Effects Circuit Info - Distortion, OD, Fuzz

[ashcatlt]

This is partly an attempt to reply to minions, and partly an attempt to get the discussion of active clipping circuits out of the Passive Clipping thread, and partly just a place to throw some info.

First, I highly suggest reading The Technology of the Fuzz Face. It's pretty much a component by component explanation of how and why the original classic distortion box does what it does.

But we were talking about parallel diode clipping, so I'm going to first post this schematic of my personal favorite:

[ashcatlt]

Sept 1, 2010 22:45:28 GMT -5 sumgai said:

ash,

Ah, the old "diodes in parallel with the amp" trick - I see!

In essence, you're correct, when the diode conducts, any signal that exceeds the forward voltage drop should go to ground. This would indeed be clipping. And it would also be the prime source of so much square wave action that it'd be incredible if it were tonally acceptable.

More to the point, the greater the voltage presented to the diode, the greater the percentage of total clipped signal. IOW, in this kind of circuit, we don't want too much voltage at the diode input, we do want to have something get through to the amp.   The math would work out something like this:

Assume 5v present at the diode input. (Not a bad figure for a preamp circuit, or the output of some stompboxes.) A silicon diode will conduct 4.35 volts worth of signal, leaving a mere 0.65 volts worth of signal to go out the jack. That's a whopping 80+% loss, no? Now if we just use our pickups themselves, and they generate only about 1 volt, then the signal that passes out of the jack will be the same, 0.65v, but that's about 65% of the total signal, yes?

Test time: Which would you rather have, a signal that's almost fully present, or a signal that's been clipped to the bone, and beyond? That said, it's still a matter of time as well, to reach the clipping point, so what the ear hears from that 80+% lost signal will be really harsh - ringing and square waves out the gazoo.

Yes, this can all be made to work, no matter what we cobble together. But despite my training (for which I paid good money), my proclivities still tend towards the KISS principle. 'Nuff said.


~!~!~!~!~!~

You said that diode are sometimes used outside of the feedback loop, and that's true, but also as you stated, the net effect is a crossover notch. Sometimes this is musically acceptable, sometimes not - depends on what order harmonic is most affected. However, this is often ameliorated by using an assymetrical diode setup. The notch might still be there, but it will be "swamped" by the remaining signal that isn't clipped as severely. Or so it appears to my ears, back when I used to breadboard this stuff all the time. (Still got those boards too, but hardly ever use 'em now. Too many other irons in the fire (and not a horse in sight that needs a good shoeing!).)
...
sumgai

Kind of going backwards here, but I was talking about something like the Rat (above) with diodes parallel to the signal, after the opamp, completely outside the feedback loop, and very much like what you're talking about in above paragraphs.

So, the way you've calculated this, it seems the 5V peak is in one direction, implying that the signal also swings to -5V on its "downward" excursion. That's a total of 10V peak-to-peak. But how can we get 10V out of a pedal with a 9V power supply? You can't. Yes, brand new batteries often get close to 10V until they drain down a bit, but most opamps won't swing all the way from rail to rail, so let's just sit with 9V supply. Better yet, since I'm currently only talking about the opamp and the diodes, let's call it a bipolar +-4.5V supply, okay?

The way I've heard it explained, the opamp (the triangle) in the Rat takes the + input, and amplifies it until it is equal to the singnal present at the - input. Notice that the output of the opamp feeds back through the gain pot to that - input.

If the gain pot (the 150k Log pot in the feedback loop) is turned so that the wiper is all the way to the left (in the schematic), there is no resistance there, so the - input gets exactly the output. There is no need to amplify the + signal, 'cause they're already equal. Gain = 1x, or 0db.

If we turn the gain toward the right (again, in the diagram), we introduce some resistance, and the - input starts to see a smaller signal. The opamp has to amplify the output to compensate, and we start to get some actual gain. Gain > 1x, or > 0db.

Let's say we find some combination of input voltage x gain factor that makes the opamp want to put out a 10V (+-5V) peak-to-peak signal. The opamp can only ever put out as much voltage as it can get from the power supply (actually, as I said, usually a little less). It really wants to get to that +-5V, and tries its darnedest, but the best it can do is +-4.5V. So what happens to that other .5 on either side? It gets clipped off.

Now it goes out to the diodes. These are connected in anti-parallel from the signal "hot" to ground, just like we were doing in JohnH's passive implementation. These are silicon diodes, so sg's calculations above apply. The total peak-to-peak output from the diodes is +-0.65V, or 1.3V altogether.

But it gets much worse! If the input voltage is 1V p2p, we're only talking about a gain of 10x. This opamp can give us a whole lot more than that! What happens if we crank it up to 100x gain? And this is generally considered to be a "medium duty" pedal. There are quite a few which go much further.

[jcgss77]

So, if I am reading into this correctly, you are saying that the way the gain pot on Rat works is it affects the negative aspect of the curve, while allowing the positive to work through the circuit unaffected? Sorry if this is common circuit knowledge, but I am trying to learn.

And the diodes will then clip the already overdriven signal, or is this just between gain stages?

Also, you state that if you hook a 9v power supply to this, it will only get 9v out of it (+-4.5v). I thought that supplies usually were a little beefier than what is stated, by about 10-20%, or is this a case of RMS voltage versus peak voltage? So I assume that you would get a little better performance if you used a bigger supply?

