Again, there are situations in which the signal clipping is eligible, for example in B and C class amplifiers. But in many application the signal must be amplified without any distortion or clipping whatsoever. Therefore, we must be able to design amplifiers with maximum unclipped amplification gain.

Let's first see the case that the load resistance is very high or no load is connected. In that case, the AC load line is almost the same as the DC load line, so we can safely work only with the DC load line. The maximum oscillation output can be from 0.5V to V_CE(cut). We avoid operating the transistor near the saturation area because in that area the output signal is highly distorted, therefore we use the number 0.5 for our calculations. So, to set the Q point, all we have to do is to find the middle. There is a simple formula which does exactly this:

Let's see an example. Suppose that we calculated the I_C(sat)=4mA and V_CE(cut)=12V. From these points we draw the DC load line as shown bellow. To set the optimum Q point, we need to know only the V_CE(cut) which is 12V:

This way we achieve maximum oscillation within the complete linear area of the transistor, without exceeding the V_CE(cut) value (12V).

Suppose now that the load resistor is not that big, and the AC load line has different slope than the DC load line. First of all, let's make something clear: Since the load is connected in parallel (AC analysis) to the collector or emitter resistor, this means that the resulting AC resistor (r_c or r_e) can only be smaller than the collector or emitter DC resistor (R_C or R_E). Therefore, it is easy to understand that the I_c(sat) current of the AC load line can only be higher than the I_C(sat) current of the DC load line. And since the AC and DC load lines intersect at the Q point, it is absolutely certain that the V_ce(cut)-AC voltage of the AC load line can only be less than the V_CE(cut)-DC voltage of the DC load line.

The previous statement makes clear that, if the AC load line is not the same as the DC load line, the output oscillation without clipping becomes smaller. As a matter of fact, the new oscillation range will be 0,5V up to Vce(cut-off)-ac. To calculate the Q point we use the same formula as before, but we replace the V_CE(cut)-DC term with the V_ce(cut)-AC:

This formula tells us that, in order to achieve the maximum unclipped output, we need to set the Q point in the middle (approximately) of the AC load line. Here is an example:

There are many ways to change the Q point, since any change on the DC bias will also change the Q point. The designer may choose to go with the trial and error method, or by solving the mathematics equations. But no matter which method is used, the designer must be able to locate the biasing part that needs to be changed, so that this change will have big effect on the Q point and small or no effect on the rest of the circuit and its characteristics.

As always, we suppose that the transistor is biased with a voltage divider, and the emitter has also a small feedback resistor. Let's remember how V_CE is calculated:

So, by changing either R_C or R_E, we can change the V_CE, thus we change the V_CEQ of the Q point. But which one to choose? The answer is simple. The capacitor C_E acts as a bridge in AC signal, which means that that the resistor R_E does not have any affect on the AC signal, and therefore has no affect on the AC load line. Therefore, we prefer to change the emitter resistor R_E, since it affects only the DC load line. By increasing the I_E (=I_C) current the V_CEQ point shifts rights. If I_E is decreased the V_CEQ point shifts left.

Here is an example of a common collector connection:

It is obvious that if R_E is changed, it will have an affect on both AC and DC load lines, since there is no bypassing capacitor across this resistor. As we know, this type of connection is also called "emitter follower", because the emitter voltage follows the base voltage:

Moreover, from the schematic we can calculate the V_CE:

We replace the term V_E in the second formula (2) from the first formula (1):

This equation tells us that we can change the V_CE and thus the V_CEQ of the Q point by changing the base voltage V_B. So, we can simply change the R_B1 or R_B2 resistor to achieve the optimum Q point. There is something that we need to take into account here. First, let's see the AC equivalent of the circuit:

As you see, R_B1 and R_B2 are still active components in the AC equivalent. These components have an affect at the input signal, since R_B1 and R_B2 will eventually define the input stage impedance. A large change on either resistor may require to re-design the circuit.

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