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Referring to section 2.3.4 A1 (pages 94-95) of the Art of Electronics 3rd Ed., I'm having difficulty understanding how to calculate distortion of a grounded emitter amplifier. The book says:

Because the gain is proportional to the drop across the collector resistor, the nonlinearity equals the ratio of instantaneous swing to average quiescent drop across the collector resistor:

$$\frac{\Delta G}{G} \approx \frac{\Delta V_{out}}{V_{drop}}$$

where Vdrop is the average, or quiescent, voltage drop across the collector resistor Rc.

Now in the example in the first paragraph of page 95, Vcc = +10V and the output is biased to half of Vcc hence Vdrop = 5V. The authors were successful at comparing measured distortions to predicted values. The results were 0.7% distortion for 0.1V output sinewave amplitude and 6.6% for 1V amplitude. I tried to "predict" the distortion but either my math or general understanding is wrong. For the first instance, I said the ratio of instantaneous swing to be 0.2V and the average quiescent drop to be 5V:

$$ \frac{\Delta V_{out}}{V_{drop}} = \frac{0.2}{5} = 0.04 $$

Or 4%.What is the mistake I am making?

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  • \$\begingroup\$ Show the schematic. In a grounded emitter amplifier, the non-linearity of the base-emitter junction will produce about 100% distortion for about 0.2 volts across that input, if you trust the standard Intercept-point maths. The Art of Electronics math is merely reverse-engineering the collector swing, to extract the delta voltage across the base-emitter junction. From what I recall, 4 milliVolts on that junction will produce 10% distortion. \$\endgroup\$ Commented Nov 6, 2018 at 5:38

2 Answers 2

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The problem occurs because \$r_e=\frac{k\:T}{q\:I_\text{E}}\$ and because the gain is \$A_V=\frac{R_\text{C}}{r_e}\$ with the emitter grounded. Ignoring the base's recombination current, \$I_\text{E}\approx I_\text{C}=\frac{V_\text{CC}-V_{\text{C}}}{R_\text{C}}\$, so:

$$A_V=\frac{R_\text{C}}{\frac{k\:T}{q\:\frac{V_\text{CC}-V_{\text{C}}}{R_\text{C}}}}=\frac{R_\text{C}}{\frac{k\:T}{q}}\cdot\frac{V_\text{CC}-V_{\text{C}}}{R_\text{C}}=\frac{V_\text{CC}-V_{\text{C}}}{\frac{k\:T}{q}}$$

Note that the collector resistor disappeared. Also, you should know that \$V_T=\frac{k\:T}{q}\approx 26\:\text{mV}\$ at room temperature. So the above equation, discounting temperature variations, becomes:

$$A_V=\frac{V_\text{CC}-V_{\text{C}}}{26\:\text{mV}}\tag{1}\label{eq1}$$

In your example, \$V_\text{CC}=10\:\text{V}\$ and the nominal \$V_\text{C}=5\:\text{V}\$, so with the signal value at its midpoint the gain is about \$A_V\approx 192.31\$. But with \$V_\text{C}=5\:\text{V}+100\:\text{mV}=5.1\:\text{V}\$ then \$A_V\approx 188.46\$ and with \$V_\text{C}=5\:\text{V}-100\:\text{mV}=4.9\:\text{V}\$ then \$A_V\approx 196.15\$. Roughly, this is about a gain change of \$\pm 4\$ riding on an average gain of about \$200\$ (rounding things up a bit.) That works out to about \$\pm 2\:\%\$.

But this is the "peak" figure, as the text mentions. The text suggests that you divide this value by 3, as a rule they provide, to get the "waveform distortion." This computes out to \$ 0.6\overline{6}\:\%\$ and they round it to \$\pm 0.7\:\%\$.

(I suspect they mean by this to imply total harmonic distortion. If so, the calculation for this is complex and beyond the scope of my answer in order to prove their rule of thumb answer. But there are also several different meanings for distortion, too. And a complete comparison of them here would be far, far beyond my desire. So you will simply have to accept their approach for these purposes. They have the tools on hand for the analysis and were able to empirically test their estimates. So let's leave it at that.)

It's simple enough for you to use the above formula, using \$V_\text{C}=5\:\text{V}-1\:\text{V}=4\:\text{V}\$ and also \$V_\text{C}=5\:\text{V}+1\:\text{V}=6\:\text{V}\$ to work out, after dividing by 3 again, their figure of \$\approx 6.6\:\%\$.

