ECE 1001 : BASIC

ELECTRONICS
Department Department of of Electronics Electronics and and Communication
Communication Engineering, Engineering, MIT, MIT, Manipal Manipal
2
1
Part – I :
Analog
Electronics
Reference:

Robert L.
Boylestad, Louis
CHAPTER-1: Nashelsky,
Electronic Devices
DIODES AND
& ​Circuit Theory,
APPLICATONS
11​th ​Edition, PHI,
2012
 differentiate between ideal
and ​practical diodes

Department of Electronics and ▪ ​Explain the concept of

Communication Engineering, MIT,
Manipal
static and dynamic
1 resistance of the diode.

Module – 1 : ▪ ​Explain various

Diodes breakdown phenomenon
observed in diodes.

Learning ▪ ​Describe the working of

Zener diode and its I-V
outcomes characteristic.

At the end of this ▪ ​Explain the operation of

module, students diode as capacitor.
will be able to:

▪ ​Explain the operation of

PN junction diode under
Department of Electronics and
different biasing ​condition. Communication Engineering, MIT,
Manipal
▪ ​Draw the I-V 3
characteristic of diode and
▪ Basic of
Semiconduct
ors
Department Department of of
Electronics Electronics and and

▪ Doping in Communication Communication

Engineering, Engineering, MIT, MIT,
Manipal Manipal

Semiconduct 2

Review
ors
Semiconductors
​
Common semiconducting materials Crystal structure

of silicon
http://fourier.eng.hmc.edu/e84/le ​ctures/ch4/node1.html
http://www.austincc.edu/HongXiao/overvie

w/basic-semi/sld007.htm
Department of Electronics and Communication Engineering, MIT, Manipal

5
Doping in Semiconductors

Schematic of a silicon crystal lattice doped with

impurities to produce n-type and p-type
semiconductor material.
[http://www.pveducation.org/pvcdrom/pn-junction/dop
ingl].
Department of Electronics and Communication Engineering, MIT, Manipal

6
Self test

1.Why silicon is preferred over germanium

for
semiconductor devices?

2.List different elemental and compound

semiconductors.
Department of Electronics and Communication Engineering, MIT, Manipal

7
P-N Junction Diode

Anode Cathode ​P N
Common practical diodes available in market

Department of Electronics and Communication Engineering, MIT, Manipal

8
P-N Junction Diode- conti...

Used in numerous applications

• Switch,
• Rectifier,
• Regulator,
• Voltage multiplier,
• Clipping,
• Clamping, etc.

Department of Electronics and Communication Engineering, MIT, Manipal

9
P-N Junction Diode under
biasing
P-N junction (a) in contact (b) formation of
depletion region

[http://www.imagesco.com/articles/photovoltaic/photo
voltaic-pg3.html].

Department of Electronics and Communication Engineering, MIT, Manipal

10
P-N Junction Diode under biasing
condition

Unbias condition

Diode under zero bias conditions

Department of Electronics and Communication Engineering, MIT, Manipal

11
Forward bias
▪ Positive of battery connected to p-type
(anode)
▪ Negative of battery connected to n-type
(cathode)

Department of Electronics and Communication Engineering, MIT, Manipal

Diode under forward biasing conditions

12
Reverse bias

▪ Positive of battery connected to n-type

material (cathode)
▪ Negative of battery connected to p-type
material (anode)
Department of Electronics and Communication Engineering, MIT, Manipal

Diode under reverse biasing conditions

13
Self test

1. The arrow direction in the diode symbol indicates

a. Direction of electron flow. b. Direction of hole flow
(Direction of conventional current) c. Opposite to the
direction of hole flow d. None of the above

2. When the diode is forward biased, it is equivalent

to
a. An off switch b. An On switch c. A high resistance
d. None of the above

Department of Electronics and Communication Engineering, MIT, Manipal

14
I-V characteristic of practical
diode
Department of Electronics and Communication Engineering, MIT, Manipal
15
Diode (mA)
symbol

