Chapter-5 Diode Circuits and Power Supplies

Diode Circuits and Power Supplies 1

Muhammed Riyas A.M,Assistant Professor,Dept. of ECE,MCET Pathanamthitta

Chapter-5

Diode Circuits and Power Supplies

Contents

Diode circuits: Series and parallel diode circuits, Clippers, Clampers, Voltage multipliers

Rectifiers: Half-wave and full wave (including bridge) rectifiers, Derivation of Vrms, Vdc, ripple factor,

peak inverse voltage, rectification efficiency in each case, capacitor filter, working and design of a

simple zener voltage regulator.

Power Supplies: Block diagram description of a DC Power supply, Principle of SMPS.

5.1 Introduction

Diode circuits to be considered perform functions such as rectification, clipping. and clamping.

These functions are possible only because of the nonlinear properties of the pn junction diode. The

conversion of an ac voltage to a dc voltage, such as for a dc power supply, is called rectification. Clipper

diode circuits clip portions of a signal that are above or below some reference level. Clamper circuits shift

the entire signal by some dc value, Zener diodes, which operate in the reverse-bias breakdown region, have

the advantage that the voltage across the diode in this region is nearly constant over a wide range of

currents. Such diodes are used in voltage reference or voltage regulator circuits.

5.2 Clippers

There are a variety of diode networks called clippers that have the ability to “clip” off a portion of the

input signal without distorting the remaining part of the alternating waveform. Depending on the orientation

of the diode, the positive or negative region of the input signal is “clipped” off. Often, dc battery is also used to

fix the clipping level. The input waveform can be clipped at different levels by simply changing the battery

voltage and by interchanging the position of various elements.

Clipper circuits are used in radars and digital computers etc. when it is desired to remove signal

voltages above or below a specified voltage level. Another application is in radio receivers for communication

circuits where noise pulses that rise well above the signal amplitude are clipped down to the desired level.

There are two general categories of clippers: series and parallel. The series configuration is defined as

one where the diode is in series with the load, while the parallel variety has the diode in a branch parallel to

the load. Here we assume that the diode is ideal. This simply means that the turn-on voltage of the diode is

any voltage greater than 0 V, and the diode resistance is 0 Ω.

5.2.1 Series Clippers

When the diode is connected in series with the load to form a clipper, the configuration is called

series clippers. Series clippers can be used to clip either the positive or negative portions of the input signal.



Series Negative Clipper

During the positive half cycle of the input signal, the diode is turned on and it acts as closed switch. Hence no voltage drops across the diode. All the input voltage appears across the load, and hence the ouptut signal voltage is same as input voltage.

During negative half cycle of the input voltage, the diode is reverse biased and acts as open switch. In such a case, no current flows through the diode and the load, and hence the output voltage across the load is zero.

Fig: Series Negative Clipper

Biased Series Negative Clippers

If we want to change/adjust the clipping level of AC voltage, then external biasing voltage must be

used. The figure given below shows a biased (series) clipper. A biasing voltage is connected in series with the

diode as shown in the figure.



Fig: Biased Series Negative Clipper

When the input signal is positive, it tries to establish the current through the diode. However, the

biasing voltage V opposes the flow of current and tries to keep the diode in “off” state. As soon as the signal

voltage rises above voltage V, the diode starts to conduct and the current flows through the load. The output

appears across the load as soon as the current flows though it. Maximum value of output voltage is Vm – V.

Series Positive Clipper

Positive clippers are used to clip positive portions of the input signal and allow the negative portions

of the signal to pass through .That means,during the positive half cycle of the input signal, diode is reverse

biased and acts as open switch. In such a case, no current flows through the diode and the load, and hence the

output voltage across the load is zero.



Fig: Series Positive Clipper

During negative half cycle of the input voltage, the diode is turned on and it acts as closed switch.

Hence no voltage drops across the diode. All the input voltage appears across the load, and hence the ouptut

signal voltage is same as input voltage.

Biased Series Positive Clippers

When the input signal is positive, the diode is reverse biased, it is considered to be in “off” state and it

acts as an open switch. As the signal voltage goes positive, it aids the battery voltage in reverse biasing the

diode and consequently the reverse bias voltage across the diode increases. For the complete positive half

cycle, the diode stays in reverse bias and acts as open switch.

Now consider the instant when the signal voltage goes negative. As soon as the signal voltage goes

negative, it starts to oppose the battery voltage and tries to forward bias the diode. However at this point, the

signal voltage is not sufficient to forward bias the diode. Once the signal voltage is greater than the battery

voltage (we assume Vm to be greater than battery voltage V) it forward biases the diode. The diode is now in

the “on” state and output voltage is developed acoss the load. Minimum value of output voltage is -(Vm – V).



Fig: Biased Series Positive Clipper

In all the above discussions, note that the diode is considered to be ideal one. In a practical diode, the

breakdown voltage will exist (0.7 V for silicon and 0.3 V for Germanium).

5.2.2 Parallel Clippers

In parallel or shunt clipping circuits, the diode is connected in parallel to the output. In this discussions, the

diode is considered to be a practical one. In a practical diode, the breakdown voltage will exist (0.7 V for silicon

and 0.3 V for Germanium).

Parallel Positive Clipper

In this diode clipping circuit, the diode is forward biased during the positive half cycle of the sinusoidal input

waveform. For the diode to become forward biased, it must have the input voltage magnitude greater than

+0.7 volts (0.3 volts for a germanium diode).

When this happens the diodes begins to conduct and all positive half cycles are bypassed through the diode to

the ground. Thus the output voltage which is taken across the diode can never exceed 0.7 volts during the

positive half cycle. During the negative half cycle, the diode is reverse biased and does not conduct. As a result

input voltage will appears at the output. In this way positive clipper passes only negative going half cycle of the

input to the output.



Fig: Parallel Positive Clipper

Parallel Negative Clipper

The diode is forward biased during the negative half cycle of the sinusoidal waveform and limits or clips it to -

0.7 volts while allowing the positive half cycle to pass unaltered when reverse biased. As the diode limits the

negative half cycle of the input voltage it is therefore called a negative clipper circuit.

Fig:Parallel Negative Clipper

Combinational Clipper



If we connected two diodes in inverse parallel as shown, then both the positive and negative half cycles would

be clipped as diode D1 clips the negative half cycle of the sinusoidal input waveform while diode D2 clips the

positive half cycle.

For ideal diodes the output waveform above would be zero. However, due to the forward bias voltage drop

across the diodes the actual clipping point occurs at +0.7 volts and -0.7 volts respectively. But we can increase

this ±0.7V threshold to any value we want up to the maximum value, (Vm) of the sinusoidal waveform either by

connecting together more diodes in series creating multiples of 0.7 volts, or by adding a voltage bias to the

diodes.

