1
\$\begingroup\$

Integrator and differentiator circuits can be made using an op-amp:

enter image description here

and a simple RC network:.

enter image description here

Is one better than the other? What are the differences between the two, if any?

\$\endgroup\$
  • 1
    \$\begingroup\$ One is fairly mathematically correct and the other is a low-pass filter that behaves somewhat like an integrator at higher frequencies. \$\endgroup\$ – Andy aka May 8 '13 at 7:07
  • \$\begingroup\$ @Andyaka I'm a little confused now. I did the RC Integrator derivation using Laplace, which which gives me a ratio of Vout/Vin=1/(sRC+1). As you mentioned its a low pass filter, as omega tends to infinity Vout/Vin tends to zero. So how can it work as an integrator at high frequencies when its Vout/Vin tends to zero as we keep on increasing the frequency? \$\endgroup\$ – user23564 May 10 '13 at 10:56
  • 1
    \$\begingroup\$ The RC circuit (without op-amp) approximates to an integrator as frequency rises and this approximation gets more accurate with higher frequencies - the sCR term becomes more dominant than 1 because s is increasing. An integrator's output will also fall to zero at infinite frequency. \$\endgroup\$ – Andy aka May 10 '13 at 11:10
  • \$\begingroup\$ @Andyaka Reading your first comment makes so much sense now. Thanks ! :) \$\endgroup\$ – user23564 May 10 '13 at 11:51
1
\$\begingroup\$

The lower circuit, RC

enter image description here

only works when:

  • The output impedance of Vin source is 0Ω

and at the same time:

  • Vin is much larger than VC . Ideally the capacitor voltage is 0V, but that defeats the whole function as an integrator. The voltage across the capacitor influences the current through the resistor.

and at the same time:

  • The load of the circuit (VC) side is ∞

These requirements are not easily to met. The latter requirement is caused by the fact that any load on the capacitor C or resistor R will change the frequency responce of the integrator.

The opamp circuit relaxes these requirements, at least for the infinite load and for the ratio between input and output voltage.

enter image description here

Because the output is driven by an opamp, the output impedance of the circuit is resonably low (in practice maybe couple tens of ohms) and therefore it is sufficient if the output load is about 100 times higher than that; say > 1000Ω A requirement that is a lot easier to meet.

Also because the opamp will try to keep its inputs at the same voltage (V- is virtual ground), the voltage across the capacitor is no longer influencing the input current of the integrator. The input impedance of the opamp circuit is entirely specified by the resistor, hence an ideal integrator.

\$\endgroup\$
  • \$\begingroup\$ Explains a lot why the op-amp is a better choice. But I don't understand a few things about the RC Integrator. Could you elaborate the Second point. Why should Vin be much greater than Vc? And why should the capacitor voltage be zero? \$\endgroup\$ – user23564 May 10 '13 at 10:46
  • \$\begingroup\$ Like it or not, you are integrating current over time. Current is the voltage across the resistor divided by the resistor value. However, ideally you want to integrate the input voltage over time and that voltage is equal to the sum of the voltage across the resistor and the voltage on the capacitor. When the voltage on the capacitor is not equal to 0V, it will influence the current through the resistor. So the integrator only works when the voltage on the capacitor is much smaller than the input voltage, ideally 0. \$\endgroup\$ – jippie May 10 '13 at 10:52
  • \$\begingroup\$ 'Voltage across the capacitor ideally zero', are you referring to a DC input voltage ? \$\endgroup\$ – user23564 May 10 '13 at 11:01
  • \$\begingroup\$ The voltage resulting from integration. \$\endgroup\$ – jippie May 10 '13 at 11:13
  • \$\begingroup\$ I understand it now,so the RC integrator is not actually an integrator but an approximation which works well when the reactance of the capacitor is high i.e. when frequency is high. Which @Andyaka pointed out in the comment above also. Thanks, I never realized it was only an approximation. \$\endgroup\$ – user23564 May 10 '13 at 11:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.