GGG micro V problem

[movinginslomo]

First is the pedal does not function correctly. When first powered on the effects works, but after a minute or less the effect dissapears. Could either of the ICs be bad? It was built stock from stock form a kit using supplied parts, minus one 10uf elctrolytic cap that was not in the kit, there was however a 47uf cap. I decided to stick with the build document and use the 10uf which I happened to have. Any help appreciated, I'm completely lost at debugging effects.

[movinginslomo]

oh also to save time, wiring is solid, no solder bridges, prefab double-sided PCB

[Luke51411]

oh also to save time, wiring is solid, no solder bridges, prefab double-sided PCB

Many times I've been sure of this only to find I forgot to solder an ic pin or had a cold joint or lifted trace or something. Do you have a multimeter? Take voltages at the ic and or transistor pins and post them here. Pictures of both sides of the board and a link to the build doc usually prove helpful.

[Govmnt_Lacky]

First... Welcome.

OK.... Lets get to the good stuff. First of all, do you own a multimeter? If not... get one. You will use it ALWAYS.

In order to get this problem solved (I read your thread on DIYSB also) we will need some basic information and most of it can be acquired with a multimeter so if you do not have one... refer to above. If you do not know how to use one, you will need to read up on it OR it will be a long process unfortunately.

We are going to need voltage measurements on your circuit at various points. Did you build this to work off of a battery? A wall power supply? Both?

Can you post up some pictures of the PCB front and back. Instructions on how to post pictures are on the sticky here...

http://www.madbeanpedals.com/forum/index.php?topic=3823.0

There is a great instruction for posting the right info over at DIYSB (might be here but I am lazy right now) and it is located as a sticky at the top of the "Building you own stompbox" thread labeled "Debugging.. What to do when it doesn't work"

Hope all this helps. Post away when you acquire the info needed.

[midwayfair]

[Moved here from other thread]

You're going to need to take voltages on all pins of all active devices in the circuit. You're looking for about 4.5V at the emitters of Q1 and Q2, and pins 1, 2, 3, 5, 6, 7 of the 4558 and pins 3, 4, 5, 12, 13, & 14 of the LM13700; you should have 0v on pin 4 of the 4558 and pin 6 of the LM13700 and +9V on pin 8 of the 4558 and pin 11 of the LM13700.

You can audio probe the base and emitter of Q1, pin 1 of the 4558, and pin 5 or 12 of the LM13700 (I'm not sure which one it actually uses). You can also probe pin 7 of the 4558, and you should also be able to hear something at the base of Q3.

I can't read schematics

Here's a great tutorial:
http://www.beavisaudio.com/techpages/SchematicToReality/

You'll have to be able to read a schematic at least a little to effectively troubleshoot this pedal or any other pedal, so let's break down the Mutron V.

Schematic
http://www.generalguitargadgets.com/pdf/ggg_microv_sc.pdf

The easiest way to break down a schematic is to locate a few starting points.
1) Where's your audio input?
2) Where's your audio output?
3) What are your voltage sources?

And some very basic knowledge of components:
1) What are the places where you will hear audio if you audio probe? For chips, you can look up their datasheet -- they'll tell you if something's an audio pin or a power pin.
2) What are working voltages? This is complicated and varies by pedals, but in a 9V circuit, 0 and 9V won't make any noise at all, and 4.5V is right in the middle and most of your audio signals will run through pins that are closer to 4.5V than 0V or 9V in general -- for op amps, most audio pins just will be 4.5V naturally.

The Micro V is more complicated than the LPB-1 in Dano's tutorial, but the easiest way to break down anything is to take it one piece at a time. So let's look at the components. Active components are transistors and chips -- stuff that requires power to do anything and adds to your signal. Passive components require no power and impede the flow of signal in some way.

Input is labeled input. That's helpful. Let's stick with the audio path for now. We can tackle the power later.

We can follow the input signal (ignore the little bendy line) to C1. That's our input capacitor. This prevents DC from the active components from getting into our pickups. Pickups don't like DC -- they make an AC signal instead. Capacitors BLOCK frequencies that are lower than a circuit amount. More capacitance means they pass a lower frequency. Pure DC is 0Hz -- that is, an infinitely low frequency. No capacitor in the world is big enough to actually allow that frequency to pass, so any capacitor will block all DC from passing. AC signal is always higher, and our audio signal is AC, so the capacitor allows our pickup's signal to pass. Hooray! We're off to the races.

