A transistor is operated in common-emitter configuration at $V_c = 2 ext{ V}$ such that a change in the base

Question

A transistor is operated in common-emitter configuration at $V_c = 2	ext{ V}$ such that a change in the base current from $100\, \mu	ext{A}$ to $200\, \mu	ext{A}$ produces a change in the collector current from $5	ext{ mA}$ to $10	ext{ mA}$. The current gain is:

Accepted Answer

1. **Identify the Given Data:**
   * Constant collector voltage, $V_c = 2	ext{ V}$
   * Initial base current, $I_{B1} = 100\, \mu	ext{A}$
   * Final base current, $I_{B2} = 200\, \mu	ext{A}$
   * Initial collector current, $I_{C1} = 5	ext{ mA}$
   * Final collector current, $I_{C2} = 10	ext{ mA}$
2. **Calculate Changes in Currents:**
   * Change in base current, $\Delta I_B = I_{B2} - I_{B1} = 200\, \mu	ext{A} - 100\, \mu	ext{A} = 100\, \mu	ext{A} = 100 	imes 10^{-6}	ext{ A}$
   * Change in collector current, $\Delta I_C = I_{C2} - I_{C1} = 10	ext{ mA} - 5	ext{ mA} = 5	ext{ mA} = 5 	imes 10^{-3}	ext{ A}$
3. **Calculate AC Current Gain ($\beta_{ac}$):** 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.
   * $\beta_{ac} = \frac{\Delta I_C}{\Delta I_B} = \frac{5 	imes 10^{-3}	ext{ A}}{100 	imes 10^{-6}	ext{ A}} = \frac{5 	imes 10^{-3}}{10^{-4}} = 50$

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