homework6sol

homework6sol - ECE164 Fall 2005 Problem Set 6 Solutions...

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Unformatted text preview: ECE164 Fall 2005 Problem Set 6 Solutions Problem 1 1. a) The complete small signal model: g m1 v gs1 g mb1 v bs1 r ds1 g m6 v gs6 g mb6 v bs6 r ds6 R r ds4 + − g m2 v gs2 g mb2 v bs2 r ds2 g m7 v gs7 g mb7 v bs7 r ds7 R r ds5 + − r ds3 v o1 v o2 v i1 v i2 1 b) Yes, this is a symmetric circuit, and thus can be split in half, with virtual differential grounds. In differential mode, the source of M1 and M2 are grounded, so the body generators are off. g m1 v gs1 r ds1 g m6 v gs6 g mb6 v bs6 r ds6 R r ds4 + − v o1 v i1 c) We can use the half circuit technique to analyze the common mode behavior. We split the r ds of M3 into two resistors of twice the value. g m1 v gs1 g mb1 v bs1 r ds1 g m6 v gs6 g mb6 v bs6 r ds6 R r ds4 + − v o1 v i1 2r ds3 2 d) Because the circuit is perfectly symmetric, A dm- cm = A cm- dm = 0. Body effect and the channel length modulation are neglected for all transistors except M3. g m1 v gs1 g m6 v gs6 R + − v o1 v i1 There is a very simple way to analyze the differential mode gain of this circuit. If we ignore all the r ds ’s, then all the current from the M1 g m generator goes through the M6 current source, which in turn gets pulled through R . This means effectively, M6 just passes the current from M1 through R , and can be ignored. So our gain is simply A dm =- g m 1 R For the common mode case, again M6 just conveys the current from M1 to R , so we have a source degenerated amplifier. A cm =- g m 1 R 1 + 2 g m 1 r ds 3 ≈ - R 2 r ds 3 2. a) From the sizes, we know the currents are symmetric. The P mobility is a fourth the N mobility, but M4 is four times larger than M3, so I D 4 = I D 5 = I D 3 . By symmetry, the current through M1 and M2 are equal, and...
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homework6sol - ECE164 Fall 2005 Problem Set 6 Solutions...

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