Question

A 741-type op-amp has a gain bandwidth product of 1 MHz. A non-inverting amplifier using this op-amp and having a voltage gain of 20 dB will exhibit a 3 dB bandwidth of:

• A. 50 kHz
• B. 100 kHz
• C. $\dfrac{1000}{17}$ kHz
• D. $\dfrac{1000}{7.07}$ kHz

Hint:

For op-amps with a known gain bandwidth product, you can calculate the bandwidth by dividing the GBP by the amplifier’s voltage gain in linear form.

Answer 1

The Correct Option is B

The gain bandwidth product (GBP) for an op-amp is the product of its voltage gain and bandwidth: \[ GBP = A_v \times f_{3dB} \] where $A_v$ is the voltage gain and $f_{3dB}$ is the 3 dB bandwidth.
We are given:
- GBP = 1 MHz (or 1000 kHz),
- Voltage gain $A_v = 20 \, \text{dB}$.
First, convert the voltage gain from dB to linear scale: \[ A_v = 10^{\frac{20}{20}} = 10 \] Now, using the gain bandwidth product formula: \[ 1000 \, \text{kHz} = 10 \times f_{3dB} \] Solving for $f_{3dB}$: \[ f_{3dB} = \dfrac{1000 \, \text{kHz}}{10} = 100 \, \text{kHz} \] Thus, the 3 dB bandwidth of the amplifier is 100 kHz.