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.
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1\$\begingroup\$ 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. \$\endgroup\$– jonkAug 29, 2020 at 1:48
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1\$\begingroup\$ \$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. \$\endgroup\$– jonkAug 29, 2020 at 2:05
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1\$\begingroup\$ @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. \$\endgroup\$– jonkAug 29, 2020 at 20:26
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1\$\begingroup\$ @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. \$\endgroup\$– jonkAug 29, 2020 at 20:38
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1\$\begingroup\$ @RuviandeCésaro If you want to see a typical equation for the saturation current, you can look here. There's a small bit of discussion there. \$\endgroup\$– jonkAug 30, 2020 at 2:40
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1 Answer
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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.
Image taken from Malvino and Bates, 8th edition Appendix B.