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DC POWER SUPPLY EE301 / UNIT

u nit

To understand the function of each stage in block diagram.

To sketch block diagram for dc power supply

To explain the function and sketch the circuit for each stage dc power supply

To draw the schematic diagram for dc power supply.

C1 C2

L

Dz

RL

R

Litar Penapis

Litar Pengatur

Voltan

Litar Pembahagi

Voltan

Pengubah

Litar Penerus

Voltan Keluaran Lelurus.

M

N

C

G

D1

D2


5.0 Introduction

Before this we have already discussed about semiconductors in electronics circuits. After recognize all the components, let discuss about basic electronics circuits that used the components. In this unit we will discuss about DC power supply.

5.1 Reasons why DC power supply needed in electronics devices:-

Electronics device used active component as diode, transistors, and etc. this active component need dc voltage to operate.

Batteries can give constant voltage and easy to carry everywhere. But using batteries the power will not last longer after a certain period. Electronics devices that using high power supply will shorter the batteries life.

Electronics devices that using high power supply will use more batteries. So, it’s not economical if we using batteries.

Electric power supply provided to public through output sockets at houses and buildings are in AC voltage and in high value. ( 1 phase = 240 V, 3 phase = 415 V )

5.2 Block Diagram For DC Power Supply

Operation of power supply circuits built using transformer, rectifier and filter. The addition parts are voltage regulator and voltage divider.Diagram 5.2.1 shows the block diagram for DC power supply.

Transformer Rectifier Filter Regulator Voltage Divider

AC voltage DC voltage


Starting with an ac voltage, a steady dc voltage is obtained by rectifying the ac voltage, then filtering to a dc level, and finally regulating to obtain a desired fixed dc voltage.

5.3 Transformer Block diagram shows dc power supply have five stages. Each stage has own function. First stage is transformer. We use step-down transformer since the voltage is decreased from primary to secondary. Transformer at primary windings will connect to 240V 50 Hz ac power supply and transformer at secondary windings will step down to fit with electronics devices.

Since transformer consists of two winding primary and secondary that have no connection, then the purpose of using transformers is to release the circuits at secondary windings from ac power supply. This release can avoid the user at secondary from electric shock at high ac voltage supply.

Transformer connected to the voltage source is called primary winding. The coil connected to the load called the secondary winding. The ratio of primary winding called Np turns to the secondary winding called Ns known as transformation ratio.

Diagram 5.2.1 : Block Diagram For DC Power Supply

Np : Ns

Vp Vs


The voltage across the primary winding to secondary windings called turns ratio.

When value of secondary winding is less than primary winding, secondary voltage also less than primary voltage. Secondary voltage can be calculated by

Example 5.3:

Given a transformer with turns ratio 4:1 has supply voltage 240 V, 50 Hz. Calculate the Vs.

Transformation ratio =

4:1

240 V50 Hz

Np : Ns

Vp Vs

½ Vs

½ Vs

Figure 5.3.2 : Middle tap transformers

Figure 5.3.1 : Transformer


Solution:

5.4 Rectifier

4:1

240 V50 Hz

V s = ¼ x 240V = 60 V

Vs = x Vp


Most of the electronics circuit required dc power supply to operate. One important application is the design of rectifier circuits. A diode rectifier forms the first stage of dc power supply. Rectification is the process of converting an alternating (dc) voltage into one that is limited to one polarity. Rectification is classified as:-

Half wave rectifier Full wave rectifier Bridge rectifier.

5.4.1 Half wave Rectifier

Operation

When input signal in positive cycle, D diode in forward bias. D operates as closed switch so that current can flow through the circuit. Voltage drop at RL is same with input signal in positive cycle magnitude if we neglect the voltage drop at the diode.

When input signal in negative cycle, D diode in reverse bias. D operates as opened switch so that current can’t flow through the circuit. Voltage drop at RL is same with input signal in negative cycle is zero.

Diagram 5.4.1 shows half wave rectifier across RL when connected to oscilloscope.

Output Voltage

Figure 5.4.1 : Half wave Rectifier

VmVk

t

D

RL

t

Vm Vm


Output voltage for half wave rectifier obtained only in positive cycle. Since current across the diode and voltage drop at diode is 0.7V (assumed silicon diode), voltage drop is :-

Frequency

Output signal frequency is similar with input signal frequency.

