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\$ ?
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\$\begingroup\$ From what you've posted, you don't need to know \$\beta\$ to answer the question. \$\endgroup\$– The PhotonFeb 27, 2017 at 4:59
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1\$\begingroup\$ I'm having a hard time understanding why or how you got \$ 98/2 \$. \$\endgroup\$– user103380Feb 27, 2017 at 5:04
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\$\begingroup\$ I would keep the image as it was appropriate to visualizing why the solution is what it is. \$\endgroup\$– The PhotonFeb 27, 2017 at 5:04
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\$\begingroup\$ @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. \$\endgroup\$– user96630Feb 27, 2017 at 5:06
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2 Answers
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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.
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2\$\begingroup\$ \$\beta=\frac{\alpha}{1-\alpha}\$. This is essentially what @2017 already calculated. \$\endgroup\$ Feb 27, 2017 at 5:09
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\$\begingroup\$ \$\beta=\frac{98}{2}=49\$. Hence, \$I_b=\frac{20}{49} mA\$. \$\endgroup\$– user96630Feb 27, 2017 at 5:22
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You don't need to know \$\beta\$ to answer the question. Just Kirchoff's Current Law.
answered Feb 27, 2017 at 5:06
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\$\begingroup\$ But using beta I get answer directly as \$20/49\$ mA. Is that wrong? \$\endgroup\$– user96630Feb 27, 2017 at 5:14
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\$\begingroup\$ @ThePhoton How can you solve the question using Kirchoff's law? \$\endgroup\$– YashasFeb 27, 2017 at 6:24
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\$\begingroup\$ @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). \$\endgroup\$ Feb 27, 2017 at 6:27
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1\$\begingroup\$ @YashasSamaga, "98 percent of the electrons injected into base [from the emitter] region reach collector", so \$I_c = 0.98 I_e\$. \$\endgroup\$ Feb 27, 2017 at 6:30