Fundamentals-of-Microelectronics-Behzad-Razavi.pdf

I v ref i copy1 i v ref v i v ref i current mirror c

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I V REF I copy1 I V REF V I V REF I Current Mirror (c) (a) (b) X M 1 GS M REF X X M REF M 1 DD DD DD copy Figure 9.35 (a) Conceptual illustration of copying a current by an NMOS device, (b) generation of a voltage proportional to square root of current, (c) MOS current mirror. (9.125) where channel-length modulation is neglected. Thus, the black box must satisfy the following input (current)/output (voltage) characteristic: (9.126) That is, it must operate as a “square-root” circuit. From Chapter 6, we recall that a diode- connected MOSFET provides such a characteristic [Fig. 9.35(b)], thus arriving at the NMOS current mirror depicted in Fig. 9.35(c). As with the bipolar version, we can view the circuit’s operation from two perspectives: (1) takes the square root of and squares the result; or (2) the drain currents of the two transistors can be expressed as (9.127) (9.128) where the threshold voltages are assumed equal. It follows that (9.129) which reduces to if the two transistors are identical. Example 9.20 The student working on the circuits in Examples 9.12 and 9.13 decides to try the MOS coun- terpart, thinking that the gate current is zero and hence leaving the gates floating (Fig. 9.36). Explain what happens.
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 447 (1) Sec. 9.2 Current Mirrors 447 I V REF I copy ? M REF M 1 X Floating Node DD Figure 9.36 Solution This circuit is not a current mirror because only a diode-connected device can establish (9.129) and hence a copy current independent of device parameters and temperature. Since the gates of and are floating, they can assume any voltage, e.g., an initial condition created at node when the power supply is turned on. In other words, is very poorly defined. Exercise Is always off in this circuit? Generation of additional copies of with different scaling factors also follows the prin- ciples shown in Fig. 9.26. The following example illustrates these concepts. Example 9.21 An integrated circuit employs the source follower and the common-source stage shown in Fig. 9.37(a). Design a current mirror that produces and from a 0.3-mA reference. I 1 0.2 mA M V DD I V V 0.5 mA 2 in2 out2 I V REF I M REF M 1 W L ( ( 3 0.3 mA M 1 V DD V in1 V out1 M 1 V DD V in1 V out1 I1 M V DD V V in2 out2 W L ( ( 2 I M W L ( ( 5 I2 2 (a) (b) 2 2 DD Figure 9.37
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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 448 (1) 448 Chap. 9 Cascode Stages and Current Mirrors Solution Following the methods depicted in Figs. 9.28 and 9.29, we select an aspect ratio of for the diode-connected device, for , and for . Figure 9.37(b) shows the overall circuit. Exercise Repeat the above example if mA. Since MOS devices draw a negligible gate current, MOS mirrors need not resort to the tech- nique shown in Fig. 9.31. On the other hand, channel-length modulation in the current-source transistors does lead to additional errors. Investigated in Problem 53, this effect mandates circuit modifications that are described in more advanced texts [1].
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