field effect transistor (FET)

1. Topic 5
Field Effect Transistors

2. Introduction
• There are two main types of FETs:-
Junction field-effect transistor (JFET) and
Metal-oxide semiconductor field-effect transistor (MOSFET).
• FETs differ from BJTs mainly in two ways:
(1) FETs are unipolar devices, they operate with only one type
of charge carrier.
(2)While the BJT is a current-controlled device, the FET is a
voltage-controlled device where voltage between two of
the terminals (gate and source) controls the current through the
device.

3. Introduction
• A major feature of FETs is that they have very high input
resistance.
• Since the input of the FETs has very high resistance, the
device draws negligible currents. As a result, little heat is
dissipated in the device.
• This is important in VLSI circuits where there are thousands of
FETs in the circuit and we want to limit the heat created from
each FET.

4. The Junction Field-Effect
Transistor (JFET)
JFET has two regions: a p-type material and an n-type material.
Note that the p-type material actually surrounds the n channel. The p-
type region is diffused in the n-type material to form a channel and it
is connected to the gate lead.
Wire leads are connected to each end of the n-channel

5. Three leads of JFET
In comparison to the BJT, the three leads
of a JFET:
• drain – similar to BJT’s collector
• gate – similar to BJT’s base
• source – similar to BJT’s emitter

6. Symbol for JFET
As in BJT, the arrow always point to n-type. In the n-channel JFET,
the gate is p-type. Hence the arrow points to the channel which is n-
type.
The JFET has two pn junctions: gate-to-source and gate-to-drain.

7. The JFET always operates with the gate-to-source pn
junction in reverse biased.
The n-channel JFET has a positive drain supply
voltage.
VDD provides a drain-to-source voltage, thus
supplying a current from the drain to the source.
VGG supplies a negative voltage to the gate and sets a
reverse-bias voltage between the gate and the
source.

8. Why reverse bias?
By reverse biasing the gate-to-source junction, a depletion
region is produced along this pn junction. The depletion
region formed spreads into the n-channel.
It acts like a resistance to current flow.
The size of the depletion region depends on the biasing
voltage VGG. The higher VGG is, the larger the depletion
region.

9. Increasing VGG cause the depletion region to grow. It spreads
further into the n channel and the width of the n channel becomes
thinner. This creates more resistance against current flow.
Similarly, decreasing VGG, decreases the size of the depletion
region. The n channel becomes thicker and hence, creates less
resistance to current flow.

10. JFET Characteristics and
Parameters
Case 1: Gate-to-source voltage is zero (VGS = 0).
To get VGS = 0, short the gate-to-source junction (set VGG = 0).
As we increase VDD, VDS increases
as well. The drain current ID also
increases with VDD. This produces the
drain characteristic curve (ID versus VDS).
Depletion region
form in the n-
channel near to the
drain side.
It is not large
enough to have any
significant effect.

11. As VDS increases, the reverse bias voltage from gate to drain (VGD)
increases and produces a depletion region large enough to offset the
increase in VDS.
The offset is enough to keep ID constant (between B and point C).
The region between points B and C is called the constant – current
region.
At point B, the drain-to-source (VDS) voltage is called the pinch-off
voltage (VP), the ID axis is labeled IDSS which stands for drain-to-
source current with gate shorted. Both values are specified in
data sheets.

12. Breakdown
• ID will remain constant until point C, where we reach
breakdown.
• Increasing VDS beyond point C will cause the device to
enter the breakdown region.
• This will damage the device, thus a JFET should never be
operated in the breakdown region.

13. Case 2: Gate – to – source voltage is nonzero
(VGS  0).
Connect a bias voltage VGG to
the gate.
This produces a family of
curves.
Each curve is produced with
one value of VGS.
Note that ID decreases as VGS
is made more negative.

14. • This occurs because the channel is narrowing. Also notice that
pinch – off occurs at different VP’s for different VGS values.
• The value of VGS that makes ID approximately zero is called the
cutoff voltage, VGS(off).
Note that the value of VGS that sets ID to zero (i.e. widening the
depletion region to a point where the channel is completely
closed) is the most negative value VGS can take.
• In summary to the discussion above, we note that the drain current
ID is controlled by VGS. This is why JFET is a voltage-controlled
device.

