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I am currently re-reading my old textbook Microelectronics by Millman and Grabel. Attached two pages. Here is a somewhat theoretical question on large signal parameters for the BJT.

Starting with Ebers-Moll equations (3-6) and (3-7) and assuming \$V_{CB} = 0\$ they easily derive the common-base forward short circuit current gain \$h_{FB}\$ which is defined as $$ h_{FB} \equiv \left. -\frac{I_C}{I_E} \right|_{V_{CB} = 0} $$ In \$h_{FB}\$ the F stands for FORWARD (normal use of transistor) and B for common-BASE. Short-circuit means that \$V_{CB} = 0\$, i.e. the collector is connected to common (base, as in this case) .

It is easy to see that \$h_{FB} = \alpha_F\$ as in equation (3-8). All good so far.

Then there is the common-emitter forward short-circuit current gain, \$h_{FE}\$ which is defined as $$ h_{FE} \equiv \left. -\frac{I_C}{I_B} \right|_{V_{CE} = 0} $$ i.e. the gain of collector current versus base current when the collector is grounded to the common emitter.

In (3-13) they define \$\beta_F = \frac{\alpha_F}{1-\alpha_F}\$ and then simply claim that \$h_{FE} = \beta_F\$. In other words, they claim that $$ h_{FE} = \frac{\alpha_F}{1-\alpha_F} $$

But when I do the math to derive \$h_{FE}\$ from Ebers-Moll, by using emitter as common and assuming \$V_{CE} = 0\$, I get $$ h_{FE} = -\frac{\alpha_F - \frac{\alpha_F}{\alpha_R}}{1 - \alpha_F + (1 -\alpha_R) \frac{\alpha_F}{\alpha_R}} \neq \frac{\alpha_F}{1-\alpha_F}. $$ What is right here? Can you derive \$h_{FE} = \beta_F\$?

Images from Microelectronics 2nd ed. by Jacob Millman & Arvin Grabel, McGraw-Hill 1987, pages 89 and 90.

enter image description here enter image description here

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    \$\begingroup\$ You get \$h_{FE} \neq \frac{\alpha_F}{1-\alpha_F}\$. OK, not that, so what do you get? Please edit it into the question. \$\endgroup\$ Mar 28 at 1:11
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    \$\begingroup\$ Going back as far as I've done (decades before my Millman from '79), \$\beta_{_\text{F}}=h_{_\text{FE}}\$. Note that the 'h' is from 'hybrid' and it refers to the early (1950's) hybrid-\$\pi\$ model, which was a later re-analysis developed from the earlier transport and injection versions published before it. (They are mathematically equivalent, though notation is different, but the hybrid-\$\pi\$ is used today as computer programs can linearize it with less effort and its modeling of the low-current variation of \$\beta\$ is also easier to handle.) \$\endgroup\$ Mar 28 at 1:24
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    \$\begingroup\$ Here's a question: How do you design an experimental setup and then rigorously process the experimental results in order to find \$h_{_\text{FE}}\$ at \$V_{_\text{CE}}=0\:\text{V}\$? I'm curious what you imagine here. \$\endgroup\$ Mar 28 at 1:27
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    \$\begingroup\$ @periblepsis: This is DC analysis, before the hybrid-pi (small signal) model is even introduced in the text. The h at this point refers to the regular hybrid two-port parameters en.wikipedia.org/wiki/…. Maybe Millman is confusing the large-signal beta and small-signal beta? I don't know how I would set it up experimentally and I don't believe it is relevant since this is just model-based theory. \$\endgroup\$
    – Dr H.
    Mar 28 at 7:53
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    \$\begingroup\$ @periblepsis now that I think about it more, I believe the problem here is that Millman erroneously applies two-port concepts (h-parameters) to the DC model, which is non-linear. Two-ports are for linear networks only, such as the hybrid-pi. \$\endgroup\$
    – Dr H.
    Mar 28 at 8:16

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They are assuming the diode currents in Figure 3-11 are negligible when the diodes are reverse biased. \$h_{FE}\$ is defined with the collector-base junction reverse biased, therefore, \$I_{CD}\$ is negligibly small, which in turn means the \$\alpha_R\$ dependent current source is turned off.

Then \$I_C\$ = \$\alpha_F * I_{ED} = -\alpha_F * I_E\$ and the rest follows directly from that.

(Unsolicited side opinion: Millman's sign convention for the currents is awful.)

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