Author Topic: What does that bit there do? Learning to understand circuits.  (Read 75830 times)

John Lyons

Re: What does that bit there do? Learning to understand circuits.
« Reply #20 on: July 27, 2006, 04:03:41 PM »
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

Basic Audio Pedals
www.basicaudio.net/

dano12

Re: Circuit 2
« Reply #21 on: July 27, 2006, 04:53:55 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.


Hiwatt25

  • Guest
Re: What does that bit there do? Learning to understand circuits.
« Reply #22 on: July 27, 2006, 09:23:21 PM »
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. 

dano12

Re: What does that bit there do? Learning to understand circuits.
« Reply #23 on: July 28, 2006, 09:10:04 AM »
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.

R.G.

Re: What does that bit there do? Learning to understand circuits.
« Reply #24 on: July 28, 2006, 10:12:18 AM »
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?
R.G.

Quick IQ Test: If anyone in a governmental position suspected that YOU had top-secret information on YOUR computer, how many minutes would you remain outside a jail cell?

dano12

Re: What does that bit there do? Learning to understand circuits.
« Reply #25 on: July 28, 2006, 10:36:12 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.

dano12

Re: What does that bit there do? Learning to understand circuits.
« Reply #26 on: July 28, 2006, 01:40:46 PM »
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.



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.



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
Buffers: http://www.customaudioelectronics.com/frequently_asked_questions.htm
Loading and Cables: http://howard.davis2.home.att.net/LoadingandCables.htm
Discrete Buffer: 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

That's it for buffers for now. I'm sure folks will have a field day with this one, finding errors and making suggestions :)

idlechatterbox

Re: What does that bit there do? Learning to understand circuits.
« Reply #27 on: July 29, 2006, 09:18:26 AM »
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

comfortably_numb

Re: What does that bit there do? Learning to understand circuits.
« Reply #28 on: July 29, 2006, 09:51:46 AM »
Good thread, I've been looking for something like this.  I'll be watching!

R.G.

buffers and impedance
« Reply #29 on: July 29, 2006, 10:37:43 AM »
Quote
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".
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.

Quote
A 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)

Quote
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.
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.

Quote
It 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.

Quote
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?
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.

R.G.

Quick IQ Test: If anyone in a governmental position suspected that YOU had top-secret information on YOUR computer, how many minutes would you remain outside a jail cell?

disantlor

Re: buffers and impedance
« Reply #30 on: July 29, 2006, 11:28:49 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.

R.G.

Re: What does that bit there do? Learning to understand circuits.
« Reply #31 on: July 29, 2006, 01:08:54 PM »
Quote
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?
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.


Quote
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.
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.

Quote
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?

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.

Quote
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.

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.
R.G.

Quick IQ Test: If anyone in a governmental position suspected that YOU had top-secret information on YOUR computer, how many minutes would you remain outside a jail cell?

disantlor

Re: What does that bit there do? Learning to understand circuits.
« Reply #32 on: July 29, 2006, 01:31:51 PM »
wow... R.G. for the win!!!  a very enlightening read and much appreciated. 

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!!

disantlor

Re: What does that bit there do? Learning to understand circuits.
« Reply #33 on: July 30, 2006, 01:55:40 PM »
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?

hubble

Re: What does that bit there do? Learning to understand circuits.
« Reply #34 on: August 01, 2006, 11:31:23 PM »
hi, what does NC mean on a shem?  if anyone can tell me id be greatful

R.G.

Re: What does that bit there do? Learning to understand circuits.
« Reply #35 on: August 02, 2006, 12:04:56 AM »
Quote
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?
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.

Quote
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?
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.
R.G.

Quick IQ Test: If anyone in a governmental position suspected that YOU had top-secret information on YOUR computer, how many minutes would you remain outside a jail cell?

tcobretti

Re: What does that bit there do? Learning to understand circuits.
« Reply #36 on: August 02, 2006, 01:52:42 AM »
Hubble:

In my experience, "NC" means no connection.  That is likely the only thing I will be able to contribute to this thread.

hubble

Re: What does that bit there do? Learning to understand circuits.
« Reply #37 on: August 02, 2006, 08:31:15 PM »
ok.  cool thanks!

captntasty

Re: What does that bit there do? Learning to understand circuits.
« Reply #38 on: August 02, 2006, 09:05:28 PM »
I so wish I could wrap my head around this stuff.  ???  :icon_confused:
It is no measure of health to be well adjusted to a profoundly sick society. - Jiddu Krishnamurti

dano12

Re: What does that bit there do? Learning to understand circuits.
« Reply #39 on: August 02, 2006, 09:11:56 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.