ch 14 - alternating currentf
TRANSCRIPT
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24 AlternatingCurrent
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24. Alternating
Currents
24.3 Transmission
of electrical energy24.4 Rectification
24.2
Transformer
24.1 Characteristicsof alternating current
period,
frequency,peak value
and root-mean
square value
For resistive load
= P₀/2
I=I₀ sin t
Advantage of ac
and high voltages
Distinguish
between peak andrms value …
Half-wave and
full-wave
http://www.magnet.fsu.edu/education/tutorials/java/ac/index.html
Principle of
operation
2
o
rms
I I
s
P
P
s
P
s
I
I
V
V
N
N
effect of single
capacitance on
smoothing
Single diode and
bridge rectifier
http://www.magnet.fsu.edu/education/tutorials/java/ac/index.htmlhttp://www.magnet.fsu.edu/education/tutorials/java/ac/index.html
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Terms you are likely to encounter
http://upload.wikimedia.org/wikipedia/commons/4/49/Types_of_current_by_Zureks.svghttp://upload.wikimedia.org/wikipedia/commons/4/49/Types_of_current_by_Zureks.svg
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Alternating Current AC) Electricity
by Ron Kurtus (revised 2 June 2009)
Alternating current (AC) electricity is the type ofelectricity commonly used in homes and businessesthroughout the world. While direct current (DC)
electricity flows in one direction through a wire, ACelectricity alternates its direction in a back-and-forthmotion. The direction alternates between 50 and 60times per second, depending on the electrical system of
the country.
http://www.school-for-champions.com/science/ac.htm
http://www.school-for-champions.com/science/ac.htmhttp://www.school-for-champions.com/science/ac.htm
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Alternating Current AC) Electricity
by Ron Kurtus (revised 2 June 2009)
AC electricity is created by an AC electric generator,which determines the frequency. What is special aboutAC electricity is that the voltage can be readily changed,
thus making it more suitable for long-distancetransmission than DC electricity. But also, AC canemploy capacitors and inductors in electronic circuitry,allowing for a wide range of applications.
http://www.school-for-champions.com/science/ac.htm
http://www.school-for-champions.com/science/ac.htmhttp://www.school-for-champions.com/science/ac.htm
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24.1 Characteristics of a.c.
Current flow inone direction.
An alternating current (a.c.) is anelectric current that periodicallyreverses its direction in thecircuit, with a frequency findependent of the constants ofthe circuit.
http://www.toptenz.net/top-10-format-wars.php
timetime
currentcurrent
http://www.toptenz.net/top-10-format-wars.phphttp://www.toptenz.net/top-10-format-wars.php
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Terms a) Peak or maximumcurrent (Io, or Im) is themaximum current or
amplitude of current.b) One cycle is one
alternation.
c) Frequency (f) is the
number of cyclesoccurring per second.
d) Period (T) is the time forone complete cycle.
T = 1/f
Equations of graph?
amplitude
one cycle
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Equation of sinusoidal alternating current
a) The instantaneous current is the current at any time t is given by,
i = I0 sin ( t + )
where I0 = peak or maximum current
t = time
= 2f = angular frequency
= phase
f = frequency of a.c.
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A.C. flowing in a resistor,
The voltage from the sourceis
v = Vo sin t
Current though the load,
Where I = Vo /R
The current and voltage arein phase.
t I R
vi
o sin
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Power
The instantaneous power developed
across the resistor is
P = i2R
= P0 sin 2t
where P0 = Io²R
=peak or maximum power
http://www.phys.unsw.edu.au/PHYS1111/acc/acc.html
http://www.ngsir.netfirms.com/englishhtm/Rms.htm
http://www.phys.unsw.edu.au/PHYS1111/acc/acc.htmlhttp://www.ngsir.netfirms.com/englishhtm/Rms.htmhttp://www.ngsir.netfirms.com/englishhtm/Rms.htmhttp://www.phys.unsw.edu.au/PHYS1111/acc/acc.html
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Mean power
The mean power is equal
to half the maximum power.
2
2
.sin
2
2
0
2
o
o
T
o
P
RI
dt t I T R P
Io²
Io²/2
The equation i2 = Io2 sin2 t is
also a sinusoidal curve with a
mean value Io2/2.
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Review: Mean value
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Exercise 24.2
Determine the mean of the curvesshown.
Soln
a) = 0
b)
= Po/2
0
Po
Power
Po/2
a)
b)
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Exercise 24.3
Sketch a graph of an alternatingcurrent of amplitude 20 A, 50 Hz anda phase of /2 radian. What is theequation of the graph?
time/ms
current
5 10 15 20 25
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Pulse generator[oscillator]
Provides alternatingcurrent of different
frequency andamplitude.
