DIODE APPLICATIONS

Chapter 2

ELECTRONIC DEVICES EMT116/3

OBJECTIVES Explain and analyze the operation of both half and full

wave rectifiers

Explain and analyze filters and regulators and their characteristics

Explain and analyze the operation of diode limiting and clamping circuits

Explain and analyze the operation of diode voltage multipliers

Interpret and use a diode data sheet

Troubleshoot simple diode circuits

2

CONTENT

2-1 Half-wave rectifier

2-2 Full-wave rectifier

2-3 Power supply filters and regulators

2-4 Diode limiting and clamping circuits

2-5 Voltage multipliers

2-6 Diode data sheet

2-7 Troubleshooting

3

Diode Revisited

ability to conduct current in one direction and block current in other direction

When it is forward biased, it will pass current

When it is reverse biased, the current is blocked

used in circuit called RECTIFIER (ac dc)

4

The basic function of a DC power supply is to convert an AC voltage to a constant DC voltage (AC DC)

Either half or full-waverectifier convert ac input voltageto a pulsating dc voltage.

Eliminates the fluctuations- produce smooth dc voltage

Maintains a constant dcvoltage

The basic DC power supply

A half wave rectifier(ideal) allows conduction for only 180°or half of a complete cycle.

During first one cycle:

Vin goes positive – diode FB– conduct current

Vin goes negative – diode RB– no current- 0V

The output frequency is the same as the input (same shape).

Ideal diode model

2-1 Half-Wave Rectifier

ac source load resistor

6

Vp (peak voltage) : Voltage value of a waveform at its max +ve or –ve points

Vpp (peak-to-peak voltage) : 2Vp

Vrms : V root mean square. Sine wave value that indicates its heating effect / effective valueVavg : Measure on dc voltmeter = VDC. Equals to the area under the curve over a full cycle

2p

rms

VV =

πp

AVG

VV =

7

Average value of the half-wave output voltage

[ ]

[ ]

π

π

ππ

θπ

θθπ

θθππ

ππ

π

Vp

Vp

Vp

VpdVp

dVpAreaVAVG

=

+=

−−−=

−==

==

∫

∫

]11[2

)0cos(cos2

cos2

sin2

sin21

2

00

0

8

Example 1 (a):

What is the average value of the half-wave rectified voltage in figure above?

Example 1 (b):

What is the average value of the half-wave rectified voltage with a peak value of 200 V?

9

Practical Diode barrier potential of 0.7V (Si) taken into account.

During +ve half-cycle Vin must overcome Vpotential for forward bias.

Example 1: Calculate the peak o/p voltage, Vp(out)?The peak o/p voltage:

VVV inpoutp 7.0)()( −=

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

VVV

VVV inpoutp

30.47.05

7.0)()(

=−=

−=

10

Example 2:

Sketch the output V0 and determine the output level voltage for thenetwork in above figures.

11

Peak inverse voltage (PIV) is the maximum voltage across the diode when it is in reverse bias.

The diode must be capable of withstanding this amount of voltage.

)(inpVPIV =

Peak Inverse Voltage (PIV)

12

A transformer is a device that changes ac electric power at one voltage level to ac electric power at another voltage level through the action of a magnetic field.

Transformer consist of:

(1) Primary winding (input winding)

(2) Secondary winding (output winding)

(3) Magnetic core

If the secondary has more turns than the primary, the output voltage across the secondary will be higher and the current will be smaller. If the secondary has fewer turns than the primary, the output voltage across the secondary will be lower and the current will be higher.

Φ

The general arrangement of

a transformer

Transformer

There are three types of transformers: step-up, step-down, and isolation. These

components are described as follows:

1. The step-up transformer provides a secondary voltage that is greater than the primary voltage. Ex: a step-up transformer may provides a 240 Vac output with a 120 Vac input.

2. The step-down transformer provides a secondary voltage that is less than the primary voltage. Ex: a step-down transformer may provides a 30 Vac output with a 120 Vac input.

3. An isolation transformer provides an output voltage that is equal to the input voltage. This type of transformer is used to isolate the power supply electrically from the ac power line.

+

-

NP +

-

NS

120 Vac 240 Vac

+

-

NP +

-

NS

120 Vac 30 Vac

+

-

NP +

-

NS

120 Vac 120 Vac

1:2 4:1 1:1

Step-up Step-down Isolation(a) (b) (c)

Fig.2.6

The turns ratio of a transformer is equal to the voltage ratio of the component and since, the voltage ratio is the inverse of the current ratio. By formula:

sec

secsec

II

VV

NN pri

pripri

==

Where :NSec = the number of turns in the secondaryNPri = the number of turns in the primaryVSec = the secondary voltageVPri = the primary voltageISec = the secondary currentIPri = the primary current

By the equation above can be stated that:

Step-down transformer secondary current is greater than its primary current (ISec > IPri).

Step-up transformer secondary current is less than its primary current(IPri>ISec).

