Session 2
Session 2
R.M.K COLLEGE OF ENGINEERING AND
TECHNOLOGY
DEPARTMENT OF ECE
EC8252-ELECTRONIC DEVICES
SECOND SEMESTER-I YEAR- (2020-2024 BATCH): SECTION A & B
Mrs.P.Sivalakshmi M.E
AP/ECE
SESSION:19
DATE: 28.05.2021
UNIT 3 FIELD EFFECT TRANSISTORS
Output or Drain characteristics curve of JFET
• As the reverse gate-source voltage is increased, the cross-sectional area of the channel
decreases. This in turn decreases the drain current.
• At some reverse gate-source voltage, the depletion layers extend completely across the channel.
In this condition, the channel is cut off and the drain current reduces to zero.
• The gate voltage at which the channel is cutoff (i.e. channel becomes non-conducting) is called
gate-source cut off voltage VGS (off).
Transfer characteristics curve of JFET
Pinch off Voltage (VP)
A single ended geometry junction FET is shown here, in which diffusion is done from one side only the substrate is of p-type
material which is epitaxial grown on an n-type channel. Thus a P-type gate is then diffused into the n-channel. The diffused gate
is of very low resistivity material, allowing the depletion region to spread mostly into the N-Type channel. In this device, the
slab of n type semiconductor is sandwiched between two layers of P-type materials forming two PN junction.
The gate reverse voltage that removes all the free charge from the channel is called the pinch-off voltage VP.
Let us assume that the P-type region is doped with NA acceptors per cubic meter, n-type region is doped with ND
donors per cubic meter and junction formed is abrupt.
Moreover, if the accepter impurity density is assumed to be much large than donor density, then the depletion region
width in the p-region will be much smaller than the depletion width of the N region (i.e.) NA >> ND , then WP<<WN and WP=W.
We know that the relationship between potential and charge density is given by;
d2V / dX2 = -q ND/ɛ
The electric potential is equal to the potential energy
per unit of charge. So as the value of the charge
increases, the potential energy of that charge density
also increases.
Integrating the above equation subject to boundary condition we get x to w
dv/dx = -[q ND/ ɛ ]*(x-w)
Integrating again
V = -(q ND/ɛ )* (x2/2 – w x)
V = -[q ND/2ɛ](x2- 2wx)
At x=w, V=VB which is the junction or barrier potential
VB = qNDW2/2ɛ
W2 = 2ɛ /qND *VB
W = {2ɛ/ qND *VB}1/2
As the barrier potential represents, a reverse voltage, it is lowered
by an applied forward voltage V(x) at x.
since; VB = VP-V(x)
Hence;
W(x) = a-h(x) = {2ɛ/ qNd*Vp-V(x)}1/2
Where ;
ɛ - dielectic constant of channel material
q - magnitude of electronic charge
V0 – junction contact potential at x
V(x) – is the applied potential across space charge region
a - metallurgical distance between the substrate and p+
a-h(x) – is the W(x) of depletion region into channel at point x
• If the drain current is zero h(x) and v(x) are independent to x and hence b(x) or h(x) = h. It we substitute
h(x)=h=0
a = {(2ɛ/ qNd) * Vp}1/2
a2 = (2ɛ/ qNd)* Vp
|Vp| = (q N d/2ɛ) *a2
W(x) = a-h(x) = {2ɛ/ qNd*[Vp-V(x)]}1/2
Problems
Expression for Drain Current (ID)
The relation between IDSS and VP is shown in Fig. We note that gate-source cut off voltage [i.e. VGS (off)] on
the transfer characteristic is equal to pinch off voltage VP on the drain characteristic i.e.
The transfer characteristic of JFET shown in Fig. a is part of a parabola. A rather complex mathematical
analysis yields the following expression for drain current
Where
ID= drain current at given VGS
IDSS= shorted – gate drain current
VGS= gate–source voltage
VGS (off)= gate–source cut off voltage
Problems
1. Figure shows the transfer characteristic curve of a JFET. Write the equation for drain current
Solution
2.A JFET has the following parameters: IDSS = 32 mA ; VGS (off) = – 8V ; VGS= – 4.5 V. Find the value of drain
current.
