In a common emitter transistor amplifier, signal voltage across the collector resistance of 2kΩ is 2volt . What should be the value of resistance RB in series with VB B voltage supply of 2volt in the input circuit, if the D.C. base current is 10 times the signal current in RB ? (. Given: base emitter voltage textVB E=0.6V , current amplification factor β a c ≈ β d c = 100)

Question

In a common emitter transistor amplifier, signal voltage across the collector resistance of 2kΩ is 2volt . What should be the value of resistance RB in series with VB B voltage supply of 2volt in the input circuit, if the D.C. base current is 10 times the signal current in RB ? (. Given: base emitter voltage textVB E=0.6V , current amplification factor β a c ≈ β d c = 100) (A) 7kΩ  (B) 26kΩ  (C) 10kΩ  (D) 14kΩ

Tardigrade Admin · Accepted Answer

( i C ) AC =(( V text out ) AC / R C )=(2 volt /2 k Ω)=1 mA ( i B ) AC =(( i C ) AC /β ac )=(1 mA /100)=0.01 mA ( i B ) DC =10( i B ) AC =0.1 mA =10-4 A RB=(V/iB)=(2-0.6/10-4)=14 k Ω

$7 k Ω$

$26 k Ω$

$10 k Ω$

$14 k Ω$

$(i_{C})_{A C} = \frac{( V _{out} ) _{A C}}{R _{C}} = \frac{2 v o lt}{2 k Ω} = 1 m A$
$(i_{B})_{A C} = \frac{( i _{C} ) _{A C}}{β _{a c}} = \frac{1 m A}{100} = 0.01 m A$
$(i_{B})_{D C} = 10 (i_{B})_{A C} = 0.1 m A = 1 0^{- 4} A$
$R_{B} = \frac{V}{i _{B}} = \frac{2 - 0.6}{1 0 ^{- 4}} = 14 k Ω$

7kΩ

26kΩ

10kΩ

14kΩ

(iC​)AC​=RC​(Vout ​)AC​​=2kΩ2volt​=1mA (iB​)AC​=βac​(iC​)AC​​=1001mA​=0.01mA (iB​)DC​=10(iB​)AC​=0.1mA=10−4A RB​=iB​V​=10−42−0.6​=14kΩ

$7 k Ω$

$26 k Ω$

$10 k Ω$

$14 k Ω$

$(i_{C})_{A C} = \frac{( V _{out} ) _{A C}}{R _{C}} = \frac{2 v o lt}{2 k Ω} = 1 m A$
$(i_{B})_{A C} = \frac{( i _{C} ) _{A C}}{β _{a c}} = \frac{1 m A}{100} = 0.01 m A$
$(i_{B})_{D C} = 10 (i_{B})_{A C} = 0.1 m A = 1 0^{- 4} A$
$R_{B} = \frac{V}{i _{B}} = \frac{2 - 0.6}{1 0 ^{- 4}} = 14 k Ω$