UNIT-5

POWER SUPPLIES & PHOTO ELECTRIC DEVICES

1. Explain half wave rectifier and derive expression for efficiency and ripple factor?

Ans: HALF WAVE RECTIFIER

Construction of Half wave rectifier

It consists of

1. AC Power Source:

It Provides the AC input

voltage. It consists of a transformer with a secondary winding or any AC supply. It
is connected to anode of the diode.

2. Diode:

It is very important component in the circuit, It allows current to pass in only one direction
(forward bias) and blocks it in the opposite direction (reverse bias).In a half-wave rectifier, a single
diode is used. The cathode (negative terminal) of the diode is connected to one end of the load
resistor (RL)

The other end of the load resistor is connected back to the AC source to complete the circuit.

3. Load Resistor (RL):

It is Connected across the output of the rectifier circuit, where the rectified DC output voltage is
obtained. It represents the load that consumes the power.

Working :

 In a half-wave rectifier, only one half of the AC cycle (either positive or negative) is
allowed to pass through, while the other half is blocked. `

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  When the AC input is positive, the diode becomes
forward-biased and conducts, allowing current to flow
through the load. During the negative half-cycle, the
diode is reverse-biased and blocks the current,
resulting in zero output for that half of the cycle.

1. Average DC Load Current (IDC)

We have to determine the area over a complete cycle i.e., from 0 to 2 π

And then divided by 2 π

I L=¿I m Sinωt ¿ for 0<wt < π

I L=¿0 ¿ for 0<wt < 2 π

Where I m is peak value of current

2π
1
I DC = ∫ i ⅆ ( wt )
2π 0 L

2π
1
I DC = ∫ i sin wt dwt
2π 0 m

As no current flowing through negative half cycle

⊥m π
I DC = [−cos ωt ] 0
2π

Im
I DC = [ cos ( π )−cos ( 0 ) ]
2π

⊥m
I DC = =avⅇragⅇ value
π

E am
here I m=
Rf + R L + Rs

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2. Average DC Load Voltage (EDC)

E Dc =I DC R L

⊥m
E Dc = R
π L

E am
( )
R f + R L+ R s
E Dc = RL
π

Here

RL is load resistance, R s is winding resistance , R f is forward resistance .

Here R s , R f are very small compared to RL

Em
E Dc =
π

3. R.M.S value of Load Current (IRMS)

√
π
~ 1
⊥ Rms = ∫
2π 0
(
2
I m Sinwt ) ⅆω t

√ [ ]
π
1 1−cos ( 2 ωt )
I Rms= ∫
2π 0 2
ⅆω t

√ { }
π
1 ωt sin 2 ωt
I Rms= −
2π 2 4 0

Im
√ 1 π
2π 2 ()
Im
I Rms=
2

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 4. DC Power Output (PDC)

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P DC =E DC I DC =I D R L C

[ ]
2
Im
DC power output I 2D R L =
C RL
π

2
Em RL
P Dc = 2
π [ R f + R L+ R s ]
2

5. AC Power Input (PAC)

P AC =I 2Rms [ RL + Rf + R s ]

Im
I Rms=
2

2
Im
P AC = [ R L + Rf + Rs ]
4

6. Rectifier Efficiency (η)

DC output power
η=
AC Input power

P DC
η=
P AC

2
Im
2
RL
π
η=
I 2m
4 f
[ R + R L + RS ]

()
2
2
RL
π
η=
[ R f + R2 + R s ]

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 0.406
η=
I+
[
Rf + R s
RL ]
7. Ripple Factor (γ )
RMS value of AC component of output
γ=
Averag e∨DC component of output

I RMs =√ I 2ac + I 2Dc

I ac =√ I 2RMs −I 2Dc

I ac
γ=
I DC

γ=
√I 2
RMs −I 2Dc
I DC

γ=
√( I Rms 2
I DC )−1

Im
I Rms=
2

~ Im
⊥ DC =
π

√( )
Im 2
2
γ= −1
Im
π

γ=
π2
4 √−1

√ 1⋅ 4674

γ =1 ⋅211

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Disadvantages of half wave rectifier circuit