15. Relation between pinch-off voltage, VP, and
cutoff voltage, VGS(off)
(i) VP is the value of VDS when the drain current becomes
constant. It is always measured at VGS = 0,
(ii) When VGS is nonzero, pinch-off occurs for VDS values less
than VP.
VGS(off) and VP are always equal in magnitude but opposite
in sign. Thus, knowing one, we have the other.
Data sheets will generally list only one of the two. For
example, if VGS(off) = −5 V, then VP = + 5 V.

16. Example
Determine the minimum value of VDD required to
put the JFET below in the constant-current region
of operation. The cutoff voltage, VGS(off) = − 4 V
and the IDSS = 12 mA.

17. Solution
Since VGS(off) = – 4V, thus VP = 4V.
The minimum value of VDS for the JFET to enter the constant current
region is
VDS = VP = 4 V
In the constant current area with VGS = 0 V,
ID = IDSS = 12 mA
The drop across the drain resistor is
VR(D) = IDRD = (12 mA) (560 Ω) = 6.72 V
Using KVL around the drain circuit
VDD = VDS + VR(D) = 4 V + 6.72 V = 10.72 V
Thus, VDD must be 10.72 V for the device to enter the constant current area,
i.e. to make VDS = VP.

18. Example
A particular p-channel JFET has a
VGS(off) = +4 V. What is ID when VGS = + 6 V?
Solution
Recall a p-channel JFET requires a positive gate-to-
source voltage. The more positive VGS, the less the
drain current.
When VGS = 4 V, ID = 0 (cutoff).
Any further increase in VGS keeps the JFET in cut off,
so ID remains at 0.

19. JFET Transfer Characteristic
For an n-channel JFET, VGS(off) is negative. The relation between
VGS and ID is known as the transfer characteristic curve (taken from
the drain characteristic curve ):

20. equation for the JFET transfer
characteristic curve
• The equation for the JFET transfer characteristic curve is:
ID = IDSS
• Thus, if IDSS and VGS(off) are known, ID can be determined for any
VGS.
• Notice that the transfer characteristic curve is parabolic.
Because of this, JFET is referred to as a square-law device.
2
GS(off)
GS
1









V
V

21. Example
Determine the drain current for VGS = 0 V, −1 V and –4 V for a 2N5459
JFET. Refer to the data sheet below.

22. Solution
From the data sheet, we find that IDSS = 9 mA and VGS(off) =
−8 V (maximum).
Thus, we can say that for VGS = 0,
ID = IDSS = 9mA
For VGS = –1 V, we use the equation shown above:
ID = (9 mA) = 6.89 mA
For VGS = –4V :
ID = 2.25 mA
2
8
1
1 








V
V

23. JFET Forward Transconductance, gm
The ID vs VGS curve is also known as a transconductance curve.
The slope of the curve is known as the JFET forward transconductance:
gm = ΔID / ΔVGS (S).
Usually, it is very small so that it is measured in S.
Data sheets normally show the value of gm at VGS = 0 V (gm0). This value is
enough to calculate it for all values of VGS:
gm = gm0
The gm0 value can be calculate from
gm0 =









GS(off)
GS
1
V
V
GS(off)
DSS
2
V
I

24. Input Resistance
Since a JFET operates with the gate-to-source junction
reverse biased, the input resistance is very high.
This is one advantage of JFET over BJT. The input
resistance can be calculated from
RIN = |VGS / IGSS|

25. Example
Determine the input resistance of the 2N5457 JFET.
Solution
The specification sheet for the 2N5457 JFET lists a maximum
gate reverse current (IGSS) of − 1 nA under the following
conditions:
T = 25 C, VDS = 0 V,VGS = − 15 V
By using Ohm’s law, the input (gate) impedance is
|VGS / IGSS| = 15 G.

26. JFET Biasing
Just like the BJTs, we need to establish the correct
dc gate-to-source voltage to get the desired value of
drain current.
(i) Self-biased
(ii) Voltage-divider biased
Let us establish two facts about JFET operation:
(1) In any JFET circuit, all the source current passes
through the device to the drain circuit, i.e., IS = ID.
(2)This is because there is no significant gate current
because of the high input resistance,
IG = 0.

27. Self – Bias
• The self-bias circuit replaces the gate supply (−VGG)
with a gate resistor RG and a source resistor RS.
• The gate is returned to ground via RG. Let the
potential at the gate be VG.
RG does not affect the biasing of JFET. It is required for the ac
operation of the JFET. When an ac signal enters the gate, it will
be grounded if there is no RG. The presence of RG provides a
resistance to the path of the ac signal to the ground so that most
of the ac signal will go to the gate of the JFET.
• The resistor RS added in the source circuit helps to
produce a potential at the source VS.

