode, and the diode only needs its forward voltage across it to move that amount of current. The opamp doesn't know anything has changed. All it knows is that all of a sudden it doesn't have to move its output as far to pull the current stress off its - input.
The diodes are run in constant current mode, the current always being equal to the input current, Iin = Vin/Rin.
And it works the same for both polarities.
Questions?
http://www.elixant.com/~stompbox/smfforum/index.php?topic=38581.0
I keep posting this because it has good stuff in it. There have been many discussions about this topic and it always facinates me. I think I see it as the "Heart" of the distortion, although there are many other factors involved. I seem to prefer distortions that clip 2 or 3 times.
Of course the op amp also gets into the act. :icon_cool:
QuoteOf course the op amp also gets into the act.
Eventually.
Here's one I get asked about a lot- order of components in series. Normally it makes no difference if you have a resistor and cap in series, to reverse them. But sometimes it does, due to the high pass filter effect. For example see tonepad's clever layout for Dist+ and Micro amp using the same board:
http://www.tonepad.com/getFile.asp?id=6
The non inverting input of the op amp to ground (and a cap) sets the gain in both of these. It goes through the gain pot and a capacitor.
It would seem much easier to put the pot at the end, with one end to chassis ground- then only one wire from the board would be needed. That's the Dist+ way. But MXR reversed the order of the cap and resistor (one resistor electronically, 2 resistors physically, one is a limiting resistor) on the Micro amp, and seems that tonepad did the same thing.
Maybe R.G. can explain when you can and can't reverse series components better than I can.
QuoteMaybe R.G. can explain when you can and can't reverse series components better than I can.
You can ALWAYS change the order of series components as long as there is no contact to the middle connections. In the case of the D+ board there, the order of C4, R6 and R7 is immaterial. It should be done any way that makes your wiring work easily.
The impedance of any two components in series is always just the sum of their impedances. There is no high pass effect to be had with a cap and a resistor in any order if you measure only at the two ends of the string. Obviously, you can't reverse polarized components in a series string without changing things, but if you keep them in the same polarity, you can put them in any order with no changes outside the string.
thanks, that is what I thought. (it's been a long day, 13 hours of work so far and more to come. Eating Gyro dinner at my PC now).
It's odd that MXR and everyone who makes MXR layouts builds them the hard way when it's so much easier to change the micro amp to the dist+ style. Maybe tonepad can change their layout, will remove two parts and a wire.
have fun and get a real job so you don't burn out like me ;)
To distill some of the clipping information so far, perhaps we should use the MXR Distortion + as an example to walk through. This one uses diodes for clipping at the output of the opamp. (Later we can look at diode clipping in the feedback loop of the opamp)
Here we go:
(http://www.beavisaudio.com/techpages/HIW/MXRDist+.gif)
SW1: Power on/off, just like the previous circuits
R1 and R2: These form a voltage divider that presents the non-inverting input of U1 with 4.5 volts. Pin 7 of the opamp is the V+ connection, and it gets 9 volts. This sets the bias point for the opamp.
C1: We haven't seen this one in the previous schematics—a small value cap from the top of the middle of the voltage divider to ground. Purpose?
C2, C3 and R4: What is this? It looks a bit different than our previous examples of the input stage that forms a high-pass filter and stabilizes the bias point. Hmm.... Is C2/R4 the high-pass filter, and if so, what does C3 do?
U1: The LM741 single op-amp provides the gain for the circuit.
General Question 1: For the Distortion + specifically, and pedals in general: does the opamp perform any of the clipping? Or is it all the diodes, or a combination of both? Is opamp clipping desirable? What are the general design parameters for mixing opamp clipping with discrete diode clipping?
R9/C4/R5/R6: Ok, it looks like this is the feedback loop. The signal passes from the output of the opamp, through R9 back into the inverting input of the opamp. Here's my guess at what is happening. The more of the output signal that is fed back into the inverting input, the less gain the opamp is going to produce. The R6 pot bleeds more of this inverted signal off to ground as you turn it up, hence more gain. Is this correct, and what is the purpose of R5 and C4?
C5 and R7: C5 blocks DC voltage from leaving the opamp stage. R7? Not sure....
C6: Small cap to ground: what does this do?
R8: This controls the amount of output signal presented to the output of the circuit. Like the previous circuits in this thread, the pot controls the ratio of output signal vs. attenuation to ground that the output sees.
D1 and D2: R.G. talked about a diode as a non-linear resistor. Up to a certain amount of input voltage and the diode doesn't pass any through to ground. However, once the input gets up to the diode's turnover voltage, the diode passes the voltage on to ground. So D1 is going to clip off the top of the AC waveform once the opamp has amplified the input voltage to a certain point. D2 performs the same function on the bottom end of the AC waveform. Since we are using the same type of diode on each side of the waveform, the clipping occurs at the same level for positive and negative swings—hence we have symmetrical clipping. The 1N270 part specified is a germanium diode—these have a softer clipping sound. Go back and re-read R.G. explanation of the different types of knees in clipping diodes for more info.
Question 2: What does it sound like if we only clip off the top, or just the bottom? In other words, would it be of any use to play around with having only D1 or D2 in circuit?
Question 3: There was a posting a while back from a fellow who had created a clipping circuit that had a very odd shape. I think it was like devil horns, or some similar analogy—does anyone recall this?
As always, questions and answers welcome....
First, might as well get the pot values correct. Gain is 500KC and Volume is 50KA.
Notice anything about the biasing set-up, value wise?
And yup, the D+ distorts without diodes. My guess, it's a combination of the 741's specs and the biasing current. It's part of this ckts character, you probably wouldn't want too much OA clipping in other ckts.
Quote from: Fret Wire on August 08, 2006, 11:28:08 AM
First, might as well get the pot values correct. Gain is 500KC and Volume is 50KA.
1Meg and 10k are the values listed on 3 of the 4 schemes I was working from. Including the RGKeen/Jack Orman version. While the original used a reverse log pot, most of the schematics say its ok to use the values listed.
Quote from: Fret Wire on August 08, 2006, 11:28:08 AM
Notice anything about the biasing set-up, value wise?
Yep, another 1M resistor off the middle of the divider. What does it do?
Quote from: dano12 on August 08, 2006, 11:38:05 AM
1Meg and 10k are the values listed on 3 of the 4 schemes I was working from. Including the RGKeen/Jack Orman version. While the original used a reverse log pot, most of the schematics say its ok to use the values listed.
Yep, another 1M resistor off the middle of the divider. What does it do?
That's the embarrassing thing about the D+, a simple ckt, and yet most of the schematics are wrong, pot value wise. Since you're doing a "technology of the D+" so to speak, why keep spreading the same mistaken pot values and tapers?
