ENEE 303H Fall 2009 HW5 YUH revised version 10/31/09
1. (a) example of using 2 power supplies
VDD
Rd
0
Cc2
Ri
Cc1
M1
0
Vi
0
Rg
Cc3
Rs
CL
RL
0
-VSS
0
0
vsig  vi
Rg
Ri  Rg
, Rsig  Ri // Rg
(b)(d)
(Vsig-Vg)/Rsig
sCgs(Vg-Vo)
Rsig
Cgs
G
S
Vo
sCL*Vo
Vo/RL'
Cgd
Vsig
sCgd*Vg
CL
RL'
gm*vgs
gmb*vbs
B
D
0
0
RL'  RL // ro
Now the controlled current source gmb*vbs can be replaced by a
resistor 1/gmb since the voltage across it is vbs (source
absorption).
(Vsig-Vg)/Rsig
sCgs(Vg-Vo)
Rsig
Cgs
G
S
Vo
sCL*Vo
Vo/RL'
Cgd
sCgd*Vg
Vsig
CL
RL'
gm*vgs
B
D
0
0
RL'  RL // ro //
at Gate:
1
1
, YL'  '
gm b
RL
vsig  vg
Rsig


 sC gs vg  vo  sC gd vg , so vg 
vsig  sC gs Rsig vo


s C gs  C gd Rsig  1

1 
at Source: g m vg  vo  sC gs vg  vo  vo  sCL  ' 
RL 









 gm  sC gs vsig  sC gs Rsig vo   vo sCL  YL'  g m  sC gs sCgs  Cgd Rsig  1
 g  sC v  v sC  Y  g  sC sC  C R  1 g  sC sC R
 g m  sC gs vg  vo sCL  YL'  g m  sC gs
m
gs
sig
o
'
L
L
m
gs
gs
Denominator =
Y
'
L
 

gd
sig


m
gs
gs sig vo
 

 g m  s CL  Cgs  Cgs  Cgd RsigYL'  g mCgd Rsig  s 2 Rsig CL Cgs  Cgd  CgsCgd
C 

gm
1  s gs 
gm 
 gm 
vo

'
vsig
CL  C gs  C gs  C gd Rsig YL  g mC gd Rsig
Rsig CL C gs  C gd  C gsC gd
1 s
 s2
'
YL  g m
YL'  g m



Mid-band gain AM 

YL'
gm
 gm

YL'



< 1
If HP2  HP1 ( HP1 is dominant), then
(see p.593 of Sedra/Smith)
HP1 
Y

 gm
CL  Cgs  Cgs  Cgd RsigYL'  g mCgd Rsig
'
L


C  C gs  C gs  C gd Rsig YL'  g mC gd Rsig
 HP 2  L
Rsig C L C gs  C gd   C gsC gd 
 





Solve with MATLAB:
syms s Rsig Vsig gm Cgs Cgd CL YL
% here YL=1/RL', RL'=RL//ro//(1/gmb)
L=[s*(Cgs+Cgd)+1/Rsig -s*Cgs;…
-gm-s*Cgs gm+s*Cgs+s*CL+YL]
R=[Vsig/Rsig;0]
H=L\R % solve [L]*[vg;vo]=[R]
vo=H(2)
vo =
Vsig*(gm+s*Cgs)/(gm*s*Rsig*Cgd+gm+s^2*Cgs*Rsig*Cgd+s*Cgs+s^2*CL*
Rsig*Cgs+s^2*CL*Rsig*Cgd+s*CL+YL*s*Rsig*Cgs+YL*s*Rsig*Cgd+YL)
So vo/Vsig is obtained.
Solve with Mathematica:
To make vo in terms of Vsig and vbs, the process is similar, we
just need to modify the node equation at source.
(Vsig-Vg)/Rsig
sCgs(Vg-Vo)
Rsig
Cgs
G
S
Vo
sCL*Vo
Vo/RL'
Cgd
sCgd*Vg
Vsig
CL
RL'
gm*vgs
gmb*vbs
D
0
RL'  RL // ro and YL' 
at Gate:
vsig  vg
Rsig
1
RL'


