Handout 2 , Asst Prof M Anis Ch, Electronic Devices and Circuits
Lecture 2
EE-215 Electronic Devices and Circuits
Asst Prof Muhammad Anis Ch
Modeling the Diode Forward Characteristic
The Exponential Model
The Constant-Voltage Drop Model
The Ideal Diode Model
The Small-Signal Model
The Exponential Model
provides the most accurate description of diode operation
in the forward-bias region
is the most difficult to use
because of its severely non-linear nature
v
Exponential model ⟹ i = IS e V T
for this circuit
VD
ID = IS e V T
and using KVL
VDD = ID R + VD
VDD −VD
or ID = R
if IS for the diode is given,
we have two equations in two unknowns ID and VD
VD
VDD −VD
ID = IS e V T and ID = R
if IS for the diode is given,
we have two equations in two unknowns ID and VD
These two equations can be solved using either
Graphical Analysis
or Iterative Analysis
Graphical Analysis using the Exponential Model
VD
VDD −VD
ID = IS e V Tand ID = R
this analysis is performed by
1 of 26 2/20/18, 11:47 AM
,Handout 2 , Asst Prof M Anis Ch, Electronic Devices and Circuits
plotting the two equations on the i-v plane
the point of intersection is the desired solution ( VD , ID )
the curve represents the exponential diode equation.
while the straight line represents the KVL equation
Such a straight line is called load line.
The load line intersects the diode curve at point Q
Q represents the operating point of the circuit.
The coordinates of Q give the values of ID and VD
Example 4.4 using Graphical Analysis
Determine the current ID and the diode voltage VD for the circuit in Fig. 4.10 with
VDD = 5V and R = 1kΩ. Assume that the diode has a current of 1mA at a voltage of
0.7V .
Solution:
here VDD = 5V , R = 1kΩ
and I = 1mA for V = 0.7V
V
1mA
I = IS e V T ⟹ IS = I
V = 0.7V = 6.9144 × 10−16 A
e VT e 25mV
VD
VDD −VD
As the 2 circuit equations are ID = IS e V T and ID = R
VD
5−VD
⟹ ID = 6.9144 × 10−16 e 25m and ID = 1k
VD
5−VD
ID = 6.9144 × 10−16 e 25m and ID = 1k
Plotting these 2 equations in Matlab or Octave with give the desired solution
VD , ID
2 of 26 2/20/18, 11:47 AM
, Handout 2 , Asst Prof M Anis Ch, Electronic Devices and Circuits
reading VD , ID from the plot ⟹VD = 0.7358V
and ID = 4.264mA
Iterative Analysis using the Exponential Model
the 2 circuit equations
VD
VDD −VD
ID = IS e V Tand ID = R
can also be solved using a simple iterative procedure
Modeling the Diode Forward Characteristic
VD
VDD −VD
ID = IS e V T and ID = R
Example 4.4 using Iterative Analysis
Determine the current ID and the diode voltage VD for the circuit in Fig. 4.10 with
VDD = 5V and R = 1kΩ. Assume that the diode has a current of 1mA at a voltage of
0.7V .
Solution:
to start, let VD = 0.7V
Iteration 1:
VDD −VD 5−0.7
KVL ⟹ ID = R
= 1k
= 4.3mA
to have a better estimate of VD
I I
using V2 − V1 = VT ln I21 or V2 = V1 + VT ln I21
for V1 = 0.7V and I1 = 1mA (given)
and V2 = VD =? , I2 = ID = 4.3mA
I
V2 = V1 + VT ln I21 ⟹VD = 0.7 + 25m ln 4.3m1m
= 0.736V
thus after this first iteration we have
VD = 0.736V , ID = 4.3mA
3 of 26 2/20/18, 11:47 AM
Lecture 2
EE-215 Electronic Devices and Circuits
Asst Prof Muhammad Anis Ch
Modeling the Diode Forward Characteristic
The Exponential Model
The Constant-Voltage Drop Model
The Ideal Diode Model
The Small-Signal Model
The Exponential Model
provides the most accurate description of diode operation
in the forward-bias region
is the most difficult to use
because of its severely non-linear nature
v
Exponential model ⟹ i = IS e V T
for this circuit
VD
ID = IS e V T
and using KVL
VDD = ID R + VD
VDD −VD
or ID = R
if IS for the diode is given,
we have two equations in two unknowns ID and VD
VD
VDD −VD
ID = IS e V T and ID = R
if IS for the diode is given,
we have two equations in two unknowns ID and VD
These two equations can be solved using either
Graphical Analysis
or Iterative Analysis
Graphical Analysis using the Exponential Model
VD
VDD −VD
ID = IS e V Tand ID = R
this analysis is performed by
1 of 26 2/20/18, 11:47 AM
,Handout 2 , Asst Prof M Anis Ch, Electronic Devices and Circuits
plotting the two equations on the i-v plane
the point of intersection is the desired solution ( VD , ID )
the curve represents the exponential diode equation.
while the straight line represents the KVL equation
Such a straight line is called load line.
The load line intersects the diode curve at point Q
Q represents the operating point of the circuit.
The coordinates of Q give the values of ID and VD
Example 4.4 using Graphical Analysis
Determine the current ID and the diode voltage VD for the circuit in Fig. 4.10 with
VDD = 5V and R = 1kΩ. Assume that the diode has a current of 1mA at a voltage of
0.7V .
Solution:
here VDD = 5V , R = 1kΩ
and I = 1mA for V = 0.7V
V
1mA
I = IS e V T ⟹ IS = I
V = 0.7V = 6.9144 × 10−16 A
e VT e 25mV
VD
VDD −VD
As the 2 circuit equations are ID = IS e V T and ID = R
VD
5−VD
⟹ ID = 6.9144 × 10−16 e 25m and ID = 1k
VD
5−VD
ID = 6.9144 × 10−16 e 25m and ID = 1k
Plotting these 2 equations in Matlab or Octave with give the desired solution
VD , ID
2 of 26 2/20/18, 11:47 AM
, Handout 2 , Asst Prof M Anis Ch, Electronic Devices and Circuits
reading VD , ID from the plot ⟹VD = 0.7358V
and ID = 4.264mA
Iterative Analysis using the Exponential Model
the 2 circuit equations
VD
VDD −VD
ID = IS e V Tand ID = R
can also be solved using a simple iterative procedure
Modeling the Diode Forward Characteristic
VD
VDD −VD
ID = IS e V T and ID = R
Example 4.4 using Iterative Analysis
Determine the current ID and the diode voltage VD for the circuit in Fig. 4.10 with
VDD = 5V and R = 1kΩ. Assume that the diode has a current of 1mA at a voltage of
0.7V .
Solution:
to start, let VD = 0.7V
Iteration 1:
VDD −VD 5−0.7
KVL ⟹ ID = R
= 1k
= 4.3mA
to have a better estimate of VD
I I
using V2 − V1 = VT ln I21 or V2 = V1 + VT ln I21
for V1 = 0.7V and I1 = 1mA (given)
and V2 = VD =? , I2 = ID = 4.3mA
I
V2 = V1 + VT ln I21 ⟹VD = 0.7 + 25m ln 4.3m1m
= 0.736V
thus after this first iteration we have
VD = 0.736V , ID = 4.3mA
3 of 26 2/20/18, 11:47 AM