PN
V​γ is
​ 0.6 ~ 0.7 Vfor Si
0.2 ~ 0.3 V for Ge

(μA)

I-V characteristic of Practical diode

I-V characteristic of
silicon and germanium
practical diode
http://www.technologyu
k.net/physics/electrical_ Department of Electronics and
Communication Engineering, MIT,
principles/the_diode.sht Manipal
ml 16
Silicon vs.
Germanium
Diode current equation
▪ ​ID​ is
​ diode current
▪ ​Io​ is
​ reverse saturation current
▪ ​VD​ is
​ voltage across diode
▪ ​VT​ is
​ thermal voltage = T / 11600
▪ ​η i​ s a constant = 1 for Ge and 2 for Si

eII ​
D​ = ​o (​ VV
​
​ ​T ​
Dη
eI ​
- )1 ​= ​ o VV

Dη
​

I​
-​
T​
o

▪ For positive values of V​D (forward

​ bias),
eII ​
D​ ≈
VV ​ η ​
o​ TD ​ ▪ For large negative values of V​D (reverse
​

​ ≈
bias), ​ID ​
​ –Io
Department of Electronics and Communication Engineering, MIT, Manipal
17
Effect of Temperature on the
Reverse current
Reverse current ​doubles ​for every ​10 degree rise
in temperature.

II ​
​
o2

​ ​
o1
2 (​
-​
TT ​12 ​ 10/) ​Q. A Silicon diode has a saturation current
of 1pA at 20​0​C. Determine (a) Diode bias voltage when
diode current is 3mA (b) Diode bias current when the
temperature changes to 100​0​C, for the same bias
voltage.
A.
eII ​ ⎛ │​
D ​= ​0 ​ │⎝​18
​ ​η ​
VD ⎞
VT​ ​- 1​
│​ =​ T ​= ​ =​ mV ​ ⎛
│⎠​VT​ ​ 11600 ​ 293
11600 ​ ​ 25.25 ​ =​ η
VV D​ ​ T ​​ 1ln ​
│​ ⎞ │​ = ​
│⎝+ ​I ​ID​ 0​ ​ │⎠​ 103.1 ​V 2
​ II ​02 =
​

01 )/10T(T
​ ​12 ​-

⎛ │​
= 10 ​- 12 ​ │⎝​)10100(
10
- ​2 ​

⎞ │​│⎠= 256
​ ​pA I​ ​D
⎛
= 10256 ​x ​- ​12 ​
││​
│⎝​e ​1015.322( ​x ​103.1
x -​

- )1 ⎞
​ │​
3​ │​│​⎠​= 21.7
​ ​mA ​Department of Electronics and Communication
Engineering, MIT, Manipal

Effect of Temperature on the

Reverse current

I (mA)

–75​o​C
125​o​C
o​
25​ C V (volts) r d​ ​=
∆ ​∆​V I​ D​ ​D ​r ​d =
​

∆ ​∆​V I​ D​ ​D ​r ​d =
​

I (μA)
∆ ​∆​V I​ D​ ​D ​r ​d =
​

V​
∆ ​∆​ ID​ D​
Department of Electronics and
Communication Engineering, MIT, V​
Manipal ∆ ​∆​ ID​ D​
19 V​
▪ ​Static or DC ∆ ​∆​ ID​ D​
resistance​: V​ D
∆ ​∆​ ID​ ​
• ratio of diode voltage
η​
and diode current ≈ ​ I​V D​ ​T
η​
​ ​R ​ =
IV
D​
D​ D ​ ≈ ​ I​V D​ ​T
AC resistance: η​
≈ ​ I​V D​ ​T
 η​
≈ ​ I​V D​ ​T
η​
≈ ​ I​V D​ ​T

Diode
resistances

Department of Electronics and

Communication Engineering, MIT,
Manipal

20
Diode Equivalent Circuit
▪ Used during circuit analysis
▪ Characteristic curve replaced by
straight-line segments

AK
A K ​Reverse bias

1/R​F Department of Electronics and

AK Communication Engineering, MIT,
Manipal
Forward bias
21
Forward bias
Ideal diode :
R​R =
​ ∞
R​ I-V
Vγ ​ F
characteristics
Vγ
 Ideal diode and ideal
models
[http://conceptselectroni
cs.com/diodes/diode-eq
uivalent-models/].

