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EECS 170D Homework #4
Due in class Thursday November 15
th
,
2007
Problem 1:
Assume and inverter driving an identical inverter. The following technology
parameters are given
1. λ= 0.5um
2. VDD= 3V
3.
Cox= 2
fF/um
2
4.
CJN= 0.3 fF/um
2
5.
CJP= 0.5 fF/um
2
6.
CJSWN= 0.4 fF/um
2
7.
CJSWP= 0.7
fF/um
2
8.
CGDO = 0.4 fF/um
2
9.
KPn = 75uA/V
2
10.
KPp = 25uA/V
2
11.
Φ
0
= 0.6V
12.VTOn =0.7V
13.VTOp =0.7V.
As you can derive from a good layout, the minimum drain area is W*5λ (Assume a
rectangular shape with at least one contact). With the above provided information
derive the value of CL as a function of Wn and Wp for
a) Low to High transition.
b) High to Low transition.
Hint: The Keq values for the HighLow transitions are given below so you only need
to compute the values for LowHigh.:
Solution:Assuming that area = w*5λ
Applying the equation (Keq) from the book (equation number 3.11) and using the
voltage swing indicated below we can compute the values of Keq as follows:
KeqB (VR from 0V to 1.5V) = 0.697
KeqSW (VR from 0V to 1.5V) = 0.785
KeqB (VR from 1.5V to 3V) = 0.463
KeqSW (VR from 1.5 to 3) = 0.60
CL as a function of Wn and Wp
a) CL (Low to High) = 2.CGDO.(Wn+Wp)+
0.697*(Wn*2.5*CJN)+0.785*(Wn+5)*CJSWN+
EECS170D
Fall 07
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 Spring '08
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