Field Effect Transistor (FET) 1
Feb 25, 2016
1
Field Effect Transistor (FET)
2
Introduction
Field Effect Transistor (FET)
Junction Field Effect Transistor (JFET)
Metal Oxide Semiconductor FET
(MOSFET)
Depletion TypeMOSFET
Enhancement TypeMOSFET
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Junction Field Effect Transistor (JFET) n-channel JFET p-channel JFET
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JFET Introduction
• JFET is always operated with the gate source p-n junction reversed biased.
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JFET IntroductionChannel width and thus the channel resistance can be
controlled by varying the gate voltage.
JFET biased for construction Greater VGG narrows the channel
Less VGG widens the channel Water analogy for the JFET control
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JFET Characteristics and Parameters
• For VGS = 0 v, the value of VDS at which ID becomes essentially constant is the pinch-off voltage (Vp) and is denoted as IDSS.
• Breakdown occurs at point C when ID begins to increase very rapidly with any further increase in VDS .
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VGS controls ID.
The value of VGS that makes ID approximately zero is the cutoff voltage VGS(off). The JFET must operate between VGS = 0 and VGS(off) .
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Transfer Characteristics • William Bradford Shockley derived a relationship
between ID and VGS which is known as Shockley’s equation and is given by
• The above equation suggests that when VGS = 0, ID = IDSS. When VGS = Vp, ID = 0
2
p
GSDSSD V
V1II
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Transfer curve from the drain characteristics
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ExampleThe following parameters are obtained from a certain
JFET datasheet: VP = -8 v and IDSS = 5 mA. Determine the values of ID for each value of VGS ranging from 0 v to -8 v in 1 v steps. Plot the transfer characteristic curve from these data.
Solution:2
p
GSDSSD V
V1II
mA5801mA5I
2
D
mA83.3811mA5I
2
D
11
mA81.2821mA5I
2
D
mA95.1831mA5I
2
D
mA25.1841mA5I
2
D
mA703.0851mA5I
2
D
mA313.0861mA5I
2
D
mA078.0871mA5I
2
D
mA0881mA5I
2
D
12
-8 -7 -6 -5 -4 -3 -2 -1 00
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5 x 10-3
VGS
ID
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FET Biasing• The following relations can be applied to the dc
analysis of most of the FET amplifiers:A0IG
SD II 2
p
GSDSSD V
V1II
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JFET Biasing: Fixed Bias Circuit
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JFET Biasing: Fixed Bias Circuit0IG
0RIV GGRG
Circuit for dc analysis
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Fixed Bias CircuitGS Loop:• Apply KVL
GGGS VV
• Apply the Shockley’s Equation:2
p
GSDSSD V
V1II
• Plot Shockley’s equation:
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Fixed Bias Circuit• Q-Point:
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Fixed Bias Circuit• DS Loop
DSDDDD VRIV
DDDDDS RIVV
Also note that0VS
0VVVV DSDDS
DSD VV
In addition0VVVV GSGGS
GSG VV
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Example: Determine the following for the given Fig.(a) VGSQ (b) IDQ (c) VDS (d) VD (e) VG (f) VS.Solution:(a) VGSQ = -VGG = -2 V
mA625.5821mA10
VV1II
22
P
GSDSSDQ
(b)
(c) V75.4RIVV DDDDDS
(d) VD = VDS = 4.75 V
(e) VG = VGS = -2 V(f) VS = 0 V
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JFET Biasing: Self Bias Configuration
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Self Bias Circuit: DC Analysis
Self-bias Circuit for dc analysis
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JFET Self Bias CircuitIG = 0IS = ID From GS Loop: -VGS = VRS
or VGS = -ISRS
Substituting IS = ID
VGS = -IDRS.
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JFET Self Bias CircuitShockley Equation:
2
p
SDDSS
2
p
GSDSSD V
RI1IVV1II
2
p
SDDSSD V
RI1II
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JFET Self Bias Circuit: Q-Point
Self-Bias Line:Since VGS = -IDRS .If ID = 0 then VGS = 0and ID = IDDS/2 (say), then VGS = -IDDS RS /2Superimposing this straight
line on the transfer curve, we get Q-point as shown in the Fig.
