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 RC and we are really not given enough information to do so. Let choose RC = Rload.
We don't have enough information to solve for RB – we can't use the bias stability criterion since we don't have the value of RE either. We will have to (arbitrarily) select a value of RB or RE. 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 RE that is large enough to obtain a reasonable value of VBB, Selecting RE as 400Ω will not appreciably reduce the collector current yet it will help in maintaining a reasonable value of VBB. Thus,
RB = 0.1 β RE = 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
Vout(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.
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