# Why does this resistor affect this transimpedance circuit in this way?

I am working with the following circuit

simulate this circuit – Schematic created using CircuitLab

It's a transimpedance amplifier with a clamping circuit to limit the output to a diode drop.

The idea is once the base of Q1 is at -0.7V, it turns on, shunting the photodiode current away from the op amp.

The circuit kind of works. It clamps at -.7V, but it rings and then overshoots as seen in the oscilloscope trace below (taken at the output of the op amp).

Adding a resistor in series with the photodiode fixes the problem, but slows things down a lot as seen in the circuit and scope trace below.

simulate this circuit

Why is the resistor changing the output in such a way? I was not able to derive the loop gain for this circuit.

I don't think the lower speed is because the 100 ohm series resistor is making the photodiode current shunt into its capacitance, since the time constant of 10 ohm * photodiode capacitance is much faster than the bandwidth of the amplifier.

• What problem are you trying to solve with the clamping circuit?Most photodiodes aren't going to be producing more than milliamps even when driven hard, so you probably don't need Q1's gain to enable the op-amp to handle the current. Jul 30, 2017 at 21:04
• One thing to check: When the circuit limits, does the op-amp inverting input stay at 0 V? Jul 30, 2017 at 21:06
• @thephoton the goal of the circuit is to have high gain for small light levels but to handle large light levels without the problems associated with overdriving the op amp output. I'll check the input voltage and get back to you. Jul 30, 2017 at 21:19
• Your op-amp is capable of driving 70 mA, but you're limiting at a level where the photocurrent is 7 uA. Just get rid of Q1 and use two diodes if you want a ~1.4 V limit. Jul 30, 2017 at 21:22
• @ThePhoton the amplifier still rings with 2 diodes. Also, rather than a small overshoot after the ringing pulse, it now saturates momentarily at the positive power supply. Looking at the inverting input, it looks like the loop is closed (~0V) except for the oscillations on the bottom of the clamped pulse show up. Jul 31, 2017 at 17:02