First slide

Transistors

Question

An npn transistor in a common emitter mode is used as a simple voltage – amplifier with a collector current of 4 mA. The terminal of 8 V battery is connected to the collector through a load resistance $R_{L}$ and base through load resistance $R_{B}$ . The collector emitter voltage $V_{C E} = 4 V$ , base emitter voltage $V_{B E} = 0.6 V$ and current amplification factor $β$ = 100. Find $R_{B}$ (in ohm).

Difficult

A

$18.50 \times 10^{4}$
B

$19.50 \times 10^{4}$
C

$19.50 \times 10^{6}$
D

$18.50 \times 10^{6}$

Solution

$I_{c} = 4 m A = 4 \times 10^{- 3} A, V_{c c} = 8 V, V_{C E} = 4 V, V_{B E} = 0.6 V, β = 100, R_{B} = ?$
$U \sin g K V L i n o u t e r l o o p, V_{c c} - V_{B E} - I_{b} R_{B} = 0, \Rightarrow R_{B} = \frac{V_{c c -} V_{B E}}{I_{b}},$
$A l s o, β = \frac{I_{c}}{I_{b}}, \Rightarrow I_{b} = \frac{I_{c}}{β},$
$∴ R_{B} = \frac{V_{c c -} V_{B E}}{\frac{I_{c}}{β}} = β \frac{V_{C C} - V_{B E}}{I_{c}} = 100 (\frac{8 - 0.6}{4 \times 10^{- 3}}) = 25 (\frac{7.4}{10^{- 3}})$
$= 25 \times 7.4 \times 10^{3} = 25 \times 74 \times 10^{2} = 1850 \times 10^{2} = 18.50 \times 10^{4} Ω$

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