What does the diode actually do?

Started by BAARON, September 18, 2009, 11:45:24 AM

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BAARON

What does the diode do that makes this opamp compress?  What does the output signal look like in comparison to the input?  What's the output swing like - does it reach the voltage rails, or will it not go that far?  How much does it compress, and why?

I already know how to plug this into a circuit and use it as an op-amp.  I've breadboarded it and it worked fine, though I don't know how much compression it actually added to the signal.  Now I want to know more about HOW this circuit does its job... I'm not sure whether or not I want to use it in a pedal (for a TS-style gain stage), because I don't know how much compression it would add compared to something like a Burr Brown, or whether it's well-suited to using LEDs as clipping diodes (because I don't know what it's capable of as far as the output voltage swing goes).

Any clues?  Or articles you could link to by way of explanation?
B. Aaron Ennis
If somebody makes a mistake, help them understand what went wrong.  Show them how to do it right.  Be helpful.  Don't just say "you're wrong, moron."

Joe

The diode keeps the transistor from going into saturation, which is the primary cause of the "transistor sound". Placing it within the bias network keeps it conducting, so the smaller signals pass through, but the larger ones are met with more resistance and eventually clipped entirely.

There are other ways, but the diode guarantees that the transistor can never saturate, and also reacts to the input signal differently than say a fixed series resistor would. JFETs also suffer from the saturation problem, so the diode is helpful in those circuits as well.

BAARON

So at what point does the signal clip inside the opamp?  What's the available output signal swing like?  If I'm using LEDs as clipping diodes in the non-inverting feedback loop, will the opamp be able to deliver a hot enough signal that the LEDs will clip, or will it clip earlier than that within the opamp itself because of the compression diode?
B. Aaron Ennis
If somebody makes a mistake, help them understand what went wrong.  Show them how to do it right.  Be helpful.  Don't just say "you're wrong, moron."

Joe

The output should be rail-to-rail, minus about 1/2 volt on either side. (This is a guess, I never looked at it on a scope.)


BAARON

Splendid.  I'll put in the effort of breadboarding this again (takes a while - the full circuit gets pretty massive when I'm through with it!), and A/B it against an IC version of the same circuit to see how much of a difference there is as far as compression goes.
B. Aaron Ennis
If somebody makes a mistake, help them understand what went wrong.  Show them how to do it right.  Be helpful.  Don't just say "you're wrong, moron."

Joe

"Compression" has more to do with what it does to signal peaks rather than sounding like a compressor. It should sound warmer than a regular opamp.

BAARON

Ah!  Now THAT'S a good thing to know.
B. Aaron Ennis
If somebody makes a mistake, help them understand what went wrong.  Show them how to do it right.  Be helpful.  Don't just say "you're wrong, moron."