I guess it's one of my peeves.
http://www.diystompboxes.com/smfforum/index.php?topic=47202.msg348546#msg348546
Look at the values of the divider on the DOD 250. Most ckts look like this, lower resistor values and higher value caps. The D+ has high value resistors and a low value cap.
http://www.generalguitargadgets.com/diagrams/dist_250_sc.gif
On the grey DOD 250, disregard D4, R11 (led + resistor), C5 (smoothing cap), D3 (third clipping diode), and possibly R11 (ps). AnologMike can verify that. I've never had a grey 250 opened up in my hands. An actual pedal in your hand is worth a thousand schematics. :icon_smile:
I updated the schematic with the original pot values per Fretwire's suggestion.
Unfortunately, I can no longer modify my posts, so keeping this thread updated may be kind of pointless.
Whoops, you got 500KA for the volume when it should be 50KA.
QuoteUnfortunately, I can no longer modify my posts, so keeping this thread updated may be kind of pointless
Not really, this is a nice thread. When you've got the ckt all mapped out, start a new thread with just the schematic, and it's explanation. Credit RG for his help, and link back to the original thread. That way the new thread has the correct description of the ckt without the filler, and anyone can follow the link back to the original thread, where there are some really helpful questions and answers.
OK, I'll comment on the aspects I focus on for Tone Tweaking.
R3 - sets input impedance which in turn effects the high's getting into it. 100K would probably roll some off. Pre-gain
C3 - I use this for some low freq roll off. With R3 at 1M C3 at .1uf the low roll off point is around 1.6Hz, no effect. With C3 at .001uf, the lows are rolled off around 160Hz. Pre-gain
R5 - sets max gain when Drive is at minimum resistance/max gain. In conjuction with C4 it sets the low freq roll off point for the op amp gain. With C4 at .047uf it = 720Hz, (if I got my decimal in the right place) which is sort of typical. The Drive pot lowers this as the resistance increases and gain decreases adding bass at lower drive levels.
C4 - see above. I think C4 is very important in setting the "Character" of the distortion. Halving it raises the roll off point to 1440Hz, decreases gain and makes it more Distortion like IMHO. Doubling it lowers the roll off point to 360Hz increases gain and makes if more Fuzz like IMHO.
R5 in conjunction with C4 allow the gain and roll off point to be adjusted and along with the diodes etc. determine the "Character" of the pedal. Consider using switches here.
R9 - sets gain for op amp. Some distortions use a pot here and have the R5 fixed. Larger than 1M is not usually seen as it creates to much noise. 470K and 100K are other typical values.
C5 - can be used to roll off post op amp bass, but I usually make it large enough to minimize bass loss at this point. Bass reduction before the op amp is usually preferred for smoother distortion.
D1 & D2 - Almost limitless possiblities. Again, switching recommended.
C6 - High freg roll off. as is .001uf = approx. 16Khz. .01uf = 1600hz and might be more desireable, with any value inbetween tweaked to taste.
I see this as a 2 stage distortion. At first the diodes will clip and when the Drive is high enough, the op amp will start clipping as well. Different tweakers like different op amps. Let me know if I screwed any calculations up. :icon_cool:
QuoteC3 - I use this for some low freq roll off. With R3 at 1M C3 at .1uf the low roll off point is around 1.6Hz, no effect. With C3 at .001uf, the lows are rolled off around 160Hz. Pre-gain
Another error on the schematic: C3 is .01uf.
QuoteDifferent tweakers like different op amps.
The 741, with it's low slew rate, and the biasing arrangement are definately part of the ckts "character". I get the feeling that it was taken into account way back when it was designed. Try using a TL071 and the 22k/22k/10uf divider from the 250 on the D+ with no other changes, and it feels/acts different.
Quote from: Fret Wire on August 08, 2006, 01:48:30 PM
Another error on the schematic: C3 is .01uf.
Both fixed now.
If I don't mention it, I agree with the comments.
QuoteC1: We haven't seen this one in the previous schematics—a small value cap from the top of the middle of the voltage divider to ground. Purpose?
As noted in my article on bias networks at GEO, the capacitor forms a low AC impedance from the bias point to ground. This is the same function that the JFET biasing network I was describing in reply #31 works. It shunts noise from the bias resistors to ground and holds the DC level relatively firm in the face of signal wiggling the other end of the bias resistors.
C2, C3 and R4: What is this? It looks a bit different than our previous examples of the input stage that forms a high-pass filter and stabilizes the bias point. Hmm.... Is C2/R4 the high-pass filter, and if so, what does C3 do?
C2 is a low pass filter. It's there to shunt high frequencies to ground before they get into the circuit. What's the cutoff frequency?
No way to tell. The resistance that makes this a low pass filter is back inside whatever drives this. If you feed it from a direct guitar signal, then the 1nF cap combines with the cable capacitance to shunt treble to ground within the audio band because of the highly inductive nature of guitar pickups. In that way, it smooths the distortion by cutting the size of the treble components in the incoming guitar signal. It cuts treble more with more highly inductive pickups, like humbuckers.
Quote
General Question 1: For the Distortion + specifically, and pedals in general: does the opamp perform any of the clipping? Or is it all the diodes, or a combination of both? Quote
Let's think about it.
For a diodes-to-ground clipper, there is a 10K resistor in series with the diodes. The output voltage across the diodes increases without the diodes having much to do with it until we hit the beginnings of the diode's knee. From that point on, a change of about about 200mV changes the diodes from not playing to sucking all the current they can. The opamp can only put out maybe 3.5V across that 10K resistor (4.5V minus about a volt that it can't swing near +9V or ground), so the max current the opamp could ever supply to these is about 350uA (3.5V/10K). The diodes reach their clipping at about 0.7V. The rest of the way to 3.5V changes the diode voltage not much at all. When the opamp stops providing current, the signal is already almost flat. Which contributes? I'd say the diodes are doing vastly more of the work of clipping than the opamp because the diodes are already almot flat-topping the signal when the opamp quits being able to push the diodes harder. The diode knee has aready contributed whatever softness to the edges that it can by the time the opamp can start clipping the signal, so the opamp clipping is hidden by the diode's having already limited.
QuoteIs opamp clipping desirable?
Good question. What happens when an opamp clips?
Because the opamp has a huge gain that can be spent on making the output be exactly what the feedback network and input signal say it should be, the excess gain hides any internal changes that happen as long as the opamp has excess gain to hide changes. When an opamp clips, the output circuit is no longer able to pull the output high enough (or low enough) to match what the feedback network and signal say it should be. This amounts to a lower open loop gain. And the open loop gain is what is hiding the internal loss of gain! What happens is that an opamp runs out of gain with a bang. At some signal level, it can simply no longer keep up the facade that everything is OK inside, and the output signal simply flat-tops. It happens FAST. High feedback clipping thresholds are some of the sharpest-corner clipping you can get. This makes for synthesizer-sounding buzzy edges with lots of high harmonics. On mixed-frequency signals - like two or more note chords - you get mounds of intermodulation distortion.