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SIGNAL GENERATOR
Other names: Oscillator,
low voltage a.c., pulse generator
Provides different waveforms in theform of a.c.
Allow different frequencies andamplitudes of sinusoidal wave forms tobe selected.
Variation in voltage.
3/28/2016 B. H. KHOO 16
symbol
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Root mean square current.
= Irms²R
where
Irms = root mean square current
For d.c., P = I²R
The root mean square [rms] currentis the a.c. that has the same heatingeffect in a given resistor as a direct
current (d.c.)
R I
P
Power Mean
o
2
2
2
orms
I
I
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Mathematics corner [for info]
cos 2 = cos2 - sin2
= 1 - 2 s i n2
T as
I
t t
T
I
dt t T
I
dt t I T
I
o
T o
T o
T
o
2
2
]2
2sin[
2
)2cos1(2
.sin1
2
0
2
0
2
2
0
22
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Exercise 24.4
For the alternating voltageshown, determine,
a) peak voltage,
b) frequency,c) root mean square
voltage,
d) state the equation ofthe sinusoidal voltage.
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Exercise 24.5
An alternating current of
i = [5.0/A] sin [100 t/s]
passes through a load of resistance 20 . The current is measured in
ampere and the time in second. Calculate,a) the peak current,
b) the root mean square current,
c) the frequency of the current,
d) the period,
5.0 A
t I io
sin
2
o
rms
I I
A I rms
54.32
5
= 100, =2f f = 15.9 Hz
ms
T
8.62
100
22
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Exercise 24.5
An alternating current of
i = 5.0 sin [100 t/s]
Calculate,
e) the instantaneous current at 70ms,
f) the maximum power dissipatedg) the mean power dissipated,
(Ans. a) 5.0 A, b) 3.54 A , c) 15.9 Hz,d) 62.8 ms e) 3.2 A, f) 500 Wg) 250 W)
e) I = 5 sin 100(0.07)
= 3.3 A
f) Po = Io²R = 5²20
=500W
g)
= 500/2 = 250 W
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Exercise 24.5a) Sketch the graph of i versus time.
b) Sketch the graph of theinstantaneous power versus time.
c) Sketch the graph of double the
peak current but half thefrequency.
time/ms
5
-5
06331
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How to calculate root mean square current.Start from the back of the phrase.
a) Find the sum of
square of thequantity,
b) Find the mean of a),and
c) Take the square rootof b),
22
2
2
1
1
2 .....) N
N
i
i I I I I a
2
2
2
)
)
I I c
N
I I b
rms
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Exercise 24.6Find a) the mean voltage, andb) the mean squarevoltage and c) root meansquare voltage.
(Ans. a) 3.33 V b) 33.3 V2, c) 5.77 V)
Soln.
a) 3 = 10[1]
= 3.33 V
b) 3= 100[1]
= 33.3 V2
c) Vrms = 33.3 = 5.77 V
2
10
6 84
voltage/V
t/s
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Self-test 24.1
1) What is analternating current?
2) What is the peakvoltage?
1) An alternating current (a.c.)is an electric current that
periodically reverses itsdirection in the circuit, witha frequency f independentof the constants of thecircuit.
2) Peak or maximum voltage isthe maximum voltage oramplitude of voltage.
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Self-test 24.1
3) Distinguish between thepeak and root meansquare voltage.
3) Peak voltage is the maximum voltage whileroot mean square voltage is the alternatingvoltage that has the same heating effect in a
given resistor as a direct voltage. The peak voltage is larger than rms voltage.
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Self-test 24.14) What is meant by
a) frequency,
b) the root mean squarecurrent of an a.c.?
4a) Frequency (f) is thenumber of cyclesoccurring per second.
b) The root mean square
[rms] current is the a.c.that has the sameheating effect in a givenresistor as a direct
current (d.c.)
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Self-test 24.15) For an alternating voltage,v=[20/V]sin [200t/ms]
Determine
a) the peak voltage,
b) the rms voltage,c) frequency of the alternating
voltage.
5a) 20 V
b) 20/ 2 =14.1 V
c) 2 f = 200 /10¯³
f = 100 kHz
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PYP 24.1
The magnetic flux density B of the field
due to a long straight wire is given by
An overhead power cable carries analternating current of 2000 A r.m.s. Atwhat distance would the peakmagnetic flux density due to thecurrent in the cable be 100 T?
