AET 9 Rectifiers datasheet

7/21/2019 AET 9 Rectifiers datasheet

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Rectifiers

AET 9

Courtesy of the United States Air Force


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Unit 2 - Objective

2a. Identify power supply rectifieroperating principles

OVERVIEW

1. Purpose Power Supply

Transformer

Rectifier Filter

Voltage Regulator

AC Terms


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Unit 2 - Objective


OVERVIEW

2. Characteristics Half-wave Rectifier

Full-wave Rectifier

Full-wave Bridge Rectifier

3. Operation

Half-wave Rectifier

Full-wave Rectifier



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Purpose

Before we get into rectifiers, we are going

to briefly discuss the entire power supply

TRANSFORMERAC

SOURCE RECTIFIER FILTERVOLTAGE

REGULATORTO

LOAD

The purpose of the power

supply is to convert AC voltageinto DC voltage, for use in

circuits where AC voltage is not

acceptable


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Purpose

An AC sine wave is fed into the

transformer

TRANSFORMERAC


REGULATORTO

LOAD

The purpose of the transformeris to step-up or step-down the

AC voltage or current

*


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Purpose

The transformed AC sine wave is fed into

the rectifier

TRANSFORMERAC


REGULATORTO

LOAD

The purpose of the rectifier is tochange the AC into pulsating DC

It also determines the polarity

of the output voltage

*


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Purpose

The pulsating DC is fed into the filter

TRANSFORMERAC


REGULATORTO

LOAD

The purpose of the filter is tochange the pulsating DC into a

smooth DC

*


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Purpose

The Smooth DC is fed into the voltage

regulator

TRANSFORMERAC


REGULATORTO

LOAD

The purpose of the voltageregulator is to keep the output

DC voltage constant under

changing input AC voltage or

load conditions

*


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Rectifier Operating Principles

Load

Term used to describe the actual device thatdraws current from the power supply

The load represents all of the circuits servedby the power supply

A large load means high current, and a lowcircuit resistance


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Sine Wave Terms

Sine wave Peak voltage

Peak-to peak

voltage

Effective

voltage

Alternation

Formulas

0V

POSITIVE

ALTERNATION

NEGATIVE

ALTERNATION

NEGATIVE

PEAKVOLTAGE

PEAK VOLTAGE EFFECTIVE X 1.414

CYCLE

EFFECTIVE VOLTAGE = PEAK VOLTAGE X .707

POSITIVE

PEAK

VOLTAGE


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Sine Wave Terms

Sine Wave -A voltage or current that iscontinually changing in amplitude, andperiodically changes polarity

Peak Voltage - The maximum amplitude ofthe wave on either the positive or negativealternation

Peak-to-Peak Voltage -The swing in

voltage from the maximum negative peak tothe maximum positive peak


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Sine Wave Terms

Effective Voltage (RMS) - Theamount of alternating current or voltagethat produces the same heating effect anequal amount of direct current or voltage

Alternation - The variation, eitherpositive or negative, of a waveform from

the reference (often zero volts) tomaximum and back to the reference


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Sine Wave Terms

Formulas

Epk = x Epk-pk

Epk = 1.414 x Erms Epk-pk = 2 x Epk

Epk-pk = 2.828 x Eeff

Eeff = .707 x Epk

Eeff = .3535 x Epk-pk


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Characteristics

The next circuits we discuss will include thefirst two parts of our power supply, thetransformer and the rectifier

We will cover three different types of rectifiers:

Half-wave

Full-waveFull-wave Bridge


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Characteristics

Half-wave - This circuit consists of one diode, which

will be biased by the secondary of the transformer.There is a resistor (RL) in series with the diode, and isused to develop the output voltage

Full-wave - This circuit consists of two diodes, biased

by a transformer with a center-tapped secondary.There is a resistor (RL) in series with the diodes, and isused to develop the output voltage

Full-wave Bridge - This circuit consists of four diodes,biased by the secondary of the transformer. Thelayout is normally shown in a diamond shapedarrangement. There is a resistor (RL) in series withthe diodes, and is used to develop the output voltage


