Fundamentals-of-Microelectronics-Behzad-Razavi.pdf

Cls v 2006 at 1342 439 1 sec 92 current mirrors 439 q

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[Razavi.cls v. 2006] June 30, 2007 at 13:42 439 (1) Sec. 9.2 Current Mirrors 439 Q I V CC REF REF Q 1 I copy1 Q I Q I copy2 copy3 2 3 (a) (b) Q I V CC REF REF Q 1 I copy1 Q I Q I copy2 copy3 2 3 Q I V CC REF REF Q 1 Q I Q 2 3 copy I = 3 REF (c) Figure 9.26 (a) Multiple copies of a reference current, (b) simplified drawing of (a), (c) combining output currents to generate larger copies. The key point here is that multiple copies of can be generated with minimal additional complexity because and themselves need not be duplicated. Equation (9.94) readily answers the second question as well: If ( the emitter area of ) is chosen to be times ( the emitter area of ), then . We say the copies are “scaled” with respect to . Recall from Chapter 4 that this is equivalent to placing unit transistors in parallel. Figure 9.26(c) depicts an example where - are identical to , providing . Example 9.14 A multistage amplifier incorporates two current sources of values 0.75 mA and 0.5 mA. Using a bandgap reference current of 0.25 mA, design the required current sources. Neglect the effect of the base current for now. Solution Figure 9.27 illustrates the circuit. Here, all transistors are identical to ensure proper scaling of . Q I V CC REF REF Q 1 Q Q 2 3 0.75 mA Q Q 4 5 0.25 mA 0.5 mA Figure 9.27
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 440 (1) 440 Chap. 9 Cascode Stages and Current Mirrors Exercise Repeat the above example if the bandgap reference current is 0.1 mA. The use of multiple transistors in parallel provides an accurate means of scaling the reference in current mirrors. But, how do we create fractions of ? This is accomplished by realizing itself as multiple parallel transistors. Exemplified by the circuit in Fig. 9.28, the idea is Q I V CC REF 0.25 mA REF1 Q REF2 Q REF3 I copy Q 1 X Figure 9.28 Copying a fraction of a reference current . to begin with a larger ( here) so that a unit transistor, , can generate a smaller current. Repeating the expressions in (9.89) and (9.90), we have (9.95) (9.96) and hence (9.97) Example 9.15 It is desired to generate two currents equal to 50 A and 500 A from a reference of 200 A. Design the current mirror circuit. Solution To produce the smaller current, we must employ four unit transistors for such that each carries 50 A. A unit transistor thus generates 50 A (Fig. 9.29). The current of 500 A requires 10 unit transistors, denoted by for simplicity. I V CC REF I Q 1 0.2 mA copy1 A E 4 A E I Q copy2 2 A E 10 X Figure 9.29
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 441 (1) Sec. 9.2 Current Mirrors 441 Exercise Repeat the above example for a reference current of 150 A. Effect of Base Current We have thus far neglected the base current drawn from node in Fig. 9.26(a) by all transistors, an effect leading to a significant error as the number of copies (i.e., the total copied current) increases. The error arises because a fraction of flows through the bases rather than through the collector of . We analyze the error with the aid of the diagram shown in Fig. 9.30, where and denote one unit transistor and unit transistors, Q I V CC REF REF Q 1 A E E nA I n copy I copy β copy β I Figure 9.30
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