4. Determine the value of drain current for the circuit shown in Fig.
Session 2
Session 2
Session 2

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Session 2

  • 3. R.M.K COLLEGE OF ENGINEERING AND TECHNOLOGY DEPARTMENT OF ECE EC8252-ELECTRONIC DEVICES SECOND SEMESTER-I YEAR- (2020-2024 BATCH): SECTION A & B Mrs.P.Sivalakshmi M.E AP/ECE SESSION:19 DATE: 28.05.2021 UNIT 3 FIELD EFFECT TRANSISTORS
  • 4. Output or Drain characteristics curve of JFET
  • 5. • As the reverse gate-source voltage is increased, the cross-sectional area of the channel decreases. This in turn decreases the drain current. • At some reverse gate-source voltage, the depletion layers extend completely across the channel. In this condition, the channel is cut off and the drain current reduces to zero. • The gate voltage at which the channel is cutoff (i.e. channel becomes non-conducting) is called gate-source cut off voltage VGS (off). Transfer characteristics curve of JFET
  • 6. Pinch off Voltage (VP) A single ended geometry junction FET is shown here, in which diffusion is done from one side only the substrate is of p-type material which is epitaxial grown on an n-type channel. Thus a P-type gate is then diffused into the n-channel. The diffused gate is of very low resistivity material, allowing the depletion region to spread mostly into the N-Type channel. In this device, the slab of n type semiconductor is sandwiched between two layers of P-type materials forming two PN junction. The gate reverse voltage that removes all the free charge from the channel is called the pinch-off voltage VP. Let us assume that the P-type region is doped with NA acceptors per cubic meter, n-type region is doped with ND donors per cubic meter and junction formed is abrupt. Moreover, if the accepter impurity density is assumed to be much large than donor density, then the depletion region width in the p-region will be much smaller than the depletion width of the N region (i.e.) NA >> ND , then WP<<WN and WP=W. We know that the relationship between potential and charge density is given by; d2V / dX2 = -q ND/ɛ The electric potential is equal to the potential energy per unit of charge. So as the value of the charge increases, the potential energy of that charge density also increases.
  • 7. Integrating the above equation subject to boundary condition we get x to w dv/dx = -[q ND/ ɛ ]*(x-w) Integrating again V = -(q ND/ɛ )* (x2/2 – w x) V = -[q ND/2ɛ](x2- 2wx) At x=w, V=VB which is the junction or barrier potential VB = qNDW2/2ɛ W2 = 2ɛ /qND *VB W = {2ɛ/ qND *VB}1/2 As the barrier potential represents, a reverse voltage, it is lowered by an applied forward voltage V(x) at x. since; VB = VP-V(x) Hence; W(x) = a-h(x) = {2ɛ/ qNd*Vp-V(x)}1/2
  • 8. Where ; ɛ - dielectic constant of channel material q - magnitude of electronic charge V0 – junction contact potential at x V(x) – is the applied potential across space charge region a - metallurgical distance between the substrate and p+ a-h(x) – is the W(x) of depletion region into channel at point x • If the drain current is zero h(x) and v(x) are independent to x and hence b(x) or h(x) = h. It we substitute h(x)=h=0 a = {(2ɛ/ qNd) * Vp}1/2 a2 = (2ɛ/ qNd)* Vp |Vp| = (q N d/2ɛ) *a2 W(x) = a-h(x) = {2ɛ/ qNd*[Vp-V(x)]}1/2
  • 10. Expression for Drain Current (ID) The relation between IDSS and VP is shown in Fig. We note that gate-source cut off voltage [i.e. VGS (off)] on the transfer characteristic is equal to pinch off voltage VP on the drain characteristic i.e. The transfer characteristic of JFET shown in Fig. a is part of a parabola. A rather complex mathematical analysis yields the following expression for drain current Where ID= drain current at given VGS IDSS= shorted – gate drain current VGS= gate–source voltage VGS (off)= gate–source cut off voltage
  • 12. 1. Figure shows the transfer characteristic curve of a JFET. Write the equation for drain current Solution 2.A JFET has the following parameters: IDSS = 32 mA ; VGS (off) = – 8V ; VGS= – 4.5 V. Find the value of drain current.
  • 13. 4. Determine the value of drain current for the circuit shown in Fig.