m = 2.5 for electrons and 2.7 for holes in Si.
Also, one input of the EXOR gate is tied to zero therefore 'f' will be directly transferred to the output.
The logic gates will introduce gate delays which will lead to the answer .
W = B = Bandwidth
C = 6 x 106 x 12 = 72 Mbps.
(s3 + 2s2 + 5s + 6)y(s) = (s2 + 4) x (s)
s3y(s) + 2s2y(s) + 5sy(s) + 6y(s) = s2 x (s) + 4 x (s)
Replacing s by and y(s) by y(t) and x(s) by x(t) we get
y(t) + 2y(t) + 5y(t) + 6y(t)
= x(t) + 4x(t)
+ 2 + 5 + 6 = + 4x
This is required differential equation.
Emax - Emin = 0.5 x 1 kV = 0.5 kV
2Emax = 1.5 kVi, Emax = 0.75 kV
Emin = 0.25 kV.
?
? = - 8VP - 16
? + 8VP + 16 = 0
? (VP + 4)2 = 0
? VP = -4 V.
(C) is without over modulation.
= 6.903 x 10-8
?
?
fB = - 8.6 x 1011
The threshold voltage is always negative for p-channel and hence implant is of p-type.
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