hw5 - Problem 1 A = A(1 a.f A A = a(1 a.f a f => A/A...

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Problem 1) A = A/(1+a.f) A- A = a/(1+a.f+a. f) => A/A = (a.f/(1+a.f)). f/f if a.f >> 1, the fractional change in f is not attenuated in the fractional change of closed loop gain. Otherwise, it will be attenuated. Problem 2a)
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Problem 2b) Problem 3a) Open loop gain = avn * 1/[1+ gmp/gmn*(avn/avp)] > 80 avn = gm 2 . r 02 avp = gm 3 . r 03 = 122 & gm 2 = gm 3 => avn>232 => L2 min = .6 um 3b) gm 2 /(2 π C L ) = 100 MHz => gm 2 = 18.84 mA/v 3c) gm/I D =10 => I B = I D = 1.88 mA from I D /W graph : I D2 /W 2 = 5.5 => W 2 = 342.5 um from I D /W graph : I D3 /W 3 = 1.36 => W 3 = 1385 um 3d) g element sets the output common mode voltages of the differential amplifier to the desired value V ocdes . When V oc becomes smaller than V ocdes, the current source injects an additional current to the differential branches. Since Vgs of M3 is set by I B , Vds of M3 decreases and V oc increases. If V oc becomeslarger than V ocdes, the current source drags an additional current from the differential branches. Since Vgs of M3 is set by I B , Vds of M3
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hw5 - Problem 1 A = A(1 a.f A A = a(1 a.f a f => A/A...

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