Whether that is desirable depends entirely on your musical taste. It is not all that musical, but then neither is a cow bell, and I hear those in songs too. But it's not the soft, smooth, creamy distortion that a lot of people like.
QuoteWhat are the general design parameters for mixing opamp clipping with discrete diode clipping?
The D+ *is* discrete diode clipping IMHO. The diodes clip so much before the opamp that the opamp adds little. It's not zero, granted, but little.
Quote
R9/C4/R5/R6: Ok, it looks like this is the feedback loop. The signal passes from the output of the opamp, through R9 back into the inverting input of the opamp. Here's my guess at what is happening. The more of the output signal that is fed back into the inverting input, the less gain the opamp is going to produce. The R6 pot bleeds more of this inverted signal off to ground as you turn it up, hence more gain. Is this correct, and what is the purpose of R5 and C4?
C4 blocks DC from going through R6 and R7. So for DC purposes, the gain is alway unity, and the bias voltage is not multiplied by the AC gain. This cap forces the output DC voltage to be whatever DC voltage is on the + input, as long as the opamp can possibly make that true.
R9 is the feedback resistor and R5+R6 is the value of the input gain setting resistor.
A noninverting opamp will have a gain of unity (because the output always follows that + input, henca a gain of one) plus Rf/Ri. In this case, the gain is 1+(R9/(R5+R6)). R5+R6 can be as much as 1,004,700 ohms or as little as 4700 ohms, giving gains of 1+ (1) up to 1+213. So changing R9 changes the gain from 2 to 214 for all signals above the C4/(R5+R6) rolloff. That changes the output signal level, and that in turn changes how soon the diodes clip from the input signal and how hard they clip. Your description of R9/R5/R6 is correct. Changing the setting of R6 changes how much output signal current gets back to the - input node, and so the gain.
QuoteC5 and R7: C5 blocks DC voltage from leaving the opamp stage. R7? Not sure....
C5- yes. R7 forms the top of the diode "voltage divider" and limits current into the diodes. Actually, R7 is a form of clipping control as well. As we saw above, R7 limits the total current that the opamp can put into the diodes before hitting opamp clipping because the output voltage can't go high or low enough.
QuoteC6: Small cap to ground: what does this do?
It's part of the diode "voltage divider" but a frequency dependent part. R7/C6 forms a low pass network that keeps the signal at the diodes from being as large when the frequency is high. So the voltage at which the diodes will clip starts decreasing at the R7/C6 rolloff frequency and you get less distortion on treble.
QuoteQuestion 2: What does it sound like if we only clip off the top, or just the bottom? In other words, would it be of any use to play around with having only D1 or D2 in circuit?
In this case, only the side with the diode is clipped - at least until you run into opamp clipping on the non-diode polarity. What happens is that one side is heavily clipped, one is not. This will produce an octave effect, as you are introducing heavy non-symmetrical distortion. The signal level increases because you now no longer have both sides limited to one diode drop, so the un-limited side can go to maybe 3.5V, and the total signal is about 4V peak to peak instead of 1.4V peak to peak.
QuoteQuestion 3: There was a posting a while back from a fellow who had created a clipping circuit that had a very odd shape. I think it was like devil horns, or some similar analogy—does anyone recall this?
Yes.
Here's another question. You know how R7 works with the diodes. What happens if you put a 1K resistor in series between R7 and the diodes, and take the signal output from the junction of R7 and this new resistor between R7 and the diodes?
Question RG: R7/C6 form a lowpass of 15.9khz. One of the popular mods for an existing D+ is to lower R7's value for more output volume, 5k is popular. The lowpass filter is now at 31.8khz, if my math is right. C6 would have to be raised from .001 to .0022uf to get the lowpass back to 14.5khz if you want more volume, but the original rolloff.
But now R7 is halved and more current is going to the diodes (Ge 1N270's). Is this going to obliterate any roundness hitting the knee, or will it even be noticeable with Ge diodes? Would halving R7 be more noticeable with si or led diodes?
Come to think of it, but the popular mod for more volume when building a D+ from scratch, is to use a 100k volume pot (leaving R7 at 10k). I can't help but think the volume pot value also interacts with the values of R7/C6. Correct?
Just a friendly bump.
There were some un answered questions and it would be great to keep this going.
Yeah, it's an old thread...but it's a good one.
John
what a great read.bump
Quote from: Fret Wire on August 08, 2006, 02:25:10 PMQuestion RG: R7/C6 form a lowpass of 15.9khz. One of the popular mods for an existing D+ is to lower R7's value for more output volume, 5k is popular. The lowpass filter is now at 31.8khz, if my math is right. C6 would have to be raised from .001 to .0022uf to get the lowpass back to 14.5khz if you want more volume, but the original rolloff.
But now R7 is halved and more current is going to the diodes (Ge 1N270's). Is this going to obliterate any roundness hitting the knee, or will it even be noticeable with Ge diodes? Would halving R7 be more noticeable with si or led diodes?
There's no simple answer. Some of all of that at the same time. Sorry - Mother Nature says so. Some things She does not necessarily make easy for us.
Yes, if there is no other change, decreasing R7 to 5K kicks the highpass rolloff up. That may or may not be a factor, since it was already near the top of the audio range anyway. Us old guys have ruined our hearing listening to loud music and can't hear all that much over 10-12kHz for those of us that can still hear at all. Many younger guys too. There may or may not be much audible change in decreasing R7. The higher diode drive will be noticeable With Ge as a little more volume, not much, but a harder sound. It is difficult to say whether the actual volume increases much or whether this is a psychoacoustic effect of the harder clipping.
If you sub in Si or LEDs, the sound will certainly change. But at least part of that is that you can't drive them as hard as you drove the Ges because there just isn't enough voltage from the opamp. Here's a secret of distortion: it's not so much
what kind of diode you use for clipping, within reason, it's
how far above its clipping level you drive it. Ge is kind of special because it's pre-clipping region is so small, comparable to its knee region. But after that, whether it's Si or LED, you can look at whether the incoming signal just barely crests the clipping region or is 10x the clipping voltage for how sharply it will clip.
Opamps do this too, but in a slightly different way. For an opamp banging against its power supply limits, the clipping is always sharp. That's because the signal that matters to the opamp is a virtual signal - it's the one inside the feedback loop that would exist if the feedback loop was opened. Opamps may have all kinds of odd distortions and nonlinearities inside the feedback loop where you can't get at them. In fact, that's WHY opamps exist - they are designed to make it easy to hide those distortions under the feedback. All amplifiers do this, but opamps are the logical conclusion, with such high gain that they can hide whatever internal signal exists.