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Operation - Half-Wave Rectifier

We apply the AC to the transformer

If a high voltage is required, a step-up transformer is

used; if a low voltage is required, a step-down

transformer is used

If we do not need a change in voltage level, a one-to-

one (1:1) transformer is used

AC INPUT

T1

1:1 CR1

RL

+

+

+ OUTPUT

VOLTAGE

0V


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The transformer also helps to isolate the DCoutput from the AC input

The phasing dots on this transformer signify

that the voltage polarities at these points are

the same

AC INPUT

T1

1:1 CR1

RL

+

+

+ OUTPUT

VOLTAGE

0V


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When a diode is forward biased, its resistancedecreases, and majority current flows

When a diode is reverse biased, its resistance

is very large, and we have no majority current

flow

AC INPUT

T1

1:1 CR1

RL

+

+

+ OUTPUT

VOLTAGE

0V


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A diode placed in series with an AC source and a loadresistor (RL) will have both forward and reverse biasapplied with every cycle

A forward biased diode allows current to flow, so wehave a pulse of DC in the output

RL represents all of the circuits that will draw currentfrom the rectifier

AC INPUT

T1

1:1 CR1

RL

+

+

+ OUTPUT

VOLTAGE

0V


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We will start with the positive alternation

The voltage at the top of T1s primary is positive

The voltage on the top of the secondary is also

positive (because of the phase dots)

The secondary is functioning as the power source forthe rectifier and the load circuits

AC INPUT

T1

1:1 CR1

RL

+

+

+ OUTPUT

VOLTAGE

0V


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The positive voltage felt on the anode of CR1 will

forward bias it

Majority current flows AGAINST the arrow

Current will flow from the bottom of the secondary,through RL, through the diode, and back to the top ofthe secondary

Since RL is very large, it will drop (develop) nearly all

the secondary voltage across RL, and it represents thepositive half of the AC input voltage

We disregard the voltage drop across the forwardbiased CR1


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When the input goes through the negative alternation,

the voltage at the top of T1s secondary is negative

This puts a negative potential on the anode of CR1,

reverse biasing it

The diodes resistance is so large that majority current

stops

AC INPUT

T1

1:1 CR1

RL

+

+

OUTPUT

VOLTAGE

0V


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Even though minority current could flow, we consider it

to be negligible, and say that the output is zero

This circuit is called a half-wave rectifier because itonly allows majority current to flow for half of the inputAC waveform

This circuit produces a positive output

If a negative output is required, we can reverse thediode, allowing current to flow in the opposite direction

The next slide compares the positive output rectifier toone with a with a negative polarity output


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In the first diagram, the positive alternation forward biases the

diode, while in the second diagram the negative alternationforward biases the diode

AC INPUT

T1

1:1 CR1

RL

+

+

+ OUTPUT

VOLTAGE

0V

AC INPUTT11:1 CR1

RL

OUTPUTVOLTAGE

0V+ + +


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The direction of the diode determines the polarity of the output,

not the phasing of the transformer As proof, our next slide will compare two circuits, with different

transformer phasing

AC INPUT

T1

1:1 CR1

RL

+

+

+ OUTPUT

VOLTAGE

0V

AC INPUTT11:1 CR1

RL

OUTPUTVOLTAGE

0V+ + +


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During the positive alternation, the transformer in the first circuit

passes a positive to the anode (forward biasing it), while in thesecond circuit, the transformer puts a negative potential on the

anode (because of the phase difference)

OUTPUT

O

OUTPUTVOLTAGE

0V

AC INPUT


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During the negative alternation, the transformer in thefirst circuit will pass a negative to the anode (reversebiasing it), while in the second circuit, the transformerwill put a positive potential on the anode (forwardbiasing it)

The diode was facing in the same direction in both ofthese circuits, but because of the phase differencethey receive different alternations

Transformer phase has no effect on output polarity


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Now that weve seen how current flows andhow we develop an output in the half wave

rectifier, lets apply an AC voltage of definite

values

These same values and formulas will work no

matter if were using a positive output or

negative output rectifier


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In this diagram, we have a positive output half-wave

rectifier with an input voltage of 200VAC (200 Veff)

