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(ELEC202)[2005](f)final~wctsangaa^_45589.pdf
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Reference Answer to ELEC 202 Fall 2005 Final Paper
Dec 2 drafted and Dec 8, 2008 updated by TA ZHAN Chenchang (Jim)
Email: [email protected]

(1)
(1a) 1 W 2
I = C ( )( V .V )
b nox no1 tn
2 L
1 210 2

24 uA = i60 / ii(V . 0.8 V )
uA V
22 o1
Vo1 = 1.2 V
V .V 6V .1.2 V
dd o1
R1 == = 200 k Ib 24 uA
(1b)
V = V .V . |V |= 0.25 V

ov 2 dd o2 tp
Vo2 = 6V . 0.75 V . 0.25 V = 5V
V .V 5V .1.2 V
o2 o1
R2 == = 158.3 k
Ib 24 uA
I = 1 C (W )( V )2
pox pov 2
b 2 L
24 uA = 1 i24 / 2 i(W ) i(0.25 V )2
uA V
2 Lp

(W )p = 32 = 64
L 2
(2)
Currents of M1 and M2 have the same value.
1 W 2
I = C ( )( V )
n ox 1 ov 1
2 L
1 W 2


I = C ( )( V )
n ox 2 ov 2

2 L
WW

() = 9(i )
L 1 L 2 Vov 2 = 3iVov 1 = 300 mV
ie . ., V .V .V = V . 0V .V = 300 mV
g 2 s2 tn g tn

V .V .V = V .V .V = 100 mV
g1 s1 tn gs1 tn
Vs1 = 200 mV
V 1 0.2
sV
R =
== 2k
I 100 uA
(3) (3a)

@ vs ,
0 . vv
i = o = s rs R R
ds
Rs
vs =. vo (1) Rd
@ vo ,
vv
o vo . s
gmvgs = gm (vi . vs ) = (. ) . Rd rds
Rs
vo + vo
Rd

s voR
gm (vi + vo ) = (. ) .
R Rr
d dds
vg gR
om md
A = =. =.
R 11 R RR
vss ds
ig + ++ 1+ gR ++
m ms
Rr rds
Rd d ds Rr dds ds r
(3c)

@ vs , v . vv
dd o s
i == rs R R
ds
Rs
v = (v . v ) (1)
s dd o
Rd @ vo , v . vv . v
dd o os
gv = g (0 . v ) = () .
mgs m s
Rd rds

Rs
v . (v . v )
o dd o
Rv . vR
s dd o d
g (0 . (v . v )) = () .
m dd o
R Rr
d dds
R 1 RR 1 R
sss s
.g .. 1+ gR ++ R +
m mss
v RRRr r ggr
o dddds ds m mds
Add = = ==
v 1 R 1 R RR 1 RR
ss dsdsdd
.. g .. 1+ gR ++ + R ++
m mss
R RrRr rr g grgr
d dds dds ds ds m mds mds
(3d) 2iI 2 200 i uA
g = b == 2/
m mAV
Vov 0.2 V

rds = 1 = 1 = 100 k
Ib 0.05 / V i200 uA
gR 2mi10 k

A =. md =. =. 6.43 V / V
+ m Rd Rs 1 10 k 1k
1 g R s ++ 1+ 2mi k ++
rds 100 k 100
rds k R
1 s 11k
+ R + s +1k +
g gr
m mds 2m 2mi100 k
Add == = 0.97 V / V
1 RR 1 10 k 1k
+ Rs + d + s +1k ++ gm gmrds gmrds 2m 2mi100 k 2mi100 k
(3e) + A 6.43
PSRR = i 10 | | 20log 10 =
20 log = i | | 16.43 dB
Add 0.97

(4)
(4a) 1 W 21 2
I = i C i(V .V ) = i1.25 mi(1.2 . 0.8) = 0.1 mA
b nox gs tn
2 L 2

2iIb =
2 0.1 i mA

Vovp =
= 0.4 V k 1.25 / 2
p mAV
Vb = 5V . 0.4 V . 0.8 V = 3.8 V
(4b)
Vovn = 1.2 V . 0.8 V = 0.4 V
Vovp = 0.4 V
Output swing is [V ,V .V ] = [0.4 V , 4.6 V ]
ovn dd ovp
(4c) 2iIb 2 0.1 i mA
g == = 0.5 /
mn mAV
Vovn 0.4 V

11
r == = 100 k
dsn nIb 0.1/ V i0.1 mA
rdsp = 1 = 1 = 100 k | p | Ib 0.1/ V i0.1 mA
dsp / i 25 V
A =. gmn (rdsn || r ) =. 0.5 mA V 50 k=. V /
(4d)
Assume the amplifier is in linear region, then

Amplitude of v i is 40mV
Amplitude of v o = |A| 40mV=25 40mV=1V ii
Vo (min) = 2.5 V .1V = 1.5 V
Vo (max) = 2.5 V +1V = 3.5 V Since they are in the output voltage range as indicated by 4b), both Mn and Mp are in active region and previous ass