A capacitor placed in parallel with the feedback resistor of an op-amp circuit causes the bandwidth to

A capacitor placed in parallel with the feedback resistor of an op-amp circuit causes the bandwidth to

• A. Decreases
• B. Increases
• C. No change
• D. Undefined

Answer: (A)

A capacitor placed across the feedback resistor makes the feedback path frequency dependent, introducing a single dominant pole in the closed-loop response. In an inverting amplifier, the feedback impedance becomes Zf = Rf // 1/(jωCf), so the closed-loop gain is Vout/Vin = -Zf/Rin = -(Rf/Rin) * 1/(1 + jωRfCf). At low frequencies the gain is roughly -Rf/Rin, but as frequency increases, the capacitor provides more feedback, and the magnitude falls off roughly as 1/ω beyond the corner frequency. The corner (bandwidth) is set by ωp = 1/(RfCf), so increasing the capacitor lowers the corner frequency and thus reduces the bandwidth. In short, adding the capacitor limits how fast the circuit can respond, making the bandwidth smaller.

A capacitor placed across the feedback resistor makes the feedback path frequency dependent, introducing a single dominant pole in the closed-loop response. In an inverting amplifier, the feedback impedance becomes Zf = Rf // 1/(jωCf), so the closed-loop gain is Vout/Vin = -Zf/Rin = -(Rf/Rin) * 1/(1 + jωRfCf). At low frequencies the gain is roughly -Rf/Rin, but as frequency increases, the capacitor provides more feedback, and the magnitude falls off roughly as 1/ω beyond the corner frequency. The corner (bandwidth) is set by ωp = 1/(RfCf), so increasing the capacitor lowers the corner frequency and thus reduces the bandwidth. In short, adding the capacitor limits how fast the circuit can respond, making the bandwidth smaller.