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Lecture 20-1
MOSFET transistor I-V characteristics
iD K 2 vGS Vt( )vDS vDS2
[ ]=
iD
vGS
vDS--
++
iD
K vGS
Vt
( )2
[ ] 1 vDS
+( )=
KW
2L------K
n=
Kn
Cox
n
=iD
K 2 vGS
Vt
( )vDS
[ ]=Linear region:
vDS
sat
vGS
Vt
=
vDS
vGS
Vt
Triode region:
vDS
vGS
Vt
0
VS1B=0
VBVS2B=0
VS1B=0
VS2B>0
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Lecture 20-4
Body Effect
Positive VSB for NMOSFETs tends to increase QB, hence decrease QI, for a
fixed VGS
VB
VGS > Vt
+
n+n+ QB0
VS>0
QI
VDS > 0
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Lecture 20-5
Body Effect
Modeled as a change in the threshold voltage as a function of VSB
The source is, by definition for NMOSFET, at a lower positive potential thanthe drain, which is why we use it as our reference voltage
Vt
Vt0
2f
VSB
+ 2f
( )+=
SPICE will calculate this variation in threshold voltage, or you can over-rideits calculation by directly specifying gamma
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Lecture 20-6
Temperature Variations
The threshold voltage varies with temperature due to carrier generation in thesubstrate --- tends to decrease with increasing temperature
~2mV for every 1C increase
K also changes with temperature due to change in mobility
Tends to dominate temperature variation for large iD
Will iD increase or decrease with temperature?
Vt Vt0 2f VSB+ 2f( )+=
T1
T2 > T1
I1
2---
nC
ox
W
L----- v
GSV
t( )
2
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Lecture 20-7
Where is drain, where is source?
S
D
GB
D
GB
S
n-channel transistor p-channel transistor
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Lecture 20-8
PMOSFETs
All of the voltages are negative Carrier mobility is about half of what it is for n channels
p+
n
S DG
B
p+
The bulk is now connected to the most positive potential in the circuit
Strong inversion occurs when the channel becomes as p-type as it was n-type
The inversion layer is a positive charge that is sourced by the larger potential
and drained at the smallest potential The threshold voltage is negative for an enhancement PMOSFET
Note that the flatband voltage (which is negative) effects now tend toincrease the PFET threshold while they decreased the NFET threshold
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Lecture 20-9
PMOS
The equations are the same, but all of the voltages are negative Triode region:
iD K 2 vGS Vt( ) vDS vDS2[ ]=
vGS
Vt
vDS
vGS
Vt
K 12---nCoxWL
-----= AV
2-------
iD is also negative --- positive charge flows into the drain
Saturation expression is the same as it is for NFETs:
iD
satK v
GSV
t( )
2[ ] 1 v
DS+( )=
+V dd
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Lecture 20-10
PMOS
Characteristic appears to be the same, except that all of the voltages arenegative
-5 -4 -3 -2 -1 0
-100
-90
-80
-70
-60
-50
-40
-30
-20
-10
0
10
W=1 micronL=1 micronsV t0= -1 volt
Kp=2e-5 (A/v2)
phi =-0.6
ND=1e15
IDS
(A)
VDS
VGS=-2.5V
VGS=-2.0V
VGS=-1.5V
VGS=-1.0V
VGS=-3.0V
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Lecture 20-11
PMOS
But it is generally displayed as:
W=1 micronL=1 micronsV t0= -1 volt
Kp=2e-5 (A/v 2)phi =-0.6
ND=1e15
-IDS(A)
-VDS
VGS=-2.5V
VGS=-2.0V
VGS=-1.5V
VGS=-1.0V
VGS=-3.0V
0 1 2 3 4 5
-10
0
10
20
30
40
50
60
70
80
90
100
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Lecture 20-12
Depletion Mode NMOSFET
Depletion mode FETs have a channel implanted such that there is conductionwith VGS=0
The operation is the same as the enhancement mode FET, but the thresholdvoltage is shifted
Vt is negative for depletion NMOS, and positive for depletion PMOS
VGS
n+n+
VS VDS
n+
p
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Lecture 20-13
Depletion Mode NMOSFET
Negative gate voltage is required to turn the channel off
W=1 micronL=1 micronsV t0= -2 volt
Kp=2e-5 (A/v2
)
IDS(mA)
VDS
VGS
=1.0V
VGS=0.0V
VGS
=-1.0V
VGS=-2.0V
VGS=2.0V
0 1 2 3 4 5
0.0
0.2
0.4
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Lecture 20-14
Depletion Mode NMOSFET
The iDS vs. vGS characteristic is still quadratic in saturation
W=1 micronL=1 micronsV t0= -2 voltKp=2e-5 (A/v
2)
IDS(m
A)
VGS
-4 -3 -2 -1 0 1 2 3 4 5
0
1
2
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Lecture 20-15
Examples
Find the largest value that RD can have before the transistor fails to operate insaturation
5V
-5V
5k
RD
Vt
2V=
Kn
20A V2
=
L 10m=
W 400m=
0=
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Lecture 20-16
Examples
Find the drain currents and voltages for both transistors
10V
10k 15k
Vt
2V=
Kn
20A V2
=
L 10m=
W 100m=
0=
10V
M2 M1
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Lecture 20-17
Examples
What is the effective resistance of the transistor in the triode region?
10V
24.8k Vt 1V=
K 0.5m A V2=
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Lecture 20-18
Examples
Select the Rs so that the gate voltage is 4V, the drain voltage is 4V and thecurrent is 1mA.
10V
RD
Vt
2V=
K 1m A V
2
= 0=
RS
10V
RG1
RG2
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Lecture 20-19
Examples
Select the Rs so that the transistor is in saturation with a drain current of1.0mA and a drain voltage of 5V
Vt 1 V=K 0.5m A V
2=
0=
RD
10V
RG1
RG2
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Lecture 20-20
Examples
Solve for the drain current and voltage
20V
32k
Vt
2 V=
K 1m A V2
=
0=
4k10M