Quote from: Fret Wire on August 08, 2006, 02:25:10 PM
Come to think of it, but the popular mod for more volume when building a D+ from scratch, is to use a 100k volume pot (leaving R7 at 10k). I can't help but think the volume pot value also interacts with the values of R7/C6. Correct?
Actually, not all that much, I think. The impedance of the R7/C6/diodes is well below 50K AND 100K over most of the audio band. The bigger pot lets any following cable capacitance have more effect, I guess. I would not look at the volume pot as a major source of tone change.
Hey folks.
I'm new to the forum and the whole electrical engineering stuff and just try to teach myself all the basics.
Last week I started to read theses Electrical Engeneering Training Series which I found in this Thread.
I just got a very simple question to it: What is the difference of Ampere and current flow? Isn't that basically the same? I'm just confused because they always use both terms.
Thank you.
Amperes are the unit of current flow.
An ampere is the flow of one coulomb per second. Right - so what's this coulomb thing? :icon_mad:
A coulomb is similar to a gallon of water. Water molecules are too tiny to count, so we simply say that one gallon is 23432832489782193478 molecules of water in a gallon, and then ignore the number from then on, only referring to gallons.
In the same way, a coulomb is the charge you get by accumulating 6.241 times ten to the eighteenth power electrons. And that many electrons flowing in one second is an ampere. I made up the number of water molecules in a gallon, but the number of electrons in a coulomb is correct to three decimal places - I looked it up.
Water is a commonly used analogy to electricity. Pressure is to water what voltage is to electricity. Rate of flow is to water what amperes is to electricity.
Thanks R.G.
I already knew the coloumb thing but the water example is pretty nice.
Thank you for your answer. I just tried to avoid learning things wrong.
And again I've got a little tiny questions which bothers me a lot.
My problem is in the formula for amperes I=E/R. For example we have a DC source with 1.5V and a 1ohm Resistor we would obviously have 1.5amps right?! But if we have for example a Resistor smaller than 1 or even no resistance the amps would obviously be higher than the Voltage? Is that actually possible? How could it be that more electrons are going from the - Terminal to the + Terminal in one second than the potential difference is? Did I get something wrong? What happens at 0ohms resistance or does the electrical engineer doesn't care about it because as soon as you have a conductor you have a resistance?
Thanks folks!
Quote from: Lenni on December 19, 2008, 05:42:37 PM
And again I've got a little tiny questions which bothers me a lot.
My problem is in the formula for amperes I=E/R. For example we have a DC source with 1.5V and a 1ohm Resistor we would obviously have 1.5amps right?! But if we have for example a Resistor smaller than 1 or even no resistance the amps would obviously be higher than the Voltage? Is that actually possible?
Sure it's possible.1.5 volts into a 0.1 ohm load would give you 15 amps, for example. Not that you would get that much current out of your typical AA battery -- batteries have an internal resistance that must be added to the circuit load to get an accurate calculation. The connecting wires in your circuit also have a resistance that must be accounted for. We usually can ignore wire resistance in our calculations, as it is so small as to be insignificant in most cases.
QuoteHow could it be that more electrons are going from the - Terminal to the + Terminal in one second than the potential difference is?
That's kind of like saying, "How can a car go 50 miles per hour when it only travelled 1 mile?"
QuoteDid I get something wrong? What happens at 0ohms resistance or does the electrical engineer doesn't care about it because as soon as you have a conductor you have a resistance?
Unitl you get into the realm of superconductors, all wires have some resistance.
Ok I guess I should answer my question a little bit different.
For sure it makes sense that 1.5V with 0.1 Ohm makes 15amps BUT (here we going to your need car example that I totally don't understand) If the + terminal has 1.5 x 6.28x10^18 less electron than the - terminal why should suddenly 15 x 6.28x10^18 run over there? puh that scares me that those things already bringin me in trouble.
Is it probably just that ampere = coloumb per second thing that makes me feel so stupid?
Quote from: Lenni on December 19, 2008, 09:03:09 PM
If the + terminal has 1.5 x 6.28x10^18 less electron than the - terminal why should suddenly 15 x 6.28x10^18 run over there?
Let's go back to water and hoses.
The water flows from the output of the pump through the hose because there is a pump pushing it. Electricity flows because there is a voltage pushing it - a battery or a generator or some such. There are not a spare coulomb or two of electrons sitting around at the negative terminal looking for something to do. The electron pump forces just enough of them to the negative side and away from the positive side to be X volts of electron pressure.
Electron pumps, like water pumps, can make the most pressure when there is least flow. So if you bottle up the pressure of a water pump, you get maximum pressure. Let some of the water out, and some of the work that was making pressure is now moving water, and the pressure drops as the flow comes up. You get maximum flow when the pump is just spraying into air, no restriction at all.
An electron pump holds the maximum pressure when there is no electron flow - no current. When you let some electrons flow, the pressure drops a bit because of losses inside the electron pump. If you short the + and - terminals together, the current flow is all that the pump can do.
An electron pump - battery or other voltage source - recirculates electrons. They flow out of the - side, through the circuit, and back into the + side. So even through many, many coulombs flow, it never runs out of electrons to pump.
So a (-) terminal doesn't have more electrons than the (+) terminal (Well, Ok, it does, but only enough to make that many volts). What it has is a pump behind it that will recirculate as many electrons as it takes. And matter contains many, many more electrons than the coulombs of electricity that flows, so you don't run out of electrons.
First of all thank you very much. That is a really nice discribtion and makes it a bit easier to understand the whole thing. BUT! It doesnt' really answers my question (actually no wonder now where I look at how I asked the question ...) anyways. I ask it again in the hopefully right way
Why does a resistor between 0-1ohm increases the courrent flow and a resistor above 1 decreases it? Shouldn't a resistor always decrease the courrent flow?
Hope I'm not to anoying for you guys!
Thanks for your help!
Quote from: Lenni on December 20, 2008, 09:29:26 PM
Why does a resistor between 0-1ohm increases the courrent flow and a resistor above 1 decreases it? Shouldn't a resistor always decrease the courrent flow?
Ah. OK, I think I see.
Your concept is incorrect. It is not true that resistors between 0 and 1 ohm increase current flow and above 1 ohm decreases it.
A resistor of 0 ohms is only possible with superconductors, which are not encountered in our kind of electronics anyway. So let's just say that a resistor of 0 ohms is impossible (for us, in practical use). In reality, even a copper bar a square meter in cross section has some resistance. It may be a micro-ohm, but there will be some resistance. In fact, let's just say that it IS a micro-ohm, a millionth of an ohm, and see what the math tells us.
Ohm's law tells us that the current which flows when there is a resistance of value R connected across a voltage V will be equal to the voltage V divided by the resistance R, or I = V/R
It also says that the voltage V which happens when we force a current I through a resistance R is V = I * R.
