halfe wave rectifier and full wave recifiers

half

1. Erbil Polytechnic University
Erbil Technology College
AITE Department
2
nd
semester
2023-2024
Electronics
Lecture#3
Diode Circuits
1
By Dr. Shelan M Mustafa
Dep of AITE : Electronics

2. By Dr. Shelan M Mustafa
2
After studying this chapter, you should be able to:
1- Draw a diagram of a half-wave rectifier and explain how it
works.
2- Describe the role of the input transformer in power
supplies.
3- Draw a diagram of a full-wave rectifier and explain how it
works.
4- Draw a diagram of a bridge rectifier and explain how it
works.
Learning outcomes
Dep of AITE : Electronics

3. Topics
Covered in
Lecture 3
By Dr. Shelan M Mustafa Dep of AITE : Electronics 3
Half-Wave
Rectifier
Transformer
Full-Wave
Rectifier
Bridge
Rectifier

4. By Dr. Shelan M Mustafa Dep of AITE : Electronics 4

5. By Dr. Shelan M Mustafa Dep of AITE : Electronics 5

6. By Dr. Shelan M Mustafa Dep of AITE : Electronics 6

7. Half-Wave Rectifier
By Dr. Shelan M Mustafa Dep of AITE : Electronics 7
•�
��=
??????
�(���)
??????
•�
�(���)=�
�(??????�)−0.7�
•�
���=�
??????�
•??????
��(�??????���)=??????
��
•The dc value of the output is the average value.
•Second approximation:
•??????
��=
??????
��
??????

8. Transformers
Dep of AITE : Electronics
By Dr. Shelan M Mustafa 8
Transformers are described
as a device that transfers
energy from one circuit to
another through mutual
induction.
The amount of energy
transferred is based
primarily on the turns ratio of
the two coils.

9. Schematic
Symbols
Dep of AITE : Electronics By Dr. Shelan M Mustafa 9

10. Turns
Ratio
•The ratio of the number of turns
on the “Primary” winding versus
the number of turns on the
“Secondary” winding determines
the output of the transformer.
By Dr. Shelan M Mustafa Dep of AITE : Electronics 10SPP
S S P
INV
NVI
==

11. Turns
Ratio -
Voltage
Dep of AITE : Electronics By Dr. Shelan M Mustafa 1115120
8
1
PP
S
S S S
NV V
VV
NV V
= = =

12. Turns
Ratio -
Current
By Dr. Shelan M Mustafa Dep of AITE : Electronics 1215
15
11
SSP
S
SP
IIN
IA
NI A
= ==

13. Turns Ratio – Voltage
and Current
By Dr. Shelan M Mustafa Dep of AITE : Electronics 13SPP
S S P
INV
N V I
==

14. Center-Tap Transformer
By Dr. Shelan M Mustafa Dep of AITE : Electronics 1410120
12
1
PP
S
S S S
NV V
VV
NV V
= = =

15. Full-
Wave
Rectifier
By Dr. Shelan M Mustafa Dep of AITE : Electronics 15
The full-wave rectifier circuit
requires a center-tap
transformer.
The full-wave rectifier circuit
uses two diodes.
The peak voltage across the
load is half of the secondary
peak voltage due to the center-
tap transformer.

16. Full-Wave Rectifier
Channel 1= V
1
Channel 2 = V
RL
16
By Dr. Shelan M Mustafa
Dep of AITE : Electronics

17. Full-Wave Rectifier
V
in V
out
17By Dr. Shelan M Mustafa Dep of AITE : Electronics

18. Full-
Wave
Rectifier
•The dc value of the output is the
average value.
•V
dc = 2V
P(out)/
•The input to each diode is half of the
secondary voltage.
•Second approximation:
V
P(out) = V
P(in) - 0.7 V
•f
out = 2f
in
•I
dc(diode)=I
dc/2
By Dr. Shelan M Mustafa Dep of AITE : Electronics 18

19. Bridge
Rectifier
•The bridge rectifier circuit does not
require a center-tap transformer.
•The bridge rectifier circuit uses four
diodes.
•Two of the four diodes are conducting
at any given time, therefore the load
voltage is 1.4V less than the peak
voltage across the secondary.
By Dr. Shelan M Mustafa Dep of AITE : Electronics 19

