The dc common emitter current gain of a n-p-n transistor is 50. The potential difference applied across the collector and emitter of a transistor used in CE configuration is, VCE = 2 V. If the collector resistance,RC = 4 k Ω, the base current (IB) and the collector current (IC ) are

Question

The dc common emitter current gain of a n-p-n transistor is 50. The potential difference applied across the collector and emitter of a transistor used in CE configuration is, VCE = 2 V. If the collector resistance,RC = 4 k Ω, the base current (IB) and the collector current (IC ) are (A) IB = 10 μ A, IC = 0.5 mA (B) IB = 0.5 μ A, IC = 10 mA (C) IB = 5 μ A, IC = 1 mA (D) IB = 1 μ A, IC = 0.5 mA

Tardigrade Admin · Accepted Answer

Ic = ( VCE/RC) = (2/4 × 103) = 0.5 × 10-3 A = 0.5 mA β =(IC/IB) ⇒ IB = (Ic/β) = (0.5× 10-3/50) = 10-5 A = 10 μ A

$I_{B} = 10 μ A, I_{C} = 0.5 m A$

$I_{B} = 0.5 μ A, I_{C} = 10 m A$

$I_{B} = 5 μ A, I_{C} = 1 m A$

$I_{B} = 1 μ A, I_{C} = 0.5 m A$

$I_{c} = \frac{V _{CE}}{R _{C}} = \frac{2}{4 \times 1 0 ^{3}}$
$= 0.5 \times 1 0^{- 3} A = 0.5 m A$
$β = \frac{I _{C}}{I _{B}}$
$\Rightarrow I_{B} = \frac{I c}{β} = \frac{0.5 \times 1 0 ^{- 3}}{50}$
$= 1 0^{- 5} A = 10 μ A$

Q. The dc common emitter current gain of a n-p-n transistor is 50. The potential difference applied across the collector and emitter of a transistor used in CE configuration is, VCE​=2V. If the collector resistance,RC=4kΩ, the base current (IB​) and the collector current (IC​) are

IB​=10μA,IC​=0.5mA

IB​=0.5μA,IC​=10mA

IB​=5μA,IC​=1mA

IB​=1μA,IC​=0.5mA

Ic​=RC​VCE​​=4×1032​ =0.5×10−3A=0.5mA β=IB​IC​​ ⇒IB​=βIc​=500.5×10−3​ =10−5A=10μA

$I_{B} = 10 μ A, I_{C} = 0.5 m A$

$I_{B} = 0.5 μ A, I_{C} = 10 m A$

$I_{B} = 5 μ A, I_{C} = 1 m A$

$I_{B} = 1 μ A, I_{C} = 0.5 m A$

$I_{c} = \frac{V _{CE}}{R _{C}} = \frac{2}{4 \times 1 0 ^{3}}$
$= 0.5 \times 1 0^{- 3} A = 0.5 m A$
$β = \frac{I _{C}}{I _{B}}$
$\Rightarrow I_{B} = \frac{I c}{β} = \frac{0.5 \times 1 0 ^{- 3}}{50}$
$= 1 0^{- 5} A = 10 μ A$