Question

Given for an FET, \(g_m = 95\ \text{mA/volt}\), total capacitance \(= 5000\ \text{pF}\). For a voltage gain of \(-30\), the bandwidth will be:

• A. \(100\ \text{kHz}\)
• B. \(630\ \text{kHz}\)
• C. \(3\ \text{MHz}\)
• D. \(19\ \text{MHz}\)

Hint:

Find the gain-bandwidth product \(g_m/(2\pi C)\), then divide by the voltage gain magnitude.

Answer 1

The Correct Option is A

Step 1: For an FET amplifier the gain and bandwidth are tied together by a fixed gain-bandwidth product. The product is set by the transconductance and the total shunt capacitance: \[\text{GBW} = \frac{g_m}{2\pi C}.\]
Step 2: Insert \(g_m = 95\ \text{mA/V} = 0.095\ \text{S}\) and \(C = 5000\ \text{pF} = 5 \times 10^{-9}\ \text{F}\): \[\text{GBW} = \frac{0.095}{2\pi (5 \times 10^{-9})} = \frac{0.095}{3.14 \times 10^{-8}} \approx 3\ \text{MHz}.\]
Step 3: The bandwidth for a stage of voltage gain magnitude \(|A_v|\) is the gain-bandwidth product divided by that gain: \[\text{BW} = \frac{\text{GBW}}{|A_v|}.\]
Step 4: With \(|A_v| = 30\): \[\text{BW} = \frac{3 \times 10^{6}}{30} \approx 1.0 \times 10^{5}\ \text{Hz} = 100\ \text{kHz}.\]
Step 5: Note the trap: \(3\ \text{MHz}\) is only the gain-bandwidth product, not the bandwidth. After dividing by the gain the answer is option (A). \[\boxed{\text{BW} \approx 100\ \text{kHz}}\]