Given for an FET, gm = 9.5 mA/volt. Total capacitance =

Given for an FET, g_m = 9.5 mA/volt. Total capacitance = 500 pF. For a voltage gain of – 30 the bandwidth will be—

Given that g_m = 9.5 mA/volt, total capacitance = 500 pF, voltage gain (A₀) = – 30, BW =? We know that

A₀ × BW =	g_m
	C

where, A₀ = gain
BW = Bandwidth
C = Total capacitance
g_m = Transconductance.

or BW =	g_m
	A₀·C

or BW =	9.5 × 10^–3
	30 × 500 × 10^–12

or BW = 100 kHz
Hence alternative (A) is the correct answer.