# 522 resistive divider biasing in order to suppress

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5.2.2 Resistive Divider Biasing In order to suppress the dependence of upon , we return to the fundamental relationship and postulate that must be set by applying a well-defined . Figure 5.15 depicts an example, where and act as a voltage divider, providing a base- emitter voltage equal to (5.29) if the base current is negligible. Thus, (5.30) a quantity independent of . Nonetheless, the design must ensure that the base current remains negligible. Q 1 V CC R I C Y R X C 1 R 2 Figure 5.15 Use of resistive divider to define . Example 5.8 Determine the collector current of in Fig. 5.16 if A and . Verify that the base current is negligible and the transistor operates in the active mode.

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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 187 (1) Sec. 5.2 Operating Point Analysis and Design 187 Q 1 V CC I C Y X = 2.5 V R C R 5 k 17 k R 2 1 8 k Figure 5.16 Example of biased stage. Solution Neglecting the base current of , we have (5.31) (5.32) It follows that (5.33) (5.34) and (5.35) Is the base current negligible? With which quantity should this value be compared? Provided by the resistive divider, must be negligible with respect to the current flowing through and : (5.36) This condition indeed holds in this example because . We also note that (5.37) and hence operates in the active region. Exercise What is the maximum value of if must remain in soft saturation? The analysis approach taken in the above example assumes a negligible base current, requiring verification at the end. But what if the end result indicates that is not negligible? We now analyze the circuit without this assumption. Let us replace the voltage divider with a Thevenin
BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 188 (1) 188 Chap. 5 Bipolar Amplifiers equivalent (Fig. 5.17); noting that is equal to the open-circuit output voltage ( when the amplifier is disconnected): Q 1 V CC R I C R X C 1 R 2 V CC Q 1 V CC R I C X C R Thev V Thev I B R 1 R 2 R 1 R 2 V CC V Thev Figure 5.17 Use of Thevenin equivalent to calculate bias. (5.38) Moreover, is given by the output resistance of the network if is set to zero: (5.39) The simplified circuit yields: (5.40) and (5.41) This result along with forms the system of equations leading to the values of and . As in the previous examples, iterations prove useful here, but the exponential dependence in Eq. (5.41) gives rise to wide fluctuations in the intermediate solutions. For this reason, we rewrite (5.41) as (5.42) and begin with a guess for . The iteration then follows the sequence . Example 5.9 Calculate the collector current of in Fig. 5.18(a). Assume and A. Solution Constructing the equivalent circuit shown in Fig. 5.18(b), we note that (5.43)

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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 189 (1) Sec. 5.2 Operating Point Analysis and Design 189 Q 1 V CC I C Y X = 2.5 V R C R 5 k R 2 1 170 k 80 k Q 1 V CC R I C X C R Thev V Thev I B I B Figure 5.18 (a) Stage with resistive divider bias, (b) stage with Thevenin equivalent for the resistive di- vider and .
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