JUNCTION FIELD EFFECT TRANSISTOR(JFET). JFET operation can be compared to a water faucet : The...
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Transcript of JUNCTION FIELD EFFECT TRANSISTOR(JFET). JFET operation can be compared to a water faucet : The...
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JUNCTION FIELD EFFECT TRANSISTOR(JFET)
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JFET operation can be compared to a water faucet :
The source of water pressure – accumulated electrons at the negative pole of the applied voltage from Drain to Source
The drain of water – electron deficiency (or holes) at the positive pole of the applied voltage from Drain to Source.
The control of flow of water – Gate voltage that controls the width of the n-channel, which in turn controls the flow of electrons in the n-channel from source to drain.
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JFETs and Their Characteristics
Fig. (a) Fig. (b)
Fig. (a) is the schematic symbol for the n-channel JFET, and Fig. (b) shows the symbol for the p-channel JFET. The only difference is the direction of the arrow on the gate lead.
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JFETs and Their Characteristics
JFETN-Channel P-
Channel
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Transfer Characteristics
The transfer characteristic of input-to-output is not as straight forward in a JFET as it was in a BJT.
In a BJT, indicated the relationship between IB (input) and IC (output).
In a JFET, the relationship of VGS (input) and ID (output) is a little more complicated:
2
P
GSDSSD )
V
V(1II
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Transfer Characteristics
From this graph it is easy to determine the value of ID for a given value of VGS.
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Plotting the Transfer Curve
Shockley’s Equation Methods. Using IDSS and Vp (VGS(off)) values found in a specification sheet, the Transfer
Curve can be plotted using these 3 steps:
Step 1:
Solving for VGS = 0V:
Step 2:
Solving for VGS = Vp (VGS(off)):
Step 3:
Solving for VGS = 0V to Vp:
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Plotting the Transfer Curve
Shorthand method
VGS ID0 IDSS
0.3VP IDSS/2
0.5 IDSS/4
VP 0mA
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JFET Symbols
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Basic Operation Characteristics And Parameters
Pinch-Off Voltage
VGS Controls ID
Cutoff Voltage JFET Transfer Characteristic JFET Forward Transconductance Input Resistance and Capacitance Drain-to-Source Resistance
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JFET BIASING Fixed– Bias Self-Bias
Setting the Q-Point of a Self-Biased JFET Midpoint Bias Graphical Analysis of a Self-Biased JFET
Voltage-Divider Bias Graphical Analysis of a JFET with Voltage-Divider Bias Q-Point Stability
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The Drain Characteristic Curve when VGS = 0 V
A
B C
Ohmicregion
Constant-current region
Breakdown
IDSS
ID
VDSVp (pinch-off voltage)
VGS = 0
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A Biased n-channel JFET
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VGS = 0V and VDS = 0V
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VGS = 0V and VDS = 2V
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VGS = 0V and VDS = 5V
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VGS = 0V and VDS = 10V
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VGS = -0.5V and VDS = 0V
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VGS = -0.5V and VDS = 2V
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VGS = -0.5V and VDS = 4V
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VGS = -0.5V and VDS = 10V
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VGS = -1V and VDS = 0V
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VGS = -1V and VDS = 2V
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VGS = -1V and VDS = 3V
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VGS = -1V and VDS = 10V
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VGS = -2V and VDS = 0V
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VGS = -2V and VDS = 2V
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VGS = -2V and VDS = 3V
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VGS = -2V and VDS = 10V
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IDSS is Drain to Source current with gate
Shorted. VGS(off) is the value of VGS that makes ID
approximately zero is the cutoff voltage. Vp, pinch-off voltage, is the value of VDS at
which ID becomes constant . VGS(off) and Vp are always equal in
magnitude but opposite in sign.
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VGS(off) = - 4 and IDSS = 12mA Determine the minimum value of VDD to put the
device in the constant current region operation.