Fig: Combinational clipper

Biased Diode Clipping Circuits

To produce diode clipping circuits for voltage waveforms at different levels, a bias voltage, V is added

in series with the diode as shown. The voltage across the series combination must be greater

than V + 0.7V before the diode becomes sufficiently forward biased to conduct. For example, if the V level is

set at 4.0 volts, then the sinusoidal voltage at the diode’s anode terminal must be greater than 4.0 + 0.7 = 4.7

volts for it to become forward biased.

Biased Positive Parallel Clippers

Here the positive terminal of the external bias is connected to the cathode of the diode. Therefore

the diode is reverse biased until the voltage across it is greater than V+.7V (here the voltage level is 4+.7= 4.7

Volt).When the input signal voltage level is greater than 4.7 volt, the diode becomes forward biased and begin

to conduct. Any voltage levels above 4.7 volt are clipped off.



Fig: Biased Positive Parallel Clipper

Biased Negative Parallel Clippers

Here the negative terminal of the external bias is connected to the anode of the diode. Therefore the

diode is reverse biased until the voltage across it is less than –V-.7V (here the voltage level is -4-.7= -4.7

Volt).When the input signal voltage level is less than 4.7 volt, the diode becomes forward biased and begin to

conduct. Any voltage levels below -4.7 volt are clipped off.

Fig: Biased Negative Parallel clipper

Biased Combinational Clippers

During the positive half cycle of the input signal,first branch is effective and others remain open.

When the voltage of the positive half cycle reaches +4.7 V, diode D1 conducts and limits the waveform at +4.7



V. Diode D2 does not conduct during this period. During the negative half cycle of the input, second branch will

effective and first branch is open. When the voltage of the negative half cycle reaches -4.7 V,

diode D2 conducts and limits the waveform at -4.7 V. Diode D1 does not conduct during this period. Therefore,

all positive voltages above +4.7 V and negative voltages below –4.7 V are automatically clipped.

Fig: Biased combinational Clipper

The advantage of biased diode clipping circuits is that it prevents the output signal from exceeding preset

voltage limits for both half cycles of the input waveform, which could be an input from a noisy sensor or the

positive and negative supply rails of a power supply.

If the diode clipping levels are set too low or the input waveform is too great then the elimination of both

waveform peaks could end up with a square-wave shaped waveform.

Drawbacks of Series and Parallel Diode Clippers

In series clippers, when the diode is in ‘OFF’ position, there will be no transmission of input signal to

output. But in case of high frequency signals transmission occurs through diode capacitance which is

undesirable. This is the drawback of using diode as a series element in such clippers.

In shunt clippers, when diode is in the ‘off condition, transmission of input signal should take place to

output. But in case of high frequency input signals, diode capacitance affects the circuit operation

adversely and the signal gets attenuated (that is, it passes through diode capacitance to ground).



5.3 Clampers

The clamping network is one that will “clamp” a signal to a different dc level. The network must have

a capacitor, a diode, and a resistive element, but it can also employ an independent dc supply to introduce an

additional shift. The magnitude of R and C must be chosen such that the time constant τ= RC is large enough

to ensure that the voltage across the capacitor does not discharge significantly during the interval the diode is

non conducting.

Clamping circuits are often used in TV receivers as dc restorers. The incoming composite video signal

is normally processed through capacitively-coupled amplifiers which eliminate the dc component thereby

losing the black and white reference levels and the blanking level. These reference levels have to be restored

before applying the video signal to the picture tube.

Note that voltage drop across the diode (.7 volt) is considered for representing the output wave forms

of the following clamper circuits and it is avoided in the discussions.

Positive Clamper

Let the input signal be Vm sint. During the negative half cycle of the input sine wave, the diode

conducts and capacitor charges to Vm (peak value of the input signal) with positive polarity at the right side of

the capacitor.Here we take the Vm as 5 volt. During the positive half cycle of the input signal, the capacitor

can’t discharge since the diode does not conduct. Thus the capacitor acts as a DC source of Vm volts connected

in series with the input signal source. Then the output voltage can be expressed as Vo= Vm+Vm sint.

Fig: Positive Clamper

Negative Clamper



During the positive half cycle of the input sine wave, the diode conducts and capacitor charges to Vm

with negative polarity at the right side of the capacitor. During the negative half cycle of the input signal, the

capacitor can’t discharge since the diode does not conduct. Thus the capacitor acts as a DC source of -Vm volts

connected in series with the input signal source. Then the output voltage can be expressed as Vo= -Vm+Vm

sint.

Fig: Negative Clamper

Biased Positive Clamper

Here we take a 3 volt battery as DC source connected in series with the diode in such a way that

positive terminal of the battery is connected to the anode terminal of the diode. During the negative half cycle

of the input sine wave, the diode conducts and capacitor charges through diode and the DC source till (Vm +3)

volts with positive polarity at the right side of the capacitor. The charging of the capacitor is limited to (Vm +3)

volts due to the presence of the DC source. The output is then represented as Vo= (Vm +3)+ Vm sint.



Fig: Biased Positive Clamper

Biased Negative Clamper

Here we take a 3 volt battery as DC source connected in series with the diode in such a way that

negative terminal of the battery is connected to the cathode terminal of the diode. During the positive half

cycle of the input sine wave, the diode conducts and capacitor charges through diode and the DC source till

(Vm +3) volts with negative polarity at the right side of the capacitor. The charging of the capacitor is extended

up to (-Vm-3) volts due to the presence of the DC source. The output is then represented as Vo= -Vm -3+ Vm

sint = -(Vm +3)+ Vm sint.

Fig: Biased Negative Clamper



5.4 Voltage Multipliers

Voltage multipliers are AC-to-DC power conversion devices, comprised of diodes and capacitors, that

produce a high potential DC voltage from a lower voltage AC source. Voltage multipliers are similar in many

ways to rectifiers in that they convert AC-to-DC voltages for use in many electrical and electronic circuit

applications such as in microwave ovens, strong electric field coils for cathode-ray tubes, electrostatic and high

voltage test equipment, etc, where it is necessary to have a very high DC voltage generated from a relatively

low AC supply.

Generally, the DC output voltage (Vdc) of a rectifier circuit is limited by the peak value of its sinusoidal

input voltage. But by using combinations of rectifier diodes and capacitors together we can effectively multiply

this input peak voltage to give a DC output equal to some odd or even multiple of the peak voltage value of the

AC input voltage.

Voltage multiplier circuits are classified as voltage doubler’s, tripler’s, or quadrupler’s, etc, depending

on the ratio of the output voltage to the input voltage. In theory any desired amount of voltage multiplication

can be obtained and a cascade of “N” doublers, would produce an output voltage of 2N.Vp volts.

