# 5330 5331 5332 example 546 determine the small signal

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(5.330) (5.331) (5.332) Example 5.46 Determine the small-signal properties of an emitter follower using an ideal current source (as in Example 5.43) but with a finite source impedance . Solution Since , we have (5.333) (5.334) (5.335) Also, , and hence (5.336) (5.337)

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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 251 (1) Sec. 5.3 Bipolar Amplifier Topologies 251 We note that approaches unity if , a condition typically valid. Exercise How are the results modified if ? The buffering capability of followers is sometimes attributed to their “current gain.” Since a base current results in an emitter current of , we can say that for a current delivered to the load, the follower draws only from the source voltage (Fig. 5.93). Thus, sees the load impedance multiplied by . Q 1 V CC v X Load i L i L β + 1 Figure 5.93 Current amplification in a follower. Emitter Follower with Biasing The biasing of emitter followers entails defining both the base voltage and the collector (emitter) current. Figure 5.94(a) depicts an example similar to the scheme illustrated in Fig. 5.19 for the CE stage. As usual, the current flowing through and is chosen to be much greater than the base current. Q 1 V CC R 1 R 2 C 1 in v R E X out V Q 1 V CC R C 1 in v R E X out V B (a) (b) B I Y Figure 5.94 Biasing a follower by means of (a) resistive divider, (b) single base resistor. It is interesting to note that, unlike the CE topology, the emitter follower can operate with a base voltage near . This is because the collector is tied to , allowing the same voltage for the base without driving into saturation. For this reason, followers are often biased as shown in Fig. 5.94(b), where is chosen much less than the voltage drop across , thus lowering the sensitivity to . The following example illustrates this point. Example 5.47 The follower of Fig. 5.94(b) employs and . Calculate the bias current and voltages if A, , and V. What happens if drops to 50? Solution To determine the bias current, we follow the iterative procedure described in Section 5.2.3. Writ-
BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 252 (1) 252 Chap. 5 Bipolar Amplifiers ing a KVL through , the base-emitter junction, and gives (5.338) which, with mV, leads to (5.339) It follows that mV. Using this value in Eq. (5.338), we have (5.340) a value close to that in (5.339) and hence relatively accurate. Under this condition, mV whereas V. Since , we expect that variation of and hence negligibly affects the voltage drop across and hence the emitter and collector currents. As a rough estimate, for , is doubled ( mV), reducing the drop across by 159 mV. That is, mA, implying that a twofold change in leads to a change in the collector current. The reader is encouraged to repeat the above iterations with and determine the exact current. Exercise If is doubled, is the circuit more or less sensitive to the variation in ?

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