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VGS(off) = - 4 and IDSS = 12mA Determine the minimum value of VDD to put the
device in the constant current region operation. Since VGS(off) = - 4V, Vp = 4V. VDS = Vp=4V, VGS = 0 V ID = IDSS = 12 mA VRD = IDRD =(12mA)(560 Ohm) = 6.72 V VDD = VDS + DRD = 4 V + 6.72 V = 10.7 V
2
P
GSDSSD )
V
V(1II
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Example of An n-channel JFET transfer characteristic curve
2
P
GSDSSD )
V
V(1II
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Questions
VGS(off) = - 8 V and IDSS = 9 mA Determine the drain current, ID for VGS = 0 V,
VGS = - 1 V, and VGS = -4 V
using this formula
2
P
GSDSSD )
V
V(1II
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Answers ID = IDSS (1 - ( VGS/VGS(off)) )^2For VGS = 0 VID = 9 mA (1 - ( 0/-8) )^2 = 9 mA
For VGS = -1 VID = 9 mA (1 - ( -1/-8) )^2 = 6.89 mA
For VGS = -4 VID = 9 mA (1 - ( -4/-8) )^2 = 2.25 mA
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JFET Forward Transconductance
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Transconductance formulas
gm0 = (2Idss/( |Vgs(off)| )) gm = gm0 (1-Vgs/ Vgs(off)) Unit for transconductance, gm, is
siemens (S)
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Fixed-Bias
Investigating the input loop
IG=0A, therefore
VRG=IGRG=0V
Applying KVL for the input loop,
-VGG-VGS=0
VGG= -VGS
It is called fixed-bias configuration due to VGG is a fixed
power supply so VGS is fixed
The resulting current,
2)1(P
GSDSSD V
VII
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Investigating the graphical approach. Using below tables, we can draw the graph
VVGSGS IIDD
00 IIDSSDSS
0.3V0.3VPP IIDSSDSS/2/2
0.50.5 IIDSSDSS/4/4
VVPP 0mA0mA
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The fixed level of VGS has been superimposed as a vertical
line at
At any point on the vertical line, the level of VG is -VGG--- the
level of ID must simply be determined on this vertical line.
The point where the two curves intersect is the common
solution to the configuration – commonly referrers to as the
quiescent quiescent or operating point.
The quiescent level of ID is determine by drawing a
horizontal line from the Q-point to the vertical ID axis.
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Output loop
DDDDDS RIVV
VVS 0
SDDS VVV
SDSD VVV 0SV
DSD VV
SGGS VVV
SGSG VVV 0SV
GSG VV
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Determine VGSQ, IDQ, VDS, VD, VG, VS
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Exercise
Determine IDQ, VGSQ, VDS, VD, VG and VS
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Self Bias
The self-bias configuration eliminates the need for two dc supplies.
The controlling VGS is now determined by the voltage across the resistor RS
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For the indicated input loop:
Mathematical approach:
rearrange and solve.
SDGS RIV
2
2
1
1
P
SDDSSD
P
GSDSSD
V
RIII
V
VII
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Graphical approach
Draw the device transfer characteristic Draw the network load line
Use to draw straight line. First point, Second point, any point from ID = 0 to ID = IDSS. Choose
the quiescent point obtained at the intersection of the straight line plot and the device characteristic curve.
The quiescent value for ID and VGS can then be determined and used to find the other quantities of interest.
SDGS RIV 0,0 GSD VI
2
2
SDSSGS
DSSD
RIV
thenI
I
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For output loop
Apply KVL of output loop Use ID = IS
RDDDSDSD
SDS
DSDDDDS
VVVVV
RIV
RRIVV
)(
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Determine VGSQ, IDQ,VDS,VS,VG and VD.
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Example Determine VGSQ, IDQ, VD,VG,VS and VDS.
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Voltage-Divider Bias
The source VDD was separated into two equivalent sources to permit
a further separation of the input and output regions of the network.
IG = 0A ,Kirchoff’s current law requires that IR1= IR2 and the series
equivalent circuit appearing to the left of the figure can be used to find the level of VG.
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21
DD2G
RR
VRV
SDGGS
RSGSG
RIVV
VVV
0
Voltage-Divider Bias
VG can be found using the voltage divider rule :
Using Kirchoff’s Law on the input loop:
Rearranging and using ID =IS:
Again the Q point needs to be established by
plotting a line that intersects the transfer curve.
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Procedures for plotting
1. Plot the line: By plotting two points: VGS = VG, ID =0 and VGS = 0, ID = VG/RS
2. Plot the transfer curve by plotting IDSS, VP and calculated values of ID.
3. Where the line intersects the transfer curve is the Q point for the circuit.
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Once the quiescent values of IDQ and VGSQ are determined, the
remaining network analysis can be found.
Output loop:
2121 RR
VII DDRR
)( SDDDDDDS RIRIVV
DDDDD RIVV
SDS RIV
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Effect of increasing values of RS
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Example Determine IDQ, VGSQ, VD, VS, VDS and VDG.
