cupcake debugging - help welcome

[add4]

Hello,

So i built a cupcake a while ago and it was not working correctly ... i just found the time to get it out today and analyze a bit.

So, i started reading voltages at the opamp
at the power plug (yes i boxed before i rocked ) voltage between ground and positive side is 9.24V

at the IC
1 : 1.43
2 : oscillation between 1.90 and 1.40
3 : 1.19
4 : 0
5 : 0
6 : 1.85
7 : 1.85
8 : 2.43

so i started looking at the power stage: 2.43 at either side of R12 and the diode.

.. i don't really know what to look for next: the voltages at the input are good, but not in the power stage which would be, for me the very first place where it ca go wrong .. but i imagine it would be either grounded and 0 everywhere, or 9v everywhere..

other voltages:
Q1:
E: 0.24
C: 0.24
B: 0

Q2:
C: 2.43
C: 1.28
B: 0.25

Have you got some ideas?

thanks in advance for your time/help

[midwayfair]

You used 100K instead of 100R (100 Ohms) for R12.

[add4]

i'm going to check this.
if you're right, you get a badge for extreme debugging capabilities.


thanks for helping

[add4]

ok i just checked :

the schematic for the cupcake says 10r and it's a brown black black resistor, so i think it's correct.

but i also think there is something going on at the beginning of the power stage. it makes sense what i don't understand is why do i get 9v at the plug, and 2 at R12, (and also D2) if it's connected i should have 9v everywhere, or 2v everywhere, right?

[midwayfair]

i'm going to check this.
if you're right, you get a badge for extreme debugging capabilities.


thanks for helping

I mistyped earlier, it's supposed to be 10R and you've used something else.

Ohms law.

You have 9.14V on one side of a resistor.

You have 2.43V on the other side of the resistor.

That's a 6.7V drop across a resistor.

Either your op amp is drawing 630mA (!) and dropping almost 7V across a 10R, or your resistor is much larger than it should be.

What happens if your resistor is 10R? Well, the minimum wattage required to dissipate the heat of 63mA across 100R is over 4W, and even that would probably burn up after a few minutes. Plus your op amp, even if it were capable of drawing 630mA in this design, would probably be hot enough to burn your finger and them melt itself and probably a couple things around it as it does so. Both of those would be really noticeable problems. Since you didn't report that your pedal was destroying itself in a fiery conflagration, the logical conclusion is that the resistor is much larger than it needs to be and your op amp is drawing its usual amount of current (6.3mA assuming 6.3mA of current draw, but it could be 10K if the op amp is hardly drawing anything).

You asked why all the points connected to the power read something other than 9V? It's because you measured AFTER the resistor. Voltage drops across resistance. You probably have a little under 9V on the cathode of D2.

... Nah, I didn't do all that. It's just a very, very common mistake.

If you're looking at the resistor and seeing 10R, use your multimeter, not your eyes.

[add4]

haha .. i also assume that my battery life (using a pedal train volto to power the circuit) would drain in a few minutes.

i didn't notice anything of that so .. i'll just unsolder the resistor and measure it 
Thanks for the analysis, i know who to call next time i burn my house by messing around with 9v powered electronics circuits

[add4]

Ooooooook...
So i opened my pedal again last night and it appears that the problem i describe will occur of you soldered your 9v supply to the LED pad instead of the 9v pad....

i feel ... not very smart ... as i'm writing this

Anyway, thanks for your help !

Also i tested the pedal last night, not very loud since the kids were sleeping, but is the effect supposed to be really huge with this compressor? I mean, i definitely hear compression, and a really nice one i might add, but in my memories of the boss CS2, a compressor can sound really artificial and have a pumping effect with each note if the compression is set too high. Here, no value of the pot controlling the compression amount produced that kind of extreme sound.
is that normal for that circuit in your experience?

