By the problem description, the amplifier is to source a 35mA sinusoidal current through the load.

Note that during the positive half of the cycle, the output current is from left to right through the capacitor but, during the negative half, *the current is from right to left.*

Note also that the BJT current is (assuming the transistor is active) *always out of the emitter*.

Thus, during the negative half cycle, the current through the emitter resistor is the *sum* of the load current and BJT emitter current.

If follows that the current 'down' through the emitter resistor at the negative peak is *at least* \$35ma\$. This would be the case if the BJT were just 'turning off' at the negative peak.

Since it is already given that the emitter voltage should be \$7.22V\$ at the negative peak, and we know the resistor current is *at least* \$35mA\$ then, Ohm's Law gives

$$R_E \le \frac{7.22V}{35mA}$$

In other words, we have established an *upper bound* on the value of \$R_E\$.

We want maximum voltage swing, so we say the emitter DC voltage is
approximately 7.5V, half of our positive rail...

For sinusoidal operation and assuming the capacitor voltage is constant (the capacitor is an AC short circuit) and equal to the DC emitter voltage \$V_E\$, we have:

$$v_{O+} = V_+ - V_E $$

$$v_{O-} = -V_E \frac{R_L}{R_E + R_L} $$

Typically, one wants these to be equal in magnitude, i.e., we want the output voltage swing to be *symmetric*.

Setting \$|v_{O+}| = |v_{O-}|\$ yields the following condition for symmetric load voltage swing:

$$V_E = V_+ \frac{R_E + R_L}{R_E + 2R_L} $$

For my opinion, it is somewhat confusing to use this term "re" (and your question can serve as a "good" example). This parameter re (called "intrinsic emitter resistance") is nothing else than the inverse of the transconductance re=1/gm. I think, it is not correct to call it "resistance" because it relates the output current and the input voltage (gm=dIc/dVBE). In contrast, the classical resistor definition relates the voltage across and the current through the two-pole element.

This transconductance appears in each gain formula, for example in common emitter configuration:

**Gain A=-gm***Rc/(1+gm*Re)=-Rc/(Re+1/gm)=-Rc/(Re+re) .

The value of gm=1/re can be found by differentiating the transfer function Ic=f(Vbe) at the selected quiescent current Ic and can be found as **gm=Ic/Vt** (Vt=temperature voltage). Hence, it is a parameter which is computed at the DC bias point.

Does this answer your question?

**EDIT**

From the gain formula (with emitter resistor RE) we can derive that the gain is reduced by the factor **1/(1+gm***RE)**. From this we can again derive that the term (gm*RE) is the loop gain of the feedback system. From feedback theory we know that the disrortions are reduced also by the same factor. Rewriting this factor we get

**1/(1+gm*RE)=(1/gm)/(RE+1/gm)=re/(RE+re)**.

This prooves the result you have shown.

## Best Answer

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.