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Example Determine IDQ, VGSQ, VDS, VD and VS
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Find VDS and VGS when ID = 5mA
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Vs = ID * RS = (5mA)*(470) = 2.35 VVD = VDD – ID * RD
= 15 V – (5mA)(1k) = 15 V – 5 V = 10 VVDS = VD – VS
= 10 V – 2.35 V = 7.65 VVGS = VG – VS = 0 V – 2.35 V = - 2.35 V
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Questions
Find VDS and VGS when ID=8mA RD = 860 Ohm RS = 390 Ohm VDD = 12 V
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RD = 860 Ohm, RS = 390 Ohm, VDD = 12 V, ID=8mA
Vs = ID * RS = (8mA)*(390) = 3.12 VVD = VDD – ID * RD
= 12 V – (8mA)(860 ) = 12 V – 6.88 V = 5.12 VVDS = VD – VS
= 5.12 V – 3.12 V = 2 VVGS = VG – VS = 0 V – 3.12 V = - 3.12 V
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Setting the Q-Point of a Self-Biased JFET To determine ID for a desired value of VGS or
vice versa.
Rs = | VGS/ID |
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Midpoint Bias When VGS = VGS(off)/3.4
ID = IDSS/2
using formula below
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Graphical Analysis of a Self-Biased JFET Using VGS = -ID*Rs to: 1) Calculate VGS when ID = 0 (VGS,0) 2) Calculate VGS when ID = IDSS (VGS, IDSS) or Get IDSS from data sheet 3) Draw a line between (0,0) and (VGS, IDSS) 4) The line intersects the transfer characteristic
curve is the Q-point of the Circuit
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Load Line Calculation 1) Calculate VGS when ID = 0 (VGS,0) VGS = -ID*RS = - 0*470 = 0 (0,0)
2) Calculate VGS when ID = IDSS (VGS, IDSS)
or
Get IDSS from data sheet VGS = -ID*RS = -10mA*470 = -4.7 V (-4.7,10)
3) Draw a line between (0,0) and (-4.7,10m)
4) The line intersects the transfer characteristic curve is the Q-point of the Circuit.
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A self Bias dc load line and the Q-point
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Determine the Q-point for this figure when
IDSS = 4 mA
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1) Calculate VGS when ID = 0 (VGS,0)
VGS = -ID*RS = - 0*680 = 0
2) Calculate VGS when ID = IDSS (VGS, IDSS)
or Get IDSS from data sheet VGS = -ID*RS = - 4 mA* 680
Ohms = -2.72 V 3) Draw a line between
(0,0) and (-2.72 V, 4mA)
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Voltage-Divider Bias
IS = ID
VG = (R2/(R1+R2)) * VDD
ID = (VG – VGS)/RS
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Exercise
Determine ID and VGS
when VD = 7V, Vdd = 12 V
R1 = 6.8M Ohms R2 = 1M Ohms RD = 3.3 k Ohms RS = 1.8 k Ohms
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Answers ID = (VDD – VD)/RD =
(12 V – 7 V)/3.3k = 1.52mA
VS = IDRS = (1.52mA)(1.8K) = 2.74V
VG = (R2/(R1+R2)) VDD
= (1M / 7.8M)12V = 1.54V
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Graphical Analysis of JFET with Voltage-Divider Bias For Id = 0
Vs = Id*Rs = (0)*Rs = 0
Vgs = Vg-Vs = Vg – 0V = Vg
For Vgs = 0
Id = (Vg – Vgs)/Rs = Vg/Rs
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For Id = 0
Vgs = Vg
For Vgs = 0
Id = Vg/Rs
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Exercise #8
In a certain FET circuit, Vgs = 0V, Vdd = 15 V, Idss = 15 mA, and Rd = 470 ohms. If Rd is decrease to 330 ohms, Idss is.
A) 19.5 mA B) 10.5 mA C) 15 mA D) 1 mA
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Exercise #8 Answer
In a certain FET circuit, Vgs = 0V, Vdd = 15 V, Idss = 15 mA, and Rd = 470 ohms. If Rd is decrease to 330 ohms, Idss is.
15 mA
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Exercise #9
The JFET has Vgs(off) = -4 V.
Figure 8-57
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Exercise #10
Determine the value of Rs required for a self-biased JFET to produce a Vgs for -4 V when Id = 5 mA.