Debugging IC based OD ::schematic added::

[comfortably_numb]

I've been working on this design for a while, put it on the shelf, and recently came back to it.  I've designed my own PCB layout, which could easily be the problem, so someone please trace that for me first to make sure it looks good. 



Edit for schematic


I'm getting no sound and no LED function either. 

Battery tested at 8.0 V
Battery connected to circuit tested at 0.84 V

IC pin 1 - 2.1 mV
2 - 1.0 mV
3 - 2.0 mV
4 - 2.0 mV
5 - 2.0 mV
6 - 25 mV
7 - 2.0 mV
8 - 1.1 mV

After I tested these points I disconnected and tested the very warm battery at 6.5 V.

What gives?

[petemoore]

  Find the short in the power supply. 
  The diodes 'line' end is toward V+ ?

[JKowalski]

I think it's funny - 90% of my builds, literally, always have one single problem when I put them together. That problem is a microscopic hairline trace between V+ and GND. I don't know how! But it happens EVERY TIME. I feel for you.   

And the odd thing is, that's the only type of trace problem I have ever had on any of my pedals. And it's the worst one possible.

[comfortably_numb]

D1 isn't the issue and neither is the LED.  I ran a knife around all traces with no change.  I don't see any solder bridges anywhere, nor any touching leads.  Does anyone have any other ideas?  Is the layout correct?

Thanks all

[doitle]

Did you wire this true bypass? If so I'd wager it is wired wrong. 9V through a LED through a resistor to ground is pretty fool proof. If you are sure that your LED is facing the right way and it's not lighting up then the switch is hooked up wrong. That'd be where I'd start looking.

[sed]

Maybe you have wrong switch orientation? Is the switch PCB mounted? If so try to turn it 90 deg. If it's oriented the wrong way you will get some weird behavior imho.

Daniel

[JKowalski]

Your battery is in backward...

[frokost]

D1 is oriented the wrong way in the layout.

[R O Tiree]

... he's right, you know...

[valdiorn]

schematic looks good, but I would change out the 1Meg resistors for the voltage reference for something smaller, like 47k. Will give you a more stable reference.

[comfortably_numb]

Well there you have it then.  D1 is in correct according to the layout, the layout is just wrong.  Thank you guys for scouring this so quickly for me!  That's probably what Pete meant in the first place, I just didn't catch it.  Valdiorn, why does a lower resistance provide a more stable reference? 

Also, the switch is PCB mounted and it WAS initially installed 90 degrees out.  I realized this pretty quickly and changed it.  I do have a bypass signal, just no effect.  I'll swap that diode out and report in later.

Thank you all very much.

[Ice-9]

Glad you got it sorted, but with D1 in the wrong way round did you not notice the battery and /or D1 getting very hot.   

[frokost]

After I tested these points I disconnected and tested the very warm battery at 6.5 V.

Ice-9, he did notice, which is why everybody, starting with pete, meant it was a shorted battery.

[valdiorn]

Valdiorn, why does a lower resistance provide a more stable reference? 

It was just an idea. Basically, you want a voltage reference to look like an ideal voltage source, meaning, it won't change its output voltage if you connect a load to it. Right now it has an output resistance of 500k (2x 1Meg in parallel, as "seen from the signal"), but he's coupling the signal to it with a 1 Meg resistor, which then has a path to ground through a 10k res, a cap (which I'll assume is a perfect conductor for higher frequencies), and a 2.2 meg resistor... ok, that  a total of 3.2Meg/500k = 6,4x difference, but the normal design rule says make the input impedance at least 10x larger than the output impedance and you can ignore the output resistance of the voltage source.

Basically, that reference point is not going to be 9 volts, it's going to be
( 1 Meg | 3.2 Meg ) / ( 1 Meg + ( 1Meg | 3.2 Meg ) ) * 9 Volt = 3.891 volts, that' not 4.5 volts!
but make it 47k:
( 47k | 2.2 Meg ) / ( 47k + ( 47k | 2.2 Meg ) ) * 9 Volt = 4.467 volts, much closer, and more stable, in case you change the circuit a bit...



So, the moral of the story: using lower value resistors makes for lower output resistance of the voltage source, which means the rest of the circuit will not have as much of an effect on its stability.

[valdiorn]

whoops, I must've been high when I wrote that or sumthin'. Bacically that cap is going to negate just about everything I said about the voltage at the reference point (because that's a DC point we're talking about). However, with such high resistors you will see a ripple (a part of the signal) appear at the voltage reference. That's not good. I did some quick simulations in LTSpice and this time I'm actually right
Lower the resistors to 10k or 47k and you'll see much less signal leaking onto the reference. (doing this is just for peace of mind, or to prevent any hard-to diagnose problems that might occur later on)

Let's hope I'm not full of crap this time. I'd have edited my original post but the modify button has disappeared...