Non linearity of the single stage grounded emitter amplifier

[bob7185]

Hi, everyone I've been going over the basics reading the AoE 3rd edition by HH.  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:"... 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 ? If somebody can please help with that doubt.

[magic]

Kinda impossible to answer without more context or the same book that you are reading

[T3sl4co1l]

It's on page 94.

They're referring to the peak change in gain.  The error averaged over a full waveform is smaller, by about that much.

For example, suppose the output waveform were a triangle wave (rather than the "barn roof" it actually is).  We know the triangle wave has harmonics with amplitudes that go as 1/N^2 (for N odd).  This sums to 1.47% THD.  But the peak error is [assumption].

Hmm.

Excuse me while I step over to the blackboard and doodle something for a moment...

Ah, the closest sine wave to a triangle wave is 0.81246152 amplitude (best fit, 100 samples evenly spaced).  That is, for a triangle of peak amplitude 1 and a sine of peak 0.81..., the RMS error is minimal.

In this case, the peak error is 18.7% while the RMS is 7%, which is, very approximately, the factor of 3 mentioned.
Crude demo:
https://docs.google.com/spreadsheets/d/1gBjter_vGnuW1iC6h7usQKf3WjN0YbGgO2lV13uEJVE/edit?usp=sharing
You may also find this helpful for seeing how such a calculation can be done?

If one were to input an exponential waveform instead of a linear ramp (of some given amplitude, because again, the distortion of an exponential is proportional to its amplitude), they'd get closer to the result H&H got.  The waveform is peakier (a poorer fit to a sine), so the peak-to-RMS error can be expected to be higher, probably above 3 for the amplitude pictured.

(Ed: didn't really have to go and write a spreadsheet for that.  It's sufficient to note that the harmonics are orthogonal; the RMS test is equivalent to an inner product and the inner product of fundamental with any harmonic is zero.  So the "best fit" of course has to be the ratio of the triangle's fundamental amplitude, or 8/pi^2 ~= 0.81.  Not that these terms are probably meaningful to the beginner!)

Tim