[Ans.: 5.7 m]
Soln.
peak current Io
= 20002
= 2828 A
d = 5.7 m
d
I B
o
2
d
x x
2
)2828(10410100
76
d
I
B o
2
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PYP 24.2
Ans A
P = ½ Po =
Is independent of frequency R
V o
2
2
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24.2 Transformer
A transformer changes i.e. transforms an alternating p.d from one valueto another of greater (step-up) or smaller value (step-down) usingthe mutual induction principle.
http://www.tucsontransformer.com/img/general-electric-unit-substation-transformer-44-1012-1.htmhttp://www.tucsontransformer.com/img/general-electric-unit-substation-transformer-44-1012-1.htm
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Power Transformer
2.5 MVA General Electric Unit Substation Transformer
http://www.tucsontransformer.com/img/general-electric-unit-substation-transformer-44-1012-1.htmhttp://www.tucsontransformer.com/img/general-electric-unit-substation-transformer-44-1012-1.htmhttp://www.tucsontransformer.com/img/general-electric-unit-substation-transformer-44-1012-1.htmhttp://www.tucsontransformer.com/img/general-electric-unit-substation-transformer-44-1012-1.htm
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500 MVA Single-phaseautotransformers
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Substation Equipment:Power Transformers
http://www.osha.gov/SLTC/etools/electric_power/illustrated_glossary/substation_equipment/power_transformers.html
http://www.osha.gov/SLTC/etools/electric_power/illustrated_glossary/substation.html#Equipmenthttp://www.osha.gov/SLTC/etools/electric_power/illustrated_glossary/substation_equipment/power_transformers.htmlhttp://www.osha.gov/SLTC/etools/electric_power/illustrated_glossary/substation_equipment/power_transformers.htmlhttp://www.osha.gov/SLTC/etools/electric_power/illustrated_glossary/substation_equipment/power_transformers.htmlhttp://www.osha.gov/SLTC/etools/electric_power/illustrated_glossary/substation_equipment/power_transformers.htmlhttp://www.osha.gov/SLTC/etools/electric_power/illustrated_glossary/substation.html#Equipment
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ElectricityFlow on the
Farm
http://www.wisconsinpublicservice.com/business/farm_voltage_electricity.aspx
http://www.wisconsinpublicservice.com/business/farm_voltage_electricity.aspxhttp://www.wisconsinpublicservice.com/business/farm_voltage_electricity.aspx
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Transformer (electrical appliants)
Used in laboratory power supply.
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Description
a) A simple transformer
consists of two coils, theprimary and the secondarycoils wound over a coremade of magnetically softmaterial.
b) There is no electricalconnection between the
primary and secondarycoils, but the soft iron core
provides a magnetic link between them.
http://www.electricityforum.com/products/trans-s.htm
http://www.electricityforum.com/products/trans-s.htmhttp://www.electricityforum.com/products/trans-s.htm
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Description
a) An alternating voltage applied to the primary coil produces an a.c.through it, which produces an alternating magnetic flux in the corethreading the secondary coil.
b) An alternating voltage is induced in the secondary coil.
c) Frequency of secondary voltage is the same as the primary voltage.
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Transformer
Transformers are designed so that all the magnetic fluxproduced by the primary coil passes through thesecondary. The primary coil is connected to an a.c.source.
http://www.daviddarling.info/encyclopedia/E/electromagnetic_induction.html
CROa.c.source
primary
coil secondarycoil
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http://www.tpub.com/content/doe/h1011v4/css/h1011v4_48.htm
http://www.tpub.com/content/doe/h1011v4/css/h1011v4_48.htmhttp://www.tpub.com/content/doe/h1011v4/css/h1011v4_48.htm
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Principle [how it works]
When current rises in theprimary coil, the magneticfield through the secondarycoil due to this currentincreases.
The changing flux through thesecondary causes e.m.f to beinduced in the secondary coil
As the current reversesdirection, the emf in thesecondary reversesdirection.
the frequency of thesecondary is the same asthe primary.
http://www.daviddarling.info/encyclopedia/E/electromagnetic_induction.html
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Main causes of
energy loses
a) R esistance of coil. Power dissipated in theresistor is i2R where R is the resistance of theresistor and i the current passing through the
coil. This is reduced by using low resistancethick copper wire.
b) Eddy current. The changing flux in the core willcause and induced current called eddy currentto flow. Laminating the core reduces theenergy losses due to eddy current.
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Main causes of energy loses [ ]
c) Flux leakage occurs when the changing flux fromthe primary threads the secondary. Efficient coredesign to ensure that all the primary flux is linkedwith the secondary.
d) Hysteresis loss. Magnetization of the core isrepeatedly changed from one direction another and back again. This requires energy and causes thecore to get hot. This is reduced by using softmagnetic material for the core.
