A Bipolar Junction Transistor has - 0-98- base current is IB- 25 - A and ICBO- 200 nA-The emitter current is

Given, nbsp; Base current I_B= 25 μ A I_CBO = 200 nA α= 0.98 where, β =α1 α= 0.981 0.98 = 49 Collector current , I_C= β I_B+(1+β )I_CBO ∴ I_C = 49× 25× 10^ ...

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Standard XII

Physics

Transistor as an Amplifier

A Bipolar Jun...

Question

A Bipolar Junction Transistor has $α = 0.98$ , base current
is $I_{B} = 25 μ A$ and $I_{C B O} = 200 n A$ .
The emitter current is

1.33 mA

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1.235 mA

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1.05 mA

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1.26 mA

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Solution

The correct option is D 1.26 mA
Given,
Base current $I_{B} = 25 μ A$
$I_{C B O} = 200 n A$
$α = 0.98$
where, $β = \frac{α}{1 - α} = \frac{0.98}{1 - 0.98} = 49$
$Collector current, I_{C} = β I_{B} + (1 + β) I_{C B O}$
$∴ I_{C} = 49 \times 25 \times 10^{- 6} + (50) \times 200 \times 10^{- 9}$
$I_{C} = 1.235 \times 10^{- 3} A$
$Emitter current = I_{E} = I_{C} + I_{B}$
$= 1.235 m A + 0.025 m A$
$∴ I_{E} = 1.26 m A$

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A Bipolar Junction Transistor has α=0.98, base current is IB=25 μA and ICBO=200 nA. The emitter current is

The correct option is D 1.26 mAGiven, Base current IB=25 μA ICBO=200 nA α=0.98 where, β=α1−α=0.981−0.98=49 Collector current,IC=βIB+(1+β)ICBO ∴IC=49×25×10−6+(50)×200×10−9 IC=1.235×10−3A Emitter current=IE=IC+IB =1.235 mA+0.025 mA ∴IE=1.26 mA

A Bipolar Junction Transistor has $α = 0.98$ , base current
is $I_{B} = 25 μ A$ and $I_{C B O} = 200 n A$ .
The emitter current is