20. By Dr. Shelan M Mustafa Dep of AITE : Electronics 20

21. 21
By Dr. Shelan M Mustafa
Dep of AITE : Electronics

22. Bridge
Rectifier
•The dc value of the output is the
average value.
•V
dc = 2V
P(out)/
•Second approximation:
V
P(out) = V
P(in) - 1.4 V
•f
out = 2f
in
•I
dc(diode)=I
dc/2
By Dr. Shelan M Mustafa Dep of AITE : Electronics 22

23. Example#1
For the circuit shown below
calculate
(a) Average output voltage
(b) Average load current
(c) Average diode current
(d) Peak load current
(e) Peak diode current
(f) Power dissipation in diode
By Dr. Shelan M Mustafa Dep of AITE : Electronics 23

24. Solution#1
�
�(??????�)=2�
���=1.4142×50�=70.71�
�
�(���)=�
�(??????�)−0.7�=70.7�−0.7�=70�
By Dr. Shelan M Mustafa Dep of AITE : Electronics 24
�)�
��=
�
�(���)
??????
=
70�
3.14
=22.50�
�)??????
��=
�
��
??????
??????
=
22.5�
4.7??????Ω
=4.78????????????
�) ??????
��(�??????���)=??????
��=4.78????????????
�)??????
�=
�
�(���)
??????
??????
=
70�
4.7??????Ω
=14.89????????????
�) ??????
??????=??????
�(�??????���)�
??????=14.89????????????×0.7�=10.42??????�
�) ??????
�(�??????���)=??????
�=14.89????????????

25. Example#2
25
From figure bellow:
a. Find the peak and average voltage across each half of the secondary
b. Sketch the voltage waveform across RL.
c. Determine the peak and average current through each diode.
d. Determine the PIV for each diode.
By Dr. Shelan M Mustafa
Dep of AITE : Electronics

26. Solution#2
�
�(??????�)=2�
���=1.4142×220�=311.11�
�
�??????�(���)=
1
6
�
�(??????�)=51.85�
peak voltage across each diode �
�??????����?????? =
??????�??????�(���)
2
=25.92�
By Dr. Shelan M Mustafa Dep of AITE : Electronics 26
�
��=
2�
�(���)
??????
=
50.45�
3.14
=16.05�
??????
�=
�
�(���)
??????
??????
=
25.22�
1??????Ω
=25.22????????????
????????????�=2�
�(���)+�
??????=51.14�
�
�(���)=�
�??????����??????−0.7=25.92�−0.7�=25.22�
??????
��=
�
��
??????
??????
=
16.05�
1??????Ω
=16.05????????????
??????
�(�??????���)=
??????
�
2
=12.61????????????
??????
��(�??????���)=
??????
��
2
=8.02????????????

27. Example#3
•A full-wave bridge rectifier with a 220 V rms sinusoidal input has a
load resistor of 10 kΩ. Assuming silicon diodes, calculate
•(a) dc output voltage and dc load current
•(b) peak output voltage and peak load current
•(c) peak and average diode currents
•(d) power rating of each diode
By Dr. Shelan M Mustafa Dep of AITE : Electronics 27

28. Solution#3
�
�(??????�)=2�
���=1.4142×220�=311.1�
�
�(���)=�
�(??????�)−1.4�=311.1�−1.4�=309.72�
By Dr. Shelan M Mustafa Dep of AITE : Electronics 28
�) �
��=
2�
�(���)
??????
=
2×309.72�
3.14
=197.17�
�) ??????
��=
�
��
??????
??????
=
197.17�
10??????Ω
=19.72????????????
�) ??????
��(�??????���)=
??????
��
2
=9.85????????????
�) ??????
�=
�
�(���)
??????
??????
=
309.72�
10??????Ω
=30.09????????????
�) ??????
??????=??????
�(�??????���)�
??????=15.05????????????×0.7�=10.53??????�
�) ??????
�(�??????���)=
??????
�
2
=15.04????????????