Example 5.4.1:A half wave rectifier circuit has 20 V p-p voltages in, 50 Hz. Assumed no voltage drop, calculate:-

i) Vo for rectifier signalii) Frequency out for the signal

Solution :

i)

ii)

5.4.2 Full wave Rectifier

Vo = Vi - 0.7V

Vm

VMG

VNG

t

t

t

Vk

t

M

N

CA

B

G

D1

D2 RL

Vi = 20 Vp-p= 10 Vp

so Vo = 10 Vp

Frequency out for signal = Frequency in for signal= 50 Hz


Operation

When given dc voltage to the circuit, M and N at secondary transformer become positive and negative cycles. During the positive half of input voltage, both output voltage are positive; therefore diode D1 is forward bias and conducting and D2 is reverse bias and cut off. Current will flow along M, D1, C, A, B, G. A positive wave cycle will result at RL load.

During the negative half cycle, D1 is reverse bias and D2 is forward biased. Current will flow along N, D2, C, A, B, G. Since the direction current flow through RL is similar to the current flow through the positive cycles, so similar wave will produced.

Diagram 5.4.2 shows full wave rectifier across RL when connected to oscilloscope.

Output Voltage

Output voltage for full wave rectifier will result only in positive and negative cycles. Since current across the diode at one cycle and voltage drop at diode is 0.7V (assumed silicon diode ), voltage drop is :-

Frequency

Output signal frequency is twice with input signal frequency.

Vo = VM-G - 0.7V

Diagram 5.4.2 : Full wave Rectifier


example 5.4.2:

Given input voltage of full wave rectifier is 20 Vp-p 50 Hz. Transformation ratio 2:1. Assumed no voltage drop, calculate:-

i) Output voltage at rectifier ii) Output frequency signal

Solution:

i) Refer to figure 5.4.2

ii)

5.4.3 Bridge Rectifier

M

N

A

B

C

E

F

D1

D2D3

D4

RL

Vm

t

Vk

t

Vm = 20 Vp-p= 10 Vp

V

VmMN

=Ns

Np

VMN =Ns

NpxVm

=1

210x Vp

= 5 Vp

VMG = ½ VMN

= 2.5 Vp

then Vk = VMG

= 2.5 Vp

Output frequency signal = 2 x Input frequency signal= 100 Hz


Operation

During the positive half of the input voltage cycle, M is positive and N is negative. D1 and D3 are forward biased, D2 and D4 are reversed biased. Direction will flow along M, E, A, B, C, F N. A positive wave cycle will result at RL load.

During the negative half of the input voltage cycle, M is negative and N is positive. D2 and D4 are forward biased, D1 and D3 are reversed biased. Direction will flow along M, E, A, B, C, F N. A positive wave cycle will result at RL load. Since the direction current flow through RL

is similar to the current flow through the positive cycles, so similar wave will produced.

Diagram 5.4.3 shows output waveform rectifier across RL when connected to oscilloscope.

Output Voltage

Output voltage for bridge rectifier will result in both positive and negative cycles. Since current across the diode at one cycle and voltage drop at diode is 1.4V (assumed silicon diode ), voltage drop is :-

Frequency

Output signal frequency is twice with input signal frequency.

Vk = VM-N - 1.4V

Rajah 5.4.3 : Bridge Rectifier


Example 5.4.3:

A fullwave bridge rectifier has input voltage 20 Vp-p 50 Hz. Given transformation ratio is 2:1. Assume there are no voltage drop at diode, calculate:-

i) Output voltage signalii) Output frequency signal

Solution :

i) Refer to figure 5.4.3

ii)

Answer the question

Vi = 20 Vp-p= 10 Vp

V

VmMN

=Ns

Np

VMN =Ns

NpxVm

=1

210x Vp

= 5 Vp

then Vo = VMN

= 5 Vp

Output frequency signal = 2 x frequency input signal

= 100 Hz


1. State 3 reasons why dc power supply needed.

2. Sketch block diagram for dc power supply.

3. State 2 functions of transformers in dc power supply.

4. Describe rectifier circuit.

5. List down 3 types of rectifier circuits and sketch the circuits.


1. Three reasons why dc voltage are needed:-

a. Electric devices needed dc voltage.b. Batteries can not be able to accommodate electric devices needed.c. Electrical devices need dc power supply to operate but ac power

supplies are given to the houses.