Next we come to R1, C1, R2, C12. These do a couple things.

This is our first resistor. Resistors impede signal of all frequencies. They allow both DC and AC to pass.

Resistances are commonly evaluated in pairs -- a voltage divider. PRR has said that "everything useful is a voltage divider" which was an absolutely eye-opening realization for me. Resistors in parallel DIVIDE the signal, just like they DIVIDE their resistance in parallel.



The most common voltage divider is a volume pot.


We don't have a volume pot in this circuit, but we have plenty of voltage dividers. Our first one is created when R1 and R2 meet. We have 100K dividing against 100K. R2 is impeding signal to ground (okay, "Vref" actually, but it's a spot that doesn't make any noise, so let's call it ground for now. I'll do my best to explain it later.). It's called a load in this case, and the signal going through R1 is going to try to "drive" it. We plug that into the above formula (or, FFS, let's be realistic, normally I use a calculator for anything that's not equal values ...) and we get 50K. That's half. R1 wants to try to drive the signal past R2 but R2 is so heavy that half of your signal just decides to fuck off to ground. We can now deduce that this voltage divider will reduce our input signal to half. (Why? We'll have to find out later ...) What would have happened if R1 was 50K and R2 was 150K? Plug it into the calculator and see, and try to visualize it.

The input also contains another very common little circuit: An RC filter. Two of them, actually. That's resistor - capacitor. These are your most common, simple filters.

A resistor coming in front of a capacitor to ground will form a LOW PASS filter. You have one of those in your guitar -- your tone knob. Here it's R1 -- 100K -- and C12 -- 100pF. If I plug that into a calculator (or do it by hand -- 2piRC) (http://sim.okawa-denshi.jp/en/CRlowkeisan.htm) -- I get 15,915Hz. That's really high, not really too much guitar stuff up there. (It's there to filter out some noise. You'll see that a LOT in circuits. Go open up 10 overdrives and you'll probably see something almost exactly like that in 5 of them.)

C1 and R2 form a HIGH PASS filter. This is, quite literally, the most common filter found in pedals. It's in every single pedal in some manner, in two important places: The input and output. A high pass filter cuts bass. (Passes highs.) If we plug C1 and R2's values into a calculator (http://sim.okawa-denshi.jp/en/CRhikeisan.htm) we'll see that its cutoff is about 16Hz. That's below the range of human hearing, and way below the range of the guitar, so it's passing all our bass. There's another important one at the output of the effect -- C7 and R7. You can plug those into the calculator if you want.

Finally we reach our first active component. Q1. This is a transistor. Transistor circuits are actually a sneaky form of voltage divider -- your input comes in at the base (the middle pin on most), and then it modulates the supply voltage of the entire circuit (9V) to produce an amplified copy of the signal. You can amplify the voltage of the original signal (make it LOUDER) by taking the output from the collector or amplify the current by taking the output from the emitter. You then use resistors at the collector and emitter to limit  Let's look more closely at Q1.

Remember how I said that 9V is one of those places in the circuit that you can't hear anything? Well, the collector of Q1 is connected right to the 9V. The schematic doesn't really go anywhere at that point. We can't go any further and we can't hear anything there (go ahead and listen with your audio probe - nuffin), so we can't be amplifying the voltage. Let's look at the emitter -- the schematic continues from there. We also have a resistor, R3, which keeps our signal from just going straight to ground. (And more importantly, you need some resistance there to keep the transistor from just conducting all of +9V through it and burning up the transistor. There has to be some resistance at at least the collector or emitter. Here it's just at the emitter, sometimes it's both, sometimes it's just the collector.)

This is called a BUFFER. Why use a buffer? This makes a very ROBUST signal that isn't louder. Your pickups aren't very good at making a strong signal. That's why we amplify them -- they can't make a speaker make noise. In this specific pedal's case, it's also because we're going to make the signal do two different things.

The circuit gets a little more complicated here, so let's continue tackling the audio first. We're looking to get to our end point, which is the output. It's a mostly straight line by heading toward those little triangle things.

