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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

Dr H.
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    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. – Kuba hasn't forgotten Monica Mar 28 '23 at 01:11
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    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.) – periblepsis Mar 28 '23 at 01:24
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    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. – periblepsis Mar 28 '23 at 01:27
  • @Kubahasn'tforgottenMonica: OK it is edited now. – Dr H. Mar 28 '23 at 07:46
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    @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 https://en.wikipedia.org/wiki/Two-port_network#Hybrid_parameters_(h-parameters). 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. – Dr H. Mar 28 '23 at 07:53
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    @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. – Dr H. Mar 28 '23 at 08:16
  • Isn't beta = dic/dib? – Antonio51 Mar 28 '23 at 08:34
  • @Antonio51 yes in a small-signal model. This is a large-signal (dc, no differentials) model. – Dr H. Mar 28 '23 at 08:37
  • @DrH. Right. I just forget. – Antonio51 Mar 28 '23 at 08:46
  • AFAIK, \$ \beta_0 \equiv h_{FE} = \frac{\alpha_F}{1-\alpha_F} \$. Just different names. All _static_ CE parameters to name correctly. If you do the math and don't find that result, you got something wrong or you missed some step. There're no approximations. See e.g. Sze if you want a more in-depth discussion, but it says the same. – edmz May 02 '23 at 19:18

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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.)

Eeyn
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