QB_PH_SEC8_015: Operational Amplifiers & Analog Circuits

An operational amplifier has an open-loop gain of $A = 10^5$ and an open-loop unity-gain bandwidth of $f_t = 1.0\text{ MHz}$. If it is configured as an inverting amplifier with closed-loop gain $|A_{v}| = 9$, what is the closed-loop 3-dB bandwidth (in kHz)? (Round off to the nearest integer)

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Gain-Bandwidth Product (GBW)**: For an internally compensated op-amp, the open-loop gain rolls off at $-20\text{ dB/decade}$ with constant gain-bandwidth product: $\text{GBW} = f_t = 1.0\text{ MHz}$. 2. **Noise Gain Definition**: Feedback factor $\beta = \frac{R_1}{R_1 + R_f}$. The loop bandwidth is governed by the noise gain: $1/\beta = 1 + \frac{R_f}{R_1} = 1 + |A_v| = 1 + 9 = 10$ 3. **Closed-Loop Bandwidth Calculation**: $f_{-3\text{dB}} = \frac{f_t}{1/\beta} = \frac{1.0 \times 10^6\text{ Hz}}{10} = 100 \times 10^3\text{ Hz} = 100\text{ kHz}$ 4. **Official Answer**: 100 kHz.