Audio path
Your audio signal continues through C2 and gets picked up at the + pin of the LM13700 chip. If we open our datasheet and look at what the pins mean in the LM13700, we can see that this is an audio pin, a "non-inverting" pin. "Inverting" refers to the audio phase of the signal. Here the signal is in phase. And we can also see that the signal "comes out" at the spot labeled 5, 12 (check the datasheet again ... audio outputs) and then continues on ...

But wait. C3 ALSO is picked up in the same spot. And that one goes to the same spot C2's signal went through! The signals are in phase (remember, non-inverting) and then they head to pin 2 of a simple dual op amp, which our datasheet says is an INVERTING pin. The signals mix together -- in-phase signals don't cancel. (Out of phase signals cancel.) But the phase of both signals is inverted by the time it gets here.

But wait, there's more. C4 and R5 ALSO are picked up near the same spot as C2 and C3 were. And this one ... heads to the output of our mixed signals. This C4/R5 signal is out of phase. This cancels some signal at certain frequencies. This would become way too long to be useful if I went into micro detail on every filter in the pedal so let's leave it at that for now.

But wait, there's more. Our signal is going to a fourth place. Somewhere that doesn't go toward the output. C9 and R10 head toward the really special part of the pedal.

The Envelope Detector
An envelope detector converts an AC signal (that's our guitar signal!) into a DC voltage. Why would you want to do that? Because lots of components function on DC voltages. Envelopes are super fun. You can make a transistor alter the resistance between two outside pins by applying a DC voltage to its center pin. You can light up an LED. This pedal is using the transistor method to vary the resistance between pins 1 and 16 ("amp bias input" according to the datasheet) and the bias voltage of the circuit.

[interlude: The bias voltage

There's an important spot in any circuit that uses op amps called the "Vb" or "bias voltage" or sometimes just "4.5V" (in a 9V circuit). It's the spot where the voltage is HALFWAY between the supply voltage (9V) and the lowest voltage (usually ground -- or it could be a negative voltage). Every op amp needs a spot where that voltage happens to reference, to make the chip happy and functional.

In many pedals it's just two resistors, one connected to +9V, the other to ground, of equal or approximately equal values.

Here it's more complicated. If we scour the schematic, we don't see any spot where there are two equal resistors. But there is a spot that has nothing to do with the audio path, is connected only to +9V and ground, and where two resistors connected to those two voltages meet: the center pin of Q2. And if you look at it, you'll see it's another buffer like Q1 was -- the collector is connected to +9V, and there's a resistor between the emitter and ground. What's going on here? It's making a buffered DC voltage. There's a little flag on its emitter titled "Vb". We see this flag all over the circuit, meaning all those points are connected together. But this is where it originates.

That voltage doesn't have any audio. It's silent. Our guitar signal is AC voltage, waffling around that 4.5V point. That's really important! It's the CENTER of our audio signal. If we send all our audio signal to that center point, it has no amplitude, which means it's also silent! So that goes back to why I said talking about R2 that we can pretend "Vb" is a ground point. You might see it referred to as "virtual ground" or "AC ground" or "audio ground."

It's buffered because that makes it very strong, just like the buffer for our audio signal. Lots of things in the circuit are going to want to use it, so we want to make sure there's enough current for them all to do so. There wouldn't be if it were just resistors. That's too complicated for this post, so you'll have to take it as read.

Now, your op amp wants to reference that voltage to amplify the signal. By changing how much it's "allowed" to access that voltage you can change how much it is allowed to amplify the signal, or how much of your signal is "lost" to ground.

End interlude]

IC1B is a full wave rectifier. This is a pretty simple one involving diodes. Diodes are a semiconductor, like ICs and Transistors, but much simpler. They have two points, a + (anode) end and - (cathode) end. If you put positive voltage on the + end, it will conduct to the - end. If you put negative voltage on the - end, it'll go to the + end. If you put negative voltage on the + end, it won't go anywhere.

