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 terminal of a 8 V battery is connected to the collector through a load resistance RL and to the base through a resistance RB . The collector-emitter voltage VCE=4V base-emitter voltage VBE=0.6 V and base current amplification factor β d . c .=100 . Find the values of RL and RB .

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 terminal of a 8 V battery is connected to the collector through a load resistance RL and to the base through a resistance RB . The collector-emitter voltage VCE=4V base-emitter voltage VBE=0.6 V and base current amplification factor β d . c .=100 . Find the values of RL and RB . (A) RL=1kΩ , RB=185 kΩ  (B) RL=2kΩ , RB=150 kΩ  (C) RL=1 kΩ , RB=240 kΩ  (D) RL=3kΩ , RB=185kΩ

Tardigrade Admin · Accepted Answer

See figure. Potential difference across

R_{L}

I_{c} R_{L} = 8 V - V_{CE}

= 8 V - 4 V = 4 V

Now

I_{C} R_{L} = 4 V

R_{L} = \frac{4}{4 \times 1 0 ^{- 3}} = 1 0^{3} Ω = 1 k Ω

Further, for base-emitter equation,

V_{CC} = I_{B} R_{B} + V_{BE}

or

I_{B} R_{B} =

Potential difference across

R_{B}

= V_{CC} - V_{BE} = 8 - 0.6 = 7.4 V

Again

I_{B} = \frac{I _{C}}{β} = \frac{4 \times 1 0 ^{- 3}}{100} = 4 \times 1 0^{- 5} A

R_{B} = \frac{7.4}{4 \times 1 0 ^{- 5}} = 1.85 \times 1 0^{5} Ω = 185 k Ω

$R_{L} = 1 k Ω$ , $R_{B} = 185 k Ω$

$R_{L} = 2 k Ω, R_{B} = 150 k Ω$

$R_{L} = 1 k Ω, R_{B} = 240 k Ω$

$R_{L} = 3 k Ω, R_{B} = 185 k Ω$

RL​=1kΩ , RB​=185kΩ

RL​=2kΩ,RB​=150kΩ

RL​=1kΩ,RB​=240kΩ

RL​=3kΩ,RB​=185kΩ

$R_{L} = 1 k Ω$ , $R_{B} = 185 k Ω$

$R_{L} = 2 k Ω, R_{B} = 150 k Ω$

$R_{L} = 1 k Ω, R_{B} = 240 k Ω$

$R_{L} = 3 k Ω, R_{B} = 185 k Ω$