Basic electronics MODULE_2_Diode App.pptx

Department of Electronics and Communication Engineering, MIT, Manipal
Module – 2 :
Applications of Diodes
Chapter 1: Diodes and Applications
1
Reference:
Robert L. Boylestad, Louis Nashelsky, Electronic Devices &
Circuit Theory, 11th Edition, PHI, 2012
Part – I : Analog Electronics

Application of Diodes
At the end of this module, students will be able to:
 Explain need for AC to DC conversion
 Discuss basic DC power supply unit.
 Discuss and analyze the working of a various rectifier circuits.
 Explain how capacitor filter can be used to minimize the ac
component.
2

CONTENT
3
 Introduction
 Half wave rectifier (HWR)
 Full wave rectifiers:
1) Center tapped FWR
2) Bridge Rectifier
 Capacitor filter

INTRODUCTION
4
 What is an AC and an DC signal?
 Eg. of AC signal
Define
 Average value
 RMS or effective value
Electricity Distribution in INDIA: AC signal of 230V, 50HZ.
Necessity of DC power: Many electronic gadgets
)
2
(
sin
)
( ft
A
t
Vin 

)
50
2
(
sin
2
230
)
( t
t
Vin 

Fig. 1: AC signal with A=230V, f=50Hz
Note: The average or DC value of this signal is equal to zero.

INTRODUCTION
5
DC power supply
Fig. 2 : Block Diagram of Basic DC power supply

Activity
1. List in a table the names of at least six
products/applications that we use in daily life that
require dc power supply along with the range of
values.
INTRODUCTION
2. List the appliances or products around us that need
power supply. Classify them under the umbrella of dc
or ac power supply that is used for its working.
7

CONTENT
8
 Introduction: Basic DC power supply
 Full wave rectifiers:
1) Center tapped FWR
2) Bridge Rectifier

• Diode passes only for half of the signal time
period Hence the name HWR.
HALF WAVE RECTIFIER (HWR)
Fig 7: Circuit of HWR
10

Working HWR
Fig 8: Equivalent Circuit of HWR, when node A is positive w.r.t node B
Fig 9: Equivalent Circuit of HWR, when node A is negative w.r.t node B
Note: Current through load exist only for one half cycle
11

HALF WAVE RECTIFIER
Fig 10: Input and rectified output with ideal diode
Fig 11: Secondary input and rectified output with practical diode
Simulation of HWR
12

RECTIFIER
Performance of Rectifiers is measured using the following
parameters:
• DC voltage
• Peak Inverse Voltage (PIV)
• Ripple factor
• Efficiency
1
2
2


dc
rms
V
V
















L
R
rms
V
L
R
dc
V
2
2

13
dc
V

HALF WAVE RECTIFIER
Assume ideal diodes
 During positive cycle i = Im sin(ωt)
Peak current
 During negative half cycle, i = 0
Average value of load current in half wave rectifier is
non zero
L
R
m
V
F
R
L
R
V
m
V
m
I 








m
I
t
d
i
dc
I 


2
0
)
(
2
1
0
,
0 
 F
R
V
14

Half Wave Rectifier
 Average output voltage is
 RMS value of load current in half wave rectifier is:
 RMS output voltage is
L
dc
dc R
I
V 
2
2
1
2
0
)
(
2
2
1 m
I
t
d
i
rms
I 













L
rms
rms R
I
V 
15

 PIV : should be greater than Vm , peak of secondary
voltage.
 Ripple factor is:
 Efficiency:
Half Wave Rectifier
21
.
1
1
2
2 














m
V
m
V
%
6
.
40
2
4
2
2



















L
R
rms
V
L
R
dc
V
16

ACTIVITY: Do it yourself
2. what happens when the diode connection is
reversed? Draw the input and output waveform. Will
the values of PIV, ripple factor and efficiency for this
changed circuit change?
Self Test
1. HWR is used to rectify the AC signal which has peak
value of 25V. Which all diodes can be selected whose
PIV rating is
(a) 5V (b) 15V (c) 30V (d) both a and b
HWR
17

Advantages of HWR
• Simple circuit
• Single diode
• PIV rating is Vm
Disadvantages of HWR
• High ripple factor
• Low efficiency
Half Wave Rectifier
18