When your signal gets to pin 6 ("inverting input" according to the 4558's datasheet), any time negative-going voltage appears on the - side of D1, it's allowed to continue on in the circuit. Any time positive-going voltage appears in the circuit, it goes through the op amp. It gets inverted (again, "inverting input") and comes out where it meets D2. Since it's inverted, what was once a positive voltage is now a negative-going voltage and can pass through D2. And since they're both negative voltages now, they don't cancel out.

This is our rectified voltage. I'm going to call it a "control voltage."

By the way, you can listen to that. No, really, listen to it -- you can audio probe the anode of D2. It'll sound like an octave fuzz. That's one way octave effects are created with analog signals. Neat, huh?

R12 and R13 are part of a sneaky voltage divider -- they are actually a gain control for the inverting op amp stage. Resistors in series add their values. When you turn the sensitivity pot, the resistance goes up or down. The inverting op amp stage's volume is calculated by the total value of R12+R13 divided by R10. It's never less than a little over 1x, and it gets over 30x stronger.

[Remember how we cut our input signal by half? This is actually part of the reason. This means that at minimum sensitivity, the pedal can read the incoming guitar signal as being weaker than it was originally, without needing to do weird things in this place. It's just one way to design the pedal, and it helps lessen distortion in other places. Anyway.]

Our guitar signal isn't a constant source of voltage. For that reason, envelopes work by averaging the signal and smoothing it out by storing some voltage on a capacitor.

Our voltage right now is negative, so we're going to need the - side of a capacitor to store it. Moving along we find C11, which looks funny -- its positive side is connected to GROUND. Ground is usually the lowest voltage in the circuit. What the hell? Well, it turns out that ground is often more positive than the control voltage.

R15 allows some DC to drain off at a constant rate. You can actually calculate the time if you want -- this is another LC filter, and it uses a similar calculation.

At last we come to Q3. This is the transistor we're going to wiggle to change the resistance between its pins. The collector is connected to our bias pins of the LM13700. The emitter is connected, through a small resistor, to the reference voltage we created at Q2. By applying a negative voltage at the base of the transistor, we're going to change how much gain the LM13700 is allowed to make.

Putting it all together
When the gain of the LM13700 changes, it also changes the frequency range of the signal that went in at pins 3/14 of the LM13700. This is what creates the filter sweep.

In the end, it's actually fairly simple ...

Hopefully this will be useful to you while troubleshooting the pedal.

[movinginslomo]

oh also to save time, wiring is solid, no solder bridges, prefab double-sided PCB

Many times I've been sure of this only to find I forgot to solder an ic pin or had a cold joint or lifted trace or something. Do you have a multimeter? Take voltages at the ic and or transistor pins and post them here. Pictures of both sides of the board and a link to the build doc usually prove helpful.

actually I did forget to solder some IC pins on the socket then realized and removed the chip and soldered the errant pins, would that cause issues?

[jimilee]

I've been building for awhile now, very confident of my skills, most fire right up, no issues. Four builds back or so, apparently, after pulling my hair out, I re flowed everything and that fixed my issue, just saying, if you haven't tried it, it'd be worth a shot. Other thank that, how long does it take before it starts working again?


Sent from my iPhone using Tapatalk

[movinginslomo]

I've been building for awhile now, very confident of my skills, most fire right up, no issues. Four builds back or so, apparently, after pulling my hair out, I re flowed everything and that fixed my issue, just saying, if you haven't tried it, it'd be worth a shot. Other thank that, how long does it take before it starts working again?


Sent from my iPhone using Tapatalk

I doesn't start working again lol. It basically stays as some sort of bass filter only allow bass frequencies through. When I first tested I was like "wow this filter is not being triggered at all"

[movinginslomo]

Q1 is 1.37v
Q2 1.96v
Initially values for the IC's were at 4.5v but after 30 seconds or so of being left on:
pins 1,2,3,5,6,7 of the RC455D are at 2.1v (?!)
pins 3, 4, 5, 12, 13, & 14 of the LM13700 also hover around 2.2v
The pins that should be +9v are reading nothing

[lincolnic]

I have no advice here that hasn't already been given, but I just want to say...Jon Patton, you are a goddamn treasure.

[Luke51411]

I have no advice here that hasn't already been given, but I just want to say...Jon Patton, you are a goddamn treasure.

This.
Part of the reason I read debug threads, always something new to be learned.

