Question

An $n-p-n$ transistor in a common-emitter mode is used as a simple voltage-amplifier with a collector- current of $4 \,mA$. The terminals of a $8\, V$ battery is connected to the collector through a load-resistance $R_L$ and to the base through a resistance $R_B$. The collector-emitter voltage $V_{CE} = 4 \,V$, the base-emitter voltage $V_{BE} = 0.6 \,V$ and the current amplification factor $\beta_{dc} = 100$. Then

• A. $R_{L} = 1\, k\Omega, R_{B} = 185 \,k\Omega$
• B. $R_{L} = 1\, k\Omega = R_{B}$
• C. $R_L = 2\,k\Omega, R_B = 15\, k\Omega$
• D. $R_L = 185 \,k\Omega, R_B = 1\, k\Omega$

Answer 1

Answer: (A)

$R_{L} = 1\, k\Omega, R_{B} = 185 \,k\Omega$

Answer 2

An $n-p-n$ transistor in a common-emitter mode with connections as given is shown in figure. Collector emitter voltage, $V_{CE} = V_{CC} -I_{C} R_{L}$ $\therefore R_{L} = \frac{V_{CC} -V_{CE}}{I_{C}}$ $= \frac{8 V -4 V}{4\times 10^{-3} A}$ $= 10^{3} \Omega = 1 k\Omega$ As $\beta_{dc} = \frac{I_{C}}{I_{B}}$ $\therefore I_{B} = \frac{I_{C}}{\beta_{dc}}$ $= \frac{4\times 10^{-3}}{100}A$ $= 4\times 10^{-5}A$ Base-emitter voltage, $V_{BE} = V_{CC} - I_{B} R_{B}$ $\therefore R_{B} = \frac{V_{CC} -V_{BE}}{I_{B}}$ $= \frac{8V -0.6 V}{4\times 10^{-5} A}$ $= 1.85 \times 10^{5}\Omega$ $= 185\, k\Omega$