28. Circuit Analysis
• Since IG  0,
VG = IGRG = 0.
• Now, IS produces a voltage drop across RS.
VS = ISRS = IDRS.
• Then
VGS = VG – VS = 0 – IDRS = – IDRS.
• Now,
VD = VDD – IDRD
• Since VS = IDRS, the drain-to-source voltage is
VDS = VD – VS
= VDD – ID(RD + RS)

29. Example
Find VDS and VGS in the circuit shown.
For the particular JFET in this circuit, the
internal parameter values such as gm,
VGS(off), and IDSS are such that a drain
current, ID, of approximately 5 mA is
produced.
Another JFET, even of the same type, may
not produce the same results when
connected in this circuit due to the
variations in parameter values.

30. Solution
VS = IDRS
= (5 mA) (220 Ω) = 1.1 V
VD = VDD – IDRD
= 15 V – (5 mA) (1.0 kΩ) = 10 V
Thus,
VDS = VD – VS
= 10 V – 1.1 V = 8.9 V
Since VG = 0 V,
VGS = VG – VS
= 0 V – 1.1 V = – 1.1 V

31. Setting the Q-point of a Self-
Biased JFET
The basic approach to establishing a JFET bias point is
to determine ID for a desired value of VGS or vice versa.
Then, calculate the required value of RS by using
RS = |VGS / ID|
The values of ID and VGS can be determined in two
ways:
1. Directly from the transfer characteristic curve or
2. From the transconductance equation ID = IDSS with the
values of IDSS and VGS(off) obtained from the JFET data
sheet.

32. Example
Determine the value of RS
(at VGS = −5 V) required to
self-bias an n-channel
JFET that has the transfer
characteristic curve shown.
Solution
From the graph, ID = 6.25 mA at VGS = -5 V.
Then
RS = |VGS / ID| = 5 V / 6.25 mA = 800 Ω

33. Example
Determine the value of RS required to self-bias a p-channel
JFET with IDSS = 25 mA and VGS(off) = 15 V. VGS is to be 5 V.
Solution
Use the square-law equation:
ID = IDSS
= (25 mA)[1–(5V/15V )]2
= 11.1 mA
Now determine RS:
RS = |VGS / ID|
= 5 V / 11.1 mA = 450 Ω
2
GS(off)
GS
1









V
V

34. Midpoint Bias
It is desirable to bias a JFET near the midpoint of its transfer characteristic
curve where ID = .
Under ac signal condition, it allows the maximum amount of drain current
swing between IDSS and 0.
When ID = ,
= IDSS
0.5 ½ =
VGS = 0.29VGS(off) =
From this VGS value, the required RS can be determined.
To set the drain voltage at midpoint i.e. , select a value of RD to produce the
desired voltage drop.
Choose RG arbitrarily large to prevent loading on the driving stage in a
2
DSSI
2
DSSI
2
DSSI
2
GS(off)
GS
1









V
V









GS(off)
GS
1
V
V
4.3
GS(off)V

35. Example
Select resistor values for RD and RS for the circuit
below to set up an approximate midpoint bias. For this
particular JFET, the parameters are IDSS = 12 mA and
VGS(off) = −3 V.
Solution
For midpoint bias,
ID  IDSS/2 = 6 mA
and
VGS  VGS(off) / 3.4 = − 882 mV
Then,
RS = |VGS/ID|
= 882 mV / 6 mA = 147 Ω
From
VD = VDD − IDRD,
RD = (12 V − 6 V) / 6 mA = 1 k
R D
R S
R G
1 0 M 
V D D
+ 1 2 V

36. Graphical Analysis of a Self-Biased
JFET
• Find VGS at ID = 0,
VGS = −IDRS = (0)(470 Ω) = 0 V
ID = IDSS:
VGS = −IDRS
= −(10 mA)(470 Ω) = −4.7 V
• Draw a line (dc load line)
connecting the two
points. Wherever the
load line intersects the
characteristic curve,
we have the Q-point
of the circuit.