[nzCdog]

[Moved here from other thread]

You're going to need to take voltages on all pins of all active devices in the circuit. You're looking for about 4.5V at the emitters of Q1 and Q2, and pins 1, 2, 3, 5, 6, 7 of the 4558 and pins 3, 4, 5, 12, 13, & 14 of the LM13700; you should have 0v on pin 4 of the 4558 and pin 6 of the LM13700 and +9V on pin 8 of the 4558 and pin 11 of the LM13700.

You can audio probe the base and emitter of Q1, pin 1 of the 4558, and pin 5 or 12 of the LM13700 (I'm not sure which one it actually uses). You can also probe pin 7 of the 4558, and you should also be able to hear something at the base of Q3.

I can't read schematics

Here's a great tutorial:
http://www.beavisaudio.com/techpages/SchematicToReality/

You'll have to be able to read a schematic at least a little to effectively troubleshoot this pedal or any other pedal, so let's break down the Mutron V.

Schematic
http://www.generalguitargadgets.com/pdf/ggg_microv_sc.pdf

The easiest way to break down a schematic is to locate a few starting points.
1) Where's your audio input?
2) Where's your audio output?
3) What are your voltage sources?

And some very basic knowledge of components:
1) What are the places where you will hear audio if you audio probe? For chips, you can look up their datasheet -- they'll tell you if something's an audio pin or a power pin.
2) What are working voltages? This is complicated and varies by pedals, but in a 9V circuit, 0 and 9V won't make any noise at all, and 4.5V is right in the middle and most of your audio signals will run through pins that are closer to 4.5V than 0V or 9V in general -- for op amps, most audio pins just will be 4.5V naturally.

The Micro V is more complicated than the LPB-1 in Dano's tutorial, but the easiest way to break down anything is to take it one piece at a time. So let's look at the components. Active components are transistors and chips -- stuff that requires power to do anything and adds to your signal. Passive components require no power and impede the flow of signal in some way.

Input is labeled input. That's helpful. Let's stick with the audio path for now. We can tackle the power later.

We can follow the input signal (ignore the little bendy line) to C1. That's our input capacitor. This prevents DC from the active components from getting into our pickups. Pickups don't like DC -- they make an AC signal instead. Capacitors BLOCK frequencies that are lower than a circuit amount. More capacitance means they pass a lower frequency. Pure DC is 0Hz -- that is, an infinitely low frequency. No capacitor in the world is big enough to actually allow that frequency to pass, so any capacitor will block all DC from passing. AC signal is always higher, and our audio signal is AC, so the capacitor allows our pickup's signal to pass. Hooray! We're off to the races.

Next we come to R1, C1, R2, C12. These do a couple things.

This is our first resistor. Resistors impede signal of all frequencies. They allow both DC and AC to pass.

Resistances are commonly evaluated in pairs -- a voltage divider. PRR has said that "everything useful is a voltage divider" which was an absolutely eye-opening realization for me. Resistors in parallel DIVIDE the signal, just like they DIVIDE their resistance in parallel.



The most common voltage divider is a volume pot.


We don't have a volume pot in this circuit, but we have plenty of voltage dividers. Our first one is created when R1 and R2 meet. We have 100K dividing against 100K. R2 is impeding signal to ground (okay, "Vref" actually, but it's a spot that doesn't make any noise, so let's call it ground for now. I'll do my best to explain it later.). It's called a load in this case, and the signal going through R1 is going to try to "drive" it. We plug that into the above formula (or, FFS, let's be realistic, normally I use a calculator for anything that's not equal values ...) and we get 50K. That's half. R1 wants to try to drive the signal past R2 but R2 is so heavy that half of your signal just decides to fuck off to ground. We can now deduce that this voltage divider will reduce our input signal to half. (Why? We'll have to find out later ...) What would have happened if R1 was 50K and R2 was 150K? Plug it into the calculator and see, and try to visualize it.

The input also contains another very common little circuit: An RC filter. Two of them, actually. That's resistor - capacitor. These are your most common, simple filters.

