Electrical – How to reset an integrator ONLY with RLC and op-amps

Following the superposition theorem you can calculate Vout (node 2) as a result of Vn1 (node 1) and V1=1V:

Vout=V1 + (V1-Vn1)/s*R1C3

For V1=0 this simplifies to the well-known MILLER integrator formula Vout=-Vn1/S*R1C3.

Funny but it seems like a very similar question has been asked just recently, and a piece of a circuit I've used before may help you too. This would be my approach.

You would have to set the voltage divider resistors at the (+) input of the left OP amp so that the input pulse would cross the threshold. If you think the pulse will be less than perfect, you might also want to experiment with an high added resistance to offer some hysteresis. When the pulse goes high, the output of that first OP am will quickly switch low, and the output of the integrator will begin to rise slowly over time. When the pulse goes low, however, its output will quickly switch high, and the output of the second OP AMP will drop almost immediately, like a RESET. This is due to the much lower resistance (and hence shorter) integration timing for the RESET case. You can reverse the action to behave like your version by switching the twodiode resistors, putting the voltage divider on the (-) input of the first OP AMP, and feeding your input to its (+) input through an additional resistor. The important thing is that now you'll have a RESET after each pulse event. If you were sampling the output of the second OP AMP, and saving the highest result recorded every time the pulse went low, you will a have a good measure of the pulse duration. In addition, I'd argue that resetting AFTER the event is always better than resetting at the start of the event. The reason is that any reset is likely to take a short but still non-zero amount of time. So resetting AFTER the accumulation ensures that any reset time does NOT corrupt your pulse duration measurement.