Half-Wave Voltage Doubler

The circuit of a half-wave voltage doubler is given in figure shown below. During the positive half cycle of the

ac input, voltage, diode D1 being forward biased conducts (diode D2 does not conduct because it is reverse-

biased) and charges capacitor C1 upto peak values of secondary voltage Vm with the polarity, as marked in

figure shown below.

Fig: Half Wave Voltage Doubler

During the negative half-cycle of the input voltage diode D2 gets forward biased and conducts charging

capacitor C2. For the negative half cycle, the lower end of the transformer secondary is positive while upper

end is negative. The polarity of the capacitor C2 has also been marked in the figure. Now starting from the



bottom of the transformer secondary and moving clockwise and applying Kirchoff’s voltage law to the outer

loop we have

−푉 − 푉 + 푉 = 0

Or

푉 = 푉 + 푉 = 푉 + 푉 = 2푉 = Twice the peak value of the transformer secondary voltage. (Since Vc1 = Vm)

During the next positive half-.cycle diode D2 is reverse-biased and so acts as an open and capacitor

C2 discharges through the load If there is no load across the capacitor, C2 both capacitors stay charged – C1 to

Vm and C2 to 2Vm. If, as expected there is a load connected to the output terminals of the voltage doubler, the

capacitor C2 discharges a little bit and consequently the voltage across capacitor C2 drops slightly. The

capacitor C2 gets recharged again in the next half-cycle. The ripple frequency in this case will be the signal

frequency (that is, 50 Hz for supply mains.)

Full-Wave Voltage Doubler

The circuit diagram for a full-wave voltage doubler is given in the figure shown below. During the

positive cycle of the ac input voltage, diode D1 gets forward biased and so conducts charging the capacitor

C1 to a peak voltage Vm with polarity indicated in the figure, while diode D2 is reverse-biased and does not

conduct.During the negative half-cycle, diode D2 being forward biased conducts and charges the capacitor

C2 with polarity shown in the figure while diode D1 does not conduct. With no load connected to the output

terminals, the output voltage will be equal to sum of voltages across capacitors C1 and C2 that is, VC1 + VC2 or

(Vm + Vm) or 2 Vm. When the load is connected to the output terminals, the output voltage VL will be somewhat

less than 2 Vm.

Fig: Full Wave Voltage Doubler

Comparison between Half wave and Full Wave Voltage Doublers



Half-Wave voltage doubler

Availability of common point between input and output lines for earthing is available

Ripple content is high

Ripple frequency is the supply frequency

Voltage regulation is poor

Maximum voltage across each capacitor is 2 Vm

PIV rating of each diode is 2 Vm

Full-Wave voltage doubler

Availability of common point between input and output lines for earthing is not available.

Ripple content is low.

Ripple frequency is twice the supply frequency.

Voltage regulation is better than that of half-wave voltage doubler.

Maximum Voltage Across Each Capacitor is Vm.

PIV rating of each diode is 2 Vm.

Voltage Tripler and Quadruples

The half-wave voltage doubler, shown in the earlier figure can be extended to provide any multiple of

the peak input voltage (that is, 3 Vm, 4 Vm or 5 Vm), as illustrated in the figure shown below. It is obvious from

the pattern of the circuit connections how additional diodes and capacitors are to be connected to provide

output voltage, 5,6,7 or 8 times the peak input voltage from a supply transformer of rating only Vm, and each

diode in the circuit of PIV rating 2 Vm. If load is small and the capacitors have little leakage, extremely high dc

voltages can be obtained from such a circuit using many sections to step-up the dc voltage.

In operation capacitor C1 is charged through diode Dl to a peak value of transformer secondary

voltage, Vm during first positive half-cycle of the ac input voltage. During the negative half cycle capacitor C2 is

charged to twice the peak voltage 2 Vm developed by the sum of voltages across capacitor C1 and the

transformer secondary. During the second positive half-cycle, diode D3 conducts and the voltage across

capacitor C2 charges the capacitor C3 to the same 2 Vm peak voltage. During the negative half-cycle diodes

D2 and D4 conduct allowing capacitor C3 to charge capacitor C4to peak voltage 2 Vm. From the fogure shown

below it is obvious that the voltage across capacitor C2 is 2 Vm, across capacitors C1 and C3 it is 3 Vm and across

capacitors C2 and C4 it is 4 Vm.

If additional diodes (each diode of PIV rating 2 Vm) and capacitors (each capacitor of voltage rating 2

Vm) are used, each capacitor will be charged to 2 Vm. Measuring from the top of the transformer secondary

winding (figure below) will give odd multiples of Vm at the output, while measuring from the bottom of

transformer secondary winding will give even multiples of the peak voltage, Vm.



Fig: Voltage Tripler and Voltage Quadruplar

5.5 DC Power Supply:

The electric energy available in our country is in the form of AC voltage 230V, 50 Hz. But most of

electronic devices operate in DC power supplies. Almost all basic household electronic circuits need an

unregulated AC to be converted to constant DC. All devices will have a certain power supply limit and the

electronic circuits inside these devices must be able to supply a constant DC voltage within this limit. That

is, all the active and passive electronic devices will have a certain DC operating point (Q-point or

Quiescent point), and this point must be achieved by the source of DC power. The DC power supply is

practically converted to each and every stage in an electronic system. Thus a common requirement for all

this phases will be the DC power supply. Regulated DC power supply is an electronic circuit that is

designed to provide a constant dc voltage of predetermined value across load terminals irrespective of ac

mains fluctuations or load variations. The block diagram of a regulated DC power supply is shown in

figure.

Fig:Block Diagram of a Power Supply

The first block in the DC power supply is a transformer. A transformer is a device which transforms

high voltage AC into low voltage AC or vice versa. Our goal is to convert high voltage AC into low voltage DC. So



there is absolutely no reason to use step-up transformer. The transformer that is used in power supply is step-

down transformer, which steps down the input AC voltage. The main reason why we use transformer in the

system are as follows.

We want to reduce the voltage level which we get from the AC mains. Transformer can do the job of

reducing the voltage level in a simple and efficient manner.

The diodes used in the rectifier block cannot handle such a high level of voltage from the AC mains. So the

voltage is first stepped down by the transformer and the reduced voltage is applied to the rectifier section.

The output of the transformer is given to the input of the rectifier.

The rectifier converts the input sinusoidal voltage to a unipolar output or pulsating DC. Full wave rectifiers

are the most commonly used rectifiers in power supply. The output of the rectifier is given to input of the filter

circuit. The output after being processed by full wave rectifier is not a pure DC. The output is a pulsating DC.

