**Fig. 1 **

**Solution: **

The maximum voltage across the amplifier is 10 V since the power supply can be visualized as a 10V power supply with a ground in the center. In this case, the ground has no significance to the operation of the amplifier since the input and output are isolated from the power supplies by capacitors.

We will have to select the value for R_{C} and we are really not given enough information to do so. Let choose R_{C} = R_{load}.

We don't have enough information to solve for R_{B} – we can't use the bias stability criterion since we don't have the value of R_{E} either. We will have to (arbitrarily) select a value of R_{B} or R_{E}. If this leads to a contradiction, or “bad” component values (e.g., unobtainable resistor values), we can come back and modify our choice. Let us select a value for R_{E} that is large enough to obtain a reasonable value of V_{BB}, Selecting R_{E} as 400Ω will not appreciably reduce the collector current yet it will help in maintaining a reasonable value of V_{BB}. Thus,

R_{B} = 0.1 β R_{E} = 0.1 (200)(400) = 8 K Ω

To insure that we have the maximum voltage swing at the output, we will use

Note that we are carrying out our calculations to four places so that we can get accuracy to three places. The bias resistors are determined by

Since we designed the bias circuit to place the quiescent point in the middle of the ac load line, we can use

V_{out}(undistorted p-p) 1.8 (2.94 x 10^{-3} ) (2 K Ω || 2 K Ω ) =5.29 V

Now we can determine the gain of the amplifier itself.

Using voltage division, we can determine the gain of the overall circuit.

The value of Rin can be obtained as

Thus the overall gain of the amplifier is

This shows that the common-emitter amplifier provides high voltage gain. However, it is very noisy, it has a low input impedance, and it does not have the stability of the emitter resistor common emitter amplifier.