EE114_HW6_Solutions

EE114_HW6_Solutions - - | *The point to note here is that...

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- | *The point to note here is that the feedback in in this circuit is a negative feedback. The way we did analysis (1+T) everywhere was with information that the circuit is a negative feedback circuit, thus you make RR = -ir/it the ir/it term comes negative telling you a negative feedback. But the equations already are for negative feedback so we make RR positive by making RR = -ir/it.
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2 because Vov3 = 2 Vov1,(current through M1 and M3 is same)
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C L2 274.75 pF = C L2 g m R () 2 2 π f c2 R := f c2 5.79 MHz = f c2 f n K := f n 15.92 MHz = f n 1 2 π 1 RC 1 := Case1: dominant pole is due to CL, Non-dominant pole and crossover at c) C L1 3.64 fF = C L1 1 2 π f n R := f n 4.37 GHz = f n f c1 K := K 2.75 = K tan PM := For 70 degress phase margin, non-dominant pole must be at f c1 1.59 GHz = f c1 g m R 2 1 2 π 1 1 := Case1: dominant pole is due to C1, crossover at approximately:
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EE114_HW6_Solutions - - | *The point to note here is that...

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