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Metal-oxide semiconductor field-effect transistor (MOSFET)
Enhancement MOSFET
N-channel
Border between
Triode and Saturation
regions
VDS=VDSsat=VGS-Vt
.
Enhancement MOSFET
There are three distinct regions of operation: cutoff, triode and saturation.
For N-channel MOSFET
1) Cutoff when VGS  VT
2) When the channel is first introduced
MOSFET is ON VGS  VT
It can operate on triode or saturation.
2.1) To operate in triode: V DS is kept small  channel is continuous
VGD  VT or VGS  VDS  VT or VDS  VGS  VT
I D  kn'
W
L
1 2 

(VGS  VT )VDS  2 V DS 


If V DS is small  V 2 DS can be neglected
I D  kn'
rDS 
W
(VGS  VT )VDS Linear relation ( I D  VDS )
L
VDS
1

(Resistance)
ID
' W
kn (VGS  VT )
L
2.2) To operate MOSFET in saturation
1. VGS  VT To induce channel
2. VGD  VT (pinched- off channel)
Or VDS  VGS  VT
In saturation ID is given by (neglecting  )
ID 
1
1 W
W 
nC ox   (V GS V T ) 2  k n' 
2
2 L
L 
channel length modulation
effect
ID 
1 ' W
kn 
2 L

2
 (VGS  VT ) (1  VDS )


2
2
 (V GS V T )  K (V GS V T )

PMOS=
P-Channel MOSFET
For P- channel VT, VGS and VDS are negative
To turn transistor ON VGS  VT
To operate in triode
VGD  VT or VDS  VGS  VT
I D  kn'
W
L
1 2 

(VGS  VT )VDS  2 V DS 
To operate in saturation
VGD  VT or V DS  VGS  VT
ID 
1
 pC ox
2
W

L
1 ' W

2
 (V GS V T )  k p 
2 L


2
 (V GS V T )

Example 1
Analysis the circuit to determine the voltage at all nodes and the current through all
brunches.
Vt= 1V ,
1 ' W
kn 
2  L

2
 = 0.5mA/V

since IG=0
RG1 and RG2 are in series
RG 2
 10
RG1  RG 2
VG 
5V > VT
Thus, it is excepted that the transistor is ON.
Assume the transistor is working in saturation region
ID 
1 ' W
kn 
2 L

2
 (VGS  VT ) Check VGD  VT

I D  0.5(VGS  1) 2 (1)
VGS  VG  VS  5  6I D (2)
Substitute (2) in (1)
I D  0.5(5  6 I D  1) 2  0.5(4  6 I D ) 2
18I D  25I D  8  0
2
I D  0.98mA  VS  6I D  5.34 V
 VG Doesn't make physical sense.
I D  0.5mA  VS  6I D  3V
Thus VGS  5  3  2v
VD  10  6I D  7V  VGD  5  7  2  VT The assumption is correct
Example 2
VT=-1
1 ' W
kp 
2 L

2
 =0.5mA/V

P-channel MOSFET
VG 
3
 5  3V
23
VGS  3  5  2  VT
The transistor is ON
Assume the transistor is working in saturation
ID 
1 ' W
kp 
2 L

2
 (V GS V T )

I D  0.5 2  (1)  0.5mA
2
VD  6 I D  3V
VGD  3  3  0  VT
 Saturation which confirms that assumption is correct.
Example 3
VT=2V
1 ' W
kn 
2  L

2
 =0.1mA/V

VGD  0  VT
The transistor is working in saturation
ID 
1 ' W
kn 
2 L

2
2
 (VGS  VT )  0.1(VGS  2)

VGS  VG  VS  10  15 I D  0  10  15 I D
I D  0.1(10  15I D  2) 2  0.1(8  15I D ) 2
22.5I D  25I D  6.5  0
2
ID 
ID 
25  (25) 2  4(22.5)(6.4)
2(22.5)
25  7
 0.71mA
2(22.5)
or
I D  0.4mA
VG  10  15(0.71)  0.65V Impassible
VG  10  15(0.4)  4
VGS  4  VT
Small signal ac equivalent circuit for MOSFET
MOSFET should work in saturation region
ID 
1 ' W
kn 
2 L

2
 (VGS  VT )

VGS  VGS  v gs
(1) the DC or quiescent current ID
(2) Current component that is
iD 
iD 
1 ' W
kn 
2 L

2
 (VGS  vgs  VT )

directly
proportional to input vgs .
1 ' W 
1 W 
W 
kn   (VGS  VT ) 2  kn'   (VGS  VT )vgs  kn'   vgs
2 L
2 L
L
(1)
(2)
(3)
(3) Current proportional to the
square of the
input. This component is
undesirable
because it represents
nonlinear distortion.
To reduce the nonlinear distortion, the input signal should be kept small
vgs
2(VGS  VT )
W
id  kn' 
L
id

 (VGS  VT )vgs  g m 
vgs

W
g m  kn' 
L

' W
 (VGS  VT )  2kn 

L

 ID

For N- channel MOSFET
VGS and VT are positive and VGS > VT  g m is positive
For P- channel MOSFET
VGS and VT are negative VGS < VT  g m is positive
To include the effect of channel length modulation,
it is modeled by output resistance ro 
Where V A 
1

,
VA
ID
VA=20 to 200 V
ro is in the range 10 to 1000K 
MOSFET Amplifier
1 ' W
kn 
2 L
VT  1.5V

2
  0.5mA / V

1 ' W 
kn   (VGS  VT ) 2
2 L
I D  0.5(VD  1.5) 2
(1)
VD  10  5.6 I D
(2)
ID 
I D  0.5(8.5  5.6 I D ) 2
 36.125  48.6 I D  15.68 I D
2
0  36.125  48.6 I D  15.68 I D
2
48.6  (48.6) 2  4(15.68)(36.125)
ID 
2(15.68)
I D  1.836 mA or
I D  1.24 mA
If I D  1.836 mA VD=-0.4V impossible
If I D  1.24mA VD=3.1V
VG  VD  VGS  3.1V
W
g m  kn' 
L

 (VGS  VT )  1.6mA / V

VDS  3.1  1.5  1.6V
V0
V  Vi
 g mVi  0
'
RG
RL
RGV0  g m RG R ' LV  R ' LVi  R ' LV0
V0 R ' L  g m RG R ' L R ' L (1  g m RG )


 5.7V / V
Vi
R'L  C
R ' L  RG
R0  5.6 K // RG  5.6 K
V
i
R ' L (1  g m RG )
V0 
Vi
R ' L  RG
Ri 
Vi  V0
1  R ' L (1  g m RG ) 
i

1 

RG
RG 
R ' L  RG 
1 gm R'L
i '
R L  RG
R ' L  RG
Ri 
 148 K
1 gm R'L
MOSFET Common Source amplifier
VT  1V
1 ' W
kn 
2 L

2
  0.5mA / V

I D  0.5mA
VA  200V
VG  5V ,VD  7V ,VS  3V
VDS  VGS  VT  1V (edge of saturation)
Max. Output swing when
VDS  5.5V  Swing =4.5V
r0 
| VA |
200

 400 K
ID
0.5mA
W
g m  kn' 
L

2
 (VGS  VT )  1(1)  1mA / V

AV   g m RD // RL // r0
 (1)(3)  3V / V
Ri  10 M // 10 M  5M
Rout  r0 // RD  6k
Low compared with CE amplifier
Very very high compared with BJT output resistance.