Question

H6P1: THE NEWFET DEVICE
all questions

Answer 1

PLZZ help on the H6P2

Answer 2

1) Minimum bias voltage should be the threshold and the swing Dv; VIN = VT+Dv

2) Here, iDS1=iDS2;
1/2K(VIN-VT)^2=1/2K(VGS2-VT)^2
Therefore, VDS2=VGS2=VIN
We know that VDD=VDS1+VDS2
Minimum VDS1(=VOUT) = VIN-VT
Therefore, minimum VDD= VIN-VT+VIN =2*VIN-VT

3) gm1= diDS1/dVGS1= K*(VIN-VT)
4) same current, same bias; VGS2=VGS1=VIN
gm2= diDS1/dVGS1= K*(VIN-VT)
5) -vin
6) Rth = 1/gm=1/(K*(VIN-VT))

Answer 3

1) Dv; VIN = VT+Dv
2) 2*VIN-VT
3) gm1= diDS1/dVGS1= K*(VIN-VT)
4) K*(VIN-VT)
5) -vin
6) 1/(K*(VIN-VT)

Answer 4

@Loveboy54@sam747 i'm so gratful thnk's lot

Answer 5

welcome

Answer 6

hi can any one help me on question 5 and 6 on the midterm exam