Department of Electronics and

Communication Engineering, MIT,
Manipal

22

I-V characteristic of
Diode Equivalent Circuit
▪ As further approximation, we can neglect
the slope of the characteristic i.e., R​F =
​ 0
AK
R​F =
​ 0
AK
Forward bias R​R =
​ ∞
Vγ

Vγ
A K Reverse bias ​Department of Electronics and Communication
Engineering, MIT, Manipal

23
Diode Equivalent Circuit

▪ As third approximation, even the cut-in

voltage can be neglected (Ideal diode)

AK

R​F =
​ 0
A K Reverse bias
AK
Forward bias

R​R =
​ ∞
Department of Electronics and
Communication Engineering, MIT,
Vγ = 0 Manipal
24
Self test

1. The
break-point
Department of Electronics and
voltage of Si Communication Engineering, MIT,
Manipal

diode is 25

a. 0.2V b. 0.7V Breakdown

c. 0.8V d. 1.0V phenomenon
in diodes
2. Why would
you use silicon
diodes instead of Two breakdown
mechanisms:
germanium
diodes? • ​Avalanche
breakdown : 26

• Occurs in Lightly Avalanche

doped diodes, Breakdown
• Occurs at high
reverse Voltage.

• ​Zener
Breakdown:
• Occurs in heavily
doped diodes.
• at lower reverse
bias voltages.

Schematic of
Avalanche
phenomenon
Department of Electronics and
Communication Engineering, MIT, http://shrdocs.com/pres
Manipal
entations/12656/index.ht
ml

Department of Electronics and

Communication Engineering, MIT,
Manipal

27
Zener
Breakdown

Schematic of
Zener
phenomenon
http://shrdocs.com/pres
entations/12656/index.ht
ml

Department of Electronics and

Communication Engineering, MIT,
Manipal

28 I​ZM or I​ZMax
​
Zener Diode
P​ZM or P​ZMax
and its ​

characteristic P​ZM =
​ V​Z​.I​ZM

s I-V characteristics of
Zener diode

P ​P ​N ​N
Department of Electronics and
Communication Engineering, MIT,
Anode Cathode Manipal

29

I​ZK ​or I​Zmin

Equivalent circuit
▪ Equivalent circuits of Zener diode
N​ N
–​N N ​+ ​
Vγ
V​Z ​+ –​R​F R​ ​ ​∞ ​R​ZP
​ R≈ ​

P P P ​Forward Reverse Breakdown

▪ Note: ​RZ​ ​is usually very small, can be
neglected
Department of Electronics and Communication Engineering, MIT, Manipal

30
Diode as capacitor- Varactor
diode
Department of Electronics and Communication Engineering, MIT, Manipal

31

Ax ​ ε ​
d​ C ​ =
Self test

1. Explain the principle of PIN diode.

2.What is the difference between PN diode and

Schottky diode.

3.Which type of diode exhibits negative resistance

and why?

4. Which of the following is not an essential element

of a dc power supply

​ . Filter c. Voltage regulator d. Voltage

a. Rectifier b
amplifier

Department of Electronics and Communication Engineering, MIT, Manipal

32
Self test

5. What is true about the breakdown voltage in a

Zener diode?
a. It decreases when current increases. ​b. It
destroys the diode. c. It equals the current times the
resistance. d. It is approximately constant

6. Which of these is the best description of a Zener

diode?
a. It is a rectifier diode. b. It is a constant voltage
device. c. It is a constant current device. ​d. It works
in the forward region.