1. Low Efficiency:
2. High Ripple Factor
3. Poor Transformer Utilization:
4. Low Average Output Voltage and Current:
5. Higher Peak Inverse Voltage (PIV) Requirement:
6. Not Suitable for High Power Applications:
7. Increased Filter Requirements:

2. Explain Full wave rectifier and derive expression for efficiency and ripple factor?

Ans: It is an electronic circuit that converts an alternating current (AC) input into a direct current
(DC) output by rectifying both the positive and negative halves of the AC waveform. This is
achieved using diodes that allow current to pass only in one direction.

Mainly it consists of Transformer, Two Diodes (D1 and D2, Load Resistor (RL)

Construction:

 Transformer: The AC input voltage is applied to the primary winding of the transformer.
The secondary winding of the transformer has a center tap, dividing the secondary voltage
into two equal halves, but with opposite polarities. The center tap is grounded, which serves
as the reference point.
 Diodes:

 Diode D1: Connected between one end terminal of the transformer’s

secondary winding and the load resistor (R).
 Diode D2: Connected between the other end terminal of the transformer’s
secondary winding and the same load resistor (R).

 Load Resistor (RL): Connected across the output, with one end connected to the common
point of D1 and D2, and the other end to the center tap (ground).

Operation:

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 When the input voltage (Vin) is applied to a center-tapped full-wave rectifier, the secondary
winding of the center-tapped transformer splits this voltage into two parts: one positive and one
negative.

During the positive half-cycle of the input voltage, the ‘A’ end of the transformer becomes
positive, while the ‘B’ end becomes negative. This forward-biases diode D 1 and reverse-biases
diode D2. As a result, only D1 conducts, allowing current to flow through D 1 and the load resistor
(RL).

During the negative half-cycle of the input voltage, the ‘B’ end of the transformer becomes
positive, and the ‘A’ end becomes negative. This forward-biases diode D 2 and reverse-biases diode
D1. In this half-cycle, current flows through D2 and the load resistor (R (RL) in the same direction as
it did during the positive half-cycle.
Therefore, the current through the load resistor (R L) flows in the same direction during both
the positive and negative half-cycles of the input voltage, producing a DC output voltage (Vout = I
* RL) across the load resistor. The waveforms represent the applied input voltage, the current
through the load, and the output voltage across the load (RL).

Determination of efficiency:

1. Average DC Load Current (IDC):

Consider one cycle of load current IL from 0 to 2 π to obtain the average value which
is DC value of Load current.
IL= im Sin ωt

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But for π to 2 π the current IL is again positive Sin ωt is negative. Hence in the region π to 2 π
the positive iL can be represented as negative of im Sin ωt

I L=¿− I m Sinωt ¿ for π < wt < 2 π

I L=¿0 ¿ for 0 < wt < 2 π

Where I m is peak value of current

Iavg =IDC

2π
1
I DC = ∫ i ⅆ ( wt )
2π 0 L

2π
1
I DC = ∫ i sin wt dwt
2π 0 m

[ ]
π 2π
1
I DC =
2π
∫ im sin wt dwt +∫ −im sin wt dwt
0 π

[ ]
π 2π
im
I DC =
2π
∫ sin wt dwt +∫ −sin wt dwt
0 π

im
I DC =
2π
[ [ −cos ωt ]0 + [ −cos ωt ] π ]
π 2π

4 Im
¿
2π

2 Im
I DC =
π

2. Average DC Load Voltage (EDC) E Dc =I DC R L

2 ⊥m
E Dc = RL
π

E am
Here I m=
Rf + R L + R s

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 Eam
2( )
R f + R L+ R s
E Dc = RL
π