Commercial transformer has an efficiency of 95% to99%. An ideal transformer has an efficiency of 100%.
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eddy current
changing field
induced changing
field
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Laminated iron core
magnetic field
produced by thecurrent.
induced
current
[eddycurrent]
induced
changing
field
coilSolid iron
core
I
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Function of soft iron core :a) concentrates the magnetic flux,
b) laminated to reduce eddy current losses.
Lamination . The core of a transformer is formed of a piles of thin iron or steel stampings (thin sheets) called lamination. These are oxidizedon the surface or lightly varnished to increase the electricalresistance from one to another.
F f
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For a transformer,
where
v s = secondary voltage
v p= primary voltage
Ns = number of turns in secondary
Np = number of turns in primary.
#vs and v p must both be either peak voltage or both rms.
*For Ns> N p, it’s a step uptransformer and
if N p> Ns it’s a step downtransformer.
p
s
p
s
N N
vv
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For an ideal transformer Power output = power input
where i p = primary current
i s
= secondary current
For ideal transformer there is notenergy lost, the input energy iscompletely transformed into theoutput energy.
s
p
p
s
i
i
v
v
p P s s
vivi
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24.3 Transmission of electrical energy
Transformer play anessential part in thetransmission of electricity.
Power plants are usually placedsome distance from towns.
Electricity needs to be
transmitted over long distance.There is always power loss intransmission lines due to theirresistance (I²R).
pylon
T i i f l i l
http://www.t2.unh.edu/spring99/Image13.gif
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Transmission of electrical energy
http://www.t2.unh.edu/spring99/pg4.html
pylon
http://www.t2.unh.edu/spring99/Image13.gifhttp://www.t2.unh.edu/spring99/pg4.htmlhttp://www.t2.unh.edu/spring99/Image13.gifhttp://www.t2.unh.edu/spring99/pg4.htmlhttp://www.t2.unh.edu/spring99/Image13.gifhttp://www.t2.unh.edu/spring99/Image13.gif
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TransmissionIn Britain a network of cables, called the national grid, links all the powerstations. It allows the demand for electricity to be shared out betweenthe power stations. Most of the cables in the grid system are carriedoverhead on pylons. Underground cables are more expensive anddifficult to maintain. They are used in cities and where the scenery must
not be spoilt.
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Advantages of using electrical energy
1. Electrical energy is the easiest form of energy to transmit, and distributed by cables.
2. For many modern appliances, electrical energy is the only form of energythat can be used.
3. Electrical energy can be converted efficiently into any one of the other forms of energy.
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How electrical energy is transmitted?
Electricity is sent over long distance using cables. Transmission is doneusing alternating current at high voltages to reduces energy losses incables.
1. The voltage is step up to high voltage before transmitted frompower station.
2. This ensures that the current flowing in the cables is small andthe rate of power dissipated in cables are minimum (I²R).
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How electrical energy is transmitted?
3. Through the national grid, the voltage is lowered in stages atreceiving substations depending on the need of the customer.
4. The national grid is made up of close network of cables that joinreceiving substations
E l 24 7
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Example 24.7
The output power P and output
voltage V from a power station isconnected to a factory by cablesof total resistance R. Calculate
a) the current flowing in the
circuit,b) the power dissipated incables,
c) the power input to thefactory.
factory
R
power station
~
P, V
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Advantages of a.c. in transmission
1. Direct current are less easy to generate thanalternating currents.
2. Alternating e.m.fs are more convenient to step upand to step down.
3. Alternating current is just as suitable for heatingas direct current. The heating effect does not depend ondirection of current. e.g. (a) lighting: filament lampsdepend on the heating effect, gas discharge lamps runas well on alternating current. (b) small motors invacuum cleaners can use a.c.
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Advantages of a.c. in transmission [2]
4. Transmission using alternating current is more efficient than
d.c. transmission. This is because high voltage transmission ismore efficient than low voltage transmission.
5. In high voltage and low current transmission of electricalpower, low currents require thinner and therefore cheaper
cables.
Disadvantage.
For use of high voltage the high cost of the substation insulationneeded. Cost of transmitting a.c. is lower than direct current.
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Practical transmission system
The energy loss in the cables can be reduced in twobasic ways:
(a) By reducing the resistance of the cables.
(b) By reducing the current flowing.
Large reductions in the resistance of the cables canonly be brought about by making the cables verythick. This is not practical for several reasons.
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Practical transmission system [2]
1. Thicker cables are more expensive as more materials are required,expensive to manufacture and installed.
2. Thicker cables may not be slung from pylons.
3. It is more difficult and costly to insulate high voltage cables than to be laid underground.
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PYP 24.4
Ans: C
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24.4 Rectification1. Rectification is the conversion ofalternating current (a.c.) to directcurrent (d.c.)
