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  • A npn transistor in a CE mode is used as a simple voltage amplifier with a collector current of 4 mA. The terminal of 8V battery is connected to the collector through a load resistor of <img

A npn transistor in a CE mode is used as a simple voltage amplifier with a collector current of 4 mA. The terminal of 8V
battery is connected to the collector through a load resistor of \mathrm{R_{L}} and to the base through a resistor of RB. The collector
emitter voltage\mathrm{V_{CE}=4.0V} and base current amplification factor \mathrm{\beta_{d c}=100 .}. The values of RL is:

Option: 1

\mathrm{1K\Omega }


Option: 2

\mathrm{1.5 K\Omega}


Option: 3

\mathrm{2k\Omega}


Option: 4

\mathrm{2.5K\Omega}


Answers (1)

best_answer

The input voltage equation for the transistor shown in the

figure is given by,

\mathrm{\begin{aligned} & \Rightarrow \mathrm{I}_{\mathrm{B}} \mathrm{R}_{\mathrm{B}}=\mathrm{V}_{\mathrm{CC}}-\mathrm{V}_{\mathrm{BE}}=8-0.6=7.4 \mathrm{~V} \\ & \text { Again, } \mathrm{I}_{\mathrm{B}}=\frac{\mathrm{l}_{\mathrm{C}}}{\beta}=\frac{4 \times 10^{-3}}{100}=4 \times 10^{-5} \mathrm{~A} \end{aligned}}

The output voltage equation is given by,

\mathrm{V_{C C}=V_{C E}+I_C R_L} \therefore R_L=V_C_C-V_C_E/I_C= 4/4\times 10^3\Omega =1K\Omega

 

Posted by

shivangi.shekhar

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