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I don't understand how to determine value of resistors and capacitor in differentiator and integrator circuit of Op-Amp.

I tried to align the amplitudes of these sine waves. I keep changing their values but I can't get it right.

The best result I can get:

differentiator results

The circuit:

differentiator circuit

Okay, so that is my question about the differentiator circuit. The next question is about my integrator circuit. As with the differentiator, I don't understand how these resistors and capacitors affect each other. I tried to make the source a triangular wave and the result a sine wave. Alas, no luck.

Here is the result:

triangular result

Here is the circuit:

triangular circuit

The next one is a square wave. I managed to get it to square wave, but the result is supposed to be a triangular wave and all I get looks like a straight line.

Here is the result:

Square result

Here is the circuit:

Square circuit

How can I determine the correct resistor and capacitor values?

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These are all RC time constants, let's call them tau. (freq. = 1/(2*pi*tau))

For differential circuit tau = R1*C3, (with R1 giving you some gain.)

For the integrator tau = R6*C3, (R5 is resetting the cap with yet another time constant.)

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Grace Firsta - it depends on your requirements, that means: What do you expect from the circuits? At first, it is not possible to realize IDEAL differentiating ot integrating circuits. Both are instable - because of different reasons. As a consequence, the desired mathematical operation is possible only in a limited frequency band. Therefore, it is wise to consider the frequency domain rather than the time domain (as you did).

Example integrator: Assuming an ideal opamp (infinite gain) the transfer function for your last circuit is H(s)=(R5/R6)/(1+jwR5C3). This is the classical first-order lowpass function with gain. The 3-dB lowpass corner frequency is at fc=1/(2*Pi*R5C3). Because the iDEAL integrator transfer function is 1/jwT you must compare both functions in order to see if and under which conditions you can use the lowpass circuit for integrating purposes.

Following this goal, you will see that for frequencies which fulfill the condition jwR5C3>>1 the lowpass approaches the integrating function: H(s)=(R5/R6)/jwR5C3=1/jwR6C3. Hence, the "integrator time constant" is T=R6C3.

Hence, you have a lower frequency limit for integration: w>>1/R5C3. In addition, you have an upper limit set by the opamp itself. Remember the general condition mentioned at the beginning: Open-loop gain of the opamp (nearly) infinity (in practice: larger than 100). This defines the upper frequency limit. As a consequence, integration of an input signal is possible with sufficient accuracy within a certain frequency band only.

Final remark: As you will see, the parallel resistor R5 "disturbs" the integration process (ideal: R5 infinite), however, this resistor is necessary to provide a minimum of dc feedback for stabilizing the operational point.

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