Fundamentals-of-Microelectronics-Behzad-Razavi.pdf

544 and 545 546 we begin the iteration with an

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(5.44) and (5.45) (5.46) We begin the iteration with an initial guess mV (because we know that the voltage drop across makes less than ), thereby arriving at the base current: (5.47) (5.48) Thus, A and (5.49) (5.50) It follows that A and hence A, still a large fluctuation with respect to the first value from above. Continuing the iteration, we obtain mV, A and A. After many iterations, mV and A. Exercise How much can be increased if must remain in soft saturation? While proper choice of and in the topology of Fig. 5.15 makes the bias relatively insensitive to , the exponential dependence of upon the voltage generated by the resistive divider still leads to substantial bias variations. For example, if is higher than its nominal value, so is , thus multiplying the collector current by (for mV). In other words, a error in one resistor value introduces a error in the collector current. The circuit is therefore still of little practical value.
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 190 (1) 190 Chap. 5 Bipolar Amplifiers 5.2.3 Biasing with Emitter Degeneration A biasing configuration that alleviates the problem of sensitivity to and is shown in Fig. 5.19. Here, resistor appears in series with the emitter, thereby lowering the sensitivity to . From an intuitive viewpoint, this occurs because exhibits a linear (rather than exponential) I- V relationship. Thus, an error in due to inaccuracies in , , or is partly “absorbed” by , introducing a smaller error in and hence . Called “emitter degeneration,” the Q 1 V CC R I C Y R X C 1 R 2 R I E E P Figure 5.19 Addition of degeneration resistor to stabilize bias point. addition of in series with the emitter alters many attributes of the circuit, as described later in this chapter. To understand the above property, let us determine the bias currents of the transistor. Neglect- ing the base current, we have . Also, , yielding (5.51) (5.52) (5.53) if . How can this result be made less sensitive to or variations? If the voltage drop across , i.e., the difference between and is large enough to absorb and swamp such variations, then and remain relatively constant. An example illustrates this point. Example 5.10 Calculate the bias currents in the circuit of Fig. 5.20 and verify that operates in the forward active region. Assume and A. How much does the collector current change if is higher than its nominal value? Solution We neglect the base current and write (5.54) (5.55) Using mV as an initial guess, we have (5.56) (5.57)
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 191 (1) Sec. 5.2 Operating Point Analysis and Design 191 Q 1 V CC Y X = 2.5 V R C R R 2 1 16 k 9 k 1 k R E 100 P Figure 5.20 Example of biased stage. and hence (5.58) With this result, we must reexamine the assumption of mV. Since (5.59) (5.60) we conclude that the initial guess is reasonable. Furthermore, Eq. (5.57) suggests that a 4-mV error in leads to a error in and hence , indicating a good approximation.
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