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101 Determine the AC/open-loop voltage gain of a 2N2222A transistor, using the results of Example 101 and Table 101 Use maximum values of the h parameters, and assume that ICQ = 50 mA, RC = 1 k , RB = 100 k , and RE = 100
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The h-parameter model developed in the previous section is very useful in the small-signal analysis of various con gurations of BJT ampli ers In this section, a set of techniques will be developed to enable you to rst establish the Q point of a transistor ampli er, then construct the small-signal model, and nally use the small-signal model for analyzing the small-signal behavior of the ampli er A major portion of this section will be devoted to the analysis of the common-emitter ampli er, using this circuit as a case study to illustrate the analysis methods At the end of the section, we will look brie y at two other common ampli er circuits: the voltage follower, or common-collector ampli er; and the common-base ampli er The discussion of the common-emitter ampli er will also provide an occasion to introduce, albeit qualitatively, the important issue of transistor ampli er frequency response A detailed treatment of this last topic is beyond the intended scope of this book A complete common-emitter ampli er circuit is shown in Figure 107 The circuit may appear to be signi cantly different from the simple examples studied in the previous sections; however, it will soon become apparent that all the machinery necessary to understand the operation of a complete transistor ampli er is already available
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vout(t)
vS(t)
Figure 107 The BJT common-emitter ampli er
We shall create a small-signal linear AC equivalent-circuit model for the ampli er based on a two-port1 equivalent circuit; this equivalent circuit can then be used in connection with equivalent circuits for the load and source to determine the actual gains of the ampli er as a function of the load and source impedances Figure 108 depicts the appearance of this simpli ed representation for a transistor ampli er, where ri and ro represent the input and output resistance of
two-port circuit is a circuit that has an input and an output port, in contrast with the one-port circuits studied in 3, which had only a single port connecting the source to a load The ampli er con guration shown in Figure 107 is representative of a general two-port circuit
10
Transistor Ampli ers and Switches
iin +
ro + ri
iout
+ ~
vin
in v
vout RL
Two-port circuit
Figure 108 Equivalent-circuit model of voltage ampli er
the ampli er, respectively, and is the open-loop voltage gain of the ampli er Throughout this section, it will be shown how such a model can be obtained for a variety of ampli er con gurations You may wish to compare this model with that shown in Figure 93 in Section 91 Note that the model of Figure 108 makes use of the simple Th venin equivalent-circuit model developed in e 3 in representing the input to the ampli er as a single equivalent resistance, ri Similarly, the circuit seen by the load consists of a Th venin equivalent circuit e In the remainder of this section, it will be shown how the values of ri , ro , and may be computed, given a transistor ampli er design It is also useful to de ne the overall voltage and current gains for the ampli er model of Figure 108 as follows The ampli er voltage gain is de ned as AV = vout RL ri = vS RL + r o RS + r i iout vout /RL RS + ri = AV = iin vs /(RS + ri ) RL (1012)
while the current gain is AI = (1013)
The rst observation that can be made regarding the circuit of Figure 107 is that the AC input signal, vS (t), has been coupled to the remainder of the circuit through a capacitor, CB (called a coupling capacitor) Similarly, the load resistance, RL , has been connected to the circuit by means of an identical coupling capacitor The reason for the use of coupling capacitors is that they provide separate paths for DC and AC currents in the circuit In particular, the quiescent DC currents cannot reach the source or the load This is especially useful, since the aim of the circuit is to amplify the AC input signal only, and it would be undesirable to have DC currents owing through the load In fact, the presence of DC currents would cause unnecessary and undesired power consumption at the load The operation of the coupling capacitors is best explained by observing that a capacitor acts as an open circuit to DC currents, while if the capacitance is suf ciently large it will act as a short circuit at the frequency of the input signal Thus, in general, one wishes to make CC as large as possible, within reason The emitter bypass capacitor, CE , serves a similar purpose, by bypassing the emitter resistance RE insofar as AC currents are concerned, since the capacitor acts as a short circuit at the signal frequency On the other hand, CE is an open circuit to DC currents, and therefore the quiescent current will ow through the emitter resistor, RE Thus, the emitter resistor can be chosen to select a given Q point, but it will not appear
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