Module – 5
UNIT -5
Field Effect Transistors
Review Questions:
1. Draw the structure of JFET and discuss its working.
2. What is ‘pinch off’ voltage? How to get its value experimentally?
3. An n-JFET is operated with negative gate voltage and not with positive one. Give
reasons.
4. Give the structure of depletion MOSFET (D - MOSFET). How is D - MOSFET
different from enhancement MOSFET (E - MOSFET)?
5. Draw and discuss drain characteristics for a D-MOSFET.
6. Discuss the formation of channel in E-MOSFET emphasizing the role of inversion
layer.
7. Give self bias circuit for JFET and explain the biasing process.
8. How can we obtain negative or positive bias voltage with proper choice of
resistors in a voltage divider bias?
9. Develop a simple small signal model/equivalent circuit for FET.
10. How are DMOSFET and EMOSFET connected in a circuit to work as resistors?
11. Derive expression for voltage gain for a common source amplifier.
12. Illustrate power efficiencies of CMOS devices through a CMOS inverter circuit.
Problems:5.1 The device parameters for an n-Channel JFET are: Maximum current IDSS = 10mA,
Pinch off voltage,
Vp =
- 4V
Calculate the drain current for
(a) VGS = 0
- 1.0v
- 4V.
(b) VGS =
(c) VGS =
Solution:The expression for drain current ID, in the saturation region is,
ID
VGS
VP
I DSS 1
2
...................( A)
(a) When VGS = 0, from Eq(A) above,
ID
I DSS
10mA
(b) When VGS = -1.0V, the drain current from Eq (A) is,
ID
10 10
3
1
4
1
2
10mA 0.56
or I D 5.6 mA
(c) When VGS = -4V = Vp, then from Eq(A),
ID
I DSS 1
or , I D
0
4
4
2
5.2 A JFET produces gate current of 2nA when gate is reverse biased with 8V.
Determine The resistance between gate and source.
Solution:Since reverse gate-source voltage, VGS, of 8v produces gate current, IG of 2nA,
Therefore, gate-to-source resistance, RGS, is
RGS
VGS
IG
RGS
4000M
8V
2nA
4000M
5.3 The reverse gate voltage of JFET when changes from 4.4V to 4.2V, the drain
current changes from 2.2 mA to 2.6 mA. Find out the value of transconductance of the
transistor.
Solution:The transconductance, gm is defined as
ID
VGS
gm
Where ΔID is change in drain current when change in gate-source voltage is ΔVGS.
In the given problem,
ΔID = (2.6 – 2.2) mA
= 0.4 mA
And,
ΔVGS = (4.4 – 4.2) V
= 0.2 V
Therefore,
0.4 mA
0.2V
or , g m 2.0 m mhos.
gm
5.4 Find out the operating point current and voltage values (IDQ and VDSQ) for a self
biased JFET having the supply voltage VDD = 20V and maximum value of drain current
as 12 mA.
Solution:We know that the value of drain current at Q-point may be taken as half of the maximum
current, that is,
I DQ
I DSS
2
12mA
2
6.0 mA
In the same way, the value of drain-source voltage at Q-point may be taken as half of
supply voltage VDD. That is,
VDD 20V
2
2
or , VDSQ 10V
VDSQ
Therefore,
IDQ = 6.0 mA
VDSQ = 10.0 V
5.5 Calculate the value of source resistance RS required to self bias a n-JFET such that
VGSQ =
-
3V. The n-JFET has maximum drain-source current IDSS = 12 mA, and pinch-
off voltage, Vp =
- 6V
Solution:The drain current, ID, in a JFET, in the saturation region is,
2
ID
VGS
Vp
I DSS 1
We have, IDSS = 12 mA, VGS = -3V and Vp = -6V, Therefore,
ID
12mA 1
or , I D
3
6
2
9.0 mA
Since the voltage VGS is generated across the source resistor RS, we have,
RS
VGS
ID
RS  333
3V
 333
9 mA
5.6 For the DMOSFET circuit shown in fig., the device parameters are:
VGS(off) = -8V, IDSS = 10mA
Determine drain-to- source voltage VDS.
VDD = + 18V
ID
RD
680Ω
+
VDS
RG
10M
Solution:In a MOSFET, there is no gate current. Therefore, there is no voltage drop across
resistor RG .
Thus, VGS = 0.
Further, when VGS =0, ID = IDSS, the maximum drain current.
Summing up voltages in the output loop and using ID = IDSS,
We have,
IDSS .RD + VDS = VDD
or VDS = VDD – IDSS . RD
= 18 – 10 X 10-3 X 0.68 X 103
= 18 – 6.8
or VDS = 11.2 V
5.7 Data sheet of an EMOSFET specifies following parameters:
ID(on) = 50 mA at VGS = 6V and VT, the threshold voltage for EMOSFET, 2V.
Determine the drain current at VGS = 3V.
Solution:We first determine the conductance parameter k for the device using the relation,
I D (ON )
k
VGS
VT
2
50 mA
(6 2) 2
or , k
3.12 mA
v2
Now, the drain current, ID is expressed as,
ID
k VGS
VT
2
3.12 10 3 (3 2) 2
or , I D
3.12 mA
5.8 The drain current changes from 5 mA to 7 mA when the gate voltage is changed
from – 4.0V to – 3.7V in the amplifier circuit shown in fig.
Calculate the voltage gain of the amplifier.
+15V
6k
RD
vo
vi
RS
400Ω
RL
6k
Solution:In case, the source resistance RS is ac grounded as done in the circuit of fig. the gain of
amplifier is,
AV = gm.rD
Where gm is transconductance of the transistor and rD is effective ac resistance seen by
the drain terminal.
Now,
gm
ID
VGS
VDS
2 mA
0.3V
or , g m 6.66 mS
And,
rD = RD
RL = 6k
6k
= 3kΩ
Therefore,
AV = gm X rD = 6.66 X 10-3 X 3 X 103
or, AV = 20
5.9 Find the drain-source voltage, VDS, for the NMOS transistor circuit shown in fig. The
device parameters are: conductance parameter, k = 600μA/v2 and VT = 2V.
+15V
4M
R1
RD
2k
+
VDS
2M
k
R2
Solution:The gate current, IG, is zero in a MOSFET. Then, from voltage divider network,
R2
VGS
R1
R2
VDD
2M
4M
or , VGS
2M
15V
5V
And, as we know
ID
k VGS
VT
2
600 10 6 (5 2) 2
5.4 mA
or , I D 5.4 mA
Applying voltage summation in the output loop,
VDD = IDRD + VDS
or, VDS = VDD – IDRD = 15 – (5.4 X 10-3 X 2 X 103)
or, VDS = 4.2V
5.10 Calculate the voltage gain in the amplifier shown in fig. The transconductance of
the transistor is 4000μs. If the 400Ω source resistance is by passed by an capacitor,
how much is voltage gain now?
+20V
RD
6k
vo
vi
RL
RG
5M
k
400Ω
RS
10k
Solution:When the source resistance RS is not by passed, the voltage gain is,
g m rD
1 g m rs
AV
...........( A)
Where rD and rS are effective (ac) resistance seen by the drain and source of the
transistor.
And,
rD = RD
= 6k
RL
10 k = 3.75 kΩ
And,
rS = RS = 400Ω
Therefore
AV
AV
4000 10 6 3.75 103
1 4000 10 6 400
4 3.75
5.7
2.6
5.7
In case, RS is bypassed, the gain (rS = 0 in the Eq(A)),
AV = gm rD = 4000 X 10-6 X 3.75 X 103
or, AV = 15