COMSATS Institute of Information Technology
Virtual campus
Islamabad
Dr. Nasim Zafar
Electronics 1
EEE 231 – BS Electrical Engineering
Fall Semester – 2012
The Diode CircuitsII
Lecture No:
10
Contents:
References:
Ø Microelectronic Circuits:
Adel S. Sedra and Kenneth C. Smith.
Ø Electronic Devices and Circuit Theory:
Robert Boylestad & Louis Nashelsky ( Prentice Hall ).
Ø Introductory Electronic Devices and Circuits:
Robert T. Paynter.
Ø Electronic Devices :
3
References (Figures):
Chapter 2 Diodes:
Figures are redrawn (with some modifications) from
Introductory Electronic Devices and Circuits
By
Robert T. Paynter
The Diode Models
1. The Ideal Diode Model
The Diode:
PN Junction Diode Schematic Symbol:
Anode
Cathode
p
n
6
Diode Circuits:
anode
Reversed bias
+
+
Forward bias
cathode
The left hand diagram shows the reverse biased junction.
No current flows flows.
The other diagram shows forward biased junction.
A current flows.
ForwardBiased Diode Circuit:
R
R
I F > 0A
I F > 0A
IF
V
IF
V
+V
-V
R
R
IF
IF
8
ReverseBiased Diode Circuit:
R
R
V
0A
0A
IT
IT
V
+V
R
-V
R
9
Effect of VF:
I
VS
5V
4.3 V
VD1 = 0.7V
R1
1k
VR1 = VS − VD1 = 5V − 0.7V = 4.3V
VR1 4.3V
I=
=
= 4.3mA
R1 1kΩ
D1
Value
Ideal
Practical
VF
0 V
0.7 V
VR1
5 V
4.3 V
I
5 mA
4.3 mA
10
Example1
V
I
VS
6V
R1
10 k
VR1 = VS − VD1 = 6V − 0.7V
D1
= 5.3V
VR1 5.3V
I=
=
= 530μA
R1 10kΩ
11
Example2
I
VS
5V
R1
1.2 k
D1
R2
2.2 k
VS − VD1
I=
R1 + R2
5V − 0.7V
=
1.2kΩ + 2.2kΩ
= 1.26mA
12
Example3
R1
5.1 k
I
VS
4V
VS − VD1 − VD 2
I=
R1
D1
D2
4V − 0.7V − 0.7V
=
5.1kΩ
= 509.8μA
13
Percentage Error:
% of error =
X X'
X
100
where X = the measured value
X’ = the calculated value
14
Example4
I
R1
1.5 k
D1
R2
1.8 k
VS
10 V
I ideal =
VS
10V
=
= 3.03mA
R1 + R2 1.5kΩ + 1.8kΩ
I prac =
VS − VD1 − VD 2 10V − 0.7V − 0.7V
=
R1 + R2
1.5kΩ + 1.8kΩ
= 2.61mA
D2
% of error =
2.61mA − 3.03mA
2.61mA
15
100 = 16.1%
Power Dissipation PD(max)
16
I0 and PD(max) Relationship:
I0 =
PD (max)
VF
where I0 = the limit on the average forward current
PD(max) = the forward power dissipation rating of the
diode
VF = the diode forward voltage (0.7V for Si)
17
Forward Power Dissipation PD(max):
I
VS
10 V
D1
RL
100
Choose a diode with forward power
dissipation PD(max) at least 20%
greater than actual power dissipation.
VS − VD1 10V − 0.7V
I=
=
= 93mA
RL
100Ω
PD = VD1 I = ( 0.7V ) ( 93mA ) = 65.1mW
PD (max) = 1.2 PD = ( 1.2 ) ( 65.1mW ) = 78.12 mW (minimum)
18
Example 5.
A diode has a forward power dissipation rating of 500
mW. What is the maximum allowable value of forward
current for the device?
I0 =
PD (max)
VF
500mW
=
= 714.29mA
0.7V
I (max) = 0.8I 0 = ( 0.8 ) ( 714.29mA ) = 571.43mA
19
Complete: Model Diode Curve (Ref 3).
IF(mA)
100
IR
VZ
80
VR(V)
80
Reverse operating
region (also called
the reverse
breakdown
region)
60
Forward
operating
region
60
RZ
RZ
Complete model
Accurate model
40
VR
IR
20
40
20
0.2
0.4
0.6
VF(V)
0.8
1.0
I=0
IF
0.7 V
2.0
RB
3.0
RB
IR( A)
VF
IF
20
Another Example:
Determine voltage across diode in Fig. 2.19 (Ref. 3)
for the values of IF = 1 mA and IF = 5 mA.
Assume that the value for RB = 5 Ω .
IF = 1 mA:
VD = 0.7V + IRB = 0.7V + ( 1mA ) ( 5Ω) = 0.705V
IF = 5 mA:
VD = 0.7V + IRB = 0.7V + ( 5mA ) ( 5Ω) = 0.725V
Bulk resistance has a significant effect on voltage drop
across diode terminals when the forward current is large.
21
The Diode Models
4. PiecewiseLinear Diode Model
5. ConstantVoltage Diode Model
6. Dynamic Resistance, AC Resistance
Piecewise Linear Diode Model:
More accurate than the ideal diode model and does not rely on nonlinear
equation or graphical techniques.
Ø
Diode IV characteristic approximated
by straight line segments.
We model each section of the diode
IV characteristic with R in series with
a fixed voltage source.
Ø
ConstantVoltage Diode Model:
Ø
Ø
If VD < VD,on: The diode operates as an open
circuit.
If VD VD,on: The diode operates as a constant
Example: Diode dc Bias Calculations
VX
Ø
I X R1 VD
I X R1
IX
2.2mA for VX
IX
0.2mA for VX 1V
IX
VT ln
IS
3V
This example shows the simplicity provided by a constant
voltage model over an exponential model.
Ø
Using an exponential model, iteration is needed to solve for
current. Using a constantvoltage model, only linear equations
need to be solved.