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I've been going over the basics reading the Art of Electronics (3rd ed.) by Horowitz & Hill. I am currently reading about the shortcomings of the single-stage grounded-emitter amplifier. Precisely, their non-linearity. I'm kinda lost in that part where the authors say (on p. 94):

Because this represents the extreme variation of gain (i.e., at the peaks of the swing), the overall waveform "distortion" (usually stated as the amplitude of the residual waveform after subtraction of the strictly linear component) will be smaller by roughly a factor of 3.

Where does that 3 come from? How did they derive it?

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    \$\begingroup\$ Geez. I just said the same thing Aof E does at the start of their section on the shortcomings, written in a comment I made today to my own answer that I provided yesterday. The short answer to your question is that they used an instrument which examines the signal, intimately, to provide a "distortion result." (You'd need calculus to replicate this here using only theory to do it.) They are saying that the instantaneous % change in gain is about equal to the % change in the output with respect to the collector resistor's voltage drop. \$\endgroup\$ – jonk Dec 2 at 18:22
  • \$\begingroup\$ But that is larger at the peaks where the collector resistor voltage drop is much smaller and it is smaller at the valleys. So, over the sine wave (using something simpler to analyze than music would be), the average distortion is about 3 times smaller than the worst case (at the peak.) Again, if you really want this, and it requires calculus, I can illustrate how that figure is arrived at. But I think the authors considered it beyond the scope to prove in the book and distracts away from their teaching process. (Students would lose track and therefore not learn the important things there.) \$\endgroup\$ – jonk Dec 2 at 18:26
  • \$\begingroup\$ Thanks for your response . Could you please illustrate how they arrive at the value 3?. \$\endgroup\$ – bob7185 Dec 2 at 18:37
  • \$\begingroup\$ Before I bother, did you follow their earlier words? In short, can you explain to me how they arrived at their \$-\frac{V_\text{DROP}}{V_T}\$ expression and discuss what it means to you and how it applies to this circuit? Because, if you cannot do that and can't also explain how it applies and what that means, then I don't think any calculus I'd attempt would be of any use. \$\endgroup\$ – jonk Dec 2 at 18:48
  • \$\begingroup\$ Yes I understand how they arrived at the previous expression since Gv= -Rc/re multiplying numerator and denominator by Ic you get -Vdrop/Vt , it means basically that the gain varies with collector current ,(high gain at low Vout, low gain at high Vout) thus causing non linearity (distortion) in the output waveform. It's not that I don't understand the concept it's just that I want to see in a more detailed fashion how they came up with the factor 3 \$\endgroup\$ – bob7185 Dec 2 at 19:17

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