The output contains large fluctuations in voltages. The power supply that we intend to design must not have

any variation in output voltage. The voltage that we get from full wave rectifier fluctuates between 0 V and

Vpeak, and hence it contains AC components. These AC components needs to be filtered out so as to obtain DC

voltage. The filter circuit reduces the variation in magnitude (ripples) of the rectifier output. However, the

output from the filter is still not a pure DC but filters removes the AC component in the voltage to a

considerable extent. This increases the average DC value of the output voltage.

The output of the filter is given to the voltage regulator circuit for stabilizing the magnitude of the DC output

voltage against the variations caused by changes in the load current and input voltage.

5.5.1 Characteristics of a power supply

There are various factors that determine the quality of the power supply like the load voltage, load current,

voltage regulation, source regulation, output impedance, ripple rejection, and so on. Some of the

characteristics are briefly explained below:

Load Regulation – The load regulation or load effect is the change in regulated output voltage when the

load current changes from minimum to maximum value.

퐿표푎푑푅푒푔푢푙푎푡푖표푛 = 푉 − 푉

푉 – Load Voltage at no load

푉 – Load voltage at full load.

From the above equation we can understand that when 푉 occurs the load resistance is infinite, that is,

the out terminals are open circuited. 푉 occurs when the load resistance is of the minimum value

where voltage regulation is lost.

%퐿표푎푑푅푒푔푢푙푎푡푖표푛 = 푉 − 푉 /푉 ∗ 100



Minimum Load Resistance – The load resistance at which a power supply delivers its full-load rated current

at rated voltage is referred to as minimum load resistance.

푀푖푛푖푚푢푚퐿표푎푑푅푒푠푖푠푡푎푛푐푒 = 푉 퐼⁄

The value of 퐼 , full load current should never increase than that mentioned in the data sheet of the

power supply.

Source/Line Regulation – In the block diagram, the input line voltage has a nominal value of 230 Volts but

in practice, here are considerable variations in ac supply mains voltage. Since this ac supply mains voltage is

the input to the ordinary power supply, the filtered output of the bridge rectifier is almost directly

proportional to the ac mains voltage.

The source regulation is defined as the change in regulated output voltage for a specified rage of lie voltage.

Output Impedance – A regulated power supply is a very stiff dc voltage source. This means that the output

resistance is very small. Even though the external load resistance is varied, almost no change is seen in the

load voltage. An ideal voltage source has an output impedance of zero.

Ripple Rejection – Voltage regulators stabilize the output voltage against variations in input voltage. Ripple is

equivalent to a periodic variation in the input voltage. Thus, a voltage regulator attenuates the ripple that

comes in with the unregulated input voltage. Since a voltage regulator uses negative feedback, the distortion is

reduced by the same factor as the gain.

5.6 Rectifiers:

Diode is a unidirectional device. It provides high resistance in one direction and a low resistance in the

other direction. Therefore it can be used as a rectifier for converting AC to DC.There are two types of rectifiers.

Half wave rectifier

Full wave rectifier

5.6.1 Half wave rectifier:

Half wave rectifier is the simplest form of the rectifier. The circuit arrangement is shown in fig(a).the

primary side is connected to the power mains(230 V,50 Hz).An AC voltage induces across the secondary

winding, which is equal to less than or greater than the primary voltage depending upon the turns ratio of

the transformer. This will be the input of the rectifier.



When the transformer secondary voltage is on its positive half cycle, the diode is forward biased and

current flows through the load RL, developing a voltage across it. When the negative voltage is on its

negative half cycle, the diode is reverse biased and no current flows through the RL.Thus there will be no

voltage across the output terminal during negative half cycle. The fig (b) shows the input and output wave

forms of a half wave rectifier. Here only positive voltages are obtained at the output and all negative half

cycles are suppressed. Thus we get a pulsating DC at the output.

Peak Inverse Voltage (PIV)

In the rectifier circuit, during negative half cycle of the secondary voltage, the diode is reverse biased.

As there is no voltage across the load resistor RL during this half cycle. The whole secondary voltage will

come across the diode. When the secondary voltage reaches its maximum value Vm in the negative half

cycle, the voltage across the diode in reverse bias is also maximum. This maximum voltage is known as

peak inverse voltage (PIV).It is the maximum voltage the diode must withstand during the reverse bias

half cycle of the input. Therefore the diode is used in the rectifier circuit must have the higher PIV rating

than this voltage.



Disadvantages:

Low output because only one half cycle delivers the output. AC components are more in the output.

5.6.1.1 Performance of Half wave Rectifier:

DC Output:

The DC output voltage or a current of the rectifier is the average value of the output voltage or current.

Ie,

The average value of a sine wave over one complete cycle is zero because it is symmetrical. But in the case of a half wave rectifier output, it consists of only alternate cycles as shown in output wave form. Hence there will be some DC current flowing through the RL.

Let,

Alternating voltage appearing across the secondary winding of the transformer V = V Sinωt

Current flowing through the load resistor i = I Sinωt

Diode forward resistance r = 0

From wave form, it is clear that current flowing through the load only in the period 0 to 휋.

i.e., i = I Sinωt ∶ for0 ≤ ωt ≤ π

i = 0 ∶ forπ ≤ ωt ≤ 2π Here area = ∫ ImSinωtd(ωt) = I [−Cosωt]

= 2 I

Time period = 2 π

Now, I = 2I /2π

I = I /π Voltage across RL V = I R

= I /πR

While diode forward resistance r is taken into account

I = V (R + r )⁄

DC voltage across RL can be written as

V = V R /π(R + r )

I = Areaunderonecompletecycle

Timeperiod



= V π(1 + r /R )⁄

Since, r ≪R

RMS Value:

Root mean square (RMS) is the effective value of the current flowing through the load and is given by

In the case of half wave rectifier,

Hence,

This is the rms value of the total current which include DC value and AC components. In the output of the rectifier, the instantaneous value of AC fluctuation is the difference of the instantaneous total value and the DC value.

Instantaneous AC value is given by

i = i − I

Here the rms value of AC component is given by,

i = I − I

V = V π⁄

퐼 =퐴푟푒푎푢푛푑푒푟푡ℎ푒푠푞푢푎푟푒푑푟푒푐푡푖푓푖푒푑푤푎푣푒표푣푒푟푎푐푦푐푙푒

푇푖푚푒푝푒푟푖표푑

I = 1

2π iπ

d(ωt)

=1

2π I Sin ωtd(ωt)

=I2π

(1− Cos2ωt)

2d(ωt)

=I

2π× 2ωt − Sin2ωt

2π0

Irms = Im / 2

Vrms = Vm / 2



Ripple Factor:

The purpose of the rectifier is to convert AC voltage to DC. But not type of rectifier converts AC to pure DC. It produces pulsating DC. This residual pulsation is called ripple. Ripple factor indicates the effectiveness of the rectifier in converting AC to perfect DC. It is the ratio of the ripple voltage to the DC voltage.