The secondary is also 200VAC, or 282.8 Vpk

The positive alternation passes, leaving a positive

voltage output pulse of 282.8 Vpk

We disregard the voltage drop across the diode in our

calculations

AC INPUTCR1

RL

AC INPUT

0V

200 VAC

60 Hz

200 VAC (EFF)OR 282V PEAK

PEAK

AVERAGE

282V

0V

+282V

VOLTAGE = +282V

VOLTAGE = +90V


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We can measure the voltage peak across RL

using an

oscilloscope

However, if we were to use a voltmeter, we would read

the average voltage

The average voltage of an unfiltered half-wave rectifier

is .318 x Peak output voltage

Therefore, a DC voltmeter would read about +90VDC

(.318 x +282.8 Vpk = +89.93VDC)

AC INPUTCR1

RL

AC INPUT

0V

200 VAC

60 Hz


PEAK

AVERAGE

282V

0V

+282V

VOLTAGE = +282V

VOLTAGE = +90V


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The output voltage of the half-wave rectifier is in theform of pulses

In a half-wave rectifier, we get one pulse of outputvoltage for every cycle of input AC; this means that thefrequency of the output is the same as the input

The output frequency of a rectifier is given in Pulses-per-Second (PPS); we call it the ripple frequency(here, the input frequency is 60Hz, and the ripplefrequency is 60PPS)

AC INPUTCR1

RL

AC INPUT

0V

200 VAC

60 Hz


PEAK

AVERAGE

282V

0V

+282V

VOLTAGE = +282V

VOLTAGE = +90V


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Notice that the output voltage in this circuit goes from

zero to 282V and back to zero

We call this peak voltage change of 282V the ripple

amplitude

The lower the ripple frequency, and the higher the

ripple amplitude, the more difficult it will be to change

this pulsating DC to a smooth DC

AC INPUTCR1

RL

AC INPUT

0V

200 VAC

60 Hz


PEAK

AVERAGE

282V

0V

+282V

VOLTAGE = +282V

VOLTAGE = +90V

I t i S


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Interim Summary

Half-wave Rectifier Has one diode

Direction of diode determines polarity of output voltage

Output peak voltage determined by turns ratio oftransformer

Output ripple frequency (in PPS) equals input

frequency (in Hz)

Output ripple amplitude is the same as the transformersecondary voltage

Average DC output voltage equals .318 times the peak


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Operation - Full-Wave Rectifier

We apply the AC to the transformer

The phase dots signify that the voltage polarities at

these points are the same

The most noticeable difference is the center tap on the

secondary of the transformer

AC INPUT

T1

1:1 CR1

RLCR2

OUTPUT

VOLTAGE

+ +


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The center-tapped transformer divides the secondary

voltage into two equal amplitude voltages that will beopposite in polarity with respect to the tap (ground)

The transformers center tap is also tied to one end ofRL providing a current path through the load resistor

AC INPUT

T1

1:1 CR1

RLCR2

OUTPUT

VOLTAGE

+ +


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The circuit could be drawn like this, creatingtwo half-wave rectifiers, producing an outputpulse for each of the alternations

Both of the pulses will be the same polarityand amplitude

AC INPUT

T1

1:1 CR1

RL

CR2

OUTPUTVOLTAGE


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The positive voltage at the top of thesecondary (with respect to ground) will forwardbias CR1

The negative voltage at the bottom of thesecondary (with respect to ground) will reversebias CR2