2. block diagram for dc power supply are given below.

3. Two functions of transformers in dc power supply:-

a. Step down transformerb. As different in primary circuit and secondary circuit.

4. Rectifier is a circuit that using a diode ar more than a diode to change from ac voltage to dc voltage.

5. three types of rectifier :-

a. Half wave rectifier

b. Full wave rectifier

Transformer Rectifier Filter Regulator Voltage Divider

Ac voltage. Dc voltage.

DRL

M

N

CA

B

G

D1

D2 RL


c. Bridge rectifier

5.5 Filter

M

N

A

B

C

E

F

D1

D2D3

D4

RL


The filtered output of figure 5.5.1 has dc value and some ac variation (ripple) after passing through filtered circuit. We used filter circuit to get the smooth output voltage. Figure 5.5.1 shows output waveform before and after passing filter circuit.

The output resulting from rectifier is a pulsating dc voltage but the pulsating dc voltage from rectifier is not good enough. A filter circuit is necessary to provide steadier dc voltage (figure 5.5.2). Although a battery has essentially a constant or dc output voltage, the dc voltage derived from ac source signal by rectifying and filtering will have some ac variation (ripple) as shows in figure 5.5.3.

Good filter can reduce Vr p-p values that obtain from ripple dc voltage. The types of filters always used are:-

Capacitor filter RC filter LC filter filter

5.5.1 Capacitor Filter

Rectifier Circuit

Filter circuit

V

t

V

t

V

tDc voltage Ripple dc voltage

Figure 5.5.1 : Output waveforms before and after passing filter circuit

( Vr )p-p

Va.t.

Va.t.

V

t

V

t

Figure 5.5.2: pure Dc voltage Figure 5.5.3: ripple Dc voltage


A very popular filter circuit is the capacitor-filter circuit and connected in series to a load RL shown in figure 5.5.4.

Capacitor is a passive element designed to store energy in its electric field. When capacitor is connected to voltage source, capacitor will store the charge across it.

Refer to figure above, ID is the current flow from rectifier circuit that obtain voltage drop through RL. ID also provides the charge to capacitor C1 because it connected in series with RL. Notice that output waveform across RL is essentially dc output voltage from half wave rectifier like in figure 5.5.5

Rectifier filter

C1 RL Vk

ID

Figure 5.5.4 : Capacitor filter

Vi

20

t

15

10

5

0 62 84 10

15Vp-p

Vo2015

B Q


When 20Vp voltage drop pass through RL at starting input half wave positive cycle, C1 also get 20Vp charge as in figure showed at output wave from A to B. Then, when VRL decrease until 0, C1 will discharge.

Time taken of C1 to discharge is longer; written as time range in figure 5.5.5, that is from B to C, in 4 ms, not quick as decreasing in input voltage to 0 which that is 1ms only.

But before C1 finish discharging, until point P, there is another beat of input voltage for half second positive cycle resist passes through RL, that make C1 once again charging 20 Vp ( to point Q ).

This happened every time at half cycle input voltage. C1 will discharge from Q to R, but at X it charging back from third beat.

As a result from this process, output voltage produced now only changes from 5V to 20V ( changes only 15Vp-p ), not change to much as before filter out, that is from 0 to 20V (changes 20Vp-p).

Any voltage changes after this network is called ripple voltage. This is still not pure dc voltage since it still has beating. It resist because capacitance value we used is not a correct value. Correct value of capacitor will decrease ripple voltage to minimum voltage, until it correct value for a pure dc voltage.

Figure 5.5.6 explains how high capacitance value will decrease the ripple voltage. This means the higher capacitance value will take longer time to discharge.

Vk20

BC10.1uF

C21uF

Ripple voltage

Time range RLC1

10

5

0t

62 84 10A C

P

R

X

Figure 5.5.5 : output waveforms of capacitor filter


Figure 5.5.6 explains by using capacitor filter C1 that is 0.1uF, time taken to discharge is until point C. Ripple voltage will be in bigger value. By using capacitor filter C2 that is 1uF ,ripple voltage value will decrease a bit because time range for that capacitor is longer. ( until point Z ).

