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. $20\,mA, 10\,mA$

Answer 1

Answer: (C)

$11\,mA, 1\,mA$

Answer 2

Here, $I_c = 10\, mA$, as $90\%$ of the holes reach the collector, so the collector current, $I_C = 90\%$ of $I_E = \frac{90}{100}I_E$ or $I_E = \frac{100}{90}I_C$ $= \frac{100}{90}\times 10$ $=11\,mA$ Now, base current, $I_B = I_E - I_C$ $=11 - 10$ $= 1\,mA$