37. Voltage – Divider Bias
• The voltage at the source VS must be more
positive than the voltage at the gate VG in order to
keep the gate-to-source junction reverse biased.
• The source voltage is VS = ISRS. The voltage at
the gate is
• Thus,
VGS = VG − VS = VG − ISRS
• Using ID = IS, we get:
ID = (VG − VGS) / RS
12
2
DDG
RR
R
VV



38. Example
• Determine ID and VGS for the JFET with voltage-
divider bias shown. For this particular JFET, the
internal parameters are such that VD  7 V.
Solution
ID = (VDD – VD)/RD
= (12 V – 7 V) / 3.3 kΩ = 1.52 mA
VS = IDRS
= (1.52 mA) (2.2 kΩ) = 3.34 V
VG = [R2/(R1 + R2)]VDD
= [(1 MΩ)/(7.8 MΩ)] 12 V = 1.54 V
VGS = VG – VS
= 1.54 V – 3.34 V = –1.8V

39. Graphical Analysis of a JFET with
Voltage-Divider Bias
• The approach is similar to that in self-bias. In this case,
however, when ID = 0, VGS is not zero because the voltage-
divider produces a voltage at the gate independent of the
drain current.
• For ID = 0,
VGS = VG
• The next point taken to determine the dc load line in at VGS
= 0,
ID =
• The generalized dc load line is as shown.
S
G
R
V

41. Stability
The transfer characteristics of a JFET can differ considerably from
one device to another and this affects the Q-point stability.
A voltage-divider bias is more stable compared to a self-biased
circuit. This is because the slope of the dc load line in a voltage-
divider bias is much smaller.
Although VGS varies quite a bit for both self-bias and voltage-
divider bias, ID is much more stable with the voltage-divider bias.
Note that, by stable, we mean that the dependency of ID on the
range of Q-points is reduced.

42. Consider the n-channel JFET spec sheet :
• From the values listed, it is possible to plot two transfer
characteristic curves

45. The Metal – Oxide Semiconductor
Field-Effect Transistor
(MOSFET)
• Main drawback to JFET operation - JFET gate-to-source
must be reverse-biased in order to control the effective
size of the channel.
• This type of operation is referred to as depletion-mode
operation.
• A MOSFET is a device that can operate in the
enhancement mode and the depletion-mode
• There are two basic types of MOSFETs:
(1) Depletion (D) MOSFETs and
(2) Enhancement (E) MOSFETs.

46. • The construction of the MOSFET differs from the JFET in that it has
no pn junctions.
• The gate of the MOSFET is insulated from the channel by a very
thin layer of silicon dioxide (SiO2). The gate terminal is made of a
metal conductor.
• Thus, going from gate to the channel, you have metal, oxide and
semiconductor layers, which is where the term MOSFET comes
from.
Substrate - main body of the
MOSFETs.
For the n-channel D-
MOSFET: the substrate is p-
type, the channel is n – type.

47. D-MOSFET
• The center line in the circle represents the channel. The arrow, as usual,
points toward the n-type.
• An n-channel MOSFET operates in the depletion mode (similar to that of
JFET) when a negative gate-to-source voltage is applied.
• When a positive gate-to-source voltage is applied, the n-channel MOSFET
operates in the enhancement- mode.
• The D-MOSFET can operate both in the enhancement and depletion
modes.

48. Depletion Mode
• Negative voltage to the gate:
negative charges on the gate repel
the conduction electrons from the
channel, leaving the positive ions in
their place.
• That decreases the conductivity of
the channel.
• The greater the negative voltage on
the gate, the greater the depletion of
n-channel electrons.
• At a sufficiently large gate-to-source
voltage VGS(off), the channel is
completely depleted and ID becomes
zero.
• Just like the n-channel JFET, the n-
channel D-MOSFET conducts drain
current for gate-to-source voltages
between VGS(off) and zero.

49. Enhancement Mode
• In the enhancement mode, a
positive gate voltage applied to
the D-MOSFET effectively
widens the channel and reduces
its resistance.

50. Enhancement MOSFET
An E-MOSFET does not actually
have a channel. It depends on the
gate voltage to form a channel
between the source and drain
terminals.
The substrate extends completely
to the SiO2 layer.
An E-MOSFET has no depletion mode. It
operates only in the enhancement mode.
In other words, the gate-to-source
potential must always be positive.

51. n-channel E-MOSFET
• A positive gate voltage applied above a
threshold value VGS(th) induces a channel
by creating a thin layer of negative charges
in the substrate region adjacent to the SiO2
layer.
• The conductivity of the channel is
increased the gate-to-source voltage. This
pulls more electrons into the channel area.
• For any voltage below the threshold value
VGS(th), there is no channel.