In fact resistance itself is defined as the ratio of the voltage across a circuit element divided by the current through it, R = V/I. As you recognize, these are the same relationship, just solved for the three different elements. When you know any two of them, you can ALWAYS calculate the third one from the other two.
So a one ohm resistor is a component which causes one volt of difference from end to end when one ampere (that is, one coulomb per second) is flowing through it. Our copper bar with its millionth of an ohm, is really one one-millionth of a volt per ampere. That is the same as one volt per million amperes. The exact amount of current flow does not matter - only the RATIO between volts and amps.
So with a copper bar of 1/1,000,000 ohm resistance, the current that flows is the voltage V across it divided by one one-millionth, or a million amperes per volt. Let's increase that to only one one-thousandth of an ohm. Now we get a voltage of one thousand amperes per volt. OK, go to one ohm.
The current that flows is one ampere per volt. (it's still I = V/R, for the V and the R you have).
At one thousand ohms, the current is I = V/1000 or one one-thousandth of an ampere, which we call a milli-ampere.
At one milliion ohms, the current is I = V /1,000,000 = one micro-ampere.
There is no break point at 1 ohm. It's a smooth change, and you can always calculate it by I = V/R; or V = I*R; or R = V/I, which are all the same thing, algebraically.
It is possible you are confusing series and parallel circuits. What happens when we have two resistors? If I have a voltage V and a resistor R, then the current which flows when I put the resistor across the voltage is I = V/R.
But what if I now add a second resistor R2 in parallel with the voltage source and with R? The voltage across each resistor is the same because they both have their ends tied across V. So two currents flow. One is I = V/R and the other is I2=V/R2. So the total current coming out of V must be the sum of the two, I + I2. Adding more resistors in parallel increases the current flow because each resistor demands and gets its own current equal to the voltage divided by its resistance. Now, hold on for some math.
Let's define a third resistance Rp to be the resistance we get when we divide V by the sum of the two currents that come out of V with two resistors. That is, Rp = V/(I+I2). We know that I = V/R and I2 = V/R2, so we substitute that in.
Rp = V/(V/R+V/R2) There is a math trick that lets us write (V/R +V/R2) as V*(1/R+1/R2), so we can say
Rp = 1/(1/R+1/R2)
That looks clumsy, but it's the way I usually use the equation for parallel resistors. If R = V/I (that is, volts per amperes) defines a resistance, Y = I/V (that is, amperes per volt) defines a conductance, how well it conducts, not how well it resists. By adding resistors in parallel we are adding additional conductances, more paths to let current flow, and thereby make more total current flow.
In the only-two-resistors case, the algebra lets you come up with Rp= R*R2/(R+R2) which is how most people use it. The conductance view seems more intuitive to me, and it works for an indefinite number of resistors, not just two.
There is an other way to hook those two resistors up. If I connect them in series so the current must flow first through R then through R2, what happens. Well, we know that if all the current that flows must flow through both of them, then the current must be identical in both of them, right? So there is only one current I in this case, not a unique one in each resistor. The voltage in the two resistors must be
V1 = I*R1 and V2 = I*R2.
We know that since the total voltage V is all we have, that the two resistor voltages must add to be V, or V = V1 +V2. That means that V = I*R1 + I *R2. We can use the math trick that I*R1+I*R2 = I *(R1+R2) and get that V = I *(R1+R2).
From that we see that resistors in series add as ohms. So adding resistors in series with each other adds resistance and lowers the current that flows. Resistors in parallel add as conductances and let more current flow as you add more resistances.
But each separate resistor lets current flow that makes the voltage across it be equal to the current through it times the resistance. Always.
Allright now I got it! Suddenly it seems more logical. 0 ohm would actually mean unlimited current flow right? For some reason I was thinking that the current flow can't be higher than the Voltage. For example a Batterie with 1.5V can't deliver 3amps because it only has 1.5V.
Anyways! Thanks so far but you just mentioned another problem that I have.
Quote
There is an other way to hook those two resistors up. If I connect them in series so the current must flow first through R then through R2, what happens. Well, we know that if all the current that flows must flow through both of them, then the current must be identical in both of them, right?
Why does that current has to flow threw both of them? Why doesn't the first resistor decreases the current a bit and the second one a bit more. I mean the current first flows threw the first one and than threw the second one right? But why is the effect of the two resistor coming right after the electrons leaving the source? What I don't understand is why can you measure ... lets say: 3amps ... all over a circuit with 18V, a 2ohm resistor and a 4ohm resistor?
Because electricity must flow in closed loops.
An electron pump (like a battery) can produce 18V. If you put one resistor across it, some current I = 18/R flows. The current going into the resistor must be equal to the current coming out of the resistor because electricity must flow in closed loops. If we want the current I to be 3 amps, then we can calculate that R must be 6 ohms.
But the same result happens if we have one resistor of 6 ohms, three resistors of two ohms each in series, or six resistors of one ohm each.
The voltage across whatever string of resistors we put there will be the same - the 18V of our supposedly-perfect battery. Since current out has to equal current in to the battery (HAS to - otherwise the battery gets charged up with a static charge because it's exporting charge, right?), then the total current out is equal to the current in.
If we put a 1 ohm resistor across it, we get I = 18A. If we put a second 1 ohm resistor in series with the first oh, and the series combination across the battery, then the total voltage across two resistors is still the 18V. But since the resistors are identical, then there must be half the voltage across each one, right? So each one-ohm resistor has 9V across it, and we know that I = V/R = 9V/1ohm = 9A. And since what's going through the first resistor can't go anywhere else but through the second resistor, we get that the total current is just the same as the battery current of 9V.
Notice that we get that same current if we use one resistor of 2 ohms instead of two resistors of one ohm.
Now we add a third 1 ohm resistor in series. The voltage gets distributed across three identical resistors, so each one must have 1/3 of the total, or 6V across it; and the current in each one must be 6V/1ohm = 6A. That's the same result we'd get if we put one three-ohm resistor across the 18V battery.
Current always goes SOMEWHERE; Nature says it goes in closed loops except in the rare cases where it's piled up on something in the form of static electricity. In electronics, it flows in loops out of and back into voltage or current sources. Series resistors can't decrease the current a little more each step - they all have to work on the same current at the same time, because it's all they have to work with.
In your question, an 18V source has a 2 ohm and a 4 ohm resistor in series across it. From the little thought experiment above, we know I can change the 2 ohm to two one ohms in series and get the same result. I only did up to N=3 resistors, but the same applies for as many resistors as you'd like; so the 4 ohm can be replaced by four one-ohms in series. Then we have six identical one-ohms, each of which has the same identical current flowing through it, and 1/6 of the voltage across it. So each resistor gets 3V across it, (18/6) and the current through it must be 3v/1ohm = 3A. And it's the same for all six identical resistors.
