Small Signal Model
i
v
S. Sivasubramani
EE101- Small Signal - Diode
Small Signal Model
i
ID
id
iD
v
vD
vd
VD
S. Sivasubramani
EE101- Small Signal - Diode
Small Signal Model -Mathematical Analysis
VD - DC value
vd - ac signal
vD - Total signal (DC + ac signal)
Diode current and voltage are related by
id = Is (exp(
vd
) − 1)
Vt
ID = Is (exp(
VD
) − 1)
Vt
Let us apply VD ,
ID ≈ Is (exp(
S. Sivasubramani
VD
))
Vt
EE101- Small Signal - Diode
(1)
Small Signal Model -Mathematical Analysis
Let us apply vD , sum of DC and ac signal,
vD = VD + vd
The current
iD = Is (exp(
vD
) − 1)
Vt
iD ≈ Is (exp(
iD ≈ Is (exp(
iD ≈ Is exp(
S. Sivasubramani
vD
))
Vt
VD + vd
))
Vt
VD
vd
) exp( )
Vt
Vt
EE101- Small Signal - Diode
Contd..
From (1)
vd
)
Vt
If vd << Vt , it can be written using Taylor’s series
iD = ID exp(
iD = ID (1 +
vd
+ ···)
Vt
ID + id = ID +
ID vd
Vt
ID
vd
Vt
The small signal voltage and current related by linear relationship
like resistor.
id = gd vd
ID
where gd =
and is a small signal conductance in f.
Vt
id =
S. Sivasubramani
EE101- Small Signal - Diode
Small Signal Model
It can also be written as
vd = rd id
Vt
. rd is a small signal resistance in Ω.
ID
It can also be obtained
diD
rd = 1/
dvD vD =VD
where rd =
For small signal (vd < VT )
+
vD
+
vd
− iD
S. Sivasubramani
rd
− id
EE101- Small Signal - Diode
Small Signal Model - Various Elements
Resistor
R
+
vD
iD
−
R
+
id
−
vd
−
+
Voltage Source - Independent
V
iD
id
Current Source - Independent
I
−
+
vD
S. Sivasubramani
vd
+
−
EE101- Small Signal - Diode
Test
Draw the small signal equivalent circuit at VD = 1 V, ID = 1 mA.
The element N has the following i − v relationship.
iD = vD2
R
R
+
+
vi sin(ωt)
vD
VI
vd
vi sin(ωt)
N
− iD
S. Sivasubramani
rN
− id
rN =
1
1
= Ω
diD
2
|vD =VD
dvD
EE101- Small Signal - Diode
Small Signal Model - Analysis
If an input ac signal is superimposed on DC, use the following
steps to solve for voltage and current in any nonlinear circuit.
1
Calculate DC operating point without ac signal using any one
of the nonlinear techniques such as analytical method or
graphical method.
2
Draw small signal equivalent circuit.
3
As it is a linear circuit, use any linear techniques to solve for
voltage/current.
4
Add DC operating point and small signal voltage current to
find the response of total signal (DC + small signal).
Keep in mind that this method will work till vd < Vt .
S. Sivasubramani
EE101- Small Signal - Diode
Example-I
1 kΩ
+
1 sin(ωt) V
vD
5V
S. Sivasubramani
− iD
EE101- Small Signal - Diode
PSPICE Result of Total Signal
* C:\IITP\Academic_Work\Subjects_Taken\2014\EE101\PSPICE\Diode_Analysis.sch
Date/Time run: 09/15/14 17:12:36
Temperature: 27.0
(A) Diode_Analysis (active)
660.0mV
650.0mV
637.5mV
630.0mV
0s
50ms
100ms
V(R1:2,D1:2)
Time
Date: September 15, 2014
Page 1
S. Sivasubramani
Time: 17:14:25
EE101- Small Signal - Diode
Small Signal Analysis Approach
1
Find operating point. (Use any nonlinear technique)
1 kΩ
+
VD = 0.64843 V;
VD
5V
ID = 4.352 mA
− ID
2
Obtain small signal circuit
1 kΩ
rd =
+
1 sin(ωt) V
vd
rd
− id
3
Vt
25
=
= 5.7445 Ω
ID
4.352
vd =
1 × rd
= 0.0057 V
R + rd
Add them
vD = 0.64843 + 0.0057 sin(ωt) V
S. Sivasubramani
EE101- Small Signal - Diode
PSPICE of Total and Small Signal Analysis
* C:\IITP\Academic_Work\Subjects_Taken\2014\EE101\PSPICE\Diode_Analysis.sch
Date/Time run: 09/16/14 21:27:22
Temperature: 27.0
(A) Diode_Analysis (active)
660.0mV
650.0mV
637.5mV
630.0mV
0s
50ms
V(x)
100ms
V(y)+V(z)
Time
Date: September 16, 2014
Page 1
S. Sivasubramani
Time: 21:32:21
EE101- Small Signal - Diode
Example-II
1 kΩ
+
5 sin(ωt) V
vD
5V
S. Sivasubramani
− iD
EE101- Small Signal - Diode
PSPICE Result of Total Signal
* C:\IITP\Academic_Work\Subjects_Taken\2014\EE101\PSPICE\Diode_Analysis.sch
Date/Time run: 09/18/14 14:34:35
Temperature: 27.0
(A) Diode_Analysis (active)
800mV
400mV
0V
0s
50ms
100ms
V(x)
Time
Page 1
Date: September 18, 2014
S. Sivasubramani
EE101- Small Signal - Diode
Time: 14:35:33
Small Signal Analysis Approach
1
2
DC operating point (Already done.)
Get the small signal circuit
1 kΩ
rd =
+
5 sin(ωt) V
vd
rd
− id
3
Vt
25
=
= 5.7445 Ω
ID
4.352
vd =
5 × rd
= 0.00286 V
R + rd
Add operating point and small signal output
vD = 0.64843 + 0.0286 sin(ωt) V
As vd > Vt , Small signal approach failed here...
For large variation from DC operating point, any one of the
nonlinear techniques should be used.
S. Sivasubramani
EE101- Small Signal - Diode
PSPICE of Total and Small Signal Analysis
* C:\IITP\Academic_Work\Subjects_Taken\2014\EE101\PSPICE\Diode_Analysis.sch
Date/Time run: 09/18/14 14:34:35
Temperature: 27.0
(A) Diode_Analysis (active)
800mV
400mV
0V
0s
50ms
V(x)
100ms
V(y)+V(z)
Time
Page 1
Date: September 18, 2014
S. Sivasubramani
EE101- Small Signal - Diode
Time: 14:36:37