# Integrator with diode

I've the circuit shown below. It's an integrator, with a diode in parallel to the capacitor, and I'm a little confused.

My questions are:

1. If I apply at Vin a square wave with an amplitude from -5 V to +5 V, what happens to the voltage on V- node?
2. With a regular integrator at the output node I should have a triangular wave that goes from 0V to $-V_{in} \times \Large \frac{t}{RC}$, but in this case what I get at the output? Is it a "rectified"square wave?

simulate this circuit – Schematic created using CircuitLab

• Why not simulate it and find out for yourself? – EM Fields Sep 25 '16 at 10:57

1. If I apply at Vin a square wave with an amplitude from -5 V to +5 V, what happens to the voltage on V- node?

In a negative feedback arrangement the amplifier will do its best to keep the - input at the same potential as the + input. The + input is grounded and we refer to the - input as "virtual ground". The voltage there should always be zero unless we drive the output into saturation at which point the feedback won't be able to correct anymore.

1. With a regular integrator at the output node I should have a triangular wave that goes from 0V to $-V_{in} \times \Large \frac{t}{RC}$, but in this case what I get at the output? Is it a "rectified"square wave?

simulate this circuit – Schematic created using CircuitLab

Figure 2. (a) and (b) show how the circuit behaves on each alternate clock cycle. (A) and (B) show the result depending on which way the squarewave starts.

Consider the diode operation first: what output voltage will satisfy the virtual ground? Clearly when the input is positve and the output is at -0.7 V (a real diode - not an ideal one) the output will go to -0.7 to maintain 0 V at the - input.

When the input goes negative the diode will be reverse biased so the capacitor will run as a normal integrator and the output will be a triangle wave.

Finally, the output will depend on the starting conditions indicated by diagrams A and B. (Please excuse the arrows. CircuitLab doesn't have a free-style line symbol.

• In A we start with a positive-going square-wave so the output clamps at -0.7 V and the triangle wave continues from there.
• In B we start with a negative-going square-wave so the triangle wave starts and ends at 0 V.

In practice, due to diode and capacitor leakage and any non-symmetry in the square wave the output may settle down somewhere in between the two conditions.

• Thanks for accepting my answer. If you un-accept for a day or two you may attract some other answers which give a different or better insight to the question. If it answers all your questions and you think it's right then that's fine. – Transistor Sep 25 '16 at 10:17
• Note that your output is only correct for RC matched perfectly to the input frequency. If the RC product is smaller, the integrator will limit during part of the cycle. If larger, the the average value will be determined by op amp characteristics (specifically offsets.) – WhatRoughBeast Sep 25 '16 at 13:30
• @What: Thanks for that. I think I covered your first point with "unless we drive the output into saturation". Point taken about offsets. Many moons ago I did a project with a commutating auto-zero op-amp to try to eliminate such problems on an integrator fed by a current transformer trying to detect DC on power distribution transformers. It didn't work well. Hall-effect sensors weren't available (to me, at least) at the time and they would have solved the problem by avoiding the integrator. – Transistor Sep 25 '16 at 13:41