MOSFET
DC circuit analysis
Common-Source Circuit
Coupling capacitor acts as
an open circuit to dc but
allows the signal voltage
to be coupled to the gate
The source terminal is
at ground potential and
is comon to both the
input and output.
1
MOSFET
DC circuit analysis
Common-Source Circuit
Applying the voltage divider rule;
VG  VGS
 R2 
VDD
 
 R1  R2 
Assume that the MOSFET is biased
in the saturation region i.e; vGS  VTN
and vDS  vDS sat   vGS  VTN
I D  K n VGS  VTN 
2
Also;
I D  VDD  I D RD
The dc equivalent circuit;
2
MOSFET
DC circuit analysis
Common-Source Circuit
If the calculated vDS is greater than
vDS(sat), the MOSFET is biased in
the saturation region as we have
initially assumed, and our analysis
is correct.
The dc equivalent circuit;
3
MOSFET
DC circuit analysis
Common-Source Circuit
If the calculated vDS is smaller than
vDS(sat), the MOSFET is biased in
the triode region and we have to
recalculate the current using the
equation for the triode region
namely;

2
iD  K n 2vGS  VTN vDS  vDS

The dc equivalent circuit;
4
MOSFET
DC circuit analysis
Common-Source Circuit
The power dissipated in the MOSFET
is;
PT  I DVDS
The dc equivalent circuit;
5
MOSFET
DC circuit analysis
Common-Source Circuit
Example 1
Calculate the drain current,
ID and drain-to-source
voltage, VDS and the power
dissipated in the transistor
PT.
6
MOSFET
DC circuit analysis
Common-Source Circuit
Example 1 – Solution
 R2 
VDD
VG  VGS  
 R1  R2 
 160 

10
 280  160 
 3.64 V
7
MOSFET
DC circuit analysis
Common-Source Circuit
Example 1 – Solution (cont’d)
Assuming that the transistor is
in saturation mode;
I D  K n VGS  VTN 
2
2


 0.25 3.64  2
 0.669 mA
VDS  VDD  I D RD
 10  0.669 10
 3.31 V
8
MOSFET
DC circuit analysis
Common-Source Circuit
Example 1 – Solution (cont’d)
VDS sat   VSG  VTN
 3.64  2
 1.64 V
Since VDS > VDS(sat), the
transistor is indeed in
saturation mode as we
have initially assumed.
9
MOSFET
DC circuit analysis
Common-Source Circuit
Example 1 – Solution (cont’d)
The power dissipated in
the transistor;
PT  I DVDS
 0.669  3.31
 2.21 mW
10
MOSFET
DC circuit analysis
Common-Source Circuit
Example 2
Design the circuit such that;
R1 R2  200 k
I DQ  1.2 mA
VSDQ  4 V
11
MOSFET
DC circuit analysis
Common-Source Circuit
Example 2 – Solution
VSDQ  4 V  10  I DQ RD
Hence;
I DQ RD  6 V
6V
6V
RD 

 5 k
I DQ 1.2 mA
12
MOSFET
DC circuit analysis
Example 2 – Solution (cont’d)
Common-Source Circuit
I DQ  K p VSGQ  VTP 
2
Substituting values;
1.2  0.4VSGQ  1.2
2
1.2
VSGQ  1.2   3
0.4
2
VSGQ  3  1.2  2.932 V
13
MOSFET
DC circuit analysis
Common-Source Circuit
Example 2 – Solution (cont’d)
 R1 
  2.932 V
10
 R1  R2 
R1
 0.2932
R1  R2
But;
R1 R2
 200 k
R1  R2
14
11/02/2008
15
MOSFET
DC circuit analysis
Common-Source Circuit
Example 2 – Solution (cont’d)
Solving the equations gives us;
R1  283 k
R2  682 k
16
MOSFET
DC circuit analysis
Common-Source Circuit
Example 2 – Solution (cont’d)
17
MOSFET
DC circuit analysis
Common-Source Circuit
Example 3
For the circuit in figure; VTN = 1 V
and Kn = 0.5 mA/V2.
Determine VGS, ID and VDS.
18
MOSFET
DC circuit analysis
Common-Source Circuit
Example 3 – Solution
 40 
VG  10
5
 60  40 
 1 V
VS  5  I D RD
 5  103 I D
VGS  VG  VS
 4  103 I D
I D  4  VGS 10 3
19
MOSFET
DC circuit analysis
Common-Source Circuit
Example 3 – Solution (cont’d)
I D  K n VGS  VTN 
2
 0.5 10 VGS  1
3
2


