A npn 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 R B . The collector emitter voltage VC E=4 V, baseemitter voltage V BE =0.6 and the base current amplification factor β dc =100, calculate the value of RB.

Question

A npn 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 R B . The collector emitter voltage VC E=4 V, baseemitter voltage V BE =0.6 and the base current amplification factor β dc =100, calculate the value of RB. (A) R B =15 k Ω (B) R B =200 k Ω (C) R B =1 k Ω (D) R B =185 k Ω

Tardigrade Admin · Accepted Answer

By the definition <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a7a97c96b1fca1febf9095480a7fa380-.png" /> ∴ iB=(iC/βd c)=(4 m A/100)=0.04 mA From figure iB RB+VB E=VC C ∴ RB=(VC C-VB E/iB)=((8-0.6)/0.04 × 10-3)=185 k Ω

$R_{B} = 15 k Ω$

$R_{B} = 200 k Ω$

$R_{B} = 1 k Ω$

$R_{B} = 185 k Ω$

By the definition

$∴ i_{B} = \frac{i _{C}}{β _{d c}} = \frac{4 m A}{100} = 0.04 m A$
From figure
$i_{B} R_{B} + V_{BE} = V_{CC}$
$∴ R_{B} = \frac{V _{CC} - V _{BE}}{i _{B}} = \frac{( 8 - 0.6 )}{0.04 \times 1 0 ^{- 3}} = 185 k Ω$

RB​=15kΩ

RB​=200kΩ

RB​=1kΩ

RB​=185kΩ

By the definition ∴iB​=βdc​iC​​=1004mA​=0.04mA From figure iB​RB​+VBE​=VCC​ ∴RB​=iB​VCC​−VBE​​=0.04×10−3(8−0.6)​=185kΩ

$R_{B} = 15 k Ω$

$R_{B} = 200 k Ω$

$R_{B} = 1 k Ω$

$R_{B} = 185 k Ω$

By the definition

$∴ i_{B} = \frac{i _{C}}{β _{d c}} = \frac{4 m A}{100} = 0.04 m A$
From figure
$i_{B} R_{B} + V_{BE} = V_{CC}$
$∴ R_{B} = \frac{V _{CC} - V _{BE}}{i _{B}} = \frac{( 8 - 0.6 )}{0.04 \times 1 0 ^{- 3}} = 185 k Ω$