Hi Lenni,
Yep -- a truly zero-ohm object would imply infinite current as I = V/R which goes to infinity as R approaches zero. Of course, even with superconductors, the capability of a source to provide current is limited due to its own internal resistances, etc. Perfect current sources, along with frictionless surfaces and massless springs, can be found at your local Physics store... (or perhaps Acme, if you're a fan of the old Warner Brother's cartoons) :icon_lol:
When considering series circuits, the current flowing through each of the devices in the series is the same. In a series circuit, there is only one path for the flow of current. The movement of charge (i.e. the current) is the same everywhere throughout a series circuit. Charge doesn't pile-up or accumulate, nor is it used up by the resistors.
Sorry R.G. but I still don't get it.
I already "accepted" that two 1ohm resistors are actually the same than one 2ohm resistor but your description just confuses me even more.
Puh I don't know how else to answer the question.
Quote
If we put a 1 ohm resistor across it, we get I = 18A. If we put a second 1 ohm resistor in series with the first oh, and the series combination across the battery, then the total voltage across two resistors is still the 18V. But since the resistors are identical, then there must be half the voltage across each one, right? So each one-ohm resistor has 9V across it, and we know that I = V/R = 9V/1ohm = 9A. And since what's going through the first resistor can't go anywhere else but through the second resistor, we get that the total current is just the same as the battery current of 9V.
All that I can say to that is just: Why? I read it 10 times now but I don't understand it. 18/2 = 9 yeah that's easy but that resistors suddenly decreasing the voltage but not just drives me crazy. I'll think about it a bit more but I would appreciate if you could try it again.
Thanks so far!
Hi Lenni,
Seems like double-speak, doesn't it? On one hand, we say that resistors limit the flow of current and at the same time, we're saying that current through each device in a series circuit is the same...
huh??
Well, let me try to take a stab at this...
Let's say we have a battery that says "100 Volts" on it... This means that there is a potential difference of 100 volts between the side marked "+" and the side marked "-". If I build an incredibly exciting circuit with this battery and a 100 Ohm resistor, something is going to happen...
Turns out that a long time ago, George Ohm figured out an "easy" explanation which we still call "Ohm's Law."
Our circuit above isn't very exciting, but it qualifies as a circuit. It contains a complete path (from + to -) and there is a potential difference. With these criteria met, an electrical current can flow.
Current is a measurement of how many charged particles fly past a point in a second. We measure this with an Ammeter, and the unit of measurement is the Ampere (1 Coulomb of charge in 1 second).
Sticking this ammeter in our little circuit would show a measurement of 1 ampere. (I = V/R = 100 Volts / 100 Ohms = 1 Ampere).
Okay -- so we've got a circuit that does nothing, right?
Wrong!
We have 1 ampere of electrical current flowing through a resistor. It *is* in fact, doing something. Yeah -- its draining our battery, but in addition to that, it is converting one form of energy to another -- in this case, electrical energy is being converted to HEAT by the resistor. How much heat energy? Well, we've come to learn that P (power) = IV, which for our little circuit is P = 1 Ampere x 100 volts = 100 Watts. Yikes! That resistor is getting HOT! (Ever try to touch a 100Watt light bulb...)
So our "useless" circuit *is* in fact doing something...
But is the resistor "limiting" the current? In the famous words of Sarah Palin, "ya betcha it is."
As you noted, if I = V/R, and R gets very small, I has to get very big... If I reduced my resistor to 10 ohms, I would now have 10 Amperes of current. Drop the resistance to 1 ohm, and I now have 100 Amperes... Drop it to .1 ohms, and I've got 1000 Amperes (welding, anyone?)... All this, of course, assumes that our crazy 100 Volt battery can produce this much current...
So yeah -- the resistor IS limiting the current...
Furthermore, it doesn't matter where in our circuit I connect the ammeter. If I put it between the + side of the battery and the "top side" of the resistor it reads 1 Amp. If I put it between the - side of the battery and the "bottom side" of the resistor, **it still reads 1 Ampere.**
** The same current the flows into the resistor flows out of it. **
Now -- George Ohm got the crazy idea of hooking two 50 Ohm resistors in series with one another. Lo and behold, the ammeter read exactly the same thing as it did with 100 Ohms (1 Amp)... If you were to try it today, 100+ years after George Ohm tried it, it would read 1 Ampere...
So, from where our ammeter sits, there is nothing different about two series 50 Ohm resistors than one 100 Ohm one. And just like before, it doesn't matter where the ammeter is placed, we still show the same 1 ampere.
In fact, George "O" got all crazy and tried connecting the ammeter in series between the two 50 ohm resistors... Still... it read 1 Ampere!!
The reason is that the same current that "comes out of the plus side of the battery" returns to the negative side. Current doesn't leave the series circuit part way through, and no external device is adding extra current somewhere in the middle. We merely have our resistors and our battery.
So -- for our series circuit, the total resistance that the battery "sees" and the voltage of the battery determines the current that will flow through the entire circuit. The "total" current is limited by this resistance (for if it was truly zero, the current would be infinite). If the resistances are in series, the current *has* to be the same through each of them (we're not taking any current away and using it anywhere else, and nothing external is providing us with current).
Hm ok that all makes sense. Especially the part of "What leaves the batterie, goes back in the batterie". Where else should the current go right?!
I guess my actual problem is that I don't understand how the batterie "knows" how much current to send out?! because the Resistor drops the voltage? But how does it drop the voltage?
I draw a little circuit for the whole thing and how it would be logical for me. I know that it obviously has to be wrong since we know that what comes out, comes back in. I guess my question is: How does a resistor do its job and on what? Is it reducing the voltage and so less current is flowing? That was my idea on this example. On the left side of the resistor are many electrons (the little red dots) and on the other side are less so there is obviously a potential difference and on the right the whole voltage is dropped. How does that potential difference really look like? Probably we can figure out my problem with solving this wrong thinking.
(http://lh5.ggpht.com/_JS73FsIOVhw/SVKMAJl6w9I/AAAAAAAACcE/dN8ld3iogbo/s640/Untitled-1%20copy.jpg)
Hi all!
I'm new to this great board, and I started lurking this thread a couple of weeks ago. Thanks to all the infos i found here I managed to build (over my breadboard :) ) the Booster, the OpAmp buffer and the Distortion+ with satisfying results; I then moved on to the Fuzz, even thought I don't like the effect very much, I thought that could be a nice exercise to practice a bit.
Now the problems: the circuit outputs a huge signal that saturates as soon as I lightly touch my guitar's strings...
The only mods I did to the original circuit have been the substitution of the 150k and 18k resistors (R3 and R2) with some resistors in series, because I didn't have the right values at the time.
I then started to debug the circuit, and I found out one strange thing: when I measure the 330k Resistor (R1) out of the circuit, my multimeter reads the correct value, but when I put it in place, it drops to 121K and I don't know why, because I take the measure exactly at its pins... could this have something to do with the fact that it's a 0.5V resistor and not a 0.4V one?
