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I know the total current inside a BJT has to be constant all throughout the device, but the figures above seem to suggest that the total current in the emitter should be larger than the total current in the collector. Since we have extra hole current in the emitter going the opposite direction as the electron current (the conventional current for both have the same direction), shouldn't it be that the current in the emitter region is higher than the current in the collector region?

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  • \$\begingroup\$ I am not sure why you are talking about hole currents going in the opposite direction of the electron current. It's much simpler than that. The emitter current is higher than the collector but that's because the current in the base also flows through the emitter, regardless of whether you choose to look at hole or electron current. Your diagram above shows that too. I am not sure what you mean when you say "total current inside a BJT has to be constant all throughout the device" \$\endgroup\$
    – DKNguyen
    Oct 27, 2021 at 2:42
  • \$\begingroup\$ What I thought was that just as the total current is the all throughout inside a pn diode (shown in the very last figure in the link pveducation.org/pvcdrom/pn-junctions/…), the total current inside a BJT should be the same. Is this not the case since it is a three terminal device? \$\endgroup\$
    – user207787
    Oct 27, 2021 at 3:09
  • \$\begingroup\$ That entire link is talking about a PN junction, not a BJT. An NPN BJT is two PN junctions mashed together so I guess one way to think about it is some of the N junction belongs to one P and the rest of it belongs to the other P And the N has a terminal to it so current can sneak in and out of it without making it to the next P. The BJT doesn't break Kirchoff's Current Law. \$\endgroup\$
    – DKNguyen
    Oct 27, 2021 at 3:12
  • \$\begingroup\$ Node laws applies inside the BJT too. Emitter collector is greater than the collector one since it has base current too but often you simplify saying they are equivalent since base current is way lower than collector current. \$\endgroup\$ Oct 27, 2021 at 6:11

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The current in the emitter region is indeed larger than in the collector, because the emitter current has to be equal to the collector current plus the base current. That's just KCL.

Just like \$β\$ is defined as \$\frac{I_c}{I_b}\$, you can also define \$α=\frac{I_c}{I_e}\$, which is always less than one--i.e., \$I_c<I_e\$. (α is also often seen defined as \$\frac{β}{β+1}\$, which can be derived from the definitions given above and KCL.)

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You are correct, emitter current is base current + collector current. In fact, emitter current is I(emitter) = I(base) + I(collector)

And you can calculate Beta from measured values... Beta = I(collector) / I(base)

Then calculate exactly how much current is in each part of the transistor.

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  • \$\begingroup\$ And @hearth posted essentially the same answer as I was typing. \$\endgroup\$ Oct 27, 2021 at 4:07

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