= V /V

Or 훾 = I /I

= I − I /I

= I I⁄ − 1

= (퐼 /2)/(퐼 /휋) − 1

= (1.57) − 1

= 1.21

Efficiency:

Efficiency of the rectifier is a measure of conversion of AC power to the useful DC output power. It is expressed as the ratio of DC output power to AC input power.

i.e., 휂 = 푃 푃⁄

Where 푃 is the power delivered to load and 푃 is the AC input power from the secondary winding of the transformer.

Here 푃 = 퐼 × 푅

= (퐼 휋⁄ ) × 푅

푃 = 퐼 (푅 + 푟 )

= (퐼 2⁄ ) (푅 + 푟 )

후 = 퐑퐢퐩퐩퐥퐞퐯퐨퐥퐭퐚퐠퐞퐃퐂퐯퐨퐥퐭퐚퐠퐞

=rmsvalueoftheACcomponent

DCcomponent



Therefore,

휂 =(퐼 휋⁄ ) × 푅

(퐼 2⁄ ) (푅 + 푟 )

= . 406푅(푅 + 푟 )

= . 406(1 + 푟 /푅 )

휂 = .406표푟40.6%

Transformer Utilization Factor:

For designing a power supply using transformer, transformer rating must be known. In order to calculate the

transformer utilization factor, the power supplied by the transformer must be arranged in terms of the d.c

power.

푉 × 퐼 = 푃

TUF is defined as the ratio of d.c output power to a.c power supplied to it by the secondary winding,

i.e,푇푈퐹 = (푟푎푡푒푑)

As the rating of the transformer is different than the actual power delivered by the secondary,the TUF is

entirely different quantity than efficiency.

In case of a single diode half wave rectifier,the rated voltage of the secondary=Vm/√2 but the actual r.m.s

value of the current flowing through the secondary winding =Im/2 and not Im/√2.

Thus

푇푈퐹 =(퐼푚 휋⁄ )2 × 푅퐿

푉√2

× 퐼2

=2√2휋

퐼 푅푉

푉 = 퐼 (푟 + 푅 )

푇푈퐹 =2√2휋

퐼 푅퐼 (푟 + 푅 )

Hence,

푇푈퐹 =2√2휋2 = .287

Regulation of Half Wave Rectifier

Regulation is defined as the variation of output DC voltage with the changes in the output DC current.

%푅푒푔푢푙푎푡푖표푛 =푉 − 푉

푉 × 100%



Where, 푉 = no load voltage.

푉 = full load voltage.

An ideal power supply should have a 0% regulation. The output voltage should not vary with load current IL.

For a half wave rectifier, the percentage regulation is given by,

%푅푒푔푢푙푎푡푖표푛 =

푉휋 − 푉

휋 − 푉휋푅 푅 + 푅

푉휋 − 푉

휋푅 푅 + 푅

Where,

푉 =푉휋

푉 = 푉 − 퐼

%푅푒푔푢푙푎푡푖표푛 =푉휋푉휋 ⎣⎢⎢⎢⎡ 1− 1 +

푅 + 푅푅

1−푅 + 푅푅 ⎦

⎥⎥⎥⎤

%푅푒푔푢푙푎푡푖표푛 = −푅 + 푅푅

Form Factor:

Form factor is the ratio of rms value to the average value.

퐹표푟푚푓푎푐푡표푟 =푟푚푠푣푎푙푢푒

푎푣푒푟푎푔푒푣푎푙푢푒

푓표푟푚푓푎푐푡표푟 = 푉 2⁄푉 휋⁄

= 휋 2 = 1.57⁄

Peak Factor:

Peak factor is the ratio of peak value to rms value.

푃푒푎푘푓푎푐푡표푟 =푝푒푎푘푣푎푙푢푒푟푚푠푣푎푙푢푒

푓표푟푚푓푎푐푡표푟 = 푉푉 2⁄

= 2

5.6.2 Full Wave Rectifiers:

In a full wave rectifier, current flows through the load during both half cycles of the input AC supply.

Thus in full wave rectifier, alternate half cycle of the input AC are inverted to get a unidirectional output

current. Full wave rectifiers are of two types,

Centre tap rectifier



Bridge rectifier

5.6.2.1 Centre Tap Rectifier:

The circuit arrangement of a centre-tap full wave rectifier is shown in fig(a).during positive half cycle

of the secondary voltage, the diode D1 is forward biased and D2 is reverse biased. The current flows through

D1, RL and upper half of the secondary winding. During negative half cycle of the secondary voltage, current

flows through D2, RL and lower half of the secondary winding. Here current through RL during both half cycle of

input AC in the same direction. Therefore the output voltage taken across RL will be DC. The input and output

wave forms of this rectifier are shown in fig (b).

Peak Inverse Voltage (PIV)

In the centre tap full wave rectifier, during positive half cycle of the secondary voltage diode D1 conducts

and when secondary voltage attains its maximum value Vm, a voltage equal to Vm will develop across RL with

the polarity marked in fig(c).The diode D2 at that instant is reverse biased and the voltage across it will be the

sum of the voltages developed across the lower winding of the secondary (Vm) and the voltage developed

across RL=Vm. Therefore peak inverse voltage across the diode in this case will be Vm+Vm=2Vm.



i.e, PIV=2 Vm

Advantages:

High efficiency

Low ripple factor

Disadvantages:

Requires centre tap rectifier with secondary voltage twice as required for half wave or bridge rectifier.

Diodes of high PIV (2V) rating are to be used.

5.6.2.2 Bridge Rectifier:

The circuit arrangement of the bridge rectifier is show in fig (a).It contains four diodes, but avoids the

need for a centre-taped transformer. During the positive half cycle of the secondary voltage, diode D2 and D3

will be forward biased. At the same time diodes D1 and D4 are reverse biased. There fore diodes D2 and D3

conduct and current flows through D2, RL, D3 and transformer secondary. During the negative half cycle of the

secondary voltage diodes D1 and D4 will be forward biased and D2 and D3 will be reverse biased. There fore the

current flows through D1, RL ,D4 and transformer secondary. Here in both cases the current flow through load

resistor RL is in the same direction. Hence DC voltage is obtained at the output. The wave forms are shown in

fig (b).



Peak inverse voltage (PIV)

When the secondary voltage is at its positive maximum value Vm diodes D2 and D3 are forward biased

and conduct. Conducting diodes have zero resistance and zero voltage drop across them, the points A and C in

fig (a) are in same potential. Similarly points B and D .Therefore the reverse voltage coming across diodes D1

and D4 will be equal to the potential difference between the points A and D will be equal to Vm.

Advantages:

Does not require centre tap rectifier.

