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Shouldn't the Is * e^(Vbe/Vt) formula apply to the emitter current and not the collector current seeing as that junction is forward biased?

Pwestm
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  • Are you suggesting that the ebers-moll equation is incorrect in some way? – Andy aka Dec 23 '23 at 20:14
  • That's not the actual basic model for the NPN. The basic level 1 DC model is found here. (There are 3 equivalent ones.) Subsequent models (level 2 added AC and level 3 adds, for example, the Early Effect) provide improvements. Gummel-Poon makes the Early Effect more physically related (good thing) and that incorporates the Late Effect, among many other improvements. And then there is VBIC and Mextram. – periblepsis Dec 23 '23 at 20:15
  • Simply divide the $I_S$ coefficient by$\frac{\beta}{\beta +1}$ and you will get the equation for $I_E = \frac{I_S}{\frac{\beta}{\beta +1}}e^{\frac{V_{BE}}{V_T}} $ current. Or you could divide $I_S$ by beta to get the base current equation $I_B = \frac{I_S}{\beta}e^{\frac{V_{BE}}{V_T}}$. – G36 Dec 23 '23 at 20:16
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    The emitter current is the same as the collector current but with the base current added to it, –  Dec 23 '23 at 20:18

2 Answers2

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That's what the transistor effect is all about.

You modulate the current flowing between two nodes (Collector and Emitter) by modulating the current flowing between two other nodes (Base and Emitter).

The current that flows between the Base and Emitter is the input of the system.

The current that flows between the Collector and Emitter is the output of the system.

Enrico Migliore
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  • With the 1st sentence you confirm the formula mentioned by Pwestm - and with your 2nd sentence you seem to claim the opposite. What is true? – LvW Dec 24 '23 at 11:43
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It is true that the ideal transistor (bipolar junction transistor) has emitter current well described by the familiar diode equation. It is also true that typical collector current will be within one or two percent of the emitter current, in normal (non-saturated) operation.

When using a three-terminal transistor as an amplifier with emitter common, the emitter is nominally at a ground potential, which implies that the 'output' current is the collector current, and thus, a useful model of the amplifier must yield a collector-current result.

Because we can assume non saturation, and small base current, the collector current thus is approximately $$ I_c \simeq I_e = I_{sat} e^{{V_{be} \over{kT} }}$$

Whit3rd
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