 sC gs vg  vo  sC gd vg , so vg 
vsig  sC gs Rsig vo


s C gs  C gd Rsig  1

1 
at Source: g m vg  vo  g mbvbs  sC gs vg  vo  vo  sCL  ' 
RL 









 gm  sC gs vsig  sC gs Rsig vo   g mbvbs  vo sCL  YL'  gm  sC gs sCgs  Cgd Rsig  1
gm  sCgs vsig  gmbvbs  vo sCL  YL'  gm  sCgs sCgs  Cgd Rsig  1 gm  sCgs sCgsRsig vo
 g m  sC gs vg  g mbvbs  vo sCL  YL'  g m  sC gs
we get the same denominator as that of the previous case
(the only difference is now RL'  RL // ro )
C gs 

gm
g

vsig  ' m b vbs
1

s
'

gm 
YL  g m 
YL  g m
vo 
'
CL  C gs  C gs  C gd Rsig YL  g mC gd Rsig
2 Rsig C L C gs  C gd  C gsC gd
1 s

s
YL'  g m
YL'  g m

vo v
bs  0







 




C gs 

gm

vsig
1

s
g m 
YL'  g m 

CL  C gs  C gs  C gd Rsig YL'  g mC gd Rsig
Rsig CL C gs  C gd  C gsC gd
1 s
 s2
'
YL  g m
YL'  g m






 




vo v
sig  0

1 s

CL  C gs  C gs  C gd
Y
'
L
gm b
vbs
 gm
Rsig YL'  g mC gd Rsig

 gm

YL'


