# (ms-word).

```SOLUTION
EE 141 Fall 1999
Homework Set #4
Due Wed. Oct 6 1999, 5.00 pm
Problem 1)
a) Pav = 1.25 mW
refer eqn. 3.42 of textbook
b) With lower voltages one would expect power to drop. But with increasing
frequency the power becomes greater than the original power at a period of 5ns.
Problem 2)
a) VOH = Vdd
VOL = 0.9 using eqn 4.11 of textbook
b) 0 static power dissipation when Vin is low.
When Vin is high
Pav = 3.2 mW
from eqn 4.13 of textbook
c) In this problem
tpLH = 0.69 RL CL
tpHL = 0.69 (RL || RPDN ) CL
RL is the load resistance. RPDN is the resistance of the pull down network.
RL = &frac12;[(Vds/Id)vout=vol + (Vds/Id)vout = vdd/2]
= &frac12; [(Vdd – Vol) / ((Kp/2) (Vdd -|Vtp|)2) + (Vdd/2) /((Kp/2)((Vdd-Vt)Vdd
– Vdd2 /4))]
= 10951 
RPDN = &frac12;[(Vds/Id)vout=vdd + (Vds/Id)vout = vdd/2]
= &frac12; [Vdd / ((Kn/2) (Vdd -Vtn)2) + (Vdd/2) /((Kn/2)((Vdd-Vt)Vdd/2 –
Vdd2/4))]
= 9791 
tpLH = .3778 ns
tpHL = .1783 ns
d) higher. Lambda,p &gt; Lambda,n. So if effect of channel modulation is included,
it is likely that Isd,p &gt; Isd,n. So Vol has to move higher to make |Vds,p| smaller and
|Vds,n| bigger.
Problem 3)
If you solved all the parts of the problem assuming a short channel device that is
ok. The grader will be notified.
For this problem use table 3.5 and 3.6
a) S= 3.5
New speed is 1225 Mhz
Power is 35 W. (with short channel- speed= 1.225 Ghz, power = Same)
b) with U= 3.5 and S= 3.5
Speed = 350 Mhz
Power = 0.816 W
(With short channel Speed = 350Mhz, power = 0.816 W)
c) consider general scaling
Pav = 1= S/U3
U3 = 3.5
Vdd= 3.5/U = 2.3V
d) There is no change in power consumed in constant voltage scaling for short
channel devices
Problem 4)
a) tw = 0.38 r c L2
= 0.38 (10000)2 (0.08/3.6) (3.6 * 0.03 + 0.04 *2)
= 0.158 ns
b)
-
We have to find Cout as seen by the second buffer
Output cap of the 2nd stage is approx by Ctotal. The problem then becomes
We want to set u = ((Ctotal) / Ci)1/2
Total delay is tp0 u + tp0 u + tw
u = 13.7
Problem 5)
a) ((A+C)(B+D)F(A+E+D)(B+E+C))’
b)
c) There is no common Euler path. Therefore no existing solution such that the
PMOS network can be laid out in one contiguous strip.
d)
```