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Unformatted text preview: ECE ECE342 HW#12 Solutions
442 HW#13 Solutions 1.
gm1 = gm1 VOV
1m × 0.2
2ID
⇒ ID =
=
= 100µA
VOV
2
2
RO = (gm2 ro2 )ro1 Since the transistors are identical with the same current ﬂowing through both, ro1 = ro2 = ro and
gm1 = gm2 = gm ; as speciﬁed in the problem:
2
RO = gm ro = 400k Ω ⇒ ro = 20k Ω Since ro = VA L
ID , we have:
L= gm = 2µn Cox W
×
L 100µ × 20k
ID ro
=
= 0.4µm
1
VA
5µ ID ⇒ 2
W
gm
(1000µ)2
=
= 12.5
=
L
2µn Cox ID
2(400µ)(100µ) For the maximum negative swing at the output, we want the MOSFETs to be biased so that each
transistor can reach VDS = VOV = 0.2V .
⇒ VG2 = Vtn + VOV + VOV = 0.5 + 0.2 + 0.2 = 0.9V
Minimum output voltage will then be Vout = 2VOV = 0.4V .
2. From the smallsignal schematic, we can write the following KCL equations, noting that the transistor M1
reduces to just a resistance ro1 , since the control voltage vgs1 = 0, and the control voltage vgs2 = 0−vy = −vy :
ix = −(gm2 + gmb2 )vy + ⇒ vx − vy
vy
=
ro2
ro1 ro1
1
vy
ro1
≈
≈
=
vx
ro2 (1 + (gm2 + gmb2 )ro1 )
ro1 ro2 (gm2 + gmb2 )
gm2 ro2 Therefore, vy is smaller than vx by a factor of gm2 · ro2 .
3. a) For the same I in cases (a) and (b):
I= gm ( W )a
V2
1W2
L
kn VOV ⇒ OV b = W
2
2
L
VOV a
( L )b 2
For the same I, if W/L is divided by 4, then VOV is multiplied by 4, or equivalently VOV is doubled,
W
= µn Cox L VOV . Thus, gm for the circuit (b) is half of the one for circuit (a). AO = gm ro = 2ID
VA
2V L
×
=A
VOV
ID
VOV Thus, if L is multiplied by 4, and VOV is halved, then AO is doubled for circuit (b).
In summary, for circuit (b), VOV is doubled, gm is halved, AO is doubled. 1 P6. ...
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This note was uploaded on 11/06/2011 for the course ECE 342 taught by Professor Nareshshanbhag during the Spring '11 term at University of Illinois, Urbana Champaign.
 Spring '11
 NareshShanbhag
 Transistor

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