Again assuming that \$I_\text{C}\approx I_\text{E}\$ and using an emitter resistor to degrade the gain (intentionally, often quite substantially), equation \$\ref{eq1}\$ becomes:

$$A_V=\frac{1}{\frac{26\:\text{mV}}{V_\text{CC}-V_\text{C}}+\frac{R_\text{E}}{R_\text{C}}}\tag{2}\label{eq2}$$

(If you need a walk-through to see how that equation arrives, I can provide it. I'm leaving it this way so that you have to do some work to see if you can arrive at the same place on your own.)

So long as \$\frac{R_\text{E}}{R_\text{C}}\gg \frac{26\:\text{mV}}{V_\text{CC}-V_\text{C}}\$, for all \$V_\text{C}\$, then the gain is \$A_V\approx \frac{R_\text{C}}{R_\text{E}}\$.

But from the textbook's statement that the quiescent value of \$V_\text{E}=250\:\text{mV}\$ and assuming that \$I_\text{C}\approx I_\text{E}\$, we know the value of \$\frac{R_\text{E}}{R_\text{C}}=\frac{250\:\text{mV}}{5\:\text{V}}=0.05\$.

So using equation \$\ref{eq2}\$ to estimate the distortion, using the same peak values of \$V_\text{C}\$ as before, we get \$A_V\approx 18.12\$ @ \$V_\text{C}=5\:\text{V}\$, \$A_V\approx 18.08\$ @ \$V_\text{C}=5.1\:\text{V}\$, and \$A_V\approx 18.15\$ @ \$V_\text{C}=4.9\:\text{V}\$. Roughly, this is about a gain change of \$\pm 0.034\$ riding on an average gain of about \$18.1\$. That works out to about \$\pm 0.2\:\%\$. Dividing this value by 3, you get a figure of about \$0.07\:\%\$. Which is very close to the textbook's statement about the results of the analyzer's figure.

If you redo the calculations now using \$\pm 1\:\text{V}\$ variation rather than \$\pm 100\:\text{mV}\$ variation, you'll again find that the results are quite similar to their analyzer results.

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  • \$\begingroup\$ I understood how you derived equation 2. Overall, this seems to be an acceptable answer. \$\endgroup\$
    – DanielJG
    Commented Nov 7, 2018 at 14:10
  • \$\begingroup\$ @DanielJG If you can see improvements to make, let me know. I didn't want to go through the trouble of deriving complete solutions so that I could determine which of several meanings of "distortion" they'd intended. Too much work for too little gain, at this stage. I could derive a sensitivity equation for equation 2, easily, though. But if you aren't familiar with those concepts in calculus, that might not help much except show an easier way to work out the peak gain as a percent of intended gain. Anyway, always open to suggestions to improve things. \$\endgroup\$
    – jonk
    Commented Nov 7, 2018 at 16:30
  • \$\begingroup\$ @jonk I know this is an old answer but how did toy calculate the 0.034%? of a gain change? \$\endgroup\$
    – G36
    Commented Aug 18, 2020 at 15:08
  • \$\begingroup\$ @G36 I didn't write "%" there. That said, I'll have to reread this before answering. When I get time, I'll do that. \$\endgroup\$
    – jonk
    Commented Aug 18, 2020 at 16:04
  • \$\begingroup\$ @G36 Looks like I just used equation 2? 1/(.026/(10-5)+.05) = 18.115942, for example. I'm not promising I got the equation right. I'd need to take more time to double-check my processes. But just plugging in values into equation 2, those I indicate, seems to get the numbers I see. Is there a deeper question here, like perhaps I need to re-check equation 2? Or that I need to re-examine why I used 4.9, 5.0, and 5.1 volts? \$\endgroup\$
    – jonk
    Commented Aug 18, 2020 at 17:55
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You are missing the "correctional" factor of 3 that is on the bottom right side of page 94. You need to apply it to get the distortion of 0.7%.

$$ \frac{1\text{V amplitude}}{5\text{V drop across the collector resistor}* \text{factor }3}=6.7\% $$

Likewise, to get the figures of close to 0.08% and 0.74% that are present below in the same first paragraph on page 95, you have to divide the result by 2.5 (that is rounded to 3).

Art of Electronics. Exercise 2.12.

$$ \frac{1\text{V amplitude}}{5\text{V drop across the collector resistor}* \text{factor }2.5}*\frac{0.0253\text{V that is }V_{T} }{0.0253+0.25\text{V across }R_{E}}=0.735\% $$

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