# Adding resistors to a differentiating op-amp?

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1. Mar 14, 2015

### dilloncyh

I know how to derive the expression Vout = -CR*d(Vin)/dt for a circuit differentiator that consists of a capacitor and a resistor, how I can understand the attached circuit, which is part of practice exercise for my class. Why do we need to add R2 and R3? When I run the simulation Vout = -V3*R3/R1, and Vout is just the normal square wave. How to analyze this circuit?

PS: I am quite new to electronics so sorry if this question sounds stupid. thanks

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2. Mar 14, 2015

### Svein

I suspect that the frequency of your square wave is too high for the differentiator. Change C1 to 1nF and see what happens.

3. Mar 14, 2015

### dilloncyh

by changing the C1 to 1nF, V2 and Vout becomes close to zero all the time (in the nano range), but still I don't quite understand why we need R2 and R3. How to analyze the circuit and get the relationship between Vout and Vin (or V2) in terms of C1, R1, R2 and R3?

thanks

4. Mar 14, 2015

### LvW

If it is your goal to differentiate the input signal, the resistor R2 makes no sense. Just remove it. The resistor R3 is necessary for stability reasons - if the opamp would be ideal (no gain drop and no phase deviations with rising frequencies) you could make R3=0.
However, for real opamps and large frequencies (in the region where the loop gain is approximately zero) it is necessary to limit the phase contribution from the feedback factor using a resistor R3. But its value should be as small as possible (trade-off between stability aspects and disturbing influence on differentiating performance).

5. Mar 14, 2015

### Svein

R1 and R3 sets the gain around the operational amplifier. C1 and R2 is the differentiating circuit (well, to be exact R2 || R3). Your circuit has a gain of -1, so I can understand why you do not see what R1 and R3 is doing. Try an experiment: Remove R2 and R3, and connect C1 directly to R1 (and the - input of the operational amplifier).

6. Mar 14, 2015

### LvW

I am afraid that - in this case - the circuit will be unstable (works for ideal opamp only).

7. Mar 14, 2015

### Svein

I am betting on the internal impedance of the source...

8. Mar 15, 2015

### Staff: Mentor

The circuit as presented is a C-R differentiator (an approximation to a differentiator) followed by a buffer amplifier of gain = -R1 / R3

In calculating the time constant of the differentiator you use R2∥R3 because the (-) input of the op-amp is a "virtual earth" and this means R3 appears as a resistance to ground from the capacitor.

Other respondents have suggested that you modify the circuit; I can't see any reason to do so, since you apparently are required to analyze the circuit provided. Instead of altering the circuit, I suggest that you change the frequency of the driving signal to allow the capacitor time to charge and discharge appreciably during each half period of the input.

If you don't know what wave-shape to expect, try a google search on "RC differentiator circuit"

9. Mar 15, 2015

### LvW

Well - I can try to justify my recommendation as given in my response: The title of the question contains the term "differentiating opamp" - for me an indication that the opamp is intended to work not as an amplifier but as the "heart" of a differentiating block.
However, I agree with you - perhaps the questioner has the task to live with the circuit as it is and to explain what it does - and how.