2
 0.5 10 3 VGS
 2VGS  1
3


I

4

V
10
Substituting for D
GS
2
8  2VGS  VGS
 2VGS  1
Hence;
VGS  7  2.646 V
20
MOSFET
DC circuit analysis
Common-Source Circuit

Example 3 – Solution (cont’d)

2
I D  0.5 10 3 VGS
 2VGS  1


 0.5 103 2.6462  2  2.646  1
I D  1.35 mA
VS  5  I D RS
 5  1.35 1
 3.65 V
21
MOSFET
DC circuit analysis
Common-Source Circuit
Example 3 – Solution (cont’d)
VD  5  I D RD
 5  1.35  2
 2.29 V
VDS  VD  VS
 2.29  3.65
VDS  5.94 V
22
MOSFET
DC circuit analysis
Common-Source Circuit
Example 3 – Solution (cont’d)
 R2 
  5
VG  5  5
 R1  R2 
 40 
 10
5
 60  40 
 1 V
VGS  VG  VS
 1  3.65
VGS  2.65 V
23
MOSFET
DC circuit analysis
Load Line and Mode of Operation
The load line equation is;
VDS  VDD  I D RD
 5  I D 20
or;
5 VDS
ID 

20 20
mA 
From the above equation, we
obtain the following two points:
when ID = 0, VDS = 5 V and
when VDS = 0, ID = 5/20 mA
24
MOSFET
DC circuit analysis
Load Line and Mode of Operation
The two points are used to plot a load line on the iD – vDS
characteristic curves as shown.
25
MOSFET
DC circuit analysis
Load Line and Mode of Operation
The Q-point is given by the drain dc current and drain-tosource voltage and is always on the load line.
26
MOSFET
DC circuit analysis
Load Line and Mode of Operation
If vGS is less than VTN, the drain current is zero and the
transistor is in the cutoff mode.
27
MOSFET
DC circuit analysis
Load Line and Mode of Operation
If vGS is just greater than VTN, the transistor turns on.
28
MOSFET
DC circuit analysis
Load Line and Mode of Operation
As vGS increases, the Q-point moves up the load line.
29
MOSFET
DC circuit analysis
Load Line and Mode of Operation
The transition point is the boundary between saturation
and non-saturation (triode) region.
30
MOSFET
DC circuit analysis
Constant-Current Biasing
The figure shows a MOSFET amplifier
where RS is replaced with a constant
current source, IQ.
The gate-to-source
voltage, VGS adjusts
itself to correspond to
the current IQ.
IQ causes the bias to be independent on the
transistor parameters and hence stabilizes
the operating (Q) point.
31
MOSFET
DC circuit analysis
Constant-Current Biasing
The figure shows the corresponding
dc equivalent circuit.
The gate terminal is at the ground
potential because we assume zero
gate current.
IQ sets the value of VGS, VD, VDS and
hence the Q-point of the MOSFET.
The current source IQ can be constructed using
MOSFETs but this topic will not be covered in
this note.
32
MOSFET
DC circuit analysis
Constant-Current Biasing Example 4
The MOSFET parameters in the figure are;
VTN  0.8 V, k n'  80 A/V 2 , W / L  3
Determine VGS and VDS.
33
15/02/08
34
MOSFET
DC circuit analysis
Constant-Current Biasing Example 4 – Solution
The dc equivalent circuit is as follows;
Assuming the MOSFET is operating in
the saturation mode;
kn' W
2
I D   VGS  VTN 
2 L
Substituting values;
6
80

10
2
250 10 6 
 3VGS  0.8
2
VGS  2.24 V
35
MOSFET
DC circuit analysis
Constant-Current Biasing Example 4 – Solution (cont’d)
The source potential;
VS  VG  VGS  0  2.24  2.24 V
Assuming zero gate current;
VD  5  I D RD
 5  I Q RD
 5  250 10 6 10 4
VD  2.5 V
36
MOSFET
DC circuit analysis
Constant-Current Biasing Example 4 – Solution (cont’d)
The drain-to-source voltage;
VDS  VD  VS  2.5  2.24
VDS  4.74 V
37
MOSFET
DC circuit analysis
EXERCISES
Problems:
3.23;
3.25;
3.27 and
3.28
38
QUIZ
39