What is the transfer function of an ideal op-amp integrator with input resistor Rin and feedback capacitor Cf, and how does a parallel feedback resistor modify it?

What is the transfer function of an ideal op-amp integrator with input resistor Rin and feedback capacitor Cf, and how does a parallel feedback resistor modify it?

• A. Vout(s) = -(1/(Rin Cf)) (1/s) Vin(s)
• B. Vout(s) = - (Rf/Rin) Vin(s)
• C. Vout(s) = - [Rf/(Rin (1 + s Rf Cf))] Vin(s)
• D. Vout(s) = - (1/(Rin)) (s Cf) Vin(s)

Answer: (A)

An ideal op-amp integrator converts input voltage into an output that is proportional to the integral of the input, inverted because it’s an inverting configuration. With the noninverting input grounded, the inverting input is a virtual ground, so the input current through Rin equals the feedback current through the feedback impedance. For a pure capacitor in the feedback path, its impedance is Zf = 1/(sCf). The input current is Vin/Rin, and it must equal the current through the feedback network, which is Vout/Zf. This gives Vin/Rin = - Vout / Zf = - Vout s Cf. Solve for the transfer function: Vout(s)/Vin(s) = - 1 / (s Rin Cf). This is the classic ideal integrator form: the output is proportional to the time integral of the input, with a negative sign. If a resistor Rf is placed in parallel with the feedback capacitor, the feedback impedance becomes Zf = (1/(sCf)) // Rf = Rf / (1 + s Rf Cf). The transfer function becomes Vout(s)/Vin(s) = - Zf / Rin = - [Rf / (Rin (1 + s Rf Cf))]. This shows how the parallel resistor modifies the response: at low frequencies the gain tends toward -Rf/Rin (finite DC gain), while at high frequencies the behavior approaches the ideal integrator (-1/(s Rin Cf)), with a pole at s = -1/(Rf Cf).

An ideal op-amp integrator converts input voltage into an output that is proportional to the integral of the input, inverted because it’s an inverting configuration. With the noninverting input grounded, the inverting input is a virtual ground, so the input current through Rin equals the feedback current through the feedback impedance.

For a pure capacitor in the feedback path, its impedance is Zf = 1/(sCf). The input current is Vin/Rin, and it must equal the current through the feedback network, which is Vout/Zf. This gives Vin/Rin = - Vout / Zf = - Vout s Cf. Solve for the transfer function:

Vout(s)/Vin(s) = - 1 / (s Rin Cf).

This is the classic ideal integrator form: the output is proportional to the time integral of the input, with a negative sign.

If a resistor Rf is placed in parallel with the feedback capacitor, the feedback impedance becomes Zf = (1/(sCf)) // Rf = Rf / (1 + s Rf Cf). The transfer function becomes

Vout(s)/Vin(s) = - Zf / Rin = - [Rf / (Rin (1 + s Rf Cf))].

This shows how the parallel resistor modifies the response: at low frequencies the gain tends toward -Rf/Rin (finite DC gain), while at high frequencies the behavior approaches the ideal integrator (-1/(s Rin Cf)), with a pole at s = -1/(Rf Cf).