Another thing: I measure 2.60V at the Q1 base and 2V at the emitter... can someone help me please? :)
Here I am again!
I thought it could be useful to show exactly how I wired all the stuff up :) Here's the pic:
(http://img150.imageshack.us/my.php?image=hemmofuzz2rv6.png)
(if the image doesn't open here's the direct link: http://img150.imageshack.us/my.php?image=hemmofuzz2rv6.png (http://img150.imageshack.us/my.php?image=hemmofuzz2rv6.png))
Note: the pinout of the two 2n3904 seem to be E-C-B, front facing the flat side, according to my multimeter... ???
Quote from: Lenni on December 24, 2008, 02:12:39 PM
Hm ok that all makes sense. Especially the part of "What leaves the batterie, goes back in the batterie". Where else should the current go right?!
I guess my actual problem is that I don't understand how the batterie "knows" how much current to send out?! because the Resistor drops the voltage? But how does it drop the voltage?
I draw a little circuit for the whole thing and how it would be logical for me. I know that it obviously has to be wrong since we know that what comes out, comes back in. I guess my question is: How does a resistor do its job and on what? Is it reducing the voltage and so less current is flowing? That was my idea on this example. On the left side of the resistor are many electrons (the little red dots) and on the other side are less so there is obviously a potential difference and on the right the whole voltage is dropped. How does that potential difference really look like? Probably we can figure out my problem with solving this wrong thinking.
(http://lh5.ggpht.com/_JS73FsIOVhw/SVKMAJl6w9I/AAAAAAAACcE/dN8ld3iogbo/s640/Untitled-1%20copy.jpg)
Lenni,
One of the problems with your picture is that you show electrons getting "lost" in the resistor. This isn't happening. All the electrons that leave the battery return to it. Otherwise, our ammeters would show different currents at the positive and negative sides of the circuit. This doesn't happen, so all our electrons are getting back to their home base.
What you're not seeing/showing in that picture is the work that the resistor forces the electrons to do. The chemical composition of the resistor causes a conversion of some of the electrical energy into heat.
Another thing the drawing does not show is that the electrons are being "pushed" by a pressure -- we call it voltage. The resistor, by forcing some work to be done to make heat, causes a pressure drop -- or, as we see it, a voltage drop. A common analogy is that the charged particles are forced through a "narrow" area, and while the same number emerge (current is constant), some the "push" behind them is gone (i.e. voltage drop). Older textbooks referred to voltage as "Electromotive force", and you'll occasionally see Ohm's law represented as E = IR (which is how I learned it, so I'm dating myself again). Resistors cause some of the electromotive force to get used up doing things inside the resistor (heating up, among others). So, for your picture, the same number of dots emerge, they just have a little less "push" behind them. The resistor ate some of the "push" -- not any of the dots.
The battery doesn't really know how much current to send out...
A battery will have certain characteristics that are largely governed by its physical makeup and chemical composition. Depending on these factors, it will have a typical voltage (which, for purposes of illustration we will say is a "fixed" value -- although in reality it is not). Different chemistry will lead to different "fully charged" voltages between the positive and negative sides of a cell. Various other factors, including plate area, determine the battery's ability to provide current to a circuit for a period of time. AAA, AA, C, and D size batteries all show a terminal voltage difference of ~1.5 volts, but their ability to produce a given current for a given period of time varies (larger battery generally = more current for longer time).
The battery attempts to provide the current that the circuit is asking for -- and Ohm's law tells us this figure. The circuit's total "effective" resistance to ground ultimately determines how much current it requires. This is why you'll see designers saying something like, "this circuit draws 20mA of current at 12 Volts". The 12V battery may be able to provide 2,000,000,000 Amperes (a *really* freaking large battery), but our example circuit will draw 20mA from this battery or from a much smaller lantern battery or series-connected AAA batteries.
Hey Lazer,
Can you link the schematic you're talking about?
Even better, please post your question and schematic in a separate topic so it's got an appropriate title and won't either get overlooked at the bottom of this one, or be lost in the archives under a wrong topic.
Perro, R.G: I'll create a new thread with the schematic included! Thanks!
Ok That the resistor isn't eating electrons is logical (how I already posted with the schematic, I know that this way is wrong. I just don't get why ;) ). You said that the resistors just taking some of the "push" means Voltage drop right? But how can there be a voltage drop or potential drop if there is still the same amount of electrons on both sides? Isn't the difference of the amounts of electrons on two sides actually what makes the potential difference?
So resistors are basically lowering the "speed" or "pressure" of the electrons. That's something that I can easily understand, but what does the Voltage drop actually means than? Back to my little schematic up there: electrons coming to the first resistor, lowering the speed to 50% (Voltage drop of 50V from 100V makes 50%, tada that boy can calculate) and than takes the rest of the 50% => 0% = no speed. Do they speed up again because the battery suddenly "pulls" on them or is the speed idea just bullshit?
I was just continuing with the NEETS and suddenly I saw this expression ( 50μA ) without that it had been explained. I was pretty confused what that actually should mean? what is that &mu standing for? Does is mean &mu per Ampere?
I would really appreciate if somebody could tell me what this stands for.
Thank you!
In order to give this thread a continuation...
(http://i207.photobucket.com/albums/bb215/DWBH_bucket/HEAVYDARLING-1.gif)
... what does the marked resistor do? Some sort of 'negative feedback'? :icon_eek:
It does multiple things.
(1) it provides the steady state base bias current.
(2) it stabilizes the DC bias point thus provided by providing negative feedback from the collector. If the collector goes down due to something like the transistor getting hotter and conducting/leaking more, the voltage across the resistor goes down, and so the base current goes down, tending to raise the collector.
(3) it provides AC negative feedback: If an input signal goes up, the collector goes down; the collector going down reduces the current supplied to the base, lessening the effect of the input signal.
This is the so-called voltage feedback transistor bias circuit. It was one of the earlier versions of transistor biasing.
Quote from: Lenni on January 10, 2009, 03:57:00 PM
I was just continuing with the NEETS and suddenly I saw this expression ( 50μA ) without that it had been explained. I was pretty confused what that actually should mean? what is that &mu standing for? Does is mean &mu per Ampere?
I would really appreciate if somebody could tell me what this stands for.
Thank you!
mu (µ) is a greek symbol for "micro", meaning to multiply by 10^-6. It's similar to mega, milli, centi, and other such prefixes, most of which can be found here. (http://en.wikipedia.org/wiki/SI_prefix)
In this case, there would be 50*10^-6 (0.00005) amperes
oh ok. Yeah with a little brainwork I actually should have figured that. But what does the & symbol in there means? Is that just a mistake and a sign which shouldn't be there or is it just me?!