Diodes of low PIV rating can be used.

Disadvantages:

Needs four diodes.

5.6.2.2 Performance of Full Wave Rectifier:

DC output:

In a full wave rectifier, it utilizes both half cycles of the input AC voltage. Therefore the DC voltage or average voltage available in the output of a full wave rectifier will be double the DC voltage available in the half wave rectifier.

i.e. 2× Vdc of the half wave rectifier



푉 = 2(푉 휋⁄ )

퐼 = 2(퐼 휋⁄ )

RMS Value:

When writing the rms value of the full wave rectifier both half cycles are to be considered. Hence the rms value of the current is given by,

Hence,

Ripple Factor:

훾 = I − I /I

= (I I⁄ ) − 1

= 퐼 /√2 /(2퐼 /휋) − 1

훾 = .482

I = 1π iπ

d(ωt)

=1π I Sin ωtd(ωt)

=Iπ

(1− Cos2ωt)2

d(ωt)

=I2π

ωt − Sin2ωt

2π0

Irms = Im / √2

Vrms = Vm / √2

=I π

2π



Efficiency:

휂 = 푃 푃⁄

푃 = 퐼 × 푅

= (2퐼 휋⁄ ) × 푅

푃 = 퐼 (푅 + 푟 )

= 퐼 √2⁄ (푅 + 푟 )

휂 =(2 퐼 휋⁄ ) × 푅

퐼 √2⁄ (푅 + 푟 )

= . 812(1 + 푟 /푅 )

( Where 푅 ≫ 푟 )

휂 = .812표푟81.2%

TUF of Centre Tap Rectifier The two diode full-wave rectifier can be thought of equivalent to two half wave rectifiers feeding to a common load. Hence TUF with respect to two half wave secondary can be written as

TUF (secondary) = 2 TUF of half wave = 2× .287

=.574

푇푈퐹(푃푟푖푚푎푟푦) =(2퐼푚 휋⁄ )2 × 푅퐿

푉√2

× 퐼√2

=8퐼휋

퐼 푅푉 퐼

=8휋

= .812

Thus,

푇푈퐹 = (푇푈퐹 + 푇푈퐹 )/2

=. 812 + .574

2 = .693

TUF of Bridge Rectifier

푇푈퐹 = 푃 푃⁄



푇푈퐹 =(2퐼푚 휋⁄ )2 × 푅퐿

푉√2

× 퐼√2

=8퐼휋

퐼 푅푉 퐼

=8휋

= .812

Regulation of Full Wave Rectifier

Regulation of output signal is defined as the variation of output DC voltage with the changes in the DC load current.

%푅푒푔푢푙푎푡푖표푛 =푉 − 푉

푉 × 100%

Where, 푉 = no load voltage.

푉 = full load voltage.

An ideal power supply should have a 0% regulation. The output voltage should not vary with load current IL. The % regulation is given by,

%푅푒푔푢푙푎푡푖표푛 =

2푉휋 − 2푉

휋 − 2푉휋푅 푅 + 푅

2푉휋 − 2푉

휋푅 푅 + 푅

Where,

푉 =2푉휋

푉 =2푉휋 − 퐼

%푅푒푔푢푙푎푡푖표푛 =2푉휋

2푉휋 ⎣

⎢⎢⎢⎡ 1− 1 +

푅 + 푅푅

1−푅 + 푅푅 ⎦

⎥⎥⎥⎤

%푅푒푔푢푙푎푡푖표푛 = −푅 + 푅푅

Form Factor:

퐹표푟푚푓푎푐푡표푟 =푟푚푠푣푎푙푢푒

푎푣푒푟푎푔푒푣푎푙푢푒

푓표푟푚푓푎푐푡표푟 = (푉 √2)⁄(2푉 휋⁄ )

= 휋 2√2 = 1.11⁄

Peak Factor:



푃푒푎푘푓푎푐푡표푟 =푝푒푎푘푣푎푙푢푒푟푚푠푣푎푙푢푒

푃푒푎푘푓푎푐푡표푟 = 푉(푉 √2)⁄

= √2

5.6.Problems:

1. In a half wave rectifier the load resistance RL=1KΩ the forward resistance of the diode rd=100Ω, input

alternating voltage (Vm) =325 volt. Find (a) Peak value, average value and rms value of the output current

(b) Efficiency of the rectifier.

Solution

(a) Peak value

퐼 = 푉 (푟 + 푅 )⁄

= 325 (100 + 1000)⁄

퐼 = 295.45푚퐴

Average value

퐼 = 퐼 휋⁄

= 295.45 휋⁄

퐼 = 94.046푚퐴

Rms value

퐼 = 퐼 /2

= 295.45 2⁄

= 147.725푚퐴

(b) Efficiency

휂 = 푃 푃⁄

푃 = 퐼 × 푅

= (94.046 × 10 ) × 1000

푃 = 8.845푊

푃 = 퐼 (푅 + 푟 )

= (147.725 × 10 ) (100 + 1000)

= 24푊

휂 = 푃 푃⁄

= 8.845/24

휂 = .3685표푟36.85%



2. In a full wave centre tap rectifier, the load resistance used is of 2KΩ,the forward resistance of each diode is

400Ω,the voltage across the half of the secondary winding is 240 Sin50t.Find (a) Im, (b) Idc (c) Irms (d) ripple

factor (e) PIV.

Solution

(a) 푰풎

퐼 = 푉 (푟 + 푅 )⁄

= 240 (400 + 2000)⁄

= .1퐴푚푝

(b) 푰풅풄

퐼 = 2 퐼 휋⁄

= (2 × .1) 휋⁄

= .636퐴푚푝 = 63.6푚퐴

(c) 푰풓풎풔

퐼 = 퐼 /√2

= . 1 √2⁄

= .070퐴푚푝 = 70.72푚퐴

(d) Ripple factor

훾 = (퐼 퐼⁄ ) − 1

= (. 07072/.0632) − 1

= .501표푟50.1%

(e) PIV

푃퐼푉 = 2푉 = 2 × 240 = 480 V

5.7 Filter Circuits:

The out put voltage of both half wave and full wave rectifier is pulsating and not pure DC.If such a

pulsating DC is given to any equipment, the working of the equipment will be disturbed. This disturbance of

ripples can be avoided by using filter circuits. Many types of passive filters are in use such as,

Shunt capacitor filter.

Series inductor filter.

Chock input LC filter.

Π- Section Filter.

5.7.1 Shunt Capacitor Filter:

This type of filter consists of a large value capacitor connected across the load resistance RL as shown

in fig (a).This capacitor offers low reactance path to AC components and very high impedance to DC.So that AC

components in the rectifier output find low reactance path through the capacitor and only a very small part

flows through RL, producing a small ripple voltage at the output as shown in fig (B).