[ashcatlt]

Sept 4, 2010 15:51:36 GMT -5 jcgss77 said:

So, if I am reading into this correctly, you are saying that the way the gain pot on Rat works is it affects the negative aspect of the curve, while allowing the positive to work through the circuit unaffected? Sorry if this is common circuit knowledge, but I am trying to learn.

Don't be sorry, that's why we're here. Somebody else has your question too.

Unfortunately, this is not correct. Both the + and - inputs to the opamp take the whole signal, on both sides of 0V. The difference is that if you pump your signal into the +, it comes out the output "right side up": + on input = + on output. Pump it into the - input, and it comes out "upside down": + on input = - on output.

The opamp adds what comes in the + input to the "upside downified" - input. If the total is not 0, it amplifies the + input until it is. Or, at least, it tries.

And the diodes will then clip the already overdriven signal, or is this just between gain stages?

No, yeah, if it's clipping in the opamp, it gets clipped again (and even further) by the diodes.

Also, you state that if you hook a 9v power supply to this, it will only get 9v out of it (+-4.5v). I thought that supplies usually were a little beefier than what is stated, by about 10-20%, or is this a case of RMS voltage versus peak voltage? So I assume that you would get a little better performance if you used a bigger supply?

I'm using 9V even to keep things simple. The opamp usually won't go all the way to the rails on either side no matter where those rails sit. Has nothing to do with peak vs RMS, the PS part of this is all DC so that peak = RMS, or more precisely RMS is meaningless.

As far as better performance, that depends on what you're looking for. Part of what makes the Rat what it is is this combination of opamp and diode clipping. Some folks actually think these things sound better with half-dead batteries, to the point where they add a control to decrease the supply voltage and get even more opamp saturation!

[JohnH]

That ProCoRat is a hairy circuit!

The gain from the opamp varies from x1, up to a large value determined by the ratio of the 150k gain pot resistance, to that of the combined impedance of the 47 and 560 ohm resistors with the 2.2 and 4.7uF caps. The gain rises with frequency and gets up to 60 or 70 db. 60db is x1000 on signal voltage. So even with a small input, the opamp can be driven until it slams up and down at the limits set by the power supply. Then on each cycle, the diodes are trimming the signal to the usual +/- 0.7V or so.

I expect that circuit could make a lot of noise indeed!

John

[jcgss77]

OK, so lets see if I got this straight. The opamp is the LM308, and pins 3 & 2 are the + &- inputs. Depending on which one you use for the signal line, the opamp will amplify the signal until it is 0. What happens when it runs 0? Is this how the opamp performs the distortion effect?

Also, a signal that is neither negative or positive is DC, correct? Would this circuit do anything if you just fed DC into it? After all, aren't batteries DC? Also, what effect would using the opposite polarity (+ or -) on the opamp input have on the sound of the Rat?

[newey]

OK, so lets see if I got this straight. The opamp is the LM308,

Yes.

pins 3 & 2 are the + &- inputs.

The input goes through a transistor (nominally a BF245A) before it goes to the opamp (this is oversimplified, there's more going on there . . ) Pin 4 of the opamp goes to ground, where it finds common ground with the input -, as well as with the battery negative pole.

Pole 3 is part of the gain control circuit.

At least, that's what I'm seeing. As JohnH noted, this is a hairy circuit. First time builders should get their feet wet on a simpler circuit. That's what I'm doing, anyway.

[ashcatlt]

I was thinking of going through the circuit component by component at some point. I'll need help with that, for sure. But we need to get over this opamp thing first.

The opamp tries to amplify until the difference between the + and - inputs (pin 2 and 3 here) is 0V. This doesn't necessarily mean that the output will be 0V.

Let's turn the gain all the way down for a minute, so that there is no resistance in the feedback path from the output (pin 6) and the - input. Let's run a sine wave in there which swings between +-1V. Let's choose a frequency for this sine wave where we can ignore all the caps. In order for the difference between the + and - inputs to be 0V, we need a +-1V signal at the - input because 1 - 1 = 0, and (-1) - (-1) = 0. Since this is exactly what's going in the + input, there's no amplificiation required. The opamp's gain is 1x.

Now let's keep that same sine wave in there, and turn the gain knob up to put some resistance in that feedback path. This works with the resistors to ground (the 560 and 47 toward the bottom of the scheme) as a voltage divider. Let's set it so that we divide the output in exactly half. Half the voltage coming out of pin 6 gets back around to the - input. So now we need +-2V from the output in order to get +-1V at the - input, and the opamp amplifies enough to get us there.

Turn the gain up so it's dividing down to 1/10, and now we need +-10V at the output to get +-1V at the - input. That's 5.5V on either side beyond our +-4.5V supply, and that's where the distortion comes in. On the "upswings", the opamp tries to get to +10V, but it gets to a point where the input is .45V, and the output hits the PS ceiling at 4.5V and stays there until the input wave comes back down to .45V or less. The same on the "downswings".

The opamp would amplify a DC input just the same. This particular circuit has capacitors in series with the signal which keep DC voltage from getting into or out of the circuit, but the opamp itself will do it.

Running a signal into the - input ends up flipping the waveform over.