AC INPUT

T1

1:1 CR1

RLCR2

OUTPUT

VOLTAGE

+ +


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Ground is negative with respect to the top of the

secondary, so current flows from ground, up through

RL, through CR1 (against the arrow), and back to the

top of T1

Since CR2 is reverse biased, it is shut off during this

alternation

AC INPUT

T1

1:1 CR1

RLCR2

OUTPUT

VOLTAGE

+ +


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This develops an output voltage across RL

that is

positive with respect to ground

The peak amplitude of the output is the peak

secondary voltage because the secondary is center

tapped

We ignore the voltage drop across CR1

AC INPUT

T1

1:1 CR1

RLCR2

OUTPUT

VOLTAGE

+ +


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The negative alternation causes a negative voltage at

the top of the secondary (with respect to ground) will

reverse bias CR1

The positive voltage at the bottom of the secondary

(with respect to ground) will forward bias CR2

200 VAC

60 Hz

T1

1:1 CR1

RL

CR2+ +

O R f


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Ground is negative with respect to the bottom of the

secondary, so current flows from ground, through RL,

through CR2 (against the arrow), and back to the

bottom of T1

CR1 is reverse biased, so it is shut off

200 VAC

60 Hz

T1

1:1 CR1

RL

CR2+ +

O ti F ll W R tifi


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The current flows through RL in the same direction for

both alternations of the input

This produces two pulses of the same polarity and

amplitude

Reversing both diodes will reverse the polarity of the

output pulses

200 VAC

60 Hz

T1

1:1 CR1

RL

CR2+ +

O ti F ll W R tifi


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Since we produce two pulses for every cycle,

the ripple frequency is two times the AC input

frequency (output in PPS equals 2 x input

frequency in Hz)

Also, since we have two pulses per cycle, wellcompute the average voltage differently (the

formula will be .636 x Peak)

Remember, the peak is only of the full

secondary

O ti F ll W R tifi


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If we use our earlier example of a 200VAC (282.8 Vpk)

input, and a 1:1 transformer, we get a voltage acrossthe full secondary of 282.8 Vpk

The center tap puts of this voltage available foreach diode (141.4 Vpk)

200 VAC

60 Hz

T1

1:1 CR1

RL

CR2+ +

O ti F ll W R tifi


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We now use our formula (.636 x Epk) to get

the resulting average voltage:

.636 x 141.4 Vpk = 89.93 VDC

200 VAC

60 Hz

T1

1:1 CR1

RL

CR2+ +

O ti F ll W R tifi


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You can see that both the half wave andfull wave rectifiers can have the same

Average voltage (as long as the inputamplitude and transformer turns ratio isthe same)

When compared to a half wave rectifier,we find that the full-wave has a higherripple frequency

This allows us to easily filter the output ofa full-wave rectifier

O ti F ll W R tifi


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The ripple amplitude of the unfiltered full-wave rectifier may be smaller than that ofthe half-wave, only because the peakoutput of the full-wave rectifier is half the

peak of the secondary

In order for the half-wave and the full-wave rectifiers to have the same peak

outputs, the transformer of the full-wavemust have double the turns ratio of thehalf wave

Full Wave Bridge Rectifier


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Full-Wave Bridge Rectifier

The bridge rectifier is a type of full wave

rectifier

It differs from the previous full-wave rectifier in

that a center-tapped transformer is not used,

and it requires four diodes

200 VAC

60 Hz

1:1

CR1

RLCR2

CR4

CR3282 V

PEAK



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This rectifier, like the previous full-wave,

allows both the positive and negative

alternations of the sine wave to be used

Our next slides will show how this is possible

200 VAC

60 Hz

1:1

CR1

RLCR2

CR4

CR3282 V

PEAK



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The positive alternation forward biases CR3 and CR4

CR2 is reverse biased (has a negative voltage appliedto its anode)

CR1 is reverse biased (has a positive voltage appliedto its cathode

AQR30020010-9307-138

AC

INPUT

+282V

0V

OUTPUT

VOLTAGE0V

+282V

282V

200 VAC

60 Hz

1:1

CR1

RLCR2

+ +

+CR4

CR3282 V

PEAK



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Current flows from the transformer, through CR4 to the

ground The ground is attached to one end of RL, so current

flows up through RL From the top of RL, current flows through CR3 back to

the transformer

AQR30020010-9307-138

AC

INPUT

+282V

0V

OUTPUT

VOLTAGE0V

+282V

282V

200 VAC

60 Hz

1:1

CR1

RLCR2

+ +

+CR4

CR3282 V

PEAK



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Current flows through the diodes against the

