Question

For a transistor amplifier in common emitter configuration having load impedance of $1\, kO$ ($h_{fe} = 50$ and $h_{oe} = 25$) the current gain is

• A. -5.2
• B. -15.7
• C. -24.8
• D. -48.78

Answer 1

Answer: (D)

-48.78

Answer 2

In CE configuration, $A_{1} = \frac{-h_{fe}}{1+h_{0e}R_{L}}$ $= \frac{-50}{1+25\times10^{-6}\times 1\times 10^{3}} = -48.78$