Solution
ECE-438, MOS Transistor
Chapter 3, Digital Integrated Circuits, J. M. Rabaey, 2nd Edition
1) Given the data in the table below for a short channel NMOS transistor with
VDSAT=0.6V and kc=100PA/V2, calculate VT0, O, J, 2°Ifµ and W/L.
1
2
3
4
5
VGS
2.5
2
2
2
2
VDS
1.8
1.8
2.5
1.8
1.8
VBS
0
0
0
-1
-2
ID(PA)
1812
1297
1361
1146
1039
For Short-Channel device:
Wª
V2 º
ID kc «Vmin(VGS VT ) min » (1 OVDS ) ,
L ¬
2 ¼
where Vmin= min(VGS-VT, VDS, VDSAT).
To begin with: the operation regions need to be determined.
For any of these data to be in saturation:
VT should be: VGS – VT > VDSAT
2- VT > 0.6 => 1.4 > VT
This is a quite high value in our process. Thus, we can assume that all data are taken in
velocity saturation. We will check this assumption later.
In Velocity Sat.:
ID
kc
V
W
VDSAT (VGS VT DSAT )(1 OVDS )
L
2
Eq.1
VDSAT and VDS are constant in 1 & 2 =>
V
VGS 1 VT DSAT
I D1
2 =>
V DSAT
I D2
VGS 2 VT 2
1812
1297
2.5 VT 0.3
=> VT0 = 0.44V
2 VT 0.3
Page: 1
In 2 & 3 VDSAT and VGS are constant =>
1297
1361
1 1.8O
ŸO
1 2.5O
0.08V 1
Using data 2 & 4: VT = 0.587 V (1)
Using data 2 & 5: VT = 0.691 V (2)
Both these values satisfy VT < 1.4 V so all the data in our table were taken in velocity
saturation.
VT 0 J
VT
2I F VSB 2I F
(1) and (2) can be used along with VT0 = 0.44 V to conclude:
J
0.29V
1
2
and 2 IF
0.6 V
also using 2nd set of data
ID = 1297PA and Eq.1 =>
W
L
15
2) An NMOS device is plugged into the test configuration shown below in Figure P1.
The input Vin=2V and the current source draws a constant current of 50PA. R is a
variable resistor that can assume values between 10k: and 30k:. Transistor M1
experiences short channel effects and has following transistor parameters:
kc=110u10-6V/A2, VT=0.4 and VDSAT=0.6V. The transistor has a W/L=2.5/0.25. For
simplicity, body effect and channel length modulation can be neglected.
a) When R=10k: find the operation region, VD and VS.
b) When R=30k: again determine the operation region, VD and VS.
Figure P1
Page: 2
(a) R=10 K: therefore V D 2.5 10 u 0.05 2V
we can assume transistor is in SAT.
kc W
2
ID
VGS VT
2 L
110 u 10 6 2.5
2
50 u 10 6
u
u 2 VS 0.4
2
0.25
Therefore, VS 1.3 V.
As VGS VT 0.3 , VDS 0.7 and V DSAT 0.6 , our assumption is correct and the
device is in saturation region.
(b) R 30 K: therefore V D 2.5 30 u 0.05 1V
It is obvious that in this case the transistor is not in velocity saturation (because VS
Would be 1.22 which is greater than VD ) .
Assuming the device is in linear region
V 2
W
ID kc VDS (VGS VT DS )
L
2
2
1 VS
6
6
50 u 10
110 u 10 u 10(1 VS )(2 VS 0.9 )
2
VS 0.93
To check our assumption:
VGS 2 0.93 1.07
V DS 0.07
So our first assumption is correct and VS 0.93
Page: 3
3) Consider the circuit shown in Figure P2.
a) Write down the equations (and only those) which are needed to determine the
voltage at node X. Do NOT plug in any values yet. Neglect short channel effects
and assume that Pp = 0.
b) Draw the (approximate) load lines for both PMOS transistor and resistor. Mark
some of the significant points.
c) Determine the required width of the transistor (for L = 0.25µm) such that X
equals 1.5 V.
d) We have, so far, assumed that M1 is a long-channel device. Redraw the load
lines Assuming that M1(PMOS) is velocity-saturated. Will the voltage at X rise or
fall?
(Use Table 3-2, Page 103)
Figure P2
PMOS is always in Saturation.
a)
kpc W
VX VTp
2 L
2
2.5 VX
R
b)
VDD/R
=2.5/20k=125uA
c)
30 x10 6 W
1.5 0.4
2
L
VSD, VX
2
1
20k:
Page: 4
Ö W/L = 2.755 , W = 0.69 Pm
d)
VSD, VX
Voltage at node X would go up since the current drive of PMOS is lower.