Self Bias line
Transfer Curve(Shockley equation)
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JFET Self Bias CircuitDS Loop: Using KVL
Substituting IS = ID,
or
In addition
SSDSDDDD RIVRIV
SDDSDDDD RIVRIV
DSDDDDS RRIVV
SDS RIV
0VG
DRDDSDSD VVVVV
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JFET Self Bias Circuit: Example 1Determine the following: VGSQ , IDQ, VDS, VS, VG, and VD.Solution:Step 1: Draw the self bias line: VGS = - IDRS , When ID = 0, VGS = 0.Choosing ID = 4 mA, VGS = -4mA×1 k = -4 vThe line is drawn below:
27VGS (volts)
JFET Self Bias Circuit: Example 1Step 2: Plot the Shockley equation: (IDSS = 8mA, VP = -6v)
2
p
GSDSSD V
V1II
VGS0 -1 -3 -4 -5 -6
ID (mA) 8 5.55 2 0.88 0.22 0
I D (m
A)
-6 -5 -4 -3 -2 -1 00
1
2
3
4
5
6
7
8
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JFET Self Bias Circuit: Example 1Step 3: Show the Shockley curve and the self bias line on the same
graph paper
From the graph, VGSQ = -2.6 v, IDQ = 2.6 mA
-6 -5 -4 -3 -2 -1 00
1
2
3
4
5
6
7
8
I D (m
A)
VGS (volts)
Self bias line
Shockley Curve
Q-Point
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JFET Self Bias Circuit: Example 1Step 4: Find the remaining quantities: VDS = VDD – ID(RS + RD ) = 20 – 2.6mA( 1 k + 3.3 k) = 8.82 v
VS = IDRS = (2.6mA)(1k) = 2.6 vVG = 0 vVD = VDS + VS = 8.82 + 2.6 = 11.42 v
(or VD = VDD – IDRD = 11.42 v)
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JFET Biasing: Voltage Divider Circuit
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JFET Biasing: Voltage Divider Circuit dc analysis
VG
21
DD2G RR
VRV
Applying KVL, 0VVV RSGSG
orRSGSG VVV
But VRS = ISRS = IDRS
Therefore
SDGSG RIVV
S
GSGD R
VVI
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Voltage Divider Circuit: Q-Point
Bias Line:(i) When ID = 0VGS = VG – IDRS = VG – (0)(RS)VGS = VG
(ii) When VGS = 0
S
G
S
GSGD R
0VRVVI
S
GD R
VI
Plot this line along with the Shockley Curve, as shown in the Figure.
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JFET Biasing: Voltage Divider Circuit dc analysis
VG
0RIVRIV sSDSDDDD
sSDDDDDS RIRIVV
)RR(IVV SDDDDDS
DDDDD RIVV
SDS RIV
21
DD2R1R RR
VII
From DS Loop:
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Voltage Divider Circuit: Example
Determine the following:(a) IDQ and VGSQ.(b) VD
(c) VS
(d) VDS
(e) VDG.
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Voltage Divider Circuit: Example 1
Solution: IDSS = 8 mA, Vp = -4 v. Shockley Equation:
Bias Line:
2GS3
2
p
GSDSSD 4
V1108VV1II
VGS -4 -2 -1 0ID mA 0 2 4.5 8
v82.110270101.2
1610270RR
VRV 36
3
21
DD2G
)105.1(I82.1
RIVV3
D
SDGGS
When ID = 0, VGS = 1.82 v For VGS = 0, ID = 1.82/1.5k = 1.21 mA
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From the Figure, IDQ = 2.4 mA, VGSQ = -1.8 v
(b) VD = VDD - IDRD = 16 – (2.4mA)(2.4k) = 10.24 v(c) VS = IDRS = 16 – (2.4mA)(2.4k) = 10.24 v
(d) VDS = V DD – ID(RD + RS ) = 16 – (2.4mA)(2.4k + 1.5k) = 6.64 v
(e) VDG = VD - VG = 10.24 – 1.82 = 8.42 v
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Voltage Divider Circuit: Example 2For the given network, Detrmine(a) VG.(b) IDQ and VGSQ.(c) VD and VS.(d) VDSQ .Solution:(a)
(b) IDSS = 10mA, Vp = -3.5 v
v16.2k110k910
20k910RR
VRV21
DD1G
2GS3
2
p
GSDSSD 4
V11010VV1II
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VGS (volts) -3.5 -2 -1 0ID (mA) 0 1.8 5.1 10
Bias Line: VGS = VG – IDRS = 2.16 – ID(1.1k)When ID = 0, VGS = 2.6 vWhen VGS = 0, I = 2.16/1.1k = 2mAFrom the graph, we see thatIDQ = 3.3 mA, VGSQ = -1.5 v(c) VD = VDD – IDQRD = 20 - (3.3mA)(2.2k) = 12.74 v VS = IDRS = 3.63 v(d) VDSQ = VDD – IDQ(RD +RS ) = 9.11 v
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Metal-Oxide -Semiconductor Field Effect Transistor (MOSFET)
MOSFET
Depletion –Type MOSFET
Enhancement-Type MOSFET
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N- Channel Depletion-Type MOSFET
Construction of D-MOSFET(n-Channel)
• The foundation of this type of FET is the substrate (p-type material).