A resistor coming in front of a capacitor to ground will form a LOW PASS filter. You have one of those in your guitar -- your tone knob. Here it's R1 -- 100K -- and C12 -- 100pF. If I plug that into a calculator (or do it by hand -- 2piRC) (http://sim.okawa-denshi.jp/en/CRlowkeisan.htm) -- I get 15,915Hz. That's really high, not really too much guitar stuff up there. (It's there to filter out some noise. You'll see that a LOT in circuits. Go open up 10 overdrives and you'll probably see something almost exactly like that in 5 of them.)

C1 and R2 form a HIGH PASS filter. This is, quite literally, the most common filter found in pedals. It's in every single pedal in some manner, in two important places: The input and output. A high pass filter cuts bass. (Passes highs.) If we plug C1 and R2's values into a calculator (http://sim.okawa-denshi.jp/en/CRhikeisan.htm) we'll see that its cutoff is about 16Hz. That's below the range of human hearing, and way below the range of the guitar, so it's passing all our bass. There's another important one at the output of the effect -- C7 and R7. You can plug those into the calculator if you want.

Finally we reach our first active component. Q1. This is a transistor. Transistor circuits are actually a sneaky form of voltage divider -- your input comes in at the base (the middle pin on most), and then it modulates the supply voltage of the entire circuit (9V) to produce an amplified copy of the signal. You can amplify the voltage of the original signal (make it LOUDER) by taking the output from the collector or amplify the current by taking the output from the emitter. You then use resistors at the collector and emitter to limit  Let's look more closely at Q1.

Remember how I said that 9V is one of those places in the circuit that you can't hear anything? Well, the collector of Q1 is connected right to the 9V. The schematic doesn't really go anywhere at that point. We can't go any further and we can't hear anything there (go ahead and listen with your audio probe - nuffin), so we can't be amplifying the voltage. Let's look at the emitter -- the schematic continues from there. We also have a resistor, R3, which keeps our signal from just going straight to ground. (And more importantly, you need some resistance there to keep the transistor from just conducting all of +9V through it and burning up the transistor. There has to be some resistance at at least the collector or emitter. Here it's just at the emitter, sometimes it's both, sometimes it's just the collector.)

This is called a BUFFER. Why use a buffer? This makes a very ROBUST signal that isn't louder. Your pickups aren't very good at making a strong signal. That's why we amplify them -- they can't make a speaker make noise. In this specific pedal's case, it's also because we're going to make the signal do two different things.

The circuit gets a little more complicated here, so let's continue tackling the audio first. We're looking to get to our end point, which is the output. It's a mostly straight line by heading toward those little triangle things.

Audio path
Your audio signal continues through C2 and gets picked up at the + pin of the LM13700 chip. If we open our datasheet and look at what the pins mean in the LM13700, we can see that this is an audio pin, a "non-inverting" pin. "Inverting" refers to the audio phase of the signal. Here the signal is in phase. And we can also see that the signal "comes out" at the spot labeled 5, 12 (check the datasheet again ... audio outputs) and then continues on ...

But wait. C3 ALSO is picked up in the same spot. And that one goes to the same spot C2's signal went through! The signals are in phase (remember, non-inverting) and then they head to pin 2 of a simple dual op amp, which our datasheet says is an INVERTING pin. The signals mix together -- in-phase signals don't cancel. (Out of phase signals cancel.) But the phase of both signals is inverted by the time it gets here.

But wait, there's more. C4 and R5 ALSO are picked up near the same spot as C2 and C3 were. And this one ... heads to the output of our mixed signals. This C4/R5 signal is out of phase. This cancels some signal at certain frequencies. This would become way too long to be useful if I went into micro detail on every filter in the pedal so let's leave it at that for now.

But wait, there's more. Our signal is going to a fourth place. Somewhere that doesn't go toward the output. C9 and R10 head toward the really special part of the pedal.

The Envelope Detector
An envelope detector converts an AC signal (that's our guitar signal!) into a DC voltage. Why would you want to do that? Because lots of components function on DC voltages. Envelopes are super fun. You can make a transistor alter the resistance between two outside pins by applying a DC voltage to its center pin. You can light up an LED. This pedal is using the transistor method to vary the resistance between pins 1 and 16 ("amp bias input" according to the datasheet) and the bias voltage of the circuit.