By increasing capacitance value, ripple voltage value will decrease, until at a suitable value of capacitor that has longer time range to get pure dc voltage and constant at 20 V.

5.6.2 RC Filter

It is possible to further reduce the amount of ripple across filter capacitor by using an additional RC filter as shown in figure 5.5.2. The purpose of added RC section is to pass most of the dc component while attenuating (reducing) as much as the ac component as possible. Resistance R will reduce ripple voltage to small value. Capacitor C2 will function in filtering the ripple voltage that still left.

Time range RLC1

15

5

0 t62 84 10A C

X

Z

Time range RLC2

10

Figure 5.5.6 : ripple voltage using capacitor C1 dan C2

C1Filter Circuit

C2 RL Vo

R

Figure 5.5.2 : RC Filter


A disadvantage of RC filter is output dc voltage across RL will drop to smaller value than before.

5 Filter

Filter obtains to overcome the disadvantage of RC filter. Resistor in RC filter replaced by inductor ( L ). Ac voltage will decrease if using inductor. Inductor has small resistance to dc voltage but has large reactance to ac voltage. So output of dc voltage will not decrease when across RL, but ac voltage in ripple voltage decrease in high value when across L.

5.5.4 LC Filter LC filter obtain when combining the advantages of series inductor and

series capacitor. It used as low-pass filter.

Filter Circuit

C2 RL Vo

L

Figure 5.5.4 : LC Filter

C1Filter Circuit

C2 RL Vk

L

Figure 5.5.3 : Filter


5.6 Regulator

Another factor of important in power supply is the amount the dc output voltage changes over range of circuit operation. The voltage provided at the output under load condition (no current drawn from the supply) is reduced when load current is drawn from the supply (under load). The amount of dc


voltage changes between no load and load conditions is described by a factor called voltage regulation. % voltage regulation

atau

where VNL = no-load voltageVFL = full load voltage

3 types of voltage regulator circuit:-

zener diode voltage regulator transistor series voltage regulation IC voltage regulator ( 78XX series)

5.6.1 Zener Diode Voltage Regulator

Zener diode will operating as voltage regulator in reverse bias condition. The advantage of zener diode is zener diode can regulate the voltage if operating in zener area. In order to operate in zener


area, input voltage must be larger than zener voltage and RL cannot make zener current decreased to zero.

Transistor Series Voltage Regulation

Series element that connected to load can control the amount of the input voltage that gets to the output. Output voltage is sampled by a circuit that provides a feedback voltage to be compared to a reference voltage.

Refer to figure 5.6.2, if the output voltage decreases, the increase VBE causes Q1 to conduct more, thereby raising output voltage-maintaining the output constant. Zener diode DZ provides the reference voltage. If the output voltage increases, the decreased VBE causes transistor Q1 to conduct less, thereby reducing the output voltage-maintaining the output constant.

5.6.3 IC voltage regulator ( 78XX series)

LM 78XX series( where XX = 05, 06, 08, 10, 12, 15, 18 atau 24 ) are three-terminal voltage regulator. IC LM7805 will obtained output voltage +5 V, LM7806 will obtained output voltage +6 V and LM7824

Rectifier circuit

Filter Dz

R

RL

Diagram 5.6.1 : Zener Diode Voltage Regulator

RRectifier circuit

Filter

Q1

Dz

RL

5.6.2 Transistor Series Voltage Regulation


will obtained output voltage +24 V. Figure XX shows IC voltage regulator ( 78XX series).

Rectifier circuit

Filter

LM74051 2

3

C1

Vk

C2

Figure 5.6.3 : IC voltage regulator ( 78XX series)


5.7. Voltage Divider Circuit

In electronics system devices, especially in complicated devices, they consists of circuit stages with different dc voltage values. As example, in tv system it used more than ten circuit stages with different function and need dc voltage around 100V, 48V, 12V and etc.

By using dc power supply, all the requirements can achieve using voltage devider after the highest value was obtained. Figure 5.7.1 and figure 5.7.2 shows voltage divider circuit constant and can change.

