Question

In an NPN transistor, the collector current is 28 mA. If 80% electrons reach the collector, its base current in mA is

• A. \(7\) mA
• B. \(14\) mA
• C. \(28\) mA
• D. \(35\) mA

Hint:

Collector current is 80 percent of the emitter current. Then I_B = I_E - I_C.

Answer 1

The Correct Option is A

Step 1: Share the emitter current:
Of every 100 electrons leaving the emitter, 80 reach the collector and the other 20 leave through the base. So the base carries one quarter of what the collector carries: $\dfrac{20}{80} = \dfrac14$.

Step 2: Apply:
$I_B = \dfrac{28}{4} = 7$ mA.

Step 3: Check with beta:
$\beta = \dfrac{\alpha}{1-\alpha} = \dfrac{0.8}{0.2} = 4$, so $I_B = I_C/\beta = 28/4 = 7$ mA.

Step 4: Why alpha is not 1:
Not every electron from the emitter reaches the collector, because some recombine with holes in the thin base. The 20 percent that do not reach the collector are exactly the base current, so $I_B = 0.2\times35 = 7$ mA.

Final Answer:
Option (A). \[ \boxed{7\text{ mA (A)}} \]