My question is related to How can I set the quiescent current for a grounded-emitter amplifier and AoE3: Grounded Emitter Amplifier Distortion originated in TAOE3 on the nonlinearity of grounded emitter amplifier. Specifically on how distortion is defined and calculated.
In 2.3.4.A on page 94-95 where it states:
... It's easy to estimate the distortion ... with a grounded emitter, the incremental (small signal) gain is \$G_V = -R_C/r_e = -I_CR_C/V_T = -V_{drop}/V_T\$, where \$V_{drop}\$ is the instantaneous voltage drop across the collector resistor ...
... the nonlinearity (fractional change of gain with swing) equals the ratio of instantaneous swing to average quiescent drop across the collector resistor: \$\Delta G/G = \Delta V_{out}/\Delta V_{drop}\$, where \$V_{drop}\$ is the average, or quiescent, voltage drop across the collector resistor \$R_C\$
... 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. Note that the distortion depends on only the ratio of swing to quiescent drop, and not directly on the operating current...
... As an example, in a grounded emitter amplifier powered from +10V, biased to half the supply (i.e. \$V_{drop} = 5V\$) \$\star\$, we measured a distortion of 0.7% at 0.1V output sine-wave amplitude and 6.6% at 1V amplitude; these values are in good agreement with the predicted values ...
The explanation is somewhat confusing - my take is that the book defines 'distortion' by the ratio, i.e. \$\Delta G/G \ \ or \ \ \Delta V_{out}/\Delta V_{drop}\$, essentially a slope value from a region of the output waveform. How can the magnitude of a ratio represent distortion or nonlinearity. Shouldn't distortion measures how non-linear a graph is?
I don't understand how 0.7% (@0.1V output) or 6.6% (@1V output) came about. If output wave amplitude peak at 0.1V, the voltage drop of \$I_CR_C\$ has to be 5-0.1 = 4.9V. Then the distortion or ratio of \$\Delta V_{drop}\$ / \$V_{drop}\$ = 0.1 / 4.9 = 2%, instead of 0.7% stated in the book. I am either misunderstanding the description or misinterpreting it.
\$\star\$: "biased to half the supply" ... when the term biased appear in BJT throughout TAOE it implies biasing the base, but here specifically it is trying to say curbing the \$V_{CC}\$ supply from +10V to +5V on the collector side, perhaps with a voltage divider on the collector side before connecting to \$R_C\$. Please correct me if I'm wrong.