Chapter 1c (RECTIFIER)

BED11103; Introduction To Electronic Chapter 2. Page 1Diode Application

RECTIFIER

Diodes are used in circuits called rectifiers that convert ac voltage into dc voltage. Rectifiers in all dc

power supplies that operate from an ac voltage source. In this section, you will study the most basic

type of rectifier, the half-wave rectifier.

After completing this section, you should be able to;

Explain and analyze the operation of half-wave rectifiers

Describe a basic dc power supply and half-wave rectification

Determine the average value of a half-wave rectified voltage

Discuss the effect of barrier potential on a half-wave rectifier output

Define peak inverse voltage (PIV)

Describe the transformer-coupled half-wave rectifier.

Prepared by: HMD UniKL BMI

BED11103; Introduction To Electronic Chapter 2. Page 2Diode ApplicationThe Basic DC Power Supply

1. The dc power supply converts the standard 240V, 60 Hz ac available at wall outlets into a

constant dc voltage.

2. The dc voltage is used to power all types of electronic circuits; television receivers, stereo

systems, VCRs, CD players, and most lab. equipments.

3. Basic block diagrams for a rectifier and complete power supply are shown in Figure below.

4. The rectifier converts the ac input voltage

to a pulsating dc voltage, which is half-

wave rectified as shown in part (a).

5. A block diagram for a complete power

supply is shown in part (b).

6. The filter eliminates the fluctuations in

the rectified voltage and produces a

relatively smooth dc voltage.

7. The regulator is a circuit that maintains a

constant dc voltage for variations in the

input line voltage or in the load.

8. The load is a circuit or device for which

the power supply is producing the dc

voltage and load current.


BED11103; Introduction To Electronic Chapter 2. Page 3Diode ApplicationThe Half-Wave Rectifier

1. A diode is connected to an ac source and to a load resistor, RL, forming a half-wave rectifier.

2. During one cycle of the input voltage, when the sinusoidal input voltage (Vin) goes positive,

the diode is forward-biased and conducts current through the load resistor, as shown in part

(a).

3. The current produces

an output voltage across the load RL, which has the same shape as the positive half-cycle of

the input voltage.

4. When the input voltage goes negative during the second half of its cycle, the diode is reverse-

biased.

5. There is no current, so the voltage across the load resistor is 0 V, as shown in Figure (b).

6. The result is only

the positive half-cycles of the ac input voltage appear across the load. Since the output does

not change polarity, it is a pulsating dc voltage with a frequency of 60 Hz, as shown in part

(c).


BED11103; Introduction To Electronic Chapter 2. Page 4Diode ApplicationAverage Value of the Half-Wave Output Voltage

The average value of the half-wave rectified output voltage is the value you would measure on a dc

voltmeter. This equation shows that VAVG is approximately 31.8% of VP for a half-wave rectified

voltage.

Effect of the Barrier Potential on the Half-Wave Rectifier Output

During the practical, diode model is used with the barrier potential of 0.7 V taken into account. This

results in a half-wave output with a peak value that is 0.7 V less than the peak value of the input, as

shown in Figure 1.


V AVG=V P

π

EXAMPLE What is the average value of the half-wave rectified voltage in Figure below?

Ans: 31.83V


Peak Inverse Voltage (PIV)

The peak inverse voltage (PIV) equals the peak value of the input voltage. PIV, occurs at the peak of

each negative alternation of the input voltage when the diode is reverse-biased.

FIGURE 2 The PIV occurs at the peak of each half-cycle of the input voltage when the diode is reverse-biased. In

this circuit, the PIV occurs at the peak of each negative half-cycle.


Vp(out )=Vp( in )−0 . 7 v

Figure 1: The effect of the barrier potential

on the half-wave rectified output voltage is

to reduce the peak value of the input by

about 0.7 V.