When the diode conducts, the capacitor charges up to peak voltage Vm (point B in figure).From this

point onwards the rectifier voltage starts to fall, but the charged capacitor tries to send current back through

the diodes. Thus the diodes are reverse biased and become open circuited. The capacitor then starts to

discharge through the load. This action prevents the load voltage from falling to zero during no conduction

period of the diode. The capacitor continues to discharge until the source voltage becomes more than the

voltage across the capacitor(at point C).Then the diode begins to conduct and the capacitor again charges to

Vm (portion CD in figure).During this time, the rectifier delivers the charging current as well as load current.

Thus the capacitor removes the AC components and also improves the output voltage. The discharging rate of

the capacitor depends upon the time constant C.RL. The ripple factor in the capacitor filter is given by,

휸 = ퟏퟒ√ퟑ풇푪푹푳

Half Wave Rectifier with Capacitor Filter:



Full Wave Centre-tap Rectifier with Capacitor Filter:

Full Wave Bridge Rectifier with Capacitor Filter:



5.7.2 Ripple factor of the full wave rectifier with a capacitor filter

From the figure, it is obvious that discharging of the capacitor is given by,

푄 = 퐼 × 푇

The charge acquired by capacitor is,

푄 = 푉 × 퐶



We know that the charge acquired by the capacitor during the charging is equal to the charge lost during

discharge.

푄 = 푄

푉 × 퐶 = 퐼 × 푇

푉 × 퐶 = × 푇

푉 × 퐶 = × ---------------------(1)

For a full wave rectifier the capacitor charges and discharges the charge carries in both the half cycles and the

total time period is equal to charging and discharging of charges in positive cycle plus charging and discharging

of charges in negative cycle.

T = (T1 + T2) |positive cycle + (T1 + T2) |negative cycle

Since Tdischarge >> Tcharge , charging time Tcharge can be neglected.

Therefore, 푇 = -----------------------(2)

Substituting equation (2) in equation (1)

Vrp-p = ×

For the ripple waveform,

Vr rms = √

Vrp-p = 2√3 Vr rms ------------------------(3)

Substituting equation (3) in equation (2)

2√3 Vr rms = ×

Vr rms = √

But ripple factor is given by,

γ = = √

5.8 Zener Voltage Regulator:

The function of regulator is to provide an output voltage V0 that is as constant as possible in spite of

the ripple in source voltage, Vs and the variations in the load current IL. Two parameters used to measure the

performance of voltage regulator are line regulation and load regulation.

Line regulation is the ratio of change in output voltage to the change in source voltage.

퐿푖푛푒푟푒푔푢푙푎푡푖표푛 =∆푉∆푉

Load regulation is defined as the ratio of change in output voltage to the change in load current.

퐿표푎푑푟푒푔푢푙푎푡푖표푛 =∆푉∆퐼



Working of Zener Voltage regulator

The figure shows the basic circuit arrangement for a zener voltage regulator. Here the zener diode

with a break down voltage Vz is connected in reverse bias across the load. A resistor Rs is also connected in

series. If a voltage Vs is applied to the circuit, until the voltage across the load is less than the break down

voltage Vz the zener diode does not conduct and no current flows through the zener diode. At this time the

same voltage Vs will appear across the load. When the input voltage Vs is more than Vz, the zener diode will

conduct and a current Iz will flow through it. This excess current will increase the drop across Rs and limits the

zener current from exceeding the maximum rated value.Thus the current ‘I’ from the DC supply splits to the

load and to the zener diode.

At this time when the zener diode conducts the voltage across it is equal to its break down voltage Vz

and remains fairly constant even when the current through it vary considerably. Thus a constant voltage is

maintained across the load.

A figure of merit for a voltage regulator is called the percent regulation, and is defined as

% Regulation = ( ) ( )

( )× 100

where VL(nom) is the nominal value of the output voltage.

As the percent regulation approaches zero percent, the circuit approaches that of an ideal voltage regulator.

Analysis and Design

We first determine the proper input series resistance Rs. This resistance limits the current through the

Zener diode and drops the excess voltage between Vs and VZ.

Rs = =



which assumes that the Zener resistance is zero for the ideal diode. Solving this equation for the diode

current. Iz, we get

Iz = − 퐼

where IL = Vz / RL , and the variables are the input voltage source VS and the load current IL.

For proper operation of this circuit, the diode must remain in the breakdown region and the power

dissipation in the diode must not exceed its rated value. In other words:

1. The current in the diode is a minimum, Iz(min), when the load current is a maximum, IL(max), and the

source voltage is a minimum, VS(min).

2. The current in the diode is a maximum, Iz(max), when the load current is a minimum, IL(min), and

the source voltage is a maximum, VPS(max).

Inserting these two specifications into the previous equation, we obtain

Rs = ( )

( ) ( )

and

Rs = ( )

( ) ( )

Equating these two expressions, we then obtain

[푉 ( ) − 푉 ][퐼 ( ) + 퐼 ( )] =[푉 ( ) − 푉 ][퐼 ( ) + 퐼 ( )]

Reasonably, we can assume that we know the range of input voltage, the range of output load current, and

the Zener voltage. The previous equation then contains two unknowns Iz(min) and Iz(max). Further, as a

minimum requirement, we can set the minimum Zener current to be one-tenth the maximum Zener current,

or IZ(min) =0.1 IZ(max). We can then solve for IZ(max), using the previous equation, as follows:

Iz(max) = ( ) ( ) ( )[ ( ) ]

( ) . . ( )



Use the maximum current obtained from the above equation, we can determine the maximum required

power rating of the Zener diode. Then we can determine the required value of the input resistance using

one of the previous equations.

Limitations

A basic zener voltage regulator has the following draw backs :

(i) It has low efficiency for heavy load currents. It is because if the load current is large, there will be

considerable power loss in the series limiting resistance.

(ii) The output voltage slightly changes due to zener impedance as Vout = VZ + IZ ZZ. Changes in load current

produce changes in zener current. Consequently, the output voltage also changes. Therefore, the use of this

circuit is limited to only such applications where variations in load current and input voltage are small.

5.8.1 Problems:

1. The zener diode shown in figure has a fixed voltage drop of 18V across it so long as the zener current is

maintained between 200mA and 2Amp.(i) find the value of R so that the load voltage remains 18V as

input voltage is free to vary from 22V to 28V.(ii) Find the maximum power dissipated by the zener diode.

Solution

Minimum voltage across R=22-18=4 V.

