# 5149 obtaining 5150 we also recognize that two

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(5.149) obtaining (5.150) We also recognize that two currents flow through : one originating from equal to and another equal to . Thus, the voltage drop across is given by (5.151) Since the voltage drop across and must add up to , we have (5.152) (5.153)

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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 209 (1) Sec. 5.3 Bipolar Amplifier Topologies 209 (5.154) Substituting for from (5.150) and rearranging the terms, we arrive at (5.155) As predicted earlier, the magnitude of the voltage gain is lower than for . With , we can assume and hence (5.156) Thus, the gain falls by a factor of . To arrive at an interesting interpretation of Eq. (5.156), we divide the numerator and denomi- nator by , (5.157) It is helpful to memorize this result as “the gain of the degenerated CE stage is equal to the total load resistance seen at the collector (to ground) divided by plus the total resistance placed in series with the emitter.” (In verbal descriptions, we often ignore the negative sign in the gain, with the understanding that it must be included.) This and similar interpretations throughout this book greatly simplify the analysis of amplifiers—often obviating the need for drawing small- signal circuits. Example 5.22 Determine the voltage gain of the stage shown in Fig. 5.37(a). Q 1 V CC R C out in R E (a) r π 2 Q 2 V CC v v Q 1 V CC R C out in R E r π 2 v v (b) Figure 5.37 (a) CE stage example, (b) simplified circuit. Solution We identify the circuit as a CE stage because the input is applied to the base of and the output is sensed at its collector. This transistor is degenerated by two devices: and the base-emitter junction of . The latter exhibits an impedance of (as illustrated in Fig. 5.7), leading to the simplified model depicted in Fig. 5.37(b). The total resistance placed in series with the emitter is
BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 210 (1) 210 Chap. 5 Bipolar Amplifiers therefore equal to , yielding (5.158) Without the above observations, we would need to draw the small-signal model of both and and solve a system of several equations. Exercise Repeat the above example if a resistor is placed in series with the emitter of . Example 5.23 Calculate the voltage gain of the circuit in Fig. 5.38(a). Q 1 V CC R C out in R E (a) Q 2 V CC v v Q 1 V CC R C out in R E r π 2 v v (b) Figure 5.38 (a) CE stage example, (b) simplified circuit. Solution The topology is a CE stage degenerated by , but the load resistance between the collector of and ac ground consists of and the base-emitter junction of . Modeling the latter by , we reduce the circuit to that shown in Fig. 5.38(b), where the total load resistance seen at the collector of is equal to . The voltage gain is thus given by (5.159) Exercise Repeat the above example if a resistor is placed in series with the emitter of . To compute the input impedance of the degenerated CE stage, we redraw the small-signal model as in Fig. 5.39(a) and calculate . Since , the current flowing through is equal to , creating a voltage drop of . Summing and and equating the result to , we have

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