Here R L > R f , R S

2 E am
E Dc =
π

3. R.M.S Value of Load Current (IRMS):

√
π
~ 1
⊥ Rms = ∫
2π 0
(
2
I m Sinwt ) ⅆω t

Since 2 half wave rectifiers are similar in working. Then

√ [ ]
π
2 1−cos ( 2 ωt )
I Rms=I m ∫
2π 0 2
ⅆω t

Im
I Rms=
√2

4. DC Power output (PDC)

2
P DC =E DC I DC =I D R L C

[ ]
2
2Im
DC power output I 2D R L =
C RL
π

2
4 Em R L
P Dc = 2
π [ R f + R L+ R s ]
2

5. AC Power input (PAC):

The A.C power input is given by

P AC =I Rms [ RL + Rf + R s ]
2

( )
2
Im
Here, I Rms =
√2

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 2
Im
P AC = [ R L + Rf + Rs ]
2

Substitute the value of Im in above equation

2
Eam 1
P AC = 2 [ R +R +R ]
2 L f s
[ R L + Rf + R s ]
2
Eam
P AC = ❑
2 [ R L+ R f + R s ]

6. Rectifier Efficiency (η)

DC output power
η=
AC Input power
P DC
η=
P AC

2
4 Im
2
RL
π
η=
I 2m
2 f
[ R + R L + RS ]

8 RL
η=
π [ Rf + R L + R s ]
2

Here R f , R s < R L

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η= 2
π

η=0.812

η %=81.2
7. Ripple factor:
RMS value of AC component of output
γ=
Average∨ DC component of output

I RMs =√ I 2ac + I 2Dc

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 I ac =√ I 2RMs −I 2Dc

I ac
γ=
I DC

γ=
√I 2
RMs −I 2Dc
I DC

γ=
√( I Rms 2
I DC )−1

Im
I Rms =
√2

~ 2Im
⊥ DC =
π

√( )
2
Im

γ=
√2 −1
2Im
π

γ=
√ π2
8
−1

γ =0.481

Advantages of Full wave rectifier:

1. Higher Efficiency
2. Higher Average Output Voltage
3. Greater Output Current
4. Lower Peak Inverse Voltage (PIV)
Dis advantages

1. Complexity
2. Higher Cost
3. Larger Size

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 4. Power Losses
5. Heat Dissipation
6. Not Ideal for Low Voltage Applications
Applications of full wave rectifier:

1. Power Supply Circuits

2. Battery Charging
3. Radio and Communication Devices
4. Welding Equipment
5. Signal Processing
6. DC Motor Drives
7. Electroplating
8. Audio Amplifiers

3. Explain filter circuits?

Ans:
 A filter passes one band of frequencies while rejecting another. They are
3 types
 passive RLC filters, active RC filters and switched capacitor filters.
 Passive filters work well at high frequencies, however, at low
frequencies the required inductors are larges, bulky and non-ideal.
 Active RC filters utilize op-amps together with resistors and capacitors
and are fabricated using discrete, thick film and thin-film technologies.

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 4. Explain 3 terminal fixed voltage IC regulators 78XX?

Ans:
It is an integrated circuit whose basic
purpose is to regulate the unregulated input
voltage and provide with a constant,
regulated output voltage.

Fixed Voltage Regulators

 These regulators provide a constant output
voltage. A popular example is the 7805 IC which provides a constant 5 volts output.
 A fixed voltage regulator can be a positive voltage regulator or a negative voltage regulator.
 A positive voltage regulator provides with constant positive output voltage.
 All those IC’s in the 78XX series are fixed positive voltage regulators.
 In the IC nomenclature – 78XX ; the part XX denotes the regulated output voltage the IC is
designed for. Examples:- 7805, 7806, 7809 etc.

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  A negative fixed voltage regulator is same as the positive fixed voltage regulator in design,
construction & operation.
 The only difference is in the polarity of output voltages.
 These IC’s are designed to provide a negative output voltage. Example:- 7905, 7906 and all
those IC’s in the 79XX series.

5. Explain construction and working of LED?

Ans: LED (Light Emitting Diode) is an optoelectronic device which works on the principle
of electro-luminance. Which convert electrical energy into light energy. It is specially doped p-n
junction diode made up of specific type of semiconductors.
When the light emitting diode light is forward
biased, then it emits light either in
visible region or infra red region.

