### Author Topic: What does that bit there do? Learning to understand circuits.  (Read 71860 times)

#### dano12

##### What does that bit there do? Learning to understand circuits.
« on: July 26, 2006, 10:29:32 AM »
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.

So, here's the Electro Harmonix LPB-1 Booster: This circuit provides a simple boost with a variable output control.
-LPB Booster

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
- Impedance
- Transistor Bias

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
- High Pass Filter
- Tone Control

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
- Output Impedance

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
- Negative Feedback

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

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
- Output Capacitor

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
- How Transistors Work
« Last Edit: July 26, 2006, 12:10:36 PM by dano12 »

#### KerryF

##### Re: What does that bit there do? Learning to understand circuits.
« Reply #1 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.

#### dano12

##### Re: What does that bit there do? Learning to understand circuits.
« Reply #2 on: July 26, 2006, 10:53:01 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.

#### KerryF

##### Re: What does that bit there do? Learning to understand circuits.
« Reply #3 on: July 26, 2006, 10:59:21 AM »
K thanks.  Any where the C2 should be, it says C1 again.  And I think thats a buffer (C2)

#### MartyMart

##### Re: What does that bit there do? Learning to understand circuits.
« Reply #4 on: July 26, 2006, 11:09:11 AM »
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
"Success is the ability to go from one failure to another with no loss of enthusiasm"
My Website www.martinlister.com

#### R.G.

##### Re: What does that bit there do? Learning to understand circuits.
« Reply #5 on: July 26, 2006, 11:23:32 AM »
Quote
9vDC - 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.

Quote
SW1 - 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.

Quote
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.
http://www.diystompboxes.com/smfforum/index.php?action=search2;search=volume%20control
Yep.

C1: See above.

Quote
Q1 - This is a PNP transistor. It provides the core of the booster.
Actually, it's an NPN transistor.
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 #6 on: July 26, 2006, 12:11:49 PM »
Awesome! Thanks guys--I've updated the original post to merge in the comments/explanations and also provide some more links.

#### Doug_H

• Guest
##### Re: What does that bit there do? Learning to understand circuits.
« Reply #7 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

Doug

#### dano12

##### Re: What does that bit there do? Learning to understand circuits.
« Reply #8 on: July 26, 2006, 12:22:45 PM »
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

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

Doug

#### zpyder

• Great Contributor!
• Posts: 447
• Total likes: 1
• Rc=1/(2pi*f*C) | f = 1/(1.39*R*C) | V=I*R | I=V/R
##### Re: What does that bit there do? Learning to understand circuits.
« Reply #9 on: July 26, 2006, 02:33:30 PM »
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...       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
www.mattrabe.com/ultraterrestrial Ultraterrestrial - Just doing our little part to make new rock go where it should have gone in the late-90's, instead of the bullshit you hear on the radio today.

#### WildMountain

##### Re: What does that bit there do? Learning to understand circuits.
« Reply #10 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.

#### dano12

##### Re: What does that bit there do? Learning to understand circuits.
« Reply #11 on: July 26, 2006, 02:53:31 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.

#### R.G.

##### Re: What does that bit there do? Learning to understand circuits.
« Reply #12 on: July 26, 2006, 03:57:40 PM »
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.
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?

#### newbie builder

##### Re: What does that bit there do? Learning to understand circuits.
« Reply #13 on: July 26, 2006, 04:36:07 PM »
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!
//

#### markm

##### Re: What does that bit there do? Learning to understand circuits.
« Reply #14 on: July 26, 2006, 06:22:08 PM »
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.

#### cakeworks

##### Re: What does that bit there do? Learning to understand circuits.
« Reply #15 on: July 26, 2006, 07:21:19 PM »
:oWHERE HAVE YOU BEEN ALL MY LIFE???... erm... i mean this thread
-Jack

Is that a plastic washing basket?

"Actually a Sterilite-branded storage tub.  Rubbermaid has better mojo, but it cost more" - Phaeton

#### idlechatterbox

##### Re: What does that bit there do? Learning to understand circuits.
« Reply #16 on: July 27, 2006, 09:57:52 AM »
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.

#### dano12

##### Re: What does that bit there do? Learning to understand circuits.
« Reply #17 on: July 27, 2006, 10:17:36 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.

Good feedback. Schematic updated.

#### dano12

##### Re: What does that bit there do? Learning to understand circuits.
« Reply #18 on: July 27, 2006, 03:19:45 PM »
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.

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.
« Last Edit: July 28, 2006, 09:13:37 AM by dano12 »

#### R.G.

##### Circuit 2
« Reply #19 on: July 27, 2006, 04:02:19 PM »
Quote
C3 ?
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.

Quote
Q1 - 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.

Quote
C2 - 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.

Quote
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.
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.
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?