PIV = VP(in)


Half-Wave Rectifier with Transformer-Coupled Input Voltage

Transformer coupling provides two advantages;

First, it allows the source voltage to be stepped up or stepped down as needed. Second, the ac

source is electrically isolated from the rectifier, thus preventing a shock hazard in the

secondary circuit.

Basic ac circuit’s recall that the secondary voltage of a transformer equals the turns ratio, n, times the

primary voltage. We will define the turns ratio as the ratio of secondary turns, Nsec, to the primary

turns, Npri : n =

N secNpri

Vsec = nVpri


EXAMPLE: Determine the peak value of the output voltage in the circuit if the turns ratio is 0.5.

Ans: 78V, 77.3V

BED11103; Introduction To Electronic Chapter 2. Page 7Diode ApplicationFULL-WAVE RECTIFIERSIn this section, you will use what you learned about half-wave rectification and expand it to full-wave

rectifiers.

After completing this section, you should be able to

Explain and analyze the operation of full-wave rectifiers

Discuss how full-wave rectification differs from half-wave rectification

Determine the average value of a full-wave rectified voltage

Describe the operation of a center-tapped full-wave rectifier

Explain how the transformer turns ratio affects the rectified output voltage

A full-wave rectifier allows unidirectional (one-way) current through the load during the entire 360°

of the input cycle, whereas a half-wave rectifier allows current through the load only during one-half

of the cycle.

The average value, which is the value measured on a dc voltmeter. VAVG is approximately 63.7% of Vp, for a full-wave rectified voltage.

The Center-Tapped Full-Wave Rectifier

The Center-Tapped Full-Wave Rectifier


V AVG=2V P

π

EXAMPLE: Find the average value of the full-wave rectified voltage.

Ans: 9.55V


1. A center-tapped rectifier is a type of full-wave rectifier that uses two diodes connected to the

secondary of a center-tapped transformer.

2. The input voltage is coupled through the transformer to the center-tapped secondary.

3. Half of the total secondary voltage appears between the center tap and each end of the

secondary winding as shown.

4. For a positive half-cycle of the input voltage, the polarities of the secondary voltages are as

shown in Figure (a). This condition forward-biases diode D1 and reverse-biases diode D2. The

current path is through D1 and the load resistor R1 as indicated.

5. For a negative half-cycle of the input voltage, the voltage polarities on the secondary are as

shown in Figure (b). This condition reverse-biases D1 and forward-biases D2. The current

path is through D1 and R1, as indicated. Because the output current during both the positive

and negative portions of the input cycle is in the same direction through the load, the output

voltage developed across the load resistor is a full-wave rectified do voltage, as shown.



Effect of the Turns Ratio on the Output Voltage

1. If the transformer's turns ratio is 1, the peak value of the rectified output voltage equals half

the peak value of the primary input voltage.

FIGURE: Center-tapped full-wave rectifier with a transformer turns ratio of 1. Vp(pri) is the peak value of the primary

voltage.

2. In order to obtain an output voltage with a peak equal to the input peak (less the diode drop), a

step-up transformer with a turns ratio of n = 2 must be used, as shown.



FIGURE : Center-tapped full-wave rectifier with a transformer turns ratio of 2.

In any case, the output voltage of a center-tapped full-wave rectifier is always one-half of the total

secondary voltage less the diode drop, no matter what the turn’s ratio.

The Full-Wave Bridge Rectifier

1. The bridge rectifier uses four diodes connected as shown. When the input cycle is positive as

in part (a), diodes D1 and D2 are forward-biased and conduct current in the direction shown.

2. A voltage is developed across R, that looks like the positive half of the input cycle. During this

time, diodes D3 and D4 are reverse-biased.


Vout =

V sec2 - 0.7v


3. When the input cycle is negative as in Figure (b), diodes D3 and D, are forward- bias and

conduct current in the same direction through RL as during the positive half-cycle.

4. During the negative half-cycle, D1 and D2 are reverse-biased. A full-wave rectified output

voltage appears across RL as a result of this action.