Transformers are often used for voltage change and isolation. The turns ratio, n of the primary to secondary determines the output

versus the input. The advantages of transformer coupling:

1) allows the source voltage to be stepped up or down2) the ac source is electrically isolated from the rectifier, thus

prevents shock hazards in the secondary circuit.

to couple ac input to the rectifier

VVV poutp 7.0(sec))( −=

(sec)pVPIV =priNNn sec=

prinVV =sec

Half-Wave Rectifier with Transformer-Coupled Input Voltage

If n>1, Vsec is greater than Vpri.If n<1, Vsec is less than Vpri.

If n=1, Vsec= Vpri. 16

Example 3:a) Determine the peak value of output voltage as shown in Figure below.

b) A power-supply transformer has a turns ratio of 5:1. What is the secondary voltage if the primary is connected to a 120 V rms source?

c) Determine the peak value of the output voltage for half-wave rectifier (use Silicon diode), if the given input voltage is 400 V and the turns ratio is 0.8.

OBJECTIVE

Explain & Analyze the operation of Full-Wave Rectifier. Discuss how full wave rectifier differs from half-wave rectifier Determine the average value Describe the operation of center-tapped & bridge. Explain effects of the transformers turns ratio PIV Comparison between center-tapped & bridge.

2-2 Full-Wave Rectifiers

18

Twice outputπ

pAVG

VV

2=

Introduction

A full-wave rectifier allows current to flow during both the positive and negative half cycles or the full 360º whereas half-wave rectifier allows only during one-half of the cycle.

The number of +ve alternations is twice the half wave for the same time interval.

The output frequency is twice the input frequency.

The average value – the value measured on a DC voltmeter.

63.7% of Vp

19

Coupled inputvoltage

This method of rectification employs two diodes connected to a secondary center-tapped transformer.

The i/p voltage is coupled through the transformer to the center-tapped secondary.

(i ) The Center-Tapped Full-Wave Rectifier

20

+ve half-cycle input voltage (forward-bias D1 & reverse-bias D2)-the current patch through the D1 and RL

-ve half-cycle input voltage (reverse-bias D1 & forward-bias D2)-the current patch through D2 and RL

The output current on both portions of the input cycle – same direction through the load.

The o/p voltage across the load resistors – full-wave rectifiers

(i ) The Center-Tapped Full-Wave Rectifier

21

priVV 2sec =If n=1, Vp(out)=Vp(pri) - 0.7V

2 Vp(sec)=Vp(pri)

If n=2,

7.0)()( −= pripoutp VV

- Effect of the Turns Ratio on the Output Voltage -

priVV 2sec =

In any case, the o/p voltage is always one-half of the total secondary voltageminus the diode drop (barrier potential), no matter what the turns ratio.

VVVout 7.02sec −=

(i ) The Center-Tapped Full-Wave Rectifier

22

-Peak Inverse Voltage (PIV)-

Maximum anode voltage:2(sec)

1pV

D +=2(sec)

2pV

D −=

D1: forward-bias – its cathode is at the same voltage of its anode minus diode drop; This is also the voltage on the cathode of D2.

PIV across D2 :

VV

VV

VPIV

p

pp

7.0 2

7.02

(sec)

(sec)(sec)

−=

−−

−=

VVV

VV

V

outpp

poutp

4.12

7.02

)((sec)

(sec))(

+=→

−=

We know that

Thus;

VVPIV outp 7.02 )( +=

(i ) The Center-Tapped Full-Wave Rectifier

For Figure below, determine:a) the voltage waveform for each half of secondary winding b) voltage across RL

c) PIV

Example 4:

+120V

-120V

Vin

2 : 1

24

Example 5:

Consider the circuit in Figure above.(a) What is the total peak secondary voltage?(b) Find the peak voltage across each half of the secondary.(c) Sketch the voltage waveform across RL.(e) What is the peak current through each diode?(f) What is the PIV for each diode?

120 Vrms

25

It employs four diodes arranged such that current flows in the direction through the load during each half of the cycle.

When Vin +ve, D1 and D2 FB and conduct current. A voltage across RL looks like +ve half of the input cycle. During this time, D3 and D4 are RB.

When Vin –ve, D3 and D4 are FB and conduct current. D1 and D2 are RB. • Used 4 diode:

2 diode in forward2 diode in reverse

• 2 diode always in series with load resistor during +ve and –ve half cycle .

Without diode drop (ideal diode):

With diode drop (practical diode):

(sec))( poutp VV =

VVV poutp 4.1(sec))( −=

(ii ) The Bridge Full-Wave Rectifier

26

VVPIV outp 7.0)( +=

Note that in most cases we take the diode drop into account.