Working Principle of LED

. When the diode is forward biased, then the current flows through the diode. The flow of current in
the semiconductors is caused by the both flow of holes in the opposite direction of
current and flow of electrons in the direction of the current. Hence there will be recombination due
to the flow of these charge carriers. The recombination indicates that the electrons in the conduction
band jump down to the valence band. When the electrons jump from one band to another band the
electrons will emit the electromagnetic energy in the form of photons and the photon energy is
equal to the forbidden energy gap (Eg). Eg = hf

Where h is known as a Planck constant, and f is the frequency of the emitted electromagnetic
radiation. The frequency of radiation is related to the velocity of light as a f= c / λ. where c is the
speed of light λ is denoted as a wavelength of an electromagnetic radiation and the above equation
will become as Eg = hc / λ

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I-V Characteristics of LED
There are different types of light emitting diodes are available in the market and
there are different LED characteristics which include the color light, or wavelength
radiation, light intensity. The important
characteristic of the LED is color.

The following graph shows the approximate

curves between the forward voltage and the
current. Each curve in the graph indicates the
different color. The safe forward voltage
ratings of most LEDs is from 1V to 3 V and forward current ratings is from 200 mA to 100 mA.
Light emitting diodes emit either visible light or invisible infrared light when forward biased. The
LEDs which emit invisible infrared light are used for remote controls

6. Explain construction and working of Photo diode?

Ans: A photodiode is a p-n junction device that consumes light energy to generate electric current.
It is also sometimes referred as photo-detector, photo-sensor, or light detector. Photodiodes are
specially designed to operate in reverse bias condition. Photodiode is very sensitive to light
so when light or photons falls on the photodiode it easily converts light into electric current.
Solar cell is also known as large area photodiode because it converts solar energy or light
energy into electric energy. However, solar cell works only at bright light. PIN photodiodes
are mostly used in high-speed applications.

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 Photodiode symbol

Working:
 When external light energy is supplied to the p-n junction photodiode, the valence electrons
in the depletion region gains energy.
 If the light energy applied to the photodiode is greater the band-gap of semiconductor
material, the valence electrons gain enough energy and break bonding with the parent atom.
 The valenc electron which breaks bonding with the parent atom will become free electron.
Free electrons moves freely from one place to another place by carrying the electric current.
 When the valence electron leave the valence shell an empty space is created in the valence
shell at which valence electron left.
 This empty space in the valence shell is called a hole. Thus, both free electrons and holes
are generated as pairs.
 The mechanism of generating electron-hole pair by using light energy is known as the inner
photoelectric effect.

7. Explain construction and working of LDR?

Ans: A Light Dependent Resistor (LDR) or a photo resistor is a device whose resistivity is a
function of the incident electromagnetic radiation. Hence, they are light sensitive devices. They are
also called as photo conductors, photo conductive cells or simply photocells. They are made up of
semiconductor materials having high resistance. There are many different symbols used to indicate

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a LDR, one of the most commonly used symbol is shown in the figure below. The arrow indicates
light falling on it.

Symbol

Working Principle of LDR:

Principle: photo conductivity.
Photo conductivity is an optical phenomenon in
which the materials conductivity is increased when
light is absorbed by the material.
When light falls i.e. when the photons fall on the
device, the electrons in the valence band
of the semiconductor material are excited to the conduction band. These photons in the incident
light should have energy greater than the band gap of the semiconductor material to make the
electrons jump from the valence band to the conduction band. Hence when light having enough
energy strikes on the device, more and more electrons are excited to the conduction band which
results in large number of charge carriers. The result of this process is more and more current starts
flowing through the device when the circuit is closed and hence it is said that the resistance of the
device has been decreased. This is the most common working principle of LDR.
Applications of LDR
1. Light sensors,
2. Camera light meter,
3. Street lamps, Alarm clock,
4. Burglar alarm circuits,
5. Light intensity meters,
6. Counting the packages moving on a conveyor belt.
--------------------------------

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