52. E-MOSFET
The dashed line indicates the absence of a channel.

53. MOSFET Characteristics and
Parameters
• Most of the characteristics of the JFET apply to
the MOSFET.
E.g. the transconductance equation for the D-
MOSFET is the same as that of the JFET.
• Recall, that where VGS = 0, it corresponds to IDSS
and where ID = 0, it corresponds to VGS(off).

54. • There are some differences with the E-MOSFET which only operates under
the enhancement mode. It requires a positive gate-to-source voltage. (n-
channel)
• It does not have a significant IDSS parameter, as do the JFET and the D-
MOSFET.
• Ideally, there is no drain current until VGS reaches a certain non-zero value
called the threshold voltage VGS(th).
ID = 0 for 0 < VGS < VGS(th)

55. Equation for the drain current
• The square law equation for the drain current of the E-
MOSFET is:
ID =
• The constant K depends on the particular MOSFET and is
given by
K =
• These values can be found from the data sheet by looking
at the specified value of ID called the on-state value, ID(on),
at the given value of VGS.
  2
thGSGS VVK 
 
  2
thGSGS
onD
VV
I


56. Data Sheet

57. Example
From the data sheet above, determine the drain current for
VGS = 5 V.
Solution
From the data sheet, ID(on) = 500 mA at VGS = 10 V and VGS(th)
= 1 V.
Solving for K:
K= mA/V2
Thus,
ID =
= (6.17)(5 − 1)2 = 98.7 mA

   
17.6
110
500
22



 VV
mA
VV
I
thGSGS
onD
  2
thGSGS VVK 

58. Handling precautions!!
• The layer of SiO2 that insulates the gate from the channel is extremely thin and
can be easily destroyed by static electricity. Hence, extra precautions must be
made in handling MOSFETs.
• Since the gate of a MOSFET is insulated form the channel, the input resistance
is very high. The gate leakage current (IGSS) is in the pA range (compared to the
gate reverse current for a JFET which is in the nA range). An input capacitance
results from the insulated gate structure. Excess static charge can be
accumulated because of the combination of the input capacitance with the very
high input resistance (like a RC circuit). This can result in damaging the device.
• Precautions for handling a MOS device include:
1. MOS devices should be shipped and stored in conductive foam. Do not use
styrofoam because it is the best static electricity generator ever devised.
2. All instruments and metal benches used in assembly or testing should be
connected to earth ground (third prong on 110 V wall outlets).
3. Assembler’s or handler’s wrist must be connected to earth ground with a length
of wire and a high value series resistor,
4. Never remove a MOS device (or any other device, for that matter) from the
circuit while the power is on.
5. Do not apply signals to a MOS device while the dc power supply is off.

59. MOSFET Biasing
D-MOSFET Biasing
• Recall that D-MOSFET can be
operated with either positive or
negative values of VGS.
• A simple bias method is to set VGS =
0 so that an ac signal at the gate
varies the gate-to-source voltage
above and below this 0 V point.
• A D-MOSFET with zero bias is
shown in the figure below. Since VGS
= 0, ID = IDSS as indicated. The drain-
to-source voltage is expressed as
VDS = VDD − IDSSRD

60. Example
Determine the drain-to-source voltage given that VGS(off) = − 8
V and IDSS = 12 mA.
Solution
The drain-to-source voltage is
VDS = VDD − IDSSRD
VDS = 18 V − (12 mA)(620 )
= 10.6 V

61. E-MOSFET Biasing
• Several of the biasing circuits used for
JFETs and D-MOSFETs cannot be used
to bias E-MOSFETs because the
enhancement mode of operation
requires a positive value of VGS.
• The two biasing methods for E-
MOSFET are the voltage-divider bias
and drain-feedback.

62. E-MOSFET Biasing
Since VGS must be greater than the threshold value
VGS(th), the goal is to make the gate voltage more
positive than the source by an amount exceeding
VGS(th).
• Drain-feedback bias is the E-MOSFET counterpart
of collector-feedback bias for BJT.
• The drain-feedback bias circuit has a negligible gate
current.
• Thus, there is no drop across RG.
• Hence,
VGS = VDS.

