Module: Electronics I
Module Number: 610/650221-222
Electronic Devices and Circuit Theory, 9th ed., Boylestad and Nashelsky
Philadelphia University
Faculty of Engineering
Communication and Electronics Engineering
Diode Equivalent Circuit
DC Analysis:
- Load Line Analysis:
The shown circuit is the simplest diode
configuration.
The intersection of the load line on the
characteristics
can
be
determined
by
applying kirchoff's voltage law.
E − VD − VR = 0
•
If V D is set to zero and solve for
I D we have the magnitude of I D on
the vertical axis.
E − VR = 0
⇒ ID =
•
E
|V =0V
R D
Now set I D = 0 A , the value of V D on the horizontal axis could be determined.
E − VD = 0
⇒ V D = E | I D =0 A
•
After finding the intersection values of both axes, the load line could be drawn
and then the point of operation as shown below.
Lecturer: Dr. Omar Daoud
Part I -II
Module: Electronics I
Module Number: 610/650221-222
Electronic Devices and Circuit Theory, 9th ed., Boylestad and Nashelsky
Ex. 2.1: For the given diode configuration in the figure below, determine both of the
Q-values and V R .
R
R
Ex.2: Determine the diode voltage and current in a series configuration using the
piece-wise model, if R=2kΩ, E=5V, V γ =0.6V and r f =10Ω.
R
R
R
R
Ex.3: For the same circuit shown in Fig.2, determine the diode voltage and current
in a series configuration, if the saturation current equals 0.1pA.
Lecturer: Dr. Omar Daoud
Part I -II
Module: Electronics I
Module Number: 610/650221-222
Electronic Devices and Circuit Theory, 9th ed., Boylestad and Nashelsky
Diode's Configurations:
Ex. 2.7: Determine both of V o and I D for the given circuit.
R
R
R
R
Ex. 2.8: Determine V D1 , V D2 and I D for the given circuit.
R
R
R
R
R
R
Ex. 2.9: Determine I, V 1 , V 2 and V o for the given circuit.
R
R
R
Lecturer: Dr. Omar Daoud
R
R
R
Part I -II
Module: Electronics I
Module Number: 610/650221-222
Electronic Devices and Circuit Theory, 9th ed., Boylestad and Nashelsky
Ex. 2.11: In this example there are two LED's that
can be used as a polarity detector.
Apply a
positive source voltage and a green light results.
Negative supplies result in a red light. Find the
resistor R in this combination to ensure a current
of 20mA through the ON diode. Both diodes have
a reverse breakdown voltage of 3V and an average
turn-on voltage of 2V.
What will happen if the Green LED is changed
with a Blue on of 5V turn-on voltage?
Lecturer: Dr. Omar Daoud
Part I -II
Module: Electronics I
Module Number: 610/650221-222
Electronic Devices and Circuit Theory, 9th ed., Boylestad and Nashelsky
Other Diode Types:
-
Solar Cell:
-
Photo diodes:
-
Light Emitting Diodes:
-
Schottky Barrier Diode:
-
Zener Diode:
At some point of applying the reverse-bias voltage, the break-down occurs. This
means that a very high current passes through the diode in the opposite direction
of the forward current.
This element could be designed to provide specific break-down voltages, and
then can be operated by limiting the current to a value within the capability of
the device (has a constant voltage reference; V z .).
R
Lecturer: Dr. Omar Daoud
R
Part I -II
Module: Electronics I
Module Number: 610/650221-222
Electronic Devices and Circuit Theory, 9th ed., Boylestad and Nashelsky
Diode Applications:
-
Rectifier Circuit:
•
Half-Wave Rectifications:
Lecturer: Dr. Omar Daoud
Part I -II
Module: Electronics I
Module Number: 610/650221-222
Electronic Devices and Circuit Theory, 9th ed., Boylestad and Nashelsky
Vdc ≅ 0.318(Vm -VK )
PIV ≥ Vm
•
Full-Wave Rectifications:
-
The DC level obtained from a sinusoidal input can be improved 100%
using a process called full-wave rectification.
-
Vdc ≅ 0.636(Vm -VK )
-
PIV ≥ 2Vm
-
The most familiar network for performing such a function is shown below
and known as a Bridge Network.