[interlude: The bias voltage

There's an important spot in any circuit that uses op amps called the "Vb" or "bias voltage" or sometimes just "4.5V" (in a 9V circuit). It's the spot where the voltage is HALFWAY between the supply voltage (9V) and the lowest voltage (usually ground -- or it could be a negative voltage). Every op amp needs a spot where that voltage happens to reference, to make the chip happy and functional.

In many pedals it's just two resistors, one connected to +9V, the other to ground, of equal or approximately equal values.

Here it's more complicated. If we scour the schematic, we don't see any spot where there are two equal resistors. But there is a spot that has nothing to do with the audio path, is connected only to +9V and ground, and where two resistors connected to those two voltages meet: the center pin of Q2. And if you look at it, you'll see it's another buffer like Q1 was -- the collector is connected to +9V, and there's a resistor between the emitter and ground. What's going on here? It's making a buffered DC voltage. There's a little flag on its emitter titled "Vb". We see this flag all over the circuit, meaning all those points are connected together. But this is where it originates.

That voltage doesn't have any audio. It's silent. Our guitar signal is AC voltage, waffling around that 4.5V point. That's really important! It's the CENTER of our audio signal. If we send all our audio signal to that center point, it has no amplitude, which means it's also silent! So that goes back to why I said talking about R2 that we can pretend "Vb" is a ground point. You might see it referred to as "virtual ground" or "AC ground" or "audio ground."

It's buffered because that makes it very strong, just like the buffer for our audio signal. Lots of things in the circuit are going to want to use it, so we want to make sure there's enough current for them all to do so. There wouldn't be if it were just resistors. That's too complicated for this post, so you'll have to take it as read.

Now, your op amp wants to reference that voltage to amplify the signal. By changing how much it's "allowed" to access that voltage you can change how much it is allowed to amplify the signal, or how much of your signal is "lost" to ground.

End interlude]

IC1B is a full wave rectifier. This is a pretty simple one involving diodes. Diodes are a semiconductor, like ICs and Transistors, but much simpler. They have two points, a + (anode) end and - (cathode) end. If you put positive voltage on the + end, it will conduct to the - end. If you put negative voltage on the - end, it'll go to the + end. If you put negative voltage on the + end, it won't go anywhere.

When your signal gets to pin 6 ("inverting input" according to the 4558's datasheet), any time negative-going voltage appears on the - side of D1, it's allowed to continue on in the circuit. Any time positive-going voltage appears in the circuit, it goes through the op amp. It gets inverted (again, "inverting input") and comes out where it meets D2. Since it's inverted, what was once a positive voltage is now a negative-going voltage and can pass through D2. And since they're both negative voltages now, they don't cancel out.

This is our rectified voltage. I'm going to call it a "control voltage."

By the way, you can listen to that. No, really, listen to it -- you can audio probe the anode of D2. It'll sound like an octave fuzz. That's one way octave effects are created with analog signals. Neat, huh?

R12 and R13 are part of a sneaky voltage divider -- they are actually a gain control for the inverting op amp stage. Resistors in series add their values. When you turn the sensitivity pot, the resistance goes up or down. The inverting op amp stage's volume is calculated by the total value of R12+R13 divided by R10. It's never less than a little over 1x, and it gets over 30x stronger.

[Remember how we cut our input signal by half? This is actually part of the reason. This means that at minimum sensitivity, the pedal can read the incoming guitar signal as being weaker than it was originally, without needing to do weird things in this place. It's just one way to design the pedal, and it helps lessen distortion in other places. Anyway.]

Our guitar signal isn't a constant source of voltage. For that reason, envelopes work by averaging the signal and smoothing it out by storing some voltage on a capacitor.

Our voltage right now is negative, so we're going to need the - side of a capacitor to store it. Moving along we find C11, which looks funny -- its positive side is connected to GROUND. Ground is usually the lowest voltage in the circuit. What the hell? Well, it turns out that ground is often more positive than the control voltage.

R15 allows some DC to drain off at a constant rate. You can actually calculate the time if you want -- this is another LC filter, and it uses a similar calculation.

At last we come to Q3. This is the transistor we're going to wiggle to change the resistance between its pins. The collector is connected to our bias pins of the LM13700. The emitter is connected, through a small resistor, to the reference voltage we created at Q2. By applying a negative voltage at the base of the transistor, we're going to change how much gain the LM13700 is allowed to make.

