In p-n-p transistor circuit, the collector current is 10 mA. If 90 % of the holes reach the collector, the emitter and base currents respectively are

Question

In p-n-p transistor circuit, the collector current is 10 mA. If 90 % of the holes reach the collector, the emitter and base currents respectively are (A) 10 mA, 1 mA (B) 22 mA, 11 mA (C) 11 mA, 1 mA (D) 20 mA, 10 mA

Tardigrade Admin · Accepted Answer

Here, Ic = 10 mA, as 90 % of the holes reach the collector, so the collector current, IC = 90 % of IE = (90/100)IE or IE = (100/90)IC = (100/90)× 10 =11 mA Now, base current, IB = IE - IC =11 - 10 = 1 mA

Q. In $p - n - p$ transistor circuit, the collector current is $10 m A$ . If $90%$ of the holes reach the collector, the emitter and base currents respectively are

$10 m A, 1 m A$

$22 m A, 11 m A$

$11 m A, 1 m A$

$20 m A, 10 m A$

Here, $I_{c} = 10 m A$ , as $90%$
of the holes reach the collector, so the collector current,
$I_{C} = 90%$ of $I_{E} = \frac{90}{100} I_{E}$
or $I_{E} = \frac{100}{90} I_{C}$
$= \frac{100}{90} \times 10$
$= 11 m A$
Now, base current,
$I_{B} = I_{E} - I_{C}$
$= 11 - 10$
$= 1 m A$

Q. In p−n−p transistor circuit, the collector current is 10mA. If 90% of the holes reach the collector, the emitter and base currents respectively are

10mA,1mA

22mA,11mA

11mA,1mA

20mA,10mA