Department of Electronics and Communication Engineering, MIT, Manipal

33
Exercises
1. Calculate the dynamic forward and reverse
resistance of a P - N junction diode, when the
applied voltage is 0.25V for Germanium Diode. I0 =
lμA and T = 300 K.
(Ans:rf=1.734 Ω; rr=390 MΩ)

2. A germanium diode has reverse saturation current

of 0.19μA. Assuming η =1, find the current in the
​
diode when it is forward biased with 0.3 V at 27​oC.
(Ans: 19.5mA)

3. The forward current in a Si diode is 15 mA at

27o​ ​C. If reverse saturation current is 0.24nA, what is
the forward bias voltage?
(Ans: 0.93V)

4. A germanium diode carries a current of 10mA

​
when it is forward biased with 0.2V at 27​oC. (a) Find
reverse sat current. (b) Find the bias voltage
required to get a current of 100mA.
(Ans: 4.42μA, 0.259V)

Department of Electronics and Communication Engineering, MIT, Manipal

34

Part – I : Analog
Electronics

Chapter 1: Diodes and

Applications

Module – 2 : ​Applications of
Diodes
 Reference​: ​Robert L. Boylestad,
Louis Nashelsky, Electronic Devices
& ​Circuit Theory, 11​th ​Edition, PHI,
2012

▪ Explain need for

Department Department of of
Electronics Electronics and and AC to DC
Communication Communication
Engineering, Engineering, MIT, MIT, conversion
Manipal Manipal

35
Application of ▪ Discuss basic DC
Diodes power supply unit.

At the end of this

▪ Discuss and
module, students
analyze the working
will be able to:
of a various rectifier
circuits. (HWR)

▪ Explain how ▪ Full wave

capacitor filter can
be used to minimize
rectifiers:
the ac component. 1) Center tapped
FWR 2) Bridge
Department of Electronics and Rectifier
Communication Engineering, MIT,
Manipal

36
CONTENT ▪ Capacitor
filter
▪ ​Introduction

▪ Half wave Department of Electronics and

Communication Engineering, MIT,

rectifier
Manipal

37
INTRODUCTION
▪ What is an AC and an DC signal?
▪ Eg. of AC signal
AtVin ​)( =

)2(sin ​π ​ft ​
tVin ​)( = )502(sin2230 ​π ​t
Define
▪ Average value
▪ RMS or effective value
​ 30V​, f=5
Fig. 1: AC signal with ​A=2 ​ 0Hz
Note: The ​average ​or ​DC value ​of this
signal is equal to zero.
Electricity Distribution in INDIA: AC signal
of 230V, 50HZ. Necessity of DC power:
Many electronic gadgets​Department of Electronics and
Communication Engineering, MIT, Manipal

38
INTRODUCTION
DC power supply

Fig. 2 : Block Diagram of Basic DC

power supply

Department of Electronics and Communication Engineering, MIT, Manipal

39
INTRODUCTION
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.

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.
Communication Engineering, MIT,
Manipal

41
CONTENT
Department of Electronics and
▪ Introduction: ▪ Capacitor
Basic DC filter
power supply

▪ ​Half wave Department of Electronics and

Communication Engineering, MIT,

rectifier Manipal

42
(HWR) HALF WAVE
RECTIFIER
▪ Full wave (HWR)
rectifiers:
1) Center tapped
FWR 2) Bridge
Rectifier
 44
Department of Electronics and
Communication Engineering, MIT,
Manipal

Working HWR

Fig 7: Circuit of
HWR

Fig 8: Equivalent Circuit

• Diode passes of HWR, when node A
is positive w.r.t node B
only for half of
the signal time
period Hence the
name HWR.
 Note: Current
through load exist
only for one half
cycle
45
Fig 9: Equivalent Circuit Department of Electronics and
Communication Engineering, MIT,
of HWR, when node A Manipal
is negative w.r.t node B