Department of Electronics
and Communication
Engineering,
Manipal Institute of
Half wave rectifier
Problem
1. A voltage v = 100 Sin ωt and frequency 50 Hz is applied to
half wave rectifier. If the load resistance is 2KΩ ,
calculate:
a) Peak value of load current
b) Average load current
c) Rms or effective load current
d) Average output voltage, Rms output voltage
e) AC input power, DC output power
f) Efficiency
g) Ripple factor

Department of Electronics
and Communication
Engineering,
Manipal Institute of
Half wave rectifier
2. An AC voltage of 230V, 50Hz is applied to transformer
having turns ratio 10:1. The secondary of transformer is
connected to half wave rectifier. The diode has cut-in
voltage 0.6V and forward resistance 10Ω. Determine
average and rms values of output current and voltage. What
should be the PIV rating of the diode? Let the load
resistance be 1.5KΩ.

CONTENT
 Introduction: Basic DC power supply
 Center tapped Full Wave rectifier (FWR)
21

Center Tapped FWR
Fig.12 center tapped FWR
Fig. 13: Secondary waveforms
22

Working of center tapped FWR
Fig. 14: Center tapped FWR
for node A is positive w.r.t B
Fig. 15: Center tapped FWR for
node B is positive w.r.t. A
Note: Current through load during both cycles is in same
direction (from node C to ground)
23

Center Tapped FWR
Fig. 16 : Input secondary and output waveforms
Note: The frequency of the output signal =2 times the input frequency
24

Center tapped FWR





m
V
t
d
t
m
V
dc
V
av
V
2
)
(
)
(
sin
0
1











m
L
m
L
dc
dc
I
R
V
R
V
I
2
2



2
)
(
)
sin
(
1 2
0
m
m
rms
V
t
d
t
V
V 
  



2
m
rms
I
I 
• The Average of output voltage
• RMS value of the voltage at the load is
• The Average of output current
25

 PIV : 2Vm , where Vm peak of secondary voltage
(between node A and ground or between node
B and ground)
 Efficiency:
Center tapped FWR
483
.
0
1
2
2
2 














m
V
m
V
%
2
.
81
2
8
2
2



















L
R
rms
V
L
R
dc
V
26

Center Tapped FWR
Self Test
Choose the correct answer: (T is the time period of the input signal)
1. In HWR, the diode is forward biased for what duration of the time
period?
(a) T/2 b) T/4 c) 3T/4 d) T
2. In center tapped FWR, each diode is forward biased for what duration
of the time period?
(a) T/2 b) T/4 c) 3T/4 d) T
3. In a center tapped FWR , current through load resistor flows for what
duration of the time period?
(a) T/2 b) T/4 c) 3T/4 d) T
4. The ripple factor of FWR is greater than HWR
(a) True (b) False
27

Department of Electronics and Communication Engineering, MIT, Manipal 28
Advantages of center tapped FWR over HWR
• High Efficiency
• low ripple factor
Disadvantages of center tapped FWR over HWR
• Uses 2 diodes
• Uses center tapped transformer
Comparison of HWR and FWR

Solved Exercise
1. A center-tapped FWR is supplied with 230V, 50 Hz AC mains
through a step down transformer with turns ratio equal to 10.
Find the average and RMS value of the load current, rating of
the diode used for proper working. PIV .
Given: Input AC mains RMS voltage =230V, turns ratio=10,
Hence Secondary RMS voltage =230/10=23V
Solution: ….
29

Full wave Rectifier
Problem:
 A sinusoidal voltage with 20-0-20 V applied to secondary of
the transformer used for full wave rectification. If the load
resistance is 1000Ω, calculate :
• Peak value of load current
• Average load current
• Rms load current
• Average output voltage
• Rms output voltage
• Efficiency
• Ripple factor
0.02A, 0.0127A, 0.0141A, 14.14v, 81.1%, 0.482

Bridge rectifier
Fig.17(a) : Bridge FWR
31

Bridge rectifier
Fig.17 (b): Bridge FWR
32

Working of Bridge FWR
Fig. 18: Bridge FWR when
node A is positive w.r.t B
Fig. 19: Bridge FWR when node
B is positive w.r.t. A
Note: Current through load for both cycles is in same direction
(from node C to ground)
33

Bridge FWR
Fig. 20 : Input and output waveforms of bridge rectifier
Simulation of FWR
Note: The frequency of the output signal =2 times the input frequency
34