Current through R, when 퐼 ( ) flows in 푅 is

퐼 ( ) + 퐼 ( ) =18푉18Ω + 200푚퐴 = 1200푚퐴

Therefore,

(i) 푅 =

= 3.33Ω

(ii) Maximum power dissipated = 푉 × 퐼 ( )

Given, 퐼 ( ) = 2퐴푚푝

Thus, Maximum power dissipated = 18푉 × 2퐴푚푝 = 36푊

2. For the circuit shown in figure, find the maximum and minimum value of zener diode current.



Solution

The zener diode will conduct maximum current, when the input voltage is maximum. i.e. 120 V.

The total current is the current through 5KΩ resistor and is

퐼 =120− 50

5퐾Ω = 14푚퐴

Load Current 퐼 = Ω

= 5푚퐴

Maximum zener current = 14− 5 = 9푚퐴

For minimum input voltage, zener current will be minimum. Therefore,

퐼 =80− 50

5퐾Ω = 6푚퐴

Minimum zener current = 6− 5 = 1푚퐴

3. Design a zener voltage regulator circuit that will maintain an output voltage of 20V across 1KΩ load when

the input voltage is 30-50V.Assume the zener knee current to be negligible compared to the load current.

Find the maximum power of the diode.

Solution

Load current 퐼 = Ω

= 20푚퐴;푅 = ( )

= .5퐾Ω = 500Ω

Maximum current through 푅 is

퐼 =50− 20. 5퐾Ω = 60푚퐴

We know that 퐼 = 퐼 + 퐼 . So,

퐼 = 퐼 − 퐼 = (60 − 20)푚퐴 = 40푚퐴

Maximum power rating of the zener diode is

40푚퐴 × 20푉 = 800푚푊 = .8푊



5.9 Switching Mode Power Supply (SMPS)

SMPS stands for switch mode power supply. In such a device, power handling electronic components

are continuously switching "on" and "off" with high frequency in order to provide the transfer of electric

energy via energy storage components (inductors and capacitors). By varying duty cycle, frequency or a

relative phase of these transitions an average value of output voltage or current is controlled. The operating

frequency range of commercial SMPS units varies typically from 50 kHz to several MHz.

Basic Concept

The basic concept behind a switch mode power supply or SMPS is the fact that the regulation is undertaken by

using a switching regulator. This uses a series switching element that turns the current supply to a smoothing

capacitor on and off.

Fig: Switching regulator concept used in SMPS

The time the series element is turned on is controlled by the voltage on the capacitor. If it is higher than

required, the series switching element is turned off, if it is lower than required, it is turned on. In this way the

voltage on the smoothing or reservoir capacitor is maintained at the required level.

Block Diagram of SMPS

Fig: Block Diagram of SMPS



If the SMPS has an AC input, then the first stage is to convert the input to DC by using a rectifier. A

SMPS with a DC input does not require this stage. In some power supplies, the rectifier circuit can be

configured as a voltage doubler by the addition of a switch operated either manually or automatically. This

feature permits operation from power sources that are normally at 115 V or at 230 V. The rectifier produces an

unregulated DC voltage which is then sent to a large filter capacitor.

The inverter stage converts DC, whether directly from the input or from the rectifier stage described above, to

AC by running it through a power oscillator, whose output transformer is very small with few windings at a

frequency of tens or hundreds of kilohertz. The frequency is usually chosen to be above 20 kHz, to make it

inaudible to humans. The switching is implemented as a multistage (to achieve high gain) MOSFET amplifier.

If the output is required to be isolated from the input, as is usually the case in mains power supplies, the

inverted AC is used to drive the primary winding of a high-frequency transformer. This converts the voltage up

or down to the required output level on its secondary winding. The output transformer in the block diagram

serves this purpose.

If a DC output is required, the AC output from the transformer is rectified. For output voltages above ten volts

or so, ordinary silicon diodes are commonly used. For lower voltages, Schottky diodes are commonly used as

the rectifier elements; they have the advantages of faster recovery times than silicon diodes and a lower

voltage drop when conducting.

The rectified output is then smoothed by a filter consisting of inductors and capacitors. For higher switching

frequencies, components with lower capacitance and inductance are needed.

A feedback circuit monitors the output voltage and compares it with a reference voltage, as shown in the block

diagram above. Depending on design and safety requirements, the controller may contain an isolation

mechanism (such as an opto-coupler) to isolate it from the DC output. Switching supplies in computers, TVs

and VCRs have these opto-couplers to tightly control the output voltage.

Advantages of SMPS

High efficiency: The switching action means the series regulator element is either on or off and

therefore little energy is dissipated as heat and very high efficiency levels can be achieved.

Compact: As a result of the high efficiency and low levels of heat dissipation, the switch mode power

supplies can be made more compact.

Flexible technology: Switch mode power supply technology can be sued to provide high efficiency

voltage conversions in voltage step up or "Boost" applications or step down "Buck" applications.

Disadvantages of SMPS



Noise: The transient spikes that occur from the switching action on switch mode power supplies are

one of the largest problems. The spikes can migrate into all areas of the circuits that the SMPSs power

if the spikes are not properly filtered. Additionally the spikes or transients can cause electromagnetic

or RF interference which can affect other nearby items of electronic equipment, particularly if they

receive radio signals.

External components: While it is possible to design a switch mode regulator using a single integrated

circuit, external components are typically required. The most obvious is the reservoir capacitor, but

filter components are also needed. In some designs the series switch element may be incorporated

within the integrated circuit, but where any current is consumed, the series switch will be an external

component. These components all require space, and add to the cost.

Expert design required: It is often possible to put together a switch mode power supply that works.

To ensure that it performs to the required specification can be more difficult. Ensuring the ripple and

interference levels are maintained can be particularly tricky.

Costs: Careful consideration of the costs of a switch mode power supply must be made before

designing or using one. Beyond the basic power supply, additional filtering may be required and this

can add to the cost.

Despite the disadvantages, switch mode power supply technology is the major form of power supply

technology used for a whole variety of applications especially those included in computers. For applications

where very low noise are required, linear regulator technology is still widely used.

Review Questions

Short Answer Questions

1. Explain any two clamper circuits.

2. A HWR having resistive load of 1000? , it rectifies an ac voltage of 325V peak value and the diode has

a forward resistance of 100?. Calculate a) peak, average and rms value of current , and b) efficiency.

3. Define line regulation and load regulation of a voltage regulator.

4. What are the advantages of SMPS?

5. Explain the operation of voltage multiplier.

Essay Questions

1. Explain any four parallel clipper circuits.

2. Derive the expressions for the rectification efficiency, ripple factor, transformer utilization factor,

form factor and peak factor of full wave rectifier.

3. Explain the full-wave bridge rectifier with capacitor filter. Also derive the expression for ripple factor.

4. Explain the operation of a zener diode voltage regulator.

5. With neat block diagram,explain the operation of SMPS.

Chapter-5 Diode Circuits and Power Supplies

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