• The source and drain terminals are connected through metallic contacts to n doped regions linked by an n channel.
• The gate is also connected to a metal contact surface but remains insulated from the n-channel by a SiO2 layer.
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Basic Operation and Charactersitics of N – Channel D-MOSFET
I D= I S= IDSS
VGS = 0
• When VGS = 0 and VDS is applied, the drain current ID = IDSS flows through the circuit due to the free electrons of the n-channel.
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Basic Operation and Characterstics of N-Channel D-MOSFET
• When VGS < 0, recombination between electrons and holes occurs. The more negative the bias, the higher the rate of combination. The resulting level of ID is reduced and becomes zero at pinch-off voltage. Electrons repelled by
negative Potential at gate.
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Basic Operation and Charactersitics of D-MOSFET
When VGS > 0, the gate will draw additional electrons from the p-substrate due to the reverse leakage current and the drain current increases at a rapid rate.
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Example: Sketch the transfer characteristics for an n-channel depletion type MOSFET with IDSS = 10 mA and Vp = -4 v.
Solution: 2
GS3
2
p
GSDSSD 4
V11010VV1II
VGS -4 -2 -1 0 +1ID (mA) 0 2.5 5.6 10 15.6
The curve is plotted on the next slide.
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Gate Source Voltage
Drai
n Cu
rren
t (A)
-4 -3 -2 -1 0 10
0.005
0.01
0.015
0.02
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P-Channel depletion type MOSFET
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Symbols
N-Channel P-Channel
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Example1: For the n-channel depletion type MOSFET of the Fig., determine
(a) IDQ and VGSQ.(b) VDS.Solution: Shockley Equation:
18 v
2GS3
2
p
GSDSSD
3V1106
VV1II
VGS -3 -2 -1 0 1
ID (mA) 0 0.7 2.7 6 10.7
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Bias Line:
v5.1M100M1018M10
RRVRV
21
DD2G
)750(I5.1RIVV DSDGGS
When ID = 0, VGS = 1.5,
When VGS = 0, ID = VG/RS = 1.5/750 = 2 mA
From the graph, IDQ = 3.1 mA, VGSQ = -0.8 v
VDS = VDD – ID(RD + RS) = 10.1 v
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Example2: Determine the following for the given network. (a) IDQ and VGSQ (b) VD.
Solution:(a) Shockley Equation:
2GS3
2
p
GSDSSD
8V1108
VV1II
VGS -8 -6 -5 -4 -2 0 1 2ID(mA) 0 0.5 1.125 2.00 4.5 8.00 10.125 12.5
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Bias Line:VGS = -IDRS. When VGS = 0, ID = 0.When ID = 2.5 mA (say)VGS = -2.510-3 2.4 1000 = -6V
-8 -6 -2 0 2-4.30
4
6
8
10
12
14
1.7
I D (m
A)
VGS (volts)
From the graph paperVGSQ = 4.3 V, ID = 1.7mA(b) VD = VDD – ID RD = 20 – (1.7mA)(6.2k) = 9.46 V
ShockleyEquation
Bias LineQ-Point
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Example 3: For the following network, determine (a) IDQ and VGSQ (b) VDS and VS.
Solution: Shockley Equation:
2GS3
2
p
GSDSSD
8V1108
VV1II
VGS -8 -6 -5 -4 -2 0 1 2ID(mA) 0 0.5 1.125 2.00 4.5 8.00 10.125 12.5
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Bias Line:VGS = -VSS – IDRS.When ID = 0VGS = -(-4) = 4 VWhen VGS = 0ID = -VSS/RS = 4/0.39k = 10.26 mA(a) From the graphVGSQ 0.5 V, IDQ 9mA(b) VDS = VDD – IDQ(RD + RS) = 7.69 V VS = -VGSQ = -0.5V
-8 -6 -4 -2 2 4 60.50
2
4
6
8
12
14
109
10
I D (m
A)
VGS (volts)Sh
ockley
Equati
on
Bias
Line
Q-Point
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N-Channel Enhancement Type MOSFETThe construction of an
enhancement type MOSFET is quite similar to that of the depletion type MOSFET except for the absence of a channel between the drain and source terminals.
