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Electronics and Communication Engineering - Exam Questions Papers - Discussion

Discussion Forum : Exam Questions Papers - Exam Paper 2 (Q.No. 32)
Exam Questions Papers
32.
For a npn BJT transistor fβ is 1.64 x 108 Hz. Cμ = 10-14 F; Cp = 4 x 10-13 F and DC current gain is 90. Find fT and gm (fβ = cut off frequency, Cμ = capacitance, Cp = parasitic capacitance, gm = transconductance, fT = gain BW product)
fT = 1.47 x 1010 Hz ; gm = 38 milli mho
fT = 1.64 x 108 Hz ; gm = 30 milli mho
fT = 1.47 x 109 Hz ; gm = 38 mho
fT = 1.33 x 1012 Hz; gm = 0.37 m-mho
Answer: Option
Explanation:

∴ fT = 90 x 1.64 x 108 = 1.47 x 1010 Hz

gm = 2pfT(Cμ + Cp) = 2 x p x 1.47 x 1010

= (10-14 + 4 x 10-13)

gm= 38 mμ.

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