HW7_Sol - ECE 315 Homework 7 Solutions 1(CS amplifier...

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ECE 315 Homework 7 Solutions 1. (CS amplifier parameters in different transistor parameter sets) Use Fig. 4.34 of a CS amplifier and Eqs. (4.56)-(4.71), for R D >> r o (approaching small-signal open-circuit load) and assuming V A is proportional to L (and hence V A ’=V A /L is constant), write down the small signal gain A vo as a function of the following parameters: (a) W/L and V OV , (b) W/L and I D , (c) I D and V OV ., where V OV is the gate overdrive voltage. (9 pts) To start, let’s draw the small signal model Since R D >> r o , we can assume r o ||R D ≈ r o By KCL, at node v out v out /r o + g m v gs = 0 v gs = v in v out /r o = -g m v in v out /v in = A vo = -g m r o This equation is the same as Eq. 4.68 if we assume R D >> r o We know from the equations in the book that g m can be written as a function of any two of the following three parameters: I D , V OV , or (W/L) g m = [2k n ’(W/L)I D ] 1/2 g m = 2I D /V OV g m = k n ’(W/L)V OV The purpose of this exercise is to let you understand clearly how to choose (a) W/L and operating point as DC gate bias (not popular choice of DC bias) when the technology is fixed; (b) W/L and I D bias (right way to DC bias your CMOS chip); (c) Same circuit layout but different technologies. r o = V A /I D = 2V A /k n ’(W/L)V OV (a) g m = k n ’ (W/L) V OV from Eq. 4.63 r o = V A /I D = V A ’L/ [k n ’/2 (W/L) V OV 2 ] A vo = -g m r o = -2V A ’L/V OV + v out + v gs g m v gs v in r o ||R D
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(b) g m = [2I D k n
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