5.8. Complete Circuit for DC Power Supply

The ac voltage is connected to a transformer which steps that ac voltage down to the level for desired dc output. The diode rectifier then provides a full wave rectified voltage that is initially filtered by a simple filter to produce a dc voltage. This resulting dc voltage usually has some ripple or ac voltage variation. A regulator circuit can use this dc input to provide a dc voltage that not only as much less ripple voltage but also remains the same dc value even if the

Voltage regulator

circuit

R1

R2

R3

80V

40V

20V

Voltage regulator

circuit

R1

Figure 5.71 : Constant voltage divider circuit

VR1

80V

0 - 40V

Rajah 5.72 : changes voltage divider circuit


input dc voltage varies somewhat, or the load connected to the output dc voltage changes.

C1 C2

L

Dz

RL

R

Filter Circuit Voltage Regulator

Circuit

Voltage dividerCircuit Transformer

Rectifier Circuit

Output dc voltage

M

N

C

G

D1

D2

Figure 5.8 : DC Power Supply Circuit.


Answer the question below.

1. Explain the function of filter circuit.

2. Name the basic filter circuit in the DC power supply and sketch the schematic diagram each of them.

3. Why does bigger capacitor filter can decrease ripple voltage in the circuit?

4. What is the function of regulator in Dc power supply?

5. Name 3 regulator circuits and sketch them.

6. Explain why voltage divider circuit is needed in power supply.


1. Filter circuit used to change the beat in dc voltage to ripple voltage or to pure dc voltage.

2. There are four types of filter :-

a. Capacitor filter

b. RC filter

c. filter

Rectifier circuit

C1 RL Vo

ID

C1Rectifier circuit

C2 RL Vo

R

C1Rectifier circuit

C2 RL Vo

L


d. LC filter

3. Large capacitor filter value can decrease ripple voltage in the circuit because increasing capacitor value can effect in increasing time range (RLC).

4. Regulator used to decrease the changes from 0 to at least to a minimum value.

5. Three types of voltage regulator are :-

a. Zener diode voltage regulator

b. Series transistor voltage regulator

Rectifier circuit

C2 RL Vo

L

Rectifier circuit

Filter Dz

R

RL

RRectifier circuit

Filter

Q1

Dz

RL


c. IC voltage regulator.

6. Voltage divider circuits are needed in power supply because each electronics devices need different voltage.

Rectifier circuit

Filter

LM74051 2

3

C1

Vk

C2


1. What factors determine whether transformers in step up or in step down?

2. There are 3 types of rectifier circuits. Explain why bridge rectifier more popular than the others.

3.

1N4001 diode has breakdown voltage=50 V. Explain what will happen to the circuit in first and second positive cycle.

4. A power supply has regulation=1%. If voltage without load is 30 V, how much voltage at full load.

5.

Figure above shows bridge rectifier using silicon diode. Assume VF = 0.7 V.

a. Calculte transformers ratio.

b. Calculate Vo and frequency

c. Sketch Vo wave.

d. Give suggestion on how to decrease ripple voltage.

D

1k

1N4001

V

t

70V

-70V

D1

D2D3

D4

1k

230 Vp-pf = 50Hz

23 Vp-p

PENILAIAN KENDIRIPENILAIAN KENDIRI


1. Factor that determines whether transformers in step up or in step down is ratio at primary windings and secondary windings. If primary windings are more than secondary windings, transformer is in step down. If secondary windings are more than primary windings, transformer is in step up.

2. Bridge rectifier more popular than two other rectifiers because :-

a. Output voltage in full wave rectifier is larger than output voltage half wave rectifier.

b. Middle tap transformer tap more expensive than normal transformer. c. Output signal frequency is more higher than input signal frequency.

That makes filter process become better.

3. Forward biased and reversed biased will flow. At half positive cycle, diode is in reversed biased. Then, when input voltage is more than 50 V, breakdown diode and reversed biased will flow. In half negative cycle, diode is in forward biased. Forward biased heavily flow.

4.

5. a.

% regulation =

1 =

= 29.7 V

Ratio==

=

=

Then, the ratio = 10 : 1

MAKLUMBALASPENILAIAN KENDIRI

MAKLUMBALASPENILAIAN KENDIRI


b.

c.

d. To decrease the ripple voltage, we connect filter circuit at output.

Output voltage ,Vo = – 2 ( 0.7 V )

= 11.5 Vp – 1.4 V= 10.1 Vp

Frequency = 50 Hz x 2= 100 Hz

10.1Vp

V

t (ms)2010

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