0V (ideal diode)

For each diode, VVPIV outp 7.0)( +=

For ideal diode, PIV = Vp(out)

(ii ) The Bridge Full-Wave Rectifier

27

100 V

10 kΩ

Example 6:

Figure above shows the full-wave rectifier (practical model) with the transformer have a 10Vrms secondary voltage.a) Sketch Vsecb) Sketch Vp(out) c) What PIV rating required for each diode?

a) What PIV rating is required for the diodes in a bridge rectifier that produces an average output voltage of 50 V? (Ans : 78.53V for ideal diodes)

b) The rms output voltage of a bridge rectifier is 20 V. What is the peak inverse voltage across the diodes? (Ans : 28.3 V for ideal condition)

Example 7:

28

VVPIV

VV

VV

V

outp

outpAVG

poutp

7.02

2

7.02

)(

)(

(sec))(

+=

=

−=

πVVPIV

VV

VVV

outp

outpAVG

poutp

7.0

24.1

)(

)(

(sec))(

+=

=

−=

π

(sec)

)(

(sec))( 7.0

p

outpAVG

poutp

VPIV

VV

VVV

=

=

−=

π

HALF-WAVE RECTIFIER CENTER-TAPPED FULLWAVE RECTIFIER

BRIDGE FULL-WAVE RECTIFIER

Summary

29

2-3 Power Supply Filters AndRegulators

OBJECTIVE

Explain & analyze the operation & characteristic of power supply filters & Regulators

Explain the purpose of a filter Describe the capacitor-input filter Define ripple voltage & calculate the ripple voltage Discuss surge current in capacitor-input filter Discuss voltage regulation & integrated circuit regulator

30

Power Supply Filters To reduce the fluctuations in the output voltage of half / full-wave

rectifier – produces constant-level dc voltage. It is necessary – electronic circuits require a constant source to

provide power & biasing for proper operation.

Filters are implemented with capacitors.

Regulators Voltage regulation in power supply done using integrated circuit

voltage regulators.

To prevent changes in the filtered dc voltage/ to fix output dc voltage due to variations in input voltage or load.

Introduction

31

In most power supply – 60 Hz ac power line voltage is converted to constant dc voltage.

60Hz pulsating dc output must be filtered to reduce the large voltage variation.

Small amount of fluctuation in the filter o/p voltage - ripple

ripple

Introduction

32

For half-wave rectifier:

+ve first quarter cycle –diode is FB-capacitor is charging within 0.7V of i/p peak

I/p decreased below its peak-capacitor retains its charge-diode become RB (cathode is more +ve than the anode)

Capacitor can discharge through load resistance – at rate by the RLC time constant (>> time constant, << capacitor will discharge)

Next cycle-diode is FB when i/p voltage exceeds the Vc by 0.7V

A capacitor-input filter will charge and discharge such that it fills in the “gaps” between each peak. This reduces variations of voltage. This voltage variation is called ripple voltage.

Capacitor-Input Filter

load

capacitor

33

Ripple Voltage: the variation in capacitor voltage due to the charging and discharging.

The advantage of a full-wave rectifier over a half-wave is quite clear. The capacitor can more effectively reduce the ripple when the time between peaks is shorter.

Easier to filter-shorted time betweenpeaks.

-smaller ripple.

Capacitor-Input Filter

34

DC

ppr

VV

r )(=

Ripple factor: indication of the effectiveness of the filter

Vr(pp) = peak to peak ripple voltage; VDC = VAVG = average value of filter’soutput voltage.

Lower ripple factor better filter[can be lowered by increasing the value of filter capacitor or increasing the load resistance]

[half-wave rectifier]

For the full-wave rectifier:

)(

)()(

211

1

rectpL

AVGDC

rectpL

ppr

VCfR

VV

VCfR

V

−==

≅ Vp(rect) = unfiltered

peak.

Capacitor-Input Filter

35

Example 8:

(a) Determine the ripple factor for the filtered bridge rectifier with a load as indicated in figure above

(b) A certain full wave rectifier has a peak output voltage of 50V with capacitor filter input = 70µF. Calculate the peak to peak ripple and DC output voltage across 700Ω load resistance.

36

Surge Current in the Capacitor-Input Filter:Being that the capacitor appears as a short during the initial charging, the current through the diodes can momentarily be quite high. To reduce risk of damaging the diodes, a surge current limiting resistor is placed in series with the filter and load.

FSM

psurge I

VVR

4.1(sec) −= IFSM = forward surgecurrent ratingspecified on

diode data sheet.

The min. surgeResistor values:

Capacitor-Input Filter

37

Connected to the output of a filtered & maintains a constant output voltage (or current) despite changes in the input, load current or temperature.

Combination of a large capacitor & an IC regulator – inexpensive & produce excellent small power supply

Popular IC regulators have 3 terminals:(i) input terminal(ii) output terminal(iii) reference (or adjust) terminal

Type number: 78xx (xx –refer to output voltage)i.e 7805 (output voltage +5.0V); 7824 (output voltage +24V)

IC Regulators

38

Gnd

output

Connected to the outputof filtered rectifier

Bridge-full waverectifier filter regulators

Regulation is the last step in eliminating the remaining ripple and maintaining the output voltage to a specific value. Typically this regulation is performed by an integrated circuit regulator. There are many different types used based on the voltage and current requirements.

FIGURE 2-23 A basic +5.0V regulated power supply

IC Regulators

39

Percent Regulations

How well the regulation is performed by a regulator is measured by it’s regulation percentage. There are two types of regulation, line and load. Line and load regulation percentage is simply a ratio of change in voltage (line) or current (load) stated as a percentage.

%100Regulation Load

%100Regulation Line

−=

∆∆

=

FL

FLNL

in

out

VVV

VV

VNL :out put voltage with no loadVFL :output voltage with full load

41