AC Amplification

AC Amplification
a single transistor AC amplification circuit is given as shown in the figure.



If the capacitances of the coupling capacitors and the emitter by-pass capacitor are large enough with respect to the frequency of the AC signal in the circuit is high enough, these capacitors can all be approximated as short circuit. Moreover, note that the AC voltage of the voltage supply is zero, it can be treated the same as the ground. Now the AC behavior of the transistor amplification circuit can be modeled by the following small signal equivalent circuit:



AC Input Impedance: For AC signals, the input of the amplification circuit is shown below, where is the internal resistance of the signal source, and the input impedance of the circuit is the three resistances , and in parallel:







where is the resistance of the PN-junction) between the base and the emitter of the transistor,






as discussed before:






AC Output Impedance: This is simply the resistance of the resistor







AC Amplification Gain: Given the AC input voltage , the base voltage can be found (voltage divider) to be







and the base current is






The collector current is and the output voltage is






Here the negative sign indicates the fact that is out of phase with , as






The voltage gain is therefore






In particular, if the input resistance is much larger than the internal resistance of the voltage source, i.e.,






and the output resistance is much smaller than the load resistance, i.e.,






then the gain can be approximated as






Example 1:



This figure shows a common-emitter amplification circuit of npn transistor. Assume , , , (assuming ). Also we assume the capacitors are large enough so that they can be considered as short circuit for the AC signals.


Find base current:






Find DC load line: when , , when ,






The DC load line is determined by these two points
Find DC operating point : The intersection of the DC load line and the curve corresponding to is the DC operating point with and (i.e., ).
Find AC load line: The AC load is . The AC load line is a straight line passing the DC operating point with its slope equal to . The intersections of AC load line with and axes can be found by













Find input voltage and current: Assume input voltage is and , the overall base voltage is , and the corresponding base current can be found from the input characteristics to be between 20 and 60 . Also note that .
Find AC output voltage: This can be found graphically from the output i-v characteristics, based on , to be , and the current is . Note that the output is in opposite phase (180 phase shift) with the input.
Find the voltage gain:








For a transistor to work properly, its DC operating point has to be set right, otherwise distortion may be caused, as shown below. So to avid distortion, the dynamic range should be maximized by setting the DC operating point at the middle point of the load line.



The circuit above can also be analyzed using the small-signal model.



Same as before, , and we have the following DC variables:





















The AC variables:






The voltage gain is:






The input resistance is , the output resistance is .


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