2. A rectifier is a conductor which islargely unidirectional.
3. Ideal rectifier or diode.
a) Must have a zero resistance whenthe current flow in one directionand
b) Must have an infinite resistancewhen the current flows inopposite direction.
current
voltage
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Forward biased
• Direction of conventional current isthe same as direction of arrow of diode.
• Rectifier conducts and has a zeroresistance.
•A real diode has low resistance
http://www.gadgetjq.com/tach_install.htm
d b d
http://www.gadgetjq.com/tach_install.htmhttp://www.gadgetjq.com/tach_install.htm
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Reversed biased
Direction of conventionalcurrent is opposite to that of thearrow of the diode.
The diode is non-conducting,and has an infinite resistance.
A real diode has a highresistance and negligible current
flows.
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Why do we need to rectify a.c.?
D.C is required for
a) battery charging
b) operating of CRO
c) operation of GM tube.
d) operation of X-ray tube
e) operation of radio receivers and transmitters.
H lf W R tifi ti
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Half Wave Rectification
During the first half cycle when P ispositive, the diode is.......................biased and is ................................
During the second half cycle when Q isnegative, the diode is ........................biased, and is ..................................
http://www.antonine-education.co.uk/physics_a2/options/module_9/Topic_3/topic_3.htm
forward
conducting
reversed
non-conducting
P
Q
http://www.antonine-education.co.uk/physics_a2/options/module_9/Topic_3/topic_3.htmhttp://www.antonine-education.co.uk/physics_a2/options/module_9/Topic_3/topic_3.htm
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Half Wave Rectification [2]
The output is a pulsatingunidirectional (direct) current.The rectifier conducts only during.................. the cycle.
The disadvantage is that only.................. cycle contributes tothe rectification.
This is adequate for a crude circuit, forexample the low voltage fan motorfor a hair dryer.
half
half
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Full wave Rectification
In the case above we see that both forward and reverse half cyclesare rectified.
· Two half-wave rectifiers are placed back to back.
· The load is connected to a centre tapping of the transformer.
· This is called a centre-tap full-wave rectifier.
· It always needs a transformer with a centre tap.
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BridgeRectifier The arrows show the forward andreverse half cycles:
www.antonine-education.co.uk/.../TOPIC_3.HTM
http://www.eleinmec.com/article.asp?18
X
Y
Ho it orks?
http://www.antonine-education.co.uk/.../TOPIC_3.HTMhttp://www.antonine-education.co.uk/.../TOPIC_3.HTM
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How it works?
During the first half cycle, when terminal X of the supplyis positive, diodes ........ and ........ are conducting, anddiodes .......... and ............. are reversed biased.
During the second half cycle, when terminal Y is positive,
diodes .......... and ......... are conducting, and diodes.......... and .......... are reversed biased.
In both half cycle. the current through the load are in the................. direction.
B D
A C
B A C
D
same
How it works?
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How it works?
Power utilised is ...................... that achieved with halfwave rectification.
The output is .............................. with an average voltageof
= 2/3 Vo
where Vo is the peak voltage.
doubled
doubled
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Alternative diagram
http://ocw.weber.edu/automotive-technology/ausv-1320-automotive-electronics/12-diodes/rectification
Smoothing
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SmoothingThe pulsating output produced by
both half-wave and full-wave
rectifiers can be made moresteady (smooth) by puttingsuitable capacitor in parallelwith the load.
http://en.wikipedia.org/wiki/Bridge_rectifier
http://en.wikipedia.org/wiki/File:Diode_bridge_smoothing.svghttp://en.wikipedia.org/wiki/File:Diode_bridge_smoothing.svghttp://en.wikipedia.org/wiki/Bridge_rectifierhttp://en.wikipedia.org/wiki/Bridge_rectifierhttp://en.wikipedia.org/wiki/File:Diode_bridge_smoothing.svghttp://en.wikipedia.org/wiki/File:Diode_bridge_smoothing.svg
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Smoothing [2]When the current flows through the load in bothhalf cycles the capacitor charges, and when thevoltage across the load decreases the capacitor
discharges.If the time constant CR is large the capacitorrecharges before it has completely discharges i.e.use a capacitor with large capacity.
The output is ripple voltage at twice the inputfrequency.
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Summary1. Alternating currents can be rectified using
diodes;
2. A single diode will carry out half wave
rectification;3. Two diodes connected to a centre tapped
transformer well carry out full wave rectification;
4. Four diodes in a bridge circuit form a bridge
rectifier.5. Capacitors are used to smooth rectified AC.