When VGS = 0, ID = 0 because the n-channel is absent.
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Basic Operation and Characteristics of an n-Channel E-MOSFET
When VGS > 0 & VDS > 0,A depletion region is creatednear the SiO2 layer void ofholes.As VGS increases, theconcentration of electronsnear the SiO2 increases and there is some flow between drain and source.The level of VGS that results in thesignificant increase in ID iscalled the Threshold Voltage (VT).
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Basic Operation and Characteristics of an n-Channel E-MOSFET
If VGS > VT is constant and VDS is increased, ID will Increase and will reach saturation.
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Drain Characteristics of an n-channel enhancement-type MOSFET
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Transfer characteristics for n-channel enhancement type MOSFET from the drain characteristics.
2TGSD VVkI where 2)Th(GS)on(GS
)on(D
VVI
k
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p-Channel enhancement-type MOSFET
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Symbols
61
Feedback Biasing of n-Channel e-MOSFET
Equations:GSDS VV DDDDGS RIVV
From the above equations, we get 0IDDGS D
VV 0VD
DDD GSR
VI
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Feedback Biasing of n-Channel e-MOSFET
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Example: Determine IDQ and VDSQ for the enhancement-type MOSFET of the following.
Solution: For the transfer curve
3
2
2)TH(GS)on(GS
)on(D
1024.0)38(
mA6
VVI
k
2TGSD VVkI
VGS 3 6 8 10ID 0 2.16mA 6 11.76mA
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For the network bias line:
For ID = 0, VGS = VDD = 12 v, and for VGS = 0ID = VDD /ID = 12 v / 2k = 6 mAFrom the graphVGSQ = 6.4 vIDQ = 2.75 mA
DDDDGS RIVV
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Voltage Divider Bias
21
DD2G RR
VRV
Applying KVL around the indicated loop:0VVV RSGSG
orRSGGS VVV
sDGGS RIVV
For the output section:
SDDDSSDDSDD
RDSRDD
RRIVRIVRI
VVVVSD
)RR(IVV SDDDDDS
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Example: Determine IDQ and VDSQ for the given enhancement type MOSFET.
Solution: Network:
v18M)1822(40M18
RRVRV
21
DD2G
DSDGGS Ik82.018RIVV
When VGS = 0, mA95.21k82.0
18ID
When ID = 0, VGS = 18 v
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Device:
23
22)TH(GS)on(GS
)on(D v/A1012.0510
mA3VV
Ik
2GS32
)TH(GSGSD 5V1012.0VVkI
VGS 5 10 15 20ID
(mA)0 0.48 12 27
From the graphVGSQ = 12.5 vIDQ = 6.7 mA
v4.14RRIVV DSDDDDS
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Combination NetworksExample1: Determine the levels of VD and VC for the
given network:Solution:
v62.3k24k8216k24
RRVRV
21
CC2B
v92.27.062.3VVV BEBE
mA825.1k6.192.2
RV
RV
IE
E
E
RE
E
mA825.1II EC
CSD III
v07.11)k7.2)(mA825.1(16)k7.2(I16V DD
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2
p
GSQDSSDQ V
V1II
Plot the following equation
From the plot, VGSQ = -3.7 v
NowVC = VB – VGSQ = 3.62 – (-3.7) = 7.32 v
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Example 2: Determine VD for the given network.Solution:From the JFET:VGS = -IDRS = -ID(2.4k)From this equation, the selfBias line is plotted as shown Below.
71
The resulting Q_point is at:VGSQ = -2.6 V, IDQ = 1mA
For the BJT:IE IC = ID = 1mAIB = IC/ = 1mA/80 = 12.5AVB = VCC – IBRB = 16 – 12.5A470k = 10.125 VVE = VD = VB – VBE = 10.125 – 0.7 = 9.425 V
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Example 3: For the network of Fig. (a), determine:VG , VGSQ, IDQ, IE, IB, VD and VC.Solution:VG = 3.3 VVGSQ = -1.25 VIDQ = 3.75 mAIE = 3.75 mAIB = 23.44AVD = 11.56 VVC = 15.88 V
Fig. (a)