hw5 solution spring 13 -
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HW5 Solutions Sp13
Instructor: Dr.Parker4/5/2013
Solution No.1There are many different possible solutions. Here are two of them.
a) b)
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Euler Path: CBADDEGEF Euler Path: CBADEEGEF
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Solution No.2 (schematic (a))Stick diagram for schematic (a)
vdd
out
C B A D D E G E F
gnd
Solution No.3 (schematic (a))For schematic a)
nnp
np
np
nppp
p
nn
n
WWW
WW
LLWL
WL
23.3514.219
43
43
34
=
=
=
==
ββ
ββ
If Wn=4 λ, thenWp=12.9 λ ≈ 13 λ
In the pmos side, transistors with inputs A, B, C, D and E (connected to the output node) have the same size of Wp= 13 λ. Transistors E (connected to the vdd node), F, and G have sizes of (2/3)(13 λ).
In the nmos side, if the critical path (most resistive path) is unit size, then alltransistors have the same size. Otherwise, whenever possible WGn=1/2WEn; while the othertransistors are equal in size to the transistors in the critical path. This helps to maintainbalance between the different resistive paths from output to ground.
Solution No.4 (schematic (a))
capacitances that get charged that affect the output node
ON
ON
ON ON
ON
OFF
OFF
OFFON
Note: The red circles identify only the capacitancesthat affect the output.
ON
ON
ON
OFF OFF OFF
OFF
OFFOFF
We named the diffusion capacitance Cdp for a pmos transistor with W=13 λ
We named the diffusion capacitance Cdn for an nmos transistor with W=4λ
From the pmos side: there are 6Cdp+4(2/3)Cdp(the 2/3 term comes from the fact that there are 4 diffusions that belong to transistors that are 2/3 times smaller than W=13 λ ).
From the nmos side: there are 15Cdn (all transistors are unit size)
We could also express the total # of capacitances using only “Cdn”. In that case Cdp= 13/4 Cdn, and
Total # capacitances that affects the output =6(13/4)Cdn +4(2/3)(13/4)Cdn+15Cdn
Solution No.4 (schematic (a))
ON
ONONOFF
OFF
OFF
OFF
ON
ON
Capacitances that get discharged thataffect the output node
ON
ON
ON
ON
OFF OFF
OFFOFF
OFFWe named the diffusion capacitance Cdp for a pmos transistor with W=13 λ
We named the diffusion capacitance Cdn for an nmos transistor with W=4λ
From the pmos side: there are 6Cdp+4(2/3)Cdp(the 2/3 term comes from the fact that there are 4 diffusions that belong to transistors that are 2/3 times smaller than W=13 λ ).
From the nmos side: there are 15Cdn (all transistors are unit size)
We could also express the total # of capacitances using only “Cdn”. In that case Cdp= 13/4 Cdn, and
Total # capacitances that affects the output =6(13/4)Cdn +4(2/3)(13/4)Cdn+15Cdn
The next slides show the solution for questions 2-4 for schematic (b)
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Euler Path: CBADEEGEF
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Solution No.2 (schematic (b))Stick diagram for schematic (b)
vdd
out
C B A D E E G E F
gnd
Solution No.3 (schematic (b))For schematic b)
nnp
np
np
nppp
p
nn
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WWW
WW
LLWL
WL
69.3514.219
5.33
5.33
35.3
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==
ββ
ββ
Note: We have 3.5Rn because on the critical path there are two transistors in parallel that are ON, so their contribution to the equivalent resistance of the path is ½ of Rn
If Wn=4 λ, thenWp=14.75 λ ≈ 15 λ
In the pmos side, transistors with inputs A, B, C, D and E (except the one connected to the vddnode) have the same size of Wp= 15 λ. Transistors E (connected to the vdd node), F, and G have sizes of (2/3)(15 λ).
In the nmos side, if the critical path (most resistive path) is unit size, then all transistors have the same size. Otherwise, whenever possible WGn=1/2WEn; while the othertransistors are equal in size to the transistors in the critical path. This helps to maintain balancebetween the different resistive paths from output to ground.
Solution No.4 (schematic (b))
ON
ON
ON ONOFF
OFF
OFF
OFFON
Capacitances that get charged thataffect the output node
ON
ON
ON
OFF OFF OFF
OFF
ONOFF
We named the diffusion capacitance Cdp for a pmos transistor with W=15 λ
We named the diffusion capacitance Cdn for an nmos transistor with W=4λ
From the pmos side: there are 6Cdp+4(2/3)Cdp(the 2/3 term comes from the fact that there are 4 diffusions that belong to transistors that are 2/3 times smaller than W=15 λ ).
From the nmos side: there are 15Cdn (all transistors are unit size)
We could also express the total # of capacitances using only “Cdn”. In that case Cdp= 15/4 Cdn, and
Total # capacitances that affects the output =6(15/4)Cdn +4(2/3)(15/4)Cdn+15Cdn
Solution No.4 (schematic (b))
OFF
OFF
OFF ONOFF
OFF
OFF
OFFON
Capacitances that get discharged thataffect the output node
ON
ON
ON
ON OFF OFF
OFF
ONOFF
We named the diffusion capacitance Cdp for a pmos transistor with W=15 λ
We named the diffusion capacitance Cdn for an nmos transistor with W=4λ
From the pmos side: there are 6Cdp+4(2/3)Cdp(the 2/3 term comes from the fact that there are 4 diffusions that belong to transistors that are 2/3 times smaller than W=15 λ ).
From the nmos side: there are 15Cdn (all transistors are unit size)
We could also express the total # of capacitances using only “Cdn”. In that case Cdp= 15/4 Cdn, and
Total # capacitances that affects the output =6(15/4)Cdn +4(2/3)(15/4)Cdn+15Cdn
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Solution No.5
B E I
G H
A F
D
C
D
WLT
WLCCox
roxgn
εε 0' ==
2' 1041
9.385.8 −×⋅
=oxC
( )( )( ) ( )2626 101.24101.2232 −− ×=×⋅⋅=WL
Solution No.6
a)
fFC gn 02.2=
b)
( ) fFfFC gn 347.102.232
=
=
Solution No.7( )( )( )
( )( ) ( )
( ) 1226
12
262
12
2
1226
1003.101.3
1029.2
101.23
10255.
2
10255.101.2/3103
−−
−−
−
−−
×=×⋅=⋅=
×=×
+×=
⋅+⋅=
×=×−=⋅
π
π
WxA
A
rLWA
LW
jsw
floor
difffloor
diff
( )
( ) ( )( )( )( ) 66 10171.2101.232/3102
2_
1003.101.3
−− ×=×+−=
⋅+=
×=×⋅=⋅=
π
π rLswPerimeterWxA
diff
jsw
( ) ( ) 2_ µµjbsp
swfloorjbswp
DCAACswPerimeterC ⋅++⋅=
( )( ) ( )( ) pFCD4106.757.1103.29.8.117.2 −×=++=
Solution No.8
( )
++=
= 1
212
4201.4
WlRR s
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