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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 Photon Feb 27 '17 at 4:59
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    \$\begingroup\$ I'm having a hard time understanding why or how you got \$ 98/2 \$. \$\endgroup\$ – KingDuken Feb 27 '17 at 5:04
  • \$\begingroup\$ I would keep the image as it was appropriate to visualizing why the solution is what it is. \$\endgroup\$ – The Photon Feb 27 '17 at 5:04
  • \$\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\$ – user96630 Feb 27 '17 at 5:06
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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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    \$\begingroup\$ \$\beta=\frac{\alpha}{1-\alpha}\$. This is essentially what @2017 already calculated. \$\endgroup\$ – The Photon Feb 27 '17 at 5:09
  • \$\begingroup\$ \$\beta=\frac{98}{2}=49\$. Hence, \$I_b=\frac{20}{49} mA\$. \$\endgroup\$ – user96630 Feb 27 '17 at 5:22
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You don't need to know \$\beta\$ to answer the question. Just Kirchoff's Current Law.

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  • \$\begingroup\$ But using beta I get answer directly as \$20/49\$ mA. Is that wrong? \$\endgroup\$ – user96630 Feb 27 '17 at 5:14
  • \$\begingroup\$ @ThePhoton How can you solve the question using Kirchoff's law? \$\endgroup\$ – Yashas Feb 27 '17 at 6:24
  • \$\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\$ – The Photon Feb 27 '17 at 6:27
  • \$\begingroup\$ @ThePhoton How do you get \$ I_e \$? \$\endgroup\$ – Yashas Feb 27 '17 at 6:28
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    \$\begingroup\$ @YashasSamaga, "98 percent of the electrons injected into base [from the emitter] region reach collector", so \$I_c = 0.98 I_e\$. \$\endgroup\$ – The Photon Feb 27 '17 at 6:30

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