Bridge Output Voltage

A bridge rectifier with a transformer-coupled input is shown in Figure a (below). During the positive

half-cycle of the total secondary voltage, diodes D1 and D2 are forward-biased.

The same is true when D3 and D4 are forward-biased during the negative half-cycle.


Vp(out) = Vp


1. As you can see in Figure b (below), two diodes are always in series with the load resistor

during both the positive and negative half-cycles.

2. If these diode drops are taken into account, the output voltage is,

THE DIODE DATA SHEET

A manufacturer's data sheet gives detailed information on a device.

After completing this section, you should be able to

Interpret and use a diode data sheet

Identify maximum voltage and current ratings

Determine the electrical characteristics of a diode

Analyze graphical data

Select an appropriate diode for a given set of specifications


Vp(out) = Vp (sec) –

BED11103; Introduction To Electronic Chapter 2. Page 13Diode ApplicationThese are the absolute maximum values under which the diode can be operated without damage to the

device.

An explanation of the parameters from Table 1 follows.

VRRM The maximum peaks reverse voltage that can be applied repetitively across the diode.

Notice that in this case, it is 50 V for the I N4001 and I kV for the 1N4007. This is the

same as PIV rating.

VR The maximum reverse do voltage that can be applied across the diode.

VRSM The maximum peak value of non repetitive reverse voltage that can be applied across

the diode.

Io The maximum average value of a 60 Hz rectified forward current.

IFSM The maximum peak value of non repetitive (one cycle) forward surge current.

TA Ambient temperature (temperature of surrounding air).

Tj The operating junction temperature.

Tstg The storage junction temperature.

Table 1: Maximum rating

RATING SYMBOL 1N4001 1N4002 1N4003 1N4004 1N4005 1N4006 1N4007 UNIT

Peak repetitive

reverse voltage

Working peak

reverse voltage

DC Blocking

voltage

VRRM

VRWM

VR

50 100 200 400 600 800 1000 V

Non repetitive

peak reverse

voltage

VRSM 60 120 240 480 720 1000 1200 V



rms reverse

voltage

VR(rms) 35 70 140 280 420 560 700 V

Average rectified

forward current

(single-

phase,resistive

load, 60Hz,

TA=75C)

Io 1.0 A

Non repetitive

peak surge

current ( surge

applied at rated

load condition )

IFSM 30 (for 1 cycle ) A

Operating and

store junction

temperature range

Tj, Tstg -65 to +175 C

ZENER DIODES

The Zener diode model to be employed for the “on” state will be shown in Figure 1a.

For the “off” state as defined by a voltage less than VZ but greater than 0V with the polarity indicated

in Figure 1b, the Zener equivalent is the open circuit.

Vi and R Fixed


FIGURE 1 Zener diode

equivalents for the (a) “on” and (b)

“off” states.

IZ = IR - IL

R

VV Li Where, and


1. In figure 2, the applied dc voltage is fixed, as is the load resistor. The analysis will be divide in

to 2 steps;

Step 1: Determine the state of the zener diode by removing it from the network and

calculating the voltage across the resulting open circuit.

2. If V ≥ VZ the Zener diode is “on”. ( Refer Figure 1a).

3. If V < VZ the Zener diode is “off”. ( Refer Figure 1b).

Step 2: substitute the appropriate equivalent circuit and solve for the desired unknowns.

4. For the Figure 2(a), the “on” state will equivalent to network Figure 3.

5. The Zener diode current must be determined using KCL.


V = VL =

R LViR+R L

VL =

Vz

FIGURE 2 (a) Basic Zener regulator and (b) Determining the state of the Zener diode


6. The power dissipated by the Zener diode is determined by;

7. Zener diodes are most frequently used in regulator networks or as a reference voltage. Figure

2 (a ) is a simple regulator design to maintain a fixed voltage across the load RL.

~end.


PZ = VZ IZ

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Chapter 1c (RECTIFIER)