The collector has current of the NPN transistor is equal to $10$ mA. If $90 %$ of the electrons emitted reach the collector, then

i_{E} = - 1 m A, i_{B} = 9 m A

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i_{E} = 9 m A, i_{B} = - 1 m A

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i_{E} = 1 m A, i_{B} = 11 m A

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i_{E} = 11 m A, i_{B} = 1 m A

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Solution

The correct option is D $i_{E} = 11 m A, i_{B} = 1 m A$
collector current $I_{c} = 10 m A$
$I_{c} = 90$
$I_{c} = \frac{90}{100} \times I_{E}$
$I_{E} = \frac{I_{C} \times 100}{90}$
$= \frac{10 m A \times 100}{90}$
$= 11.11 m A$
$b a s e c u r r e n t I_{B} = I_{E} - I_{C}$
$(11.11 - 10) = 1.1 m A$

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Similar questions

10 mA

90 %

(I_{E})

(I_{B})

I_{E}

1 m A

α

=

0.95

I_{e} = 1 m A; α = 0.95

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