63. E-MOSFET Biasing
For the voltage-divider bias, we get:
VDS = VDD − IDRD
where ID= .
12
2
DDGS
RR
R
VV


  2
thGSGS VVK 

64. Example
Determine VGS and VDS for the E-MOSFET circuit in figure below.
Assume this particular MOSFET has minimum values of ID(on) =
200 mA at VGS = 4 V and VGS(th) = 2 V.
Solution
Find K using the minimum value of ID(on).
K = ID(on)/(VGS–VGS(th))2 = 200 mA / (4V–2V)2
= 50 mA/V2
Now calculate ID for VGS = 3.13 V.
ID = K(VGS−VGS(th))2 = (50mA /V2)(3.13V–2V)2
= 63.8 mA
Finally, calculate VDS.
VDS = VDD – IDRD = 24 V – (63.8 mA) (200 Ω) = 11.2 V

65. FET Amplification
The transconductance of a FET is defined as,
.
In ac quantities,
.
Rearranging the terms, we have Id = gmVgs
This equation states that the output current Id equals the
input voltage Vgs multiplied by the transconductance gm.
GSV
I
g D
m



gsV
I
g d
m 

66. S
G
D
S
G D
gmVgs rd
Vgs Vds
Id

Id
Small signal model

67. VDD
RG
RD
VGG
ID
D
S
G
Vi
Vo
Io
IRD
Ii
Example

68. G
S
D
RG rd RD
Vi Vo
gmVgs
Io
Id
IRD,ac
Ii
Zi
Zo

69. Equivalent circuit that represents the relationship
Id = gmVgs.
Both internal resistances are
assumed to be large enough so that
they are open circuits.

70. A FET simplified equivalent circuit with an external ac drain
resistance .

71. The ac voltage gain is
Av = .
Since Vds = − IdRd, and Vgs= , thus
Av = −gmRd
gs
ds
V
V
m
d
g
I

72. There are two cases that Av can change:
• Case 1: If we attach a resistor to the source
of the FET, the gain is affected. It becomes:
Av=−
• Case 2: If the internal resistance is not
sufficiently greater than Rd (at least 10 times
greater), it appears in parallel to Rd, the gain is
reduced to the following:
Av=−






 sm
d
m
Rg
R
g
1






 '
'
dsd
dsd
m
rR
rR
g

73. Common-Source Amplifiers
JFET Amplification
A common-source amplifier is one with no source resistor (as far as ac
signal is concerned), so the source is connected to the (ac) ground.

74. It is biased such that the input stays within the linear region of operation.
The ac input signal causes the gate-to-source voltage to swing above and
below its Q-point value (VGSQ).
As the drain current , VR(D) , VD .
The drain current is in phase with the gate-to-source voltage. The drain-to-
source voltage is 180o out of phase with the gate-to-source voltage.

75. DC Analysis
Develop a dc equivalent circuit by replacing
all capacitors with opens. Then we
determine ID. If the circuit is biased at
midpoint of the dc load line, then ID =
IDSS/2.
OR
Solve for ID using the equation
ID = IDSS
2
)(
1







 

offGS
SD
V
RI

76. AC Equivalent Circuit
Replace the capacitors by shorts and the dc sources by a ground.

77. AC Equivalent Circuit
Since the input resistance to a FET is very high, all of the input
voltage from the ac signal source appears at the gate with very
little voltage is dropped across the internal ac source
resistance:
Vgs = Vin
The gain is given by
Av =−gmRd.
Thus, the output voltage becomes:
Vout = Vds = AvVgs
= −gmRdVin
where Rd=RD||RL and Vin = Vgs.

78. Example
What is the total output voltage at the unloaded amplifier
shown below? Assume IDSS = 770μA and VGS(off) = –3 V.

79. Solution
First find the dc output current. We need to solve the quadratic equation:
ID = IDSS
Rewriting the initial equation:
Let A = , B = , and C = 1.
Then we get
AID
2 + BID + 1 = 0
and the solution is
ID = (20.2 mA, 0.54 mA)
2
)(
1







 

offGS
SD
V
RI
01
12
)(
2
2
)(
2


















D
DSSoffGS
S
D
offGS
S
I
IV
R
I
V
R








2
)(
2
offGS
S
V
R
DSSoffGS
S
IV
R 12
)(


80. With this value for the drain current, determine the value of the drain
voltage:
VD = VDD – IDRD
= 12 V – (0.5mA) (3.3 kΩ)
= 10.2 V
Next we calculate gm as follows:
VGS = − IDRS = − (0.5 mA) (910 Ω) = –0.46 V
gm0 =
= 2 (770 μA) / 3 V = 0.51 mS
gm = =
= 432.7 μS
GS(off)
DSS
2
V
I









)(
0 1
offGS
GS
m
V
V
g 








V
V
mS
3
78.1
151.0

81. Finally, the ac output is
Vout = AvVin = −gmRDVin
= − (433 μS)(3.3 kΩ)(100 mV)
= − 143 mVrms
Thus the total output is an ac signal with an
(143mV) (1.41) = 202 mV
peak value riding on a dc level of 5.53 V.