arrows

You see that the voltage at the top of RL ispositive with respect to ground

AQR30020010-9307-138

AC

INPUT

+282V

0V

OUTPUT

VOLTAGE0V

+282V

282V

200 VAC

60 Hz

1:1

CR1

RLCR2

+ +

+CR4

CR3282 V

PEAK



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The negative alternation forward biases CR1

and CR2

CR3 and CR4 are reverse biased

200 VAC

60 Hz

1:1

CR1

R

L

CR2

AQR30 020010 -9307-1 38

+

OUTPUT

VOLTAGE

AC

INPUT

+282V

0V

CR4

CR3282 V

PEAK

OUTPUT

VOLTAGE

+

+

0V

+282V

282V



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Current flows from the transformer, through

CR1 to the ground Current flows up through RL From the top of RL, current flows through CR2

back to the transformer

200 VAC

60 Hz

1:1

CR1

R

L

CR2

AQR30 020010 -9307-1 38

+

OUTPUT

VOLTAGE

AC

INPUT

+282V

0V

CR4

CR3282 V

PEAK

OUTPUT

VOLTAGE

+

+

0V

+282V

282V



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You see that the voltage at the top of RL is positive

with respect to ground The direction of the diodes determines the polarity ofthe output voltage (direction of current through RL

200 VAC

60 Hz

1:1

CR1

R

L

CR2

AQR30 020010 -9307-1 38

+

OUTPUT

VOLTAGE

AC

INPUT

+282V

0V

CR4

CR3282 V

PEAK

OUTPUT

VOLTAGE

+

+

0V

+282V

282V



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The Bridge rectifier produces two output

pulses for each input cycle (output freq in PPS

= 2 x input freq in Hz)

200 VAC

60 Hz

1:1

CR1

R

L

CR2

AQR30 020010 -9307-1 38

+

OUTPUT

VOLTAGE

AC

INPUT

+282V

0V

CR4

CR3282 V

PEAK

OUTPUT

VOLTAGE

+

+

0V

+282V

282V



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Since both alternations produce a pulse, andthe voltage of the entire secondary is used,then the average voltage equals .636 x peakof the secondary

200 VAC

60 Hz

1:1

CR1

R

L

CR2

AQR30 020010 -9307-1 38

+

OUTPUT

VOLTAGE

AC

INPUT

+282V

0V

CR4

CR3282 V

PEAK

OUTPUT

VOLTAGE

+

+

0V

+282V

282V



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The unfiltered bridge rectifier has a higher

average voltage that either the half or full wave

rectifiers

200 VAC

60 Hz

1:1

CR1

R

L

CR2

AQR30 020010 -9307-1 38

+

OUTPUT

VOLTAGE

AC

INPUT

+282V

0V

CR4

CR3282 V

PEAK

OUTPUT

VOLTAGE

+

+

0V

+282V

282V



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The ripple frequency is higher than that of a

half wave, which makes it easier to change the

pulsating DC to smooth DC

200 VAC

60 Hz

1:1

CR1

R

L

CR2

AQR30 020010 -9307-1 38

+

OUTPUT

VOLTAGE

AC

INPUT

+282V

0V

CR4

CR3282 V

PEAK

OUTPUT

VOLTAGE

+

+

0V

+282V

282V



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The ripple amplitude of the bridge rectifier will

be the same as that of the half wave

200 VAC

60 Hz

1:1

CR1

R

L

CR2

AQR30 020010 -9307-1 38

+

OUTPUT

VOLTAGE

AC

INPUT

+282V

0V

CR4

CR3282 V

PEAK

OUTPUT

VOLTAGE

+

+

0V

+282V

282V

Rectifier Summary


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Rectifier Summary

Unit 2 - Objective


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Unit 2 Objective


1. Purpose

Power Supply

Transformer

Rectifier

Filter

Voltage Regulator

AC Terms

Unit 2 - Objective


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Unit 2 Objective


2. Characteristics

Half-wave Rectifier

Full-wave Rectifier


3. Operation Half-wave Rectifier

Full-wave Rectifier


AET 9 Rectifiers datasheet

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