Fig 10: Input and rectified output with ideal

diode
Simulation of HWR

Fig 11: Secondary input and rectified output

with practical diode
Department of Electronics and
Communication Engineering, MIT,
HALF WAVE
Manipal

46 RECTIFIER
RECTIFIER
Performance of Rectifiers is measured
using the following
parameters:
• ​DC voltage
• ​Peak Inverse Voltage (​ ​PIV)​
• ​Ripple factor
• ​Efficiency
V​
dc γ​ ​=
V​
rms

Vd​ c ​2​- 1 ​=
⎛ ││││​
│⎝​η
V rms R ​ Vdc 2
​ ​R ​ ⎞ ││││​
L​ L​2​ │⎠​Department of Electronics and
Communication Engineering, MIT, Manipal

47
HALF WAVE RECTIFIER
Assume ideal diodes
​ ​sin(ωt)
▪ During positive cycle ​i= Im
Peak current
▪ During negative half cycle, ​i ​= 0
Average value ​of load current in

half wave rectifier is ​non zero

R
V​γ = ​ 0 ​I ​m=
​ ,0 ​R ​F= ​ ​m-​ ​V ​γ
​ V ​
 R​ V Rm​
L​+​ F≈
​ ​ LI​ ​dc ​= ​2​1

π
2​
∫ ​π ​tdi ​)( ​ω 0
​
I​
= ​ mπ​
Department of Electronics and Communication Engineering, MIT, Manipal

48
Half Wave Rectifier
▪ Average output voltage is
▪ ​RMS value ​of load current in half
wave rectifier is:
▪ RMS output voltage is

RIV ​ =​ ⌈
dc ​ Ldc ​I rms
​ ​ ​
=
│​ 1​
│⌊2​ π
π​ 2
2∫ ​ tdi ​

0
)( ​ω ​⌉
1​ I​
│​│⌋​ 2= ​ m​2

RIV ​ =​
rms ​ rms L ​Department of Electronics and
Communication Engineering, MIT, Manipal

49
▪ ​PIV ​: voltage.
should be greater than ​V​m​, peak o
​ f
secondary
▪ Ripple factor is:
 Half Wave Rectifier
▪ Efficiency: ​
2
γ ​=
⎛ ││​
21.11 ​ │⎝2 ​π
⎞ │​ V m​ ​ rms ​ R
│​│​⎠​ V​m​- = η​ ​= ⎛​ ││││​│⎝​2 V 2​
V dc R ​ ⎞ ││││​
L​ L​ │⎠=

​ ​2​= %6.40 ​Department of Electronics and Communication

π4
Engineering, MIT, Manipal

50
HWR

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

ACTIVITY: Do it yourself
this ​changed
2. what happens
circuit
when the ​diode
change?​Department of
connection is Electronics and Communication
Engineering, MIT, Manipal
reversed​? Draw 51
the input and Half Wave
output Rectifier
waveform. Will
the values of
Advantages of
PIV, ​ripple factor
and efficiency for HWR
• Simple circuit Department of Electronics and
Communication Engineering, MIT,
Manipal
• Single diode 52
CONTENT
• PIV rating is
V​m
▪ Introduction:
Disadvantages
Basic DC
of HWR power supply

• High ripple
factor ▪ Half wave
rectifier
• Low (HWR)
efficiency
▪ ​Center
tapped Full
Wave
Fig.12 center
rectifier tapped FWR
(FWR)

Fig. 13: Secondary

▪ Capacitor waveforms
filter

54
Department of Electronics and
Department of Electronics and
Communication Engineering, MIT,
Communication Engineering, MIT,
Manipal
Manipal
53
Center Tapped FWR
Working of center tapped
FWR
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

Fig. 14: Center

Note: Current through load during both

cycles is in same ​direction (from node C
to ground)
Communication Engineering, MIT,
Manipal
Department of Electronics and
55 output waveforms

