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A transistor is operated in common-emitter configuration at Vc=2 VV_c = 2\text{ V} such that a change in the base current from 100 μA100\, \mu\text{A} to 200 μA200\, \mu\text{A} produces a change in the collector current from 5 mA5\text{ mA} to 10 mA10\text{ mA}. The current gain is:

A

75

B

100

C

150

D

50

Step-by-Step Solution

  1. Identify the Given Data: Constant collector voltage, Vc=2 VV_c = 2\text{ V} Initial base current, IB1=100 μAI_{B1} = 100\, \mu\text{A} Final base current, IB2=200 μAI_{B2} = 200\, \mu\text{A} Initial collector current, IC1=5 mAI_{C1} = 5\text{ mA}
  • Final collector current, IC2=10 mAI_{C2} = 10\text{ mA}
  1. Calculate Changes in Currents: Change in base current, ΔIB=IB2−IB1=200 μA−100 μA=100 μA=100×10−6 A\Delta I_B = I_{B2} - I_{B1} = 200\, \mu\text{A} - 100\, \mu\text{A} = 100\, \mu\text{A} = 100 \times 10^{-6}\text{ A} Change in collector current, ΔIC=IC2−IC1=10 mA−5 mA=5 mA=5×10−3 A\Delta I_C = I_{C2} - I_{C1} = 10\text{ mA} - 5\text{ mA} = 5\text{ mA} = 5 \times 10^{-3}\text{ A}
  2. Calculate AC Current Gain (βac\beta_{ac}): The AC current gain for a common-emitter configuration is defined as the ratio of change in collector current to the change in base current at a constant collector voltage.
  • βac=ΔICΔIB=5×10−3 A100×10−6 A=5×10−310−4=50\beta_{ac} = \frac{\Delta I_C}{\Delta I_B} = \frac{5 \times 10^{-3}\text{ A}}{100 \times 10^{-6}\text{ A}} = \frac{5 \times 10^{-3}}{10^{-4}} = 50
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