Quote from: Lenni on January 13, 2009, 04:56:09 PM
oh ok. Yeah with a little brainwork I actually should have figured that. But what does the & symbol in there means? Is that just a mistake and a sign which shouldn't be there or is it just me?!
It's a mistake. In HTML, special characters (http://www.tedmontgomery.com/tutorial/htmlchrc.html) are often proceeded by the ampersand (&) sign. The person who wrote this must've typed it wrong.
Quote from: R.G. on January 11, 2009, 07:25:06 PM
It does multiple things.
(1) it provides the steady state base bias current.
(2) it stabilizes the DC bias point thus provided by providing negative feedback from the collector. If the collector goes down due to something like the transistor getting hotter and conducting/leaking more, the voltage across the resistor goes down, and so the base current goes down, tending to raise the collector.
(3) it provides AC negative feedback: If an input signal goes up, the collector goes down; the collector going down reduces the current supplied to the base, lessening the effect of the input signal.
This is the so-called voltage feedback transistor bias circuit. It was one of the earlier versions of transistor biasing.
Isn't that form of biasing a bit unstable RG? Doesn't it vary quite a bit with temperature (even though it is Si). If one put a resistor from base to ground that would be a different method of biasing. Then the biasing and transistor would be more stable. I think the gain would drop a lot though.
That correct? More or less...
Quote from: kurtlives on February 14, 2009, 11:24:33 AM
Isn't that form of biasing a bit unstable RG? Doesn't it vary quite a bit with temperature (even though it is Si). If one put a resistor from base to ground that would be a different method of biasing. Then the biasing and transistor would be more stable. I think the gain would drop a lot though.
That correct? More or less...
All forms of biasing vary with temperature, just some less than others.
The first transistor bias was simply a resistor to V+. Ugly, drifty, especially with the germanium devices they used. Next was the voltage feedback form. The four-resistor "stabilized bias" form was one of the last.
We spent about two years one week in circuits class deriving Stability Factors, those being the sensitivity of the bias point to the temp drift of Vbe and Icbo. It's possible (eventually, especially if you have a grade point average to uphold!) to come up with equations based on circuit analysis that show the change of Ic and Vce with changes in Vbe for any resistor network which biases.
The voltage feedback setup gets more stable as the voltage across the feedback resistor increases and as the voltage across Re increases. It's possible to have it be quite stable, or to be downright skittery by the choices of values.
The principle problem with the voltage feedback setup is that it has a low input impedance.
bump.
this thread should not have been left alone for over 120days. It's got far too much valuable information.
^^ agree ^^ thanks for bumping it mate! i'm really exited to read this, i think Dano pretty much has the title of "Best noob info provider" although there are plenty of other contenders out there. anyway just wanted to say thanks (no need to reply to this), can't wait to devour the info after band practice ;D
Bump this! Nice thread- Much appreciated!
I would like to propose that someone who understands tube amps might help describe the basic functions of components for a fender champ, for instance.
I don't know much myself, but I have been fascinated to read the tube driven aesthetic and various adjectives used to describe soft clipping, asymmetrical distortion, compression, sustain, carbon comp dirt, coupling cap magic and etc, and would love to hear how the stompbox community thinks some of these can be emulated with analogue electronics.
Perhaps I could edit some of these into a sticky...
Thanks in advance- Roger
Hi Dano
I just want to say that the stuff on your site is great. It's helped me a lot.
http://www.beavisaudio.com/
Cheers
Ross
OK. Nice thread. I think this is a good, relevant question.
I wanted to relate the information in this thread to the schematic for the beginner project.
Assuming the beginner project is based on a the common emitter amplifier design (if it is not, then please forgive my stupidity) given by:
(http://t2.gstatic.com/images?q=tbn:ANd9GcSjRXE9rClw6bBkrVUc1J3vqXxjKnA8SR-NF19q03BuNscLBv5wu3VRM_m3)
Which resistors in the beginner project schematic:
(http://www.diystompboxes.com/beginner/schem.gif)
correspond to R1, R2, RL and RE in the common emitter amplifier diagram.
I'm trying to understand what I am building before I build it.
Thanks in advance for any insight.
SJ
im no expert,but heres my take.
RL is the 10k at the collector
RE is the 5k at the emitter
these are two different designs in the way they are biased.
so no R1 R2 in the circuit at the bottom.
the circuit on the top is biased by R1 R2. while the NPN boost by Gus is biased by the network tied to the base which suppose to provide better/higher input impedance (?).
i think i read that here in the forum but till now i still cant really understand it. somebody can chime in?
Thanks Bonaventura. That's what I figures as well.
Really what I want to know more about how this thing is biased with those 3 resistors. It really is sort of a different animal than the first few schematics discussed here.
I'll re-read this this thread to see if I can figure it out but if anyone else knows or has any insight, please don't hesitate to explain.
Thanks,
SJ
Hi. I am hoping that someone could provide a schematic for Hemmo's "Standard Fuzz". I am trying to follow along with Dano's explanation on page one of this thread about how each part works and what is does. Any help would be much appreciated as I am trying hard to learn how a two transistor pedal works.
Thanks
Quote from: bonaventura on June 02, 2012, 03:48:02 AM
these are two different designs in the way they are biased.
so no R1 R2 in the circuit at the bottom.
Not so. This is a game of spot the difference.
You can see that the boost circuit has an extra 22 uF and an extra 10k resistor at the transistors base.
For bias levels (i.e. DC levels) you should mentally remove all capacitors from the circuit (i.e treat them as broken wires) since capacitors block DC.
So that leaves us with the extra 10k resistor at the base. If you lowered the value of the 10k resistor all the way to zero (i.e. turned it into a wire) then you would have the same biasing scheme as the simpler circuit.
Therefore we have R1= 100k and R2 = 47k.
So the voltage at the base is ( 47 /147 ) * 9V = 2.87 V.
The difference is that the simple scheme connects that 2.87V directly to the base with a wire.
Gus' circuit connects it to the base through a "higher resistance wire" i.e. a 10k resistor. The voltage at the base will be pretty much the same.
So why use the 10k instead of a regular wire. My guess is that the 10k was chosen to give a particular low pass filtering at the input. Wired up this way the audio signal at the input sees an RC filter where the C is 0.1 uF and the R has a value of 10k + (100k || 47k) = 10k + 1/(1/100k + 1/47k) = 10k + 32k = 42k.
So the RC product for the filter is 0.1uF * 42k,
There are other ways of wiring things up to get that RC product at the input (e.g. replace the 10k resistor with a wire and use higher values of R1 and R2 instead). So maybe the 10k does something else that I can't think of, but I am pretty sure it does not effect biasing.
I ran this thru LT Spice, just for curiosity...With the 10K in place, you get a shelving filter...actually, HP.
Without it, you get a LP response.
You're correct, the biasing stays essentially the same.