What does this circuit do? Op-amp with transistors in feedback

I'm an electrical engineering student and I find this circuit in a guitar amplifier schematic, and basically, I'm just wondering what it does, and how one would go about working out what it does. Like what approach do you take when you're confronted with this kind of problem.

P.S. The voltage source supplies for the op-amp are actually 16V, not 15V.

In opamp feedback circuits, it's all about the current flowing in the feedback network, which must balance the current flowing from the input. Clearly, the transistors are intended to modify how the feedback network passes current, so the question is to figure out how they do that.

The basic feedback is provided by the string of resistors in the middle, which pass current according to the voltage difference divided by the total resistance. As long as the voltage across either of the 22K resistors is less than about 0.6V, neither transistor will turn on, and you have a basic inverting amplifier with a gain of about -10.

However, if the output voltage exceeds about +14V (note that the "–" input of the opamp is held at "virtual ground"), the lower transistor will start to turn on. This will pass extra current through the feedback network, reducing the gain of the amplifier overall. In terms of the application, this provides a "soft clipping" or "limiting" function. The other transistor conducts when the output tries to exceed -14V, making the operation symmetrical.

Note that the diodes are required in order to prevent current from flowing in the "wrong" direction through the 22K resistor and the B-C junction of the corresponding transistor.

• So this essentially makes this transistor-based amplifier behave more like a tube-based amplifier when overdriven? – Compro01 Mar 14 '13 at 20:14
• Reminds me of a logarithmic amplifier (and with that the intended distortion). – jippie Mar 14 '13 at 21:40

Clipping distortion. The amount of current going through R4 is always Vin/R4 (as the inverting input is virtual ground), and no current enters the terminals of the op amp. Without the transistors and diodes, all the current from R4 would then go through the three resistors in the middle feedback path. The transistor/diode paths just steal current from this path. In a first approximation, this would be a clip in both directions at the absolute value of Vout greater than the transistors' Vce plus a diode drop. The reality would be a softer clip as the transistors reach saturation.

The circuit will limit the output amplitude from the op-amp to about 20V peak to peak I reckon. It could be a bit higher as one of the previous commentors has suggested.

I don't see that it's main purpose is to provide a distortion feature, rather than protect any circuit that connects to its output.