 s2
 

Rsig CL C gs  C gd  C gsC gd
Y
'
L
 gm


So there exists a zero in finite frequencies for vbs=0, Z 
gm
,
Cgs
which can be found from setting the current which flow into the
load as zero.
The following s-plane plot is based on Exercise 6.34, p.639.
Note that the zero and poles are all on the negative real axis.
Im
s plane
Re
-gm/Cgs -whp2
-whp1
(c) Because Cgs, Cgd and CL form a continuous loop, the voltages
across these three capacitors are not totally independent;
therefore, the transfer function is of order 2.
(e)
Example: Exercise 6.34, p.639
clear
RL=10e3 ; CL=15e-15
Cgs=20e-15 ; Cgd=5e-15
Rsig=20e3
ro=20e3
gm=1.25e-3 ; gmb=0.2*gm
YLp=1/RL+1/ro+gmb
AM=gm/(YLp+gm) % mid-band gain
wz=gm/Cgs
s1=(CL+Cgs+(Cgs+Cgd)*Rsig*YLp+gm*Cgd*Rsig)/(YLp+gm) %
coefficient of s^1
s2=Rsig*(CL*(Cgs+Cgd)+Cgs*Cgd)/(YLp+gm) % coefficient of s^2
wroot=-roots([s2,s1,1])
whp1=wroot(2)
whp2=wroot(1)
% dominant pole approximation
wp1=1/s1
wp2=1/s2/wp1
% plot with the poles solved by s polynomial
f=logspace(6,12,1e4);
w=2*pi*f;
H=AM*(1+i*w/wz)./(1+i*w/whp1)./(1+i*w/whp2);
% H=gmb/(YLp+gm)./(1+i*w/whp1)./(1+i*w/whp2);
Hmag=20*log10(abs(H));
semilogx(f,Hmag)
title('Magnitude Response','fontsize',14)
xlabel('Frequency, Hz','fontsize',14)
ylabel('Magnitude, dB','fontsize',14)
grid on
we get
fhp1=whp1/2/pi= 849 MHz (by solving s polynomial)
fp1=wp1/2/pi= 729.5 MHz (by dominant pole approximation)
fz=wz/2/pi= 9.95 GHz
The following magnitude responses are plotted with the poles
solved by s polynomial because the two poles are not too far
away from each other (however, the numbers given by dominant
pole approximation 729.5 MHz and method of open-circuit time
constant 730 MHz are still not bad).
vbs=0 Magnitude Response
0
-10
Magnitude, dB
-20
-30
-40
-50
-60
-70
6
10
7
10
8
10
9
10
10
10
11
10
12
10
Frequency, Hz
Vsig=0 Magnitude Response
0
-20
Magnitude, dB
-40
-60
-80
-100
-120
-140
6
10
7
10
8
10
9
10
Frequency, Hz
10
10
11
10
12
10
2.(a)
M4007P
VDD
10Vdc
M4007P
M2
w
M4
x
y
z
Vw
M1
M3
M4007N
M4007N
0
10V
8V
6V
4V
2V
0V
0V
1V
V(X)
V(z)
2V
3V
4V
5V
V(w)
V_Vw
A(0,0)
B~=(VDD/2,VDD/2)
C(VDD,VDD)
6V
7V
8V
9V
10V
(b)
VDD1
10Vdc
M4007P
M4007P
M6
w2
M8
x2
z2
M5
M7
M4007N
M4007N
0
8.0V
7.0V
6.0V
5.0V
4.0V
3.0V
2.0V
0s
V(x2)
2us
V(z2)
4us
6us
8us
10us
12us
14us
16us
18us
20us
Time
Without perturbation, the latch circuit stays at point B (but
this can’t be seen in experiment).
VDD1
10Vdc
M4007P
M4007P
M6
w2
I1 = 0
I2 = 1u
TD = 10n
TR = 0
TF = 0
PW = 1n
PER = 1m
M8
x2
z2
I2
M5
M7
M4007N
M4007N
0
10V
8V
6V
4V
2V
0V
0s
50ns
V(x2)
100ns
150ns
200ns
250ns
300ns
V(z2)
Time
With the current pulse perturbation, V(x) goes to high and V(z)
goes to low very quickly via feedback loop.
10V
8V
6V
4V
2V
0V
0V
V(x2)
0.5V
V(z2)
1.0V
1.5V
2.0V
2.5V
3.0V
V(W2)
(c)
VDD3
10Vdc
M4007P
M4007P
M10
w3
M12
x3
z3
Cw
10n IC=0
M9
M11
M4007N
M4007N
0
3.5V
4.0V
4.5V
5.0V
5.5V
Initial condition: V(Cw) = 0 V
12V
8V
4V
0V
-2V
0s
5us
V(x3)
10us
15us
20us
25us
30us
V(z3)
Time
12V
8V
4V
0V
-2V
0V
0.2nV
V(x3)
0.4nV
0.6nV
0.8nV
1.0nV
V(z3)
V(w3)
V(x) stays at high and V(z) stays at low.
1.2nV
1.4nV
1.6nV
Initial condition: V(Cw) = 5 V
10V
8V
6V
4V
2V
0V
0s
5us
V(x3)
10us
15us
20us
25us
30us
V(z3)
Time
10V
8V
6V
4V
2V
0V
0V
V(x3)
0.5V
V(z3)
1.0V
1.5V
2.0V
2.5V
V(w3)
3.0V
3.5V
4.0V
4.5V
5.0V
Initial condition: V(Cw) = 5.03 V
(compare with initial condition V(Cw) = 5 V)
10V
8V
6V
4V
2V
0V
0s
5us
V(x3)
10us
15us
20us
25us
30us
V(z3)
Time
10V
8V
6V
4V
2V
0V
5.0V
V(x3)
5.5V
V(z3)
6.0V
6.5V
7.0V
7.5V
V(w3)
8.0V
8.5V
9.0V
9.5V
10.0V
Initial condition of Cw = 10 V
12V
8V
4V
0V
-2V
0s
5us
V(x3)
10us
15us
20us
25us
30us
V(z3)
Time
12V
8V
4V
0V
-2V
2.96nV
V(x3)
2.98nV
V(z3)
3.00nV
3.02nV
3.04nV
3.06nV
3.08nV
V(x3)
V(x) stays at low and V(z) stays at high.
3.10nV
3.12nV
3.14nV