82. Input Resistance
To calculate the exact value of the (very high) input
resistance of the amplifier, we use the equation:
Rin = RG||(VGS/IGSS)
where IGSS is the leakage current.

83. D – MOSFET Amplification
A zero-biased common-source n-channel D-MOSFET with an
ac source is capacitively coupled to the gate shown.
The gate is approximately at 0 Vdc and the source
terminal is at ground, thus making VGS = 0 V.
The signal voltage causes Vgs to swing above and below
its zero value, producing a swing in Id.

84. • The negative swing in Vgs produces the depletion mode and Ids
decreases. The positive swing in Vgs produces the enhancement
mode and Ids increases.
• Note the enhancement mode is on the right of the vertical axis and
the depletion mode is on the left.

85. The dc analysis of this amplifier is somewhat easier than for a
JFET because ID = IDSS at VGS = 0. Since ID is known, the analysis
involves calculating only VD.
VD = VDD − IDRD
The ac analysis is the same as for the JFET amplifier.

86. Example
The D-MOSFET shown earlier has the following values: RD = 33 , RG
= 10 M, C1 = 10 F, C2 = 10 F, RL = 8.2 k, IDSS = 200 mA and a gm
= 200 mS.
Determine both the dc drain voltage and ac output voltage. Given Vin =
500 mV.
Solution
Since the amplifier is zero – biased,
ID = IDSS =200 mA
And, therefore,
VD = VDD − IDRD
= 15 V − (200)(33) = 8.4 V
Rd = RD||RL = 32.9 
The ac output voltage is
Vout = −gmRdVin = −3.29 V

87. E-MOSFET Amplification
• The circuit shows a common-source n-channel E-MOSFET with
voltage-divider bias with an ac source capacitively coupled to the gate.
The gate is biased so that VGS > VGS(th).
• The signal voltage produces a swing in Vgs above and below its Q-
point value VGSQ. This in turn causes a swing in Id above and below its
Q-point, IDQ.

88. • The general procedure is to solve for the VGS, then get ID,
and finally VDS. That is,
K from ID(on) and corresponding VGS
ID = K(VGS – VGS(th))2
VDS = VDD − IDRD
12
2
DDGS
RR
R
VV



89. Example
A common source amplifier using an E-MOSFET is shown below. Find VGS,
ID, VDS, and the ac output voltage. Assume that ID(on) = 200 mA at VGS = 4 V,
VGS(th) = 2 V, and gm = 23 mS. Vin = 25 mV.
Solution
= (8.2 kΩ) / (55.2 kΩ) 15 V = 2.23 V
For VGS = 4 V, we get:
K =
= 200 mA /(4 V–2 V)2 = 50 mA/V2
Therefore,
ID = K(VGS – VGS(th))2 = (50 mA/V2)(2.23 V–2V)2 = 2.65 mA
VDS = VDD – IDRD = 15 V – (2.65 mA)(3.3 kΩ)
= 6.26 V
Rd = RD||RL = 3.3 kΩ||33 kΩ = 3 kΩ
The ac output is, then
Vout = AvVin = −gmRdVin = −(23 mS)(3 kΩ)(25 mV) = − 1.73 V
12
2
DDGS
RR
R
VV


 
  2
thGSGS
onD
VV
I


90. Common – Drain Amplifiers
The common-drain amplifier is similar to the common-
collector BJT amplifier in that the Vin is the same as Vout with no
phase shift.
The gain is actually slightly less than 1. Note the output is
taken from the source.

91. Common-Gate Amplifiers
The common-gate is similar to the common-base BJT amplifier in that it has a
low input resistance.
The voltage gain can be determined by the same formula as used with the
JFET common-source amplifier.
The input resistance can be determined by the formula below.
Rin(source) = 1/gm