@periblepsis yes I would be grateful if you could elaborate on your point of view –Xavier@periblepsis I'm still interested in your point of view. –internetI'll try to get straight to the point. I can always expand, later. So I'll just focus on the point being made in the following section of the 3rd edition of The Art of Electronics:
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The text above discusses a temperature change leading to a change in \$V_{_\text{BE}}\$. This occurs for several different physical and competing reasons. But the more important and dominant one is that the saturation current [a theoretical y-axis intercept value] shifts \$\propto T^3\$.
Their exact choices of \$I_{_\text{C}}\$, \$V_{_\text{BE}}\$, and \$\Delta\,V_{_\text{BE}}\$ are arbitrarily made for education purposes. (For example, for a device similar to the 2N2222 model in LTspice I'd need to use \$I_{_\text{C}}\approx 125\mu\text{A}\$ to get \$V_{_\text{BE}}=600\:\text{mV}\$ at \$27.5^\circ\text{C}\$. And in that case, I'd see \$-1.9536\:\text{mV}\$ change in \$V_{_\text{BE}}\$ for a \$+1^\circ\text{C}\$ change in temperature, from \$27^\circ\text{C}\$ to \$28^\circ\text{C}\$. The text is tossing out some broadly representative values. Not gospel!)
They've chosen a specific example requiring \$A_v=50\$ because this greatly exaggerates (or emphasizes) their point. A very large voltage gain necessarily also requires a very small voltage difference across the emitter resistor. (Discounting \$r_e^{\:'}\$ in their circuit -- which they didn't ignore, by the way -- says that \$A_v\approx\frac{V_{_\text{CC}}-V_{_\text{C}}}{V_{_\text{E}}}\$. So a high gain requires a small value for \$V_{_\text{E}}\$.) The choice of \$A_v=50\$ is no accident. It's pure malicious intent on their part.
They assumed, for purposes of their discussion, that the biasing resistor pair is arbitrarily stiff. (Writing, "since the base is held at constant voltage".) This means that they assume that the base voltage won't change, at all, and that any change to \$V_{_\text{BE}}\$ due to temperature variation causes a change to \$V_{_\text{E}}\$ and only to \$V_{_\text{E}}\$. (Reality, taking into account practical biasing, would involve changes to both \$V_{_\text{B}}\$ and \$V_{_\text{E}}\$.)
So in their example, since \$V_{_\text{BE}}\$ is assumed to decline by \$-2.1\:\text{mV}\$ per Kelvin degree increase, this means that \$V_{_\text{E}}\$ rises by the same amount. That leads to a necessary increase in the emitter current and associated increase in collector current.
Their example says \$\Delta T=20^\circ\text{C}\$, so this adds \$\frac{20\,\cdot\,2.1\:\text{mV}}{175\:\Omega}=20\cdot 12\:\mu\text{A}=240\:\mu\text{A}\$ to \$I_{_\text{C}}\$.
And that makes their point as they write a "nearly 25%" increase in the collector current.
Ebers-Moll, though, provides some negative feedback to this change. In their example, they used \$V_T=25\:\text{mV}\$. Ebers-Moll says that the transistor responds to this change such that \$\Delta V_{_\text{BE}}=V_T\cdot\ln\left(1+\frac{240\:\mu\text{A}}{1\:\text{mA}}\right)\approx 5.4\:\text{mV}\$.
So start with the \$20\cdot -2.1\:\text{mV}=-42\:\text{mV}\$ change due to temperature, suggesting that previously computed \$240\:\mu\text{A}\$ increase in \$I_{_\text{C}}\$. But this is countered by an increase in \$V_{_\text{BE}}\$ due to that increased collector current, which increases \$V_{_\text{BE}}\$ and therefore reduces \$V_{_\text{E}}\$. So instead of a \$42\:\text{mV}\$ increase, there's really more like \$36.6\:\text{mV}\$ change, instead. And this means not quite so much of a collector current change -- perhaps 20%-21% instead of 25%.
(It doesn't stop there because this Ebers-Moll effect reduces \$I_{_\text{C}}\$, which means the previously computed alteration of \$5.4\:\text{mV}\$ that assumed no Ebers-Moll effect is also wrong. Since the new current is less, the \$5.4\:\text{mV}\$ is a little less, which means \$I_{_\text{C}}\$ is a little higher. Etc. You can iterate this over and over. Or you can compute it once using an equation involving the product-log/LambertW function. But it's really not worth all that effort. One step of iteration gets close enough. And the authors of The Art of Electronics felt that even this first step of correction due to Ebers-Moll wasn't worth applying -- for the very reason of just how much I just wrote here to even discuss it.)
While Ebers-Moll helps to reduce the impact of temperature changes, in this case to perhaps about 80% of what the authors of The Art of Electronics suggested in the text, it doesn't really change their main point. The high gain requires a low voltage drop across the emitter resistor, which makes temperature changes to \$V_{_\text{BE}}\$ more pronounced by comparison, which leads to greater changes in the quiescent operating point of the circuit due to operating temperature differences. That point remains.
Obviously, reducing the voltage gain would help because this would allow a larger voltage difference across the emitter resistor and then temperature changes to \$V_{_\text{BE}}\$ would have less impact on the quiescent state of the stage.
The authors then go on to suggest two other roughly equivalent ways to get a higher AC voltage gain while still keeping a substantial voltage difference across the emitter leg. These examples are in Fig. 2.50 and Fig. 2.51.