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I know what thermal voltage is and what it does to the semiconductor, ie. it's the agitated electrons producing electric fields proportional to the absolute temperature. But, like the "r'e", the emitter resistance, its equation is thermal voltage over collector current. But doesn't it add when AC voltage is applied? If I'm applying 20 mV, shouldn't it be 20 mV + 26 mV? (maybe it's a vectorial sum) If it's taught 26mV/Ic, it should be it, I just need to understand why this is the only voltage considered, even with AC signal. Thanks in advance.

Ruvian de Césaro
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  • Your question has a simple answer to it if you are familiar with calculus and differentials. But if not, the explanation gets mired into analogies that really don't work so well and are more confusing than helpful. I have an idea how I might try, anyway. But do you feel comfortable with the concept of a differential in calculus? – jonk Aug 29 '20 at 00:14
  • The thermal voltage itself separately results from fundamental thermodynamics theory and the Boltzmann factor, not **k**. It's based on the simple ratio of the numbers of states at different temperatures and is really no more complex than ***fair dice*** ideas as is often used in elementary probability theory. Perhaps the best introduction is C. Kittel, ***"Thermal Physics"***, John Wiley & Sons, 1969. Read chapters 1-6 in particular. You will often find \$e^{-\frac{E_g}{k}\cdot\left(\frac{1}{T_1}-\frac{1}{T_0}\right)}\$, or something similar, as you read this stuff and develop your own sense. – jonk Aug 29 '20 at 00:27
  • Yes, I'm familiar with calculus and differentials. I just graduated in electronic engineering. Thanks for the reference. – Ruvian de Césaro Aug 29 '20 at 01:13
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    Then the answer is pretty simple. Look at the basic Shockley equation (diode or else the non-linear hybrid-\$\pi\$ model's equivalent for the bjt) and take its differential. Arrange it as a function of base-emitter voltage, first, and take the differential with respect to collector current. This is \$r_e\$. It's nothing more than the local slope of the non-linear equation. In practice, this matters because signal is applied in a way (CE arrangement) that modifies the base-emitter voltage leading to collector current changes. – jonk Aug 29 '20 at 01:48
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    \$r_e\$ is a useful fiction. It's just the local slope, though. (There are also bulk resistances at all three pins, which are completely separate from that, and aren't large.) Think closely, in fine detail, about what happens when you change the base-emitter voltage and how that impacts the collector current. This is important for gain calculations with grounded-AC topologies, as you cannot ignore that slope as it plays with the collector load. It's important for temperature reasons, as well. You know these things, as you must in order to graduate. Perhaps I'm not following your quandary. – jonk Aug 29 '20 at 02:05
  • Could you re-state what's bothering you in a different way? Perhaps if you write it differently, it will "sing" better in my mind and I'll be able to grasp more closely what you are struggling over. (This could be nothing more than a modest language and/or mirror neuron difference issue that needs to be bridged.) – jonk Aug 29 '20 at 02:06
  • Note that if the temperature changes -2mV/‘C it is no longer re=26/Ie – Tony Stewart EE75 Aug 29 '20 at 12:49
  • Re-stated: When there's no voltage applied it's 26mV/Ic. When there's any voltage applied it's still 26mV/Ic. Somehow doesn't matter the applied voltage, the voltage there is 26 mV, and Ic will define it's resistance. But I will try to take some derivatives to reach to that equation, thank you, jonk. Tony, yes, I'm aware of it. – Ruvian de Césaro Aug 29 '20 at 18:47
  • @RuviandeCésaro Please note that Tony's comment doesn't just incorporate changes in the thermal voltage, \$V_T\approx 26\:\text{mV}\$ at room temps, but also the ***much more important*** changes in the saturation current. If it were just \$V_T\$ then the change would be positive. But notice that Tony writes a negative relationship with temps. This is because the saturation current completely overwhelms the positive relationship of \$V_T\$, swamps it out and then totally reverses it and changes the sign. – jonk Aug 29 '20 at 20:22
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    @RuviandeCésaro I added the above note because you may otherwise be confused by Tony's comment and imagine he's speaking about just one effect when he's actually talking about an entirely different effect which is much larger and dominates the circumstances. Tony wrote "26/Ie" and you might have accidentally imagined that he was referring to the issue of the "26" mentioned there. I wanted to make sure that you don't improperly stick the wrong things in your head from his comment. – jonk Aug 29 '20 at 20:26
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    @RuviandeCésaro Typically, \$-1.6\:\frac{\text{mV}}{\text{K}} \lt \frac{\text{d}\,V_\text{BE}}{\text{d}\,\text{K}} \lt -2.4\:\frac{\text{mV}}{\text{K}}\$. This combines the more important impact of temperature on the saturation current, on the order of \$T^3\$, and the far less important impact of temperature on the thermal voltage, \$V_T\$, obviously on the order of \$T\$. Both impact the Shockley equation and the resulting \$V_\text{BE}\$. But the changes due to the saturation current's variation with temperature clearly dominate the situation. – jonk Aug 29 '20 at 20:38
  • @jonk On your last comment things became clear, I'm sure I learnt something interesting and neat. Thank you. So considering just the impact on the order of T is wrong? They taught me that it's right. – Ruvian de Césaro Aug 30 '20 at 02:26
  • So as I posted in the answer (and Malvino says that), the variation you're talking about, @jonk is about the 25 or 26 mV or the Is in the denominator? Just to get it clear. Thank you again. – Ruvian de Césaro Aug 30 '20 at 02:38
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    @RuviandeCésaro If you want to see a typical equation for the saturation current, you can [look here](https://electronics.stackexchange.com/a/427343/38098). There's a small bit of discussion there. – jonk Aug 30 '20 at 02:40
  • @jonk I noticed that you've got the "greater/smaller than" symbol the wrong way. -1.6 > -2.4. Anyway, I got your point. – Ruvian de Césaro Aug 30 '20 at 02:43
  • @RuviandeCésaro Yeah. I should have wrapped things with the absolute value or else changed those. Sorry about that. – jonk Aug 30 '20 at 02:44
  • @RuviandeCésaro Some more writing I've done is [here](https://electronics.stackexchange.com/a/478757/38098), [here](https://electronics.stackexchange.com/a/478757/38098), and [here](https://electronics.stackexchange.com/a/423470/38098). Plus [this](https://electronics.stackexchange.com/a/252199/38098) may be of minor interest. – jonk Aug 30 '20 at 02:47
  • Your first and second links are the same, @jonk. All are of interest, thank you very much. I've found the third previously and found it very stunning and wonderful. – Ruvian de Césaro Aug 30 '20 at 04:08

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I found the answer, and it really is what @jonk posted in the comments, it's the derivative of V(I). When I saw this I promptly remembered I saw the appendix on Malvino (bibliography of electronics course) something very similar. So I went there again to recheck and now it's very clear and meaningful. There's also the great trick to reuse equation B-2.

Malvino and Bates, 8th edition Appendix B

Image taken from Malvino and Bates, 8th edition Appendix B.

Ruvian de Césaro
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