Note: The frequency

of the output signal
=2 times the input
frequency

56
Department of Electronics and
Communication Engineering, MIT,
Manipal

Center
Fig. 16 : Input Tapped FWR
secondary and
Center tapped FWR
• The Average of output voltage
​ ​=
V av

​ ​= π
V dc ​
π​ ω​ ω​ 2 ​V ​
1 ⌈​│⌊∫ 0 ​ ​ ​)()(sin ​
Vm tdt ​ ⌉ ​│⌋= ​ π ​m​• The
 Average of output current
I d​ c =
​

V ​ dc
R​
= ​π
L​

2 ​V

Rm​
2​ I
L​= ​ π ​

• RMS value of the voltage at the

load is
1​
V r​ ms =
​ ​
π

∫​ 0 ​π
)()sin( ​
V ​m ​ω ​tdt ​2
V​
ω ​= ​ m​ 2 ​I r​ ms ​=
I ​m ​
2​Department of Electronics and Communication Engineering, MIT, Manipal
57

Center tapped FWR

▪​
PIV :​ (between 2​V​m ​, ​node ​where
​ peak
A and ​Vm ​ ​ground of
secondary or between voltage
node ​B and ground)
2
▪ Ripple factor is:
γ ​=
2 483.01
▪ Efficiency: ⎛​
││​ 2​
│⎝​ π
⎞ │​ V m​
│​│​⎠​ V​m-​ = ​η ​= ⎛​ ││││​│⎝​2 ​V rms ​2​R L​ ​V
dc R ​ ⎞ ││││​
L​ │⎠=

​ ​2​= %2.81 ​Department of Electronics and Communication

π8
Engineering, MIT, Manipal

58
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

Department of Electronics and Communication Engineering, MIT, Manipal

59
Comparison of HWR and FWR

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

Department of Electronics and Communication Engineering, MIT, Manipal

60
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: ​....
61
Department of Electronics and
Communication Engineering, MIT,
Manipal
 Fig.17(a) : Bridge
Department of Electronics and FWR
Communication Engineering, MIT,
Manipal

Bridge 62
rectifier
 Fig.17 (b): Bridge
Department of Electronics and FWR
Communication Engineering, MIT,
Manipal

Bridge 63
rectifier
 Working of
Bridge FWR

Note: Current
through load for
both cycles is in
same direction
(from node C to
ground)
Department of Electronics and
Communication Engineering, MIT,
64 Manipal
Fig. 18: Bridge Fig. 19: Bridge
FWR when node A FWR when node ​B
is positive w.r.t B is positive w.r.t. A
Bridge FWR

Simulation of FWR

Fig. 20 : Input and output waveforms of

bridge rectifier

Note: The frequency of the output signal =2

times the input frequency
Department of Electronics and
Communication Engineering, MIT,
Manipal
65
Bridge FWR
▪ ​PIV :​ (between ​Vm
​ ,​ where ​node
A ​Va​ nd ​mpeak
​ ​ ode of
n
secondary B).
voltage
▪ ​Other parameters same as
Center tapped FWR:
▪ Ripple factor is:
γ ​= 483.0 ​▪ Efficiency:
η ​= %2.81 ​Department of Electronics and Communication
Engineering, MIT, Manipal

66
Comparison of Rectifiers
• Advantages of HWR over FWR

• Advantages of Center tapped FWR

rectifier over HWR

• Advantages of bridge rectifier over to

centre-tap FWR

• Disadvantages of HWR over FWR

• Disadvantages of centre-tap FWR over

Bridge

• Disadvantages of bridge rectifier over

other rectifiers
Department of Electronics and Communication Engineering, MIT, Manipal
67
Comparison of Rectifiers
Parameters of
rectified signal
Bridge FWR
V​dc
V​RMS
PIV
Ripple factor
Efficiency
Frequency f​o
HWR Center-tapped
FWR
V