 PIV : Vm , where Vm peak of secondary voltage
(between node A and node B).
 Other parameters same as Center tapped FWR:
 Efficiency:
Bridge FWR
483
.
0


%
2
.
81


35

• Advantages of HWR over FWR
• Advantages of Center tapped FWR rectifier over HWR
• Advantages of bridge rectifier over to centre-tap FWR
Comparison of Rectifiers
• Disadvantages of HWR over FWR
• Disadvantages of centre-tap FWR over Bridge
• Disadvantages of bridge rectifier over other rectifiers
36

Parameters of
rectified signal
HWR Center-tapped
FWR
Bridge FWR
Vdc
VRMS
PIV
Ripple factor
Efficiency
Frequency fo

m
V

m
V
2

m
V
2
2
m
V
2
m
V
2
m
V
37
m
V m
V
2 m
V
1.21 0.483 0.483
40.6% 81.2% 81.2%
i
f i
f
2 i
f
2

Capacitor Filter
• Commonly referred as C type filter
• Key component of filter is the energy storing
elements. Example: Capacitor
• Capacitor helps to hold the output voltage to its
maximum or peak value.
• It can be used with HWR as well as with FWR
38

Capacitor Filter
Fig 21: C type filter with HWR
39

Capacitor Filter
Fig 22: C type filter with Bridge FWR
Simulation of HWR/FWR with C filter: with varying R and C
40

Capacitor Filter
Fig. 23 Filtered output waveform using C type filter
41

Ripple factor with Capacitor Filter
• For HWR
• For FWR
L
fCR
r
3
4
1

L
fCR
r
3
2
1

42

Ripple factor with Capacitor Filter
• DC value of filtered output for HWR
Note: here f is the frequency of the input signal
m
L
L
dc V
CR
f
CR
f
V
2
1
2


• DC value of filtered output for FWR
m
V
L
CR
f
L
CR
f
dc
V
4
1
4


43

Parameters of
rectified signal
HWR FWR
Vdc
Ripple factor
m
V
L
CR
f
L
CR
f
2
1
2
 m
V
L
CR
f
L
CR
f
4
1
4

L
fCR
r
3
2
1

L
fCR
r
3
4
1

44

Summary
At the end of this module, students will be able to:
• Discuss block diagram of a basic DC power supply unit.
• Explain and analyze the working of various rectifier circuits.
• Evaluate Output DC value, ripple factor, efficiency and PIV, of
different rectifier circuits.
• Explain the working of rectifier circuits with capacitor filter

Exercise Problems
1. Primary voltage is 120V, 60Hz. Turns ratio is 5:1. This
transformer supplies to bridge rectifier employing 4 identical
ideal diodes. The load resistance is 1kΩ. Calculate average
and rms load voltage, efficiency, ripple factor, PIV rating and
frequency of output waveform.
2. A bridge rectifier consisting of four identical diodes produces
a direct current of 124.49mA across 1000Ω resistive load.
Calculate the primary to secondary turns ratio of the
transformer if primary voltage is 220V.
46

Exercise Problems
3. A half wave rectifier with capacitor filter is supplied from
transformer having peak secondary voltage 20V and freq 50Hz.
The load resistance is 560Ω and capacitor used is 1000μF.
Calculate ripple factor and dc output voltage. Draw the filtered
output and label peak and dc value.
(Ans. for part a: 0.0103, 19.65V)
47

Exercise Problems
4. A full wave rectifier with capacitor filter has to supply an
average voltage of 30V to 900Ω load. Calculate the rms input
voltage and value of capacitor such that the ripple factor does
not exceed 0.05, assuming f = 50Hz.
5. (a) A half wave rectifier is fed from a transformer having
turns ratio 6:1.The primary voltage is 110V at 60Hz. It is decided
to have ripple factor of 0.03 and dc load current of 500mA. Find
the value of capacitor needed.
(b) Repeat for full wave bridge rectifier
(Ans for part a: 3.25 mF)
6. A load is to be supplied 10mA current at 5V dc, with ripple
not more than 0.2%. Calculate the value of capacitor needed
for the full wave bridge rectifier. Also, if the primary
voltage of transformer is 220V at 50Hz, calculate the turns
ratio needed. (Ans: 2.89mF, 62)
48

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