# Diode on op-amp output to drive tube

I am working on a tube based guitar amplifier, and found an interesting schematic online. There is an op-amp that has diodes across a resistor in the feedback path.

I understand what it makes the op-amp do, instantly skip over any value less then abs(0.5V). What I'm asking is if anyone knows why you would add this to a guitar amplifier to drive a tube.

I have also included a simulation of Switch SW101a, basically how much resistance to ground on the negative input terminal.

The only answer I could come up with is "sound" or it somehow helps correct for non-linear responses in the tube.

• Are you sure it's not part of distortion creation? I used to make fuzz boxes, with the input going thru diodes like that to clip the signal to fuzz it up. Nov 18, 2018 at 2:14

It's a form of deliberate distortion, which is popular in some guitar styles.

If you look at your blue simulation waveform, you can decompose it into two components:

• the original sinewave (about 3.4 Vpp)
• a square wave (about 0.8 Vpp) at the same frequency that has the same zero crossings.

This adds a series of odd harmonics to the signal, which gives it extra "buzz" or presence.

This is not the same as a clipping circuit, which only kicks in at high input signal levels. This effect applies at all signal levels. In fact, this effect is MORE pronounced at low signal levels.

• I assumed it would be something sound related. Thanks for the detailed explanation on the addition of odd harmonics. Nov 18, 2018 at 2:22
• at very low signal levels distortion is not present, but yeah at the middle signal levels it's most noticeable Nov 18, 2018 at 2:44

Your simulation output waveform is correct. Changing the resistor to 10k, with the switch, changes the gain (dV/dt) of the circuit to a much higher gain so you see the steeper part of the waveform when the output first starts going up (positive half cycle for example). Then when the output voltage increases enough, D102 diode turns on changing the gain (lower dV/dt) since it is in parallel with the 100k resistor and it just simply becomes a gain stage of two as long as the diode is on. When the output decreases enough, the diode turns off and the gain changes again, determined by the resistors, and it does the same thing to the waveform. So the output is now not so much sinusoidal in shape, like the input. So look at the harmonic content of the output waveform. The diode will always turn on at the nominal 0.6 volts. It's on resistance in parallel with the 100k simply changes the gain (lower). The change in gain changes the steepness (dV/dt) of the output waveform. So the circuit will not clip the output waveform. Note the output gain with the switch in the open position. The gain is 1.5 and the waveform looks much more sinusoidal and what is it's harmonic content? Compare the waveform harmonics with two with different gains. Note also the output never exceeds 2.0 volts to drive the next stage but you change the harmonic content. I hope this helps, have fun!