​ ​2 ​m​
mV V​2 ​m​ V mV
π​π​ π​ ​
mV m​ V ​ V​2 ​ V ​m​
2​22​ m​ m​ 1.21 0.483 0.483
40.6% 81.2% 81.2%
f ​ f​2 ​if​2 ​
i​ iD​ epartment of Electronics and Communication Engineering, MIT,
Manipal

68
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

Department of Electronics and Communication Engineering, MIT, Manipal

69
Capacitor Filter
Fig 21: C type filter with HWR

Department of Electronics and Communication Engineering, MIT, Manipal

70
Simulation of HWR/FWR with C filter​: with
varying R and C

71
Capacitor Filter
Fig 22: C type filter with Bridge
FWR

Department of Electronics and Communication Engineering, MIT, Manipal

Capacitor Filter
Fig. 23 Filtered output waveform
using C type filter
Communication Engineering, MIT,
Manipal

72
Department of Electronics and

Ripple factor with Capacitor

Filter
• For HWR

r​
=
fCR ​ r
32 1 ​ L •​ For FWR ​

=
fCR ​
34 1 ​ L​ Department of Electronics and

Communication Engineering, MIT, Manipal

73
Ripple factor with Capacitor
Filter
• DC value of filtered output for
HWR

V d​ c =
​

21

​ CRf ​CRfL​ L
+ 2​
Vm ​
• DC value of filtered output for
FWR
V​
​
dc =
41
CRf CRf ​ V​
+ 4​
​ L L ​ m​Note: here ​f ​is
 the frequency of the input signal
Department of Electronics and Communication Engineering, MIT, Manipal

74
Comparison of Rectifiers
Parameters of
rectified signal
HWR FWR
Vdc​
Ripple factor
CRf​ CRf
2​ L ​21 ​+ ​
V​ CRf​
L ​ m​4​ ​ 1
L4

+
r ​ fCR ​ r​
CRf ​L V ​
​ m​ 32 ​ L ​
=​ ​ =
1

​ ​Department of Electronics and Communication Engineering,

34 1 ​fCR L
MIT, Manipal

75
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
Department of Electronics and Communication Engineering, MIT, Manipal

Exercise Calculate average

and rms load
Problems voltage, efficiency,
ripple factor, PIV
rating and
1. Primary voltage frequency of output
is 120V, 60Hz. waveform.
Turns ratio is 5:1.
This ​transformer 2. Repeat this
supplies to bridge problem for center
rectifier employing tapped FWR.
4 identical ​ideal Comment on the
diodes. The load results comparing
resistance is 1kΩ. with results of
 Communication Engineering, MIT,
exercise 1. Manipal

77

Department of Electronics and

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)
 DIODES AND
APPLICATONS

Module – 3:
Voltage
Regulators
Department of Electronics and
Communication Engineering, MIT,
Manipal

78
Part – I : Reference:
ANALOG
ELECTRONICS Robert L.
Boylestad, Louis
Nashelsky,
Electronic Devices
& Circuit Theory,
CHAPTER-1: 11​th ​Edition, PHI,
2012
▪ Describe the
Department of Electronics and
Communication Engineering, MIT, working of Zener as
Manipal

1 voltage regulator

Module – 3:
▪ Discuss the IC
Voltage based voltage
Regulators regulator.

Learning
Outcomes:

At the end of this Department of Electronics and

Communication Engineering, MIT,
module, students Manipal

80
will be able to:
Zener voltage
regulation
 Communication Engineering, MIT,
Manipal

81

Department of Electronics and

III
=
z​+ ​L ​= in​
- ​z ​R ​s ​=
VV ​in

- ​ z​
I​ R =

VV ​LZ ​+​- ​(i) ​For Line regulation​, R​L is​ constant ​I and
​

= R​V ​
Z
L​

L​
is also constant and V​in varies
​ bet.

V​in(min) to
​ V​in(max) ​I

VV ​in

zs​
II
line regulation and load
regulation
R
Department of Electronics and Communication Engineering, MIT, Manipal
82