# Is it okay to take $\beta=I_c/I_b=98/2$ in this problem?

In an npn power transistor the collector current is 20mA. If 98 percent of the electrons injected into base region reach collector then the base current in mA is nearly ?

Here, in this problem is it okay to take $\beta=I_c/I_b=98/2=49$ ?

• From what you've posted, you don't need to know $\beta$ to answer the question. Feb 27 '17 at 4:59
• I'm having a hard time understanding why or how you got $98/2$.
– user103380
Feb 27 '17 at 5:04
• I would keep the image as it was appropriate to visualizing why the solution is what it is. Feb 27 '17 at 5:04
• @KingDuken Suppose 100 electrons are injected into base then 98 electrons flows into collector and 2 electrons as base current. That is why I took collector current as 98/2 times of base current. I am assuming a common emitter npn amplifier.
– user96630
Feb 27 '17 at 5:06

It sounds like they mean 98% of the emitter current, not the base current. That means you're given $\alpha = 0.98$. You can use that to calculate $\beta$ and find the base current.

• $\beta=\frac{\alpha}{1-\alpha}$. This is essentially what @2017 already calculated. Feb 27 '17 at 5:09
• $\beta=\frac{98}{2}=49$. Hence, $I_b=\frac{20}{49} mA$.
– user96630
Feb 27 '17 at 5:22

You don't need to know $\beta$ to answer the question. Just Kirchoff's Current Law.

• But using beta I get answer directly as $20/49$ mA. Is that wrong?
– user96630
Feb 27 '17 at 5:14
• @ThePhoton How can you solve the question using Kirchoff's law? Feb 27 '17 at 6:24
• @YashasSamaga, KCL tells you $I_b + I_e + I_c = 0$, and the problem statement tells you $I_e$ and $I_c$ (after a simple calculation). Feb 27 '17 at 6:27
• @ThePhoton How do you get $I_e$? Feb 27 '17 at 6:28
• @YashasSamaga, "98 percent of the electrons injected into base [from the emitter] region reach collector", so $I_c = 0.98 I_e$. Feb 27 '17 at 6:30