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To understand the NPN current equation derivation, I have been studying my lecture handout and the emitter current equation accordingly is found as :

\begin{equation} I_E= q n_i^2 A(\frac{D_p}{N_{dE}W_E}+\frac{D_n}{N_{aB}W_B})(e^{\frac{qV_{BE}}{KT}}-1) \end{equation}

The collector current equation: \begin{equation} I_C= q n_i^2 A(\frac{D_n}{N_{aB}W_B})(e^{\frac{qV_{BE}}{KT}}-1) \end{equation}

This already includes the effect of increasing the collector-emitter voltage in the factor: \begin{equation} \frac{1}{W_B} \end{equation}

As in, if we increase the reverse bias voltage the junction width increases (Wb decreases) which further increases the slope of concentration gradient of electrons in the base region and this also increases diffusion current.

However , in the lecture the professor changes the \begin{equation} I_C= q n_i^2 A(\frac{D_n}{N_{aB}W_B})(e^{\frac{qV_{BE}}{KT}}-1) \end{equation} to \begin{equation} I_C= I_s e^{\frac{qV_{BE}}{KT}}(1+\frac{V_{CE}}{V_A}) \end{equation} to include the Early effect or the increase in current due to changing collector-emitter voltage.

\begin{equation} I_s= q n_i^2 A(\frac{D_n}{N_{aB}W_B}) \end{equation} and -1 removed due to a higher operating point.

But why is this factor \begin{equation} (1+\frac{V_{CE}}{V_A}) \end{equation} necessary in the current equation?

Doesn't \begin{equation} I_s= q n_i^2 A(\frac{D_n}{N_{aB}W_B}) \end{equation} already cover that ?

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  • \$\begingroup\$ I see you are trying to make sense of your lecture. And it is hard to argue, through you, to your lecturer. So there is no possible way to "come to terms" unless everything said makes absolute sense in all circumstances. And they don't. I hope you will look over Gummel & Poon's 1970 paper. It unifies the high current regime, the Early Effect, and the forward transit time vs collector current into a single concept. Meanwhile, I can just suggest that the factor should be \$1+\frac{V_{_\text{BC}}}{V_{_\text{A}}}\$ and not as you show. \$\endgroup\$
    – jonk
    Jan 12, 2022 at 8:42
  • \$\begingroup\$ I am unfortunately a bit confused by the notations used in the paper "Gummel and Poon's 1970 paper" especially in eber's moll model equations (for ex the factor of 1/(1-alpha_n*alpha_i) and usage of charge in further equations , is there any easy way to understand this ? \$\endgroup\$
    – aziazu
    Jan 19, 2022 at 16:46

1 Answer 1

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Probably very very late.

Short answer: No, WB does not account for depletion region expansion

Long answer: WB is not what you think it is. In many cases, people blur the difference between the "quasineutral region of the base" and "the whole base width". The notes seem to have a mix where WB is the quasineutral region, but it is fixed with regards to VBE, page 5, top slide. So to answer your question: No, WB does not include base width modulation (BWM) because it does not change with expanding depletion width. Usually W is used to describe the quasineutral region of the base of a BJT.

Let's examine why (the answer is, way too many assumptions => easy confusion trap). base doping NB is usually taken much higher than collector doping NC, which means that most of the CB junction's (CBJ) depletion width is in the collector. This leads us to ignore the small part of it that ends up in the base region! Now you might think, what about EBJ? well, in forward active mode, that junction is forward biased, and the depletion width is shrinking (as opposed to CBJ who's in reverse bias and expanding its depletion region).

If you are still interested in the subject, I highly recommend this very well written book: Semiconductor Fundamentals by Robbert F. Pierret It is alot of reading, but put in baby steps, and everything it kept track of. It also goes back to examine questions like this one. The book gets into Eber-Moll's model (and derives it very easily), and continues from there. While the book explores BWM's effect on IC, it does not come up with the simple form you have (1 + VCE/VA)

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