An n-p-n transistor is connected in the common base configuration in a given amplifier. A load resistance of $

Question

An n-p-n transistor is connected in the common base configuration in a given amplifier. A load resistance of $800\, \Omega$ is connected in the collector circuit and the voltage drop across it is $0.8	ext{ V}$. If the current amplification factor is $0.96$ and the input resistance of the circuit is $192\, \Omega$, the voltage gain and the power gain of the amplifier will respectively be:

Accepted Answer

1. **Identify the Given Data:** 
   * Output/Load resistance, $R_{out} = 800\, \Omega$
   * Input resistance, $R_{in} = 192\, \Omega$
   * Current amplification factor (common base), $\alpha = 0.96$
2. **Calculate Voltage Gain ($A_v$):** The voltage gain is the product of the current gain and the resistance gain ($R_{out}/R_{in}$).
   * $A_v = \alpha 	imes \frac{R_{out}}{R_{in}} = 0.96 	imes \frac{800}{192} = 0.96 	imes \frac{25}{6} = 4$
3. **Calculate Power Gain ($A_p$):** The power gain is the product of voltage gain and current gain.
   * $A_p = A_v 	imes \alpha = 4 	imes 0.96 = 3.84$
4. **Conclusion:** The voltage gain is $4$ and the power gain is $3.84$. *(Note: The $0.8	ext{ V}$ drop across the load is extraneous information not required to find the gains).*

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