Lecturer: Dr. Omar Daoud
Part I -II
Module: Electronics I
Module Number: 610/650221-222
Electronic Devices and Circuit Theory, 9th ed., Boylestad and Nashelsky
Lecturer: Dr. Omar Daoud
Part I -II
Module: Electronics I
Module Number: 610/650221-222
Electronic Devices and Circuit Theory, 9th ed., Boylestad and Nashelsky
Ex. 2.16:
a- Sketch Vo and determine the dc level of the output for the network shown
below.
b- Repeat part a if the diode is replaced by a Si one.
c- Repeat both of a and b if the input voltage is increased to be 200V.
Ex. 2.17:
Determine the output waveform for the network shown below; calculate the output
dc level and the required PIV of each diode.
Lecturer: Dr. Omar Daoud
Part I -II
Module: Electronics I
Module Number: 610/650221-222
Electronic Devices and Circuit Theory, 9th ed., Boylestad and Nashelsky
-
Clipper Circuit:
Clippers are networks that diodes to "clip" away a portion of an input signal without
distorting the remaining part of the applied waveform.
Lecturer: Dr. Omar Daoud
Part I -II
Module: Electronics I
-
Module Number: 610/650221-222
Electronic Devices and Circuit Theory, 9th ed., Boylestad and Nashelsky
Clamper Circuit:
A clamper is a network constructed of a diode, a resistor and a capacitor that shifts a waveform to a different dc level without changing the
appearance of the
applied signal.
Lecturer: Dr. Omar Daoud
Part I -II
Module: Electronics I
Module Number: 610/650221-222
Electronic Devices and Circuit Theory, 9th ed., Boylestad and Nashelsky
-
Zener Diode Circuit:
•
The analysis of the Zener diode is quite similar to the semiconductor diodes.
•
The use of Zener Diode as a regulator
must follow one of the following
considerations:
1) V i and R Fixed:
R
-
RU
Determine the state of the
Zener diode by removing it
from
the
network
and
calculating the voltage across the remaining open circuit.
V = VL =
Vi R L
RL + R
 if V ≥ V z , the zener diode is ON
⇒
if V < V z , the zener diode is OFF
-
Substitute the equivalent circuit and solve for the desired unknowns.
Lecturer: Dr. Omar Daoud
Part I -II
Module: Electronics I
Module Number: 610/650221-222
Electronic Devices and Circuit Theory, 9th ed., Boylestad and Nashelsky
-
The ON state is shown in
the figure. Here,
VL = Vz
I R= I z + I L
⇒ Iz =
VR VL
−
R RL
V R Vi − V L
=
R
R
Pz = I z V z
But,
Ex.2.26: For the Zener diode network shown, determine V L ,V R , I z and P z .
R
R
R
R
R
R
R
R
Repeat you solution with R L =3kΩ.
R
R
2) Fixed V i and variable R L :
U
-
UR
RU
UR
RU
Due to the offset voltage range V z , there is a specific range of a resistor
R
R
values that will ensure the Zener is in the ON state.
-
Determining the minimum load resistance value that will turn the Zener
diode to the ON state.
VL = Vz =
Vi R L
RL + R
⇒ R L min =
Vz R
Vi − V z
⇒ I L max =
Lecturer: Dr. Omar Daoud
VL
Vz
=
R L R Lmin
Part I -II
Module: Electronics I
Module Number: 610/650221-222
Electronic Devices and Circuit Theory, 9th ed., Boylestad and Nashelsky
-
Once the diode is in the ON state, the voltage across R remains constant
and then the current through it.
V R = Vi − V z
VR
R
Iz = IR − IL
IR =
-
Since I z is limited to I zM as provided in the datasheet, the
R
R
R
R
I Lmin = I R − I zM
Vz
R L max =
I Lmin
Ex.2.27: For the shown Zener
diode network, determine the
range of R L and I L that will
R
R
R
R
result in V RL being maintained
R
R
at 10V. Then determine the
maximum wattage rating of the
diode.
3) Variable V i and fixed R L :
U
-
UR
RU
UR
RU
For fixed values of R L , the voltage V i must be sufficiently large to turn the
R
R
R
R
Zener diode ON.
VL = Vz =
Vi R L
RL + R
⇒ Vi min =
( R L + R)V z
RL
⇒ I R max = I zM + I L
Vi max = I R max R + V z
Ex.2.28: Determine the range of
values of V i that will maintain the
R
R
Zener diode in the ON state.
Lecturer: Dr. Omar Daoud
Part I -II