4) The circuit of Figure P3 is known as a source-follower configuration. It achieves a DC
level shift between the input and output. The value of this shift is determined by the
current I0. Assume device is in saturation J = 0.4, 2|If| = 0.6 V, VT0 = 0.43 V, kc = 115
µA/V2, and O = 0. The NMOS device has W/L = 5.4µ/1.2µ such that the short channel
effects are not observed.
a) Derive an expression giving Vi as a function of Vo and VT(Vo). If we neglect
body effect, what is the nominal value of the level shift performed by this circuit?
b) The NMOS transistor experiences a shift in VT due to the body effect. Find VT
as a function of Vo for Vo ranging from 0 to 1.5V with 0.25 V intervals. Plot VT vs.
Vo.
c) Plot Vo vs. Vi as Vo varies from 0 to 1.5 V with 0.25 V intervals. Plot two
curves: one neglecting the body effect and one accounting for it. How does the
body-effect influence the operation of the level converter? At Vo (body effect) =
1.5 V, find Vo (ideal) and, thus, determine the maximum error introduced by body
effect.
Figure P2
Page: 5
knc W
(Vi Vo Vt )2
2 L
a) ID
2ID
W
knc
L
Vi Vo VT
Neglecting body effect VT=VT0
Vi
2ID
Vo V T0
W
knc
L
2ID
Vt 0
W
knc
L
Level Shift
LS = 0.8V
b)
VT
VT 0 J
VT
0.43 0.4
2I F VSB 2I F
Vo 0.6 0.6
Page: 6
c)
Vo
Vi 2ID
W
knc
L
Vt
o
Plot Vi versus Vo for two different cases:
1) VT = VT0 (neglecting body effect, γ=0)
If body effect is neglected, Vi=Vo+constant (linear relation)
Vo=1.5 V ==> Vi=2.3V
2) VT= VT(Vo) (accounting for body effect, γ=0.4)
Vo=1.5V ==> Vi=2.57 V
Maximum error due to body effect=2.57-2.3=0.27 V.
Page: 7
5) The curves below in Figure P4 represent the gate voltage (VGS) vs. drain current (IDS)
of two NMOS devices, which are on the same die and operate in sub-threshold region.
Due to process variations on the same die the curves do not overlap.
Also assume that the transistors are within the same circuit configurations as Figure P5 in
If the input voltages are both Vin = 0.2V. What would be the respective durations to
discharge the load of CL = 1pF attached to the drains of these devices.
(Assume voltage charge for CL is 1V)
Figure P4
Figure P5
Page: 8
Vin = 0.2
Ö IDS = 3 e–8 A (1)
Or
Vin = 0.2
Ö IDS = 5 e–9 A (2)
'V
I
't
C
't1
1pF
1
3x10 8
33.3P s
't2
1pF
1
5x10 9
200 P s
6) Compute the gate and diffusion capacitances for transistor M1 of Figure P6.
Assume that drain and source areas are rectangular, and are 1 µm wide and 0.25
0.5 µm long.
Use the parameters of Example 3.5 to determine the capacitance values. Assume mj = 0.5
and mjsw = 0.44. Also compute the total charge stored at input node, for the following
initial conditions:
a) Vin = 2.5 V, Vout = 2.5 V, 0.5 V, and 0 V.
b) Vin = 0 V, Vout = 2.5 V, 0.5 V, and 0 V.
Figure P6
Page: 9
Cox = 6 fF/Pm2, MOS dimensions L = 0.25Pm and W = 1Pm
Cut-off : Cg = Cox.WL + 2Co.W
Linear
: Cg = Cox.WL + 2Co.W
Sat-Vel.Sat: Cg = 0.66Cox.WL + 2Co.W
Diffusion Cap (Cd) NOTE: Ldrain=0.5um (drain length) is not L for the MOSFET
Cd=Cj*Area+Cjsw*perimeter
Cd Area=W*Ldrain
C j LDWD C jsw (2LP
Drain
Drain Perimeter=W+2Ldrain
WD )
Cj0
Cj
mj
§ VDS
D ·
¨ 1-
¸m j
I0 ¹
©
C jsw
C jsw 0
D · mjsw
§ VDS
¨ 1-
¸m
I0 ¹ jsw
©
a) Vin = 2.5V,
Vout = 2.5V Vel. Saturation.
Cg = 1.62 fF , Q = 4.05 fC = 4.05 e –15 C
Cd = 0.827 fF
Vout = 0.5 V Linear Reg.
Cg = 2.12 fF , Q = 5.3 fC
Cd = 1.263 fF
Vout = 0 V Linear Reg.
Cg = 2.12 fF , Q = 5.3 fC
Cd = 1.56 fF
b) Vin = 0V => Cut off
Regardless of VDS
Cg = Cox.WL
Cg = 2.12 fF, Q =0
And Cds are the same as Part a.
Vin = 2.5V => Cd = 0.827 fF
Vout = 0.5 V => Cd = 1.263 fF
Vout = 0 V => Cd = 1.56 fF
Page: 10