[midwayfair]

There is no pot controlling the compression. There's an approximately 700 Ohm range on the 10K trimpot where the reference voltage on the source of the FET is appropriate for a positive voltage on the gate of the FET will drive the FET's resistance down. When the voltage produced by the rectifier and passed on to the FET is lower than the reference voltage, the FET's resistance doesn't move. The second FET is there to provide a steady current and the sus trimpot forms a voltage divider with that FET between the positive supply voltage and ground. (If you set the trimpot to 0, you'll get smoke.) The second FET works as a buffer of sorts.

I shouldn't make it sound like the trimpot doesn't do anything. But if you're outside of the very small operating range, then either you won't be able to produce a strong enough signal to trigger compression at all (even if you have gobs of gain), OR the FET will simply sit at a low resistance -- essentially be "stuck" in compressing. There are other limits to the design, like a longer decay time can be problematic, the lack of a buffer for the rectifier, an input-signal dependent threshold, etc.

The CS2 is an almost completely different device. It's so different that it would probably take me an hour just to type out all the ways it differs, never mind explaining how they work.

[add4]

I shouldn't make it sound like the trimpot doesn't do anything. But if you're outside of the very small operating range, then either you won't be able to produce a strong enough signal to trigger compression at all (even if you have gobs of gain), OR the FET will simply sit at a low resistance -- essentially be "stuck" in compressing. There are other limits to the design, like a longer decay time can be problematic, the lack of a buffer for the rectifier, an input-signal dependent threshold, etc.

thanks for your very interesting and educating answer as usual.

So if i'm moving the trimpot too high i'm essentially at 'full compression' all the time? it doesn't sound completely squashed to me.. at least not in a bad way at all.

should i try to lower that trimpot to recover a bit of dynamics then ?

[midwayfair]

I shouldn't make it sound like the trimpot doesn't do anything. But if you're outside of the very small operating range, then either you won't be able to produce a strong enough signal to trigger compression at all (even if you have gobs of gain), OR the FET will simply sit at a low resistance -- essentially be "stuck" in compressing. There are other limits to the design, like a longer decay time can be problematic, the lack of a buffer for the rectifier, an input-signal dependent threshold, etc.

thanks for your very interesting and educating answer as usual.

So if i'm moving the trimpot too high i'm essentially at 'full compression' all the time? it doesn't sound completely squashed to me.. at least not in a bad way at all.

should i try to lower that trimpot to recover a bit of dynamics then ?

If you just let the guitar idle, there will be a spot where it goes from being noisy to ... not noisy. If you keep turning it, you'll end up at a point where it's very quiet. The spot you want is just after it stops sounding noisy.

It's not so much that it's at "full compression." Or at least, that was an inaccurate way for me to describe it. Compression is a reduction in signal dynamics by a particular ratio. What's happening is that the FET is stuck at the resistance it would be driven down to if the control voltage were having an effect. In other words, it's not compressing ... it's just turned the input signal down (because it forms one leg of a voltage divider with the 330K that precedes it). You could conceivably still get some compression if the input signal is particularly large, but it won't be working in anything like the ideal range.

It can be challenging to set one of these up correctly, especially if you've never heard a working unit in person before, and unfortunately there's no real shortcut or trick or "right" setting for how any particular unit should be perfectly set up. I think Mark Hammer at one point found a note that suggested .7V on the source of the first FET as a starting point, which works out to approximately a 1V signal producing compression (that's .3V for the diode drop, and .7V before the positive rectified voltage gets above the source voltage and starts causing the FET's resistance to drop). On paper that makes sense. In real life, it's not the magic spot for everything!

[add4]

If you just let the guitar idle, there will be a spot where it goes from being noisy to ... not noisy. If you keep turning it, you'll end up at a point where it's very quiet. The spot you want is just after it stops sounding noisy.

well that's what i'm wondering .. i didn't hear much noise coming from the pedal... but i also always played my guitar when plugged in, no idle time. also i kept the volume pretty low.

i'll try to make noise with the pedal to see what you're talking about ..
anyway. event with the trimpot cranked, i think it sounds better than the two compressors i've tried (cs2 and dyna comp). this is actually a very nice effect according to my ears.

The biggest drawback i have so far is the treble loss, but i've heard it's not something that can be avoided with a compressor..

thanks for your very interesting insight as always