Putting it all together
When the gain of the LM13700 changes, it also changes the frequency range of the signal that went in at pins 3/14 of the LM13700. This is what creates the filter sweep.

In the end, it's actually fairly simple ...

Hopefully this will be useful to you while troubleshooting the pedal.

your support posts are epic Jon... just sayin

[midwayfair]

FYI, I was wrong on a couple voltages, in part because I missed that the zener was 8.2v and not 9.1V. There's a full list of what each pin should be in here:
http://www.generalguitargadgets.com/pdf/ggg_microv_instruct.pdf

Q1 is 1.37v
Q2 1.96v

Need some clarification -- which pins are these? If you're not sure, look up the datasheet of the transistor you used in those spots. We need the voltages on all three pins. Write them out like this:
C - x.xx
B - x.xx
E - x.xx

(That's collector, base, and emitter.)

Initially values for the IC's were at 4.5v but after 30 seconds or so of being left on:
pins 1,2,3,5,6,7 of the RC455D are at 2.1v (?!) ... pins 3, 4, 5, 12, 13, & 14 of the LM13700 also hover around 2.2v

(The 2.1V is actually really close to your Q2 voltage, though the voltages on the 4558 wouldn't normally be HIGHER than the half reference voltage the chip is referencing! That's weird but I'll worry about it later.)

Are you using a battery to power this? I ask because this could actually make sense if your battery were very drained, something like 4V total. If somehow the supply voltage is getting pulled down that far, it would have the same effect: You'd be near the minimum operating voltages of the chips; you would still get sound, but it would be EXTREMELY weak; the envelope wouldn't trigger -- which would leave the sound "stuck" in the low-frequency position (and thus sound like a "bass filter"), really it would describe all of the problems you're having.

But you said that you initially had 4.5V and then it got pulled down. That means something's pulling your power supply down. There aren't a ton of things that will do that in this pedal, but I'll give some suggestions in a minute.

The pins that should be +9v are reading nothing

I'm not sure what you mean by "nothing." Are they measuring 0V? Or is your multimeter reading "1" or not saying anything at all? (Make sure its range is set to 20V and not 2V.) Make sure your black probe is on a ground point before measuring any voltage point. You must be getting DC voltage to the effect because you have at least 2V elsewhere.

Okay, here are the components to check to make sure that your 9V supply is okay:
D3 - the 1N914. Double check its orientation. Make sure it's in the right spot on the pedal. Make sure you didn't mix it up with the Zener or one of the resistors.
D4 - make sure it's not burned up or shorted. You can safely remove this component entirely, actually - the reason is kind of complicated, but if the pedal makes a sudden current requirement of your DC adapter, this could destroy a zener diode. The envelope does just that, so it's actually a pretty good idea to remove this. Unless you're using an unregulated power supply, it won't have any effect.
C13 -- check that it's not backwards. (You'd probably know right away if it was -- it would be hot to the touch.)

If you used sockets, you can remove all the chips and transistors while you check those parts. you want to make sure you're getting 9V.

Your power supply -- make sure you're using a decent power supply that can give you at least 50mA of current. It would be unusual to be using something else, but this will give you plenty of current, way more than the pedal needs, just to ensure that it's not a power supply issue.

After you are sure your 9V supply is working, move onto the Vref. Go ahead and put Q2 back in and see if you're getting ~4.7V at the emitter. If you aren't, you'll need to check every component that's got a Vref flag on it in the schematic. Reflow solder joints etc.

After you've gotten Vref working, put Q1 back in and audio probe the base and collector. Check the voltages against the schematic. If it's good, go ahead and put the chips in one at a time. If you get a sudden change in voltages, or lose the audio signal at an output pin, check the components connected to the chips.

Pictures may help after you recheck the voltages and the above.

[movinginslomo]

Ah D3. The 1N914 looked suspect so I replaced it, the pedals seems to be working.. keeping it on with fingers crossed...

...been a minute still working fine...


..two minutes.. could this be the fix?

[movinginslomo]

K back to probing. Same problems. when used with a power adapter it runs fine, but with much crackling background noise.

Checking Q1 and Q2