UNIT-II DESIGN POWER SUPPLIES Power Supply Using Power Transistors

In this section of our studies we will be looking at the design of power supplies using power

transistors. We discussed the concepts of rectification and filtering using regular and zener

diodes in Section B, and we are going to start this section with a twist on our previous work –

adding a BJT. After this introduction using discrete transistors, we will be examining design

approaches using integrated circuits – both the 7800 series of integrated circuit regulators and the

LM317 adjustable regulator.

Power Supply Using Discrete Components

In Section B9 (3.4 of your text), we used the zener diode

as the voltage-controlling device in the design of a

regulated power supply. The figure to the right is a

modified version of Figure 8.20 in your text, where the

notation has been changed to correctly reproduce Figure

3.39. As we saw earlier, this is a fairly well behaved

circuit that supplies a nearly

constant output voltage over a wide range of currents.

However, as we always must, we can do better. A better regulation may be obtained if the zener

diode is connected to the base circuit of a power transistor in the EF (CC) configuration as shown

below (Figure 8.21 of your text).

In the configuration above, the transistor is referred to as a pass transistor. Because of the current

amplifying properties of the BJT in the EF (CC)configuration, the current through the zener

diode may be small. With the smaller current, there is little voltage drop across the diode

resistance, which allows the zener to approximate an ideal constant voltage source. The purpose

of CL in this circuit is to short out high frequency variations, so that

discrepancies in any of the regulated power supply loads will not be fed to any other loads.

If we assume the capacitor CL is open to dc and >>1, IL=IE= IB. Using KCL, the current

through the resistor Ri is the sum of the zener diode current plus the transistor base current, or

IR IZ IB

IZ I

L / .

Directly analogous to our work in Section B9, we are going to define two extremes for IZ in

terms of the input/output conditions:

1. Izmin occurs when the base current is maximum (IBmax=ILmax/ ) and the source voltage is

minimum.

2. Izmax occurs when the base current is minimum (IBmin=ILmin/ ) and the source voltage is

maximum.

Circuit analysis techniques yield an expression for the resistor of interest as in Equation 3.56,

where IL is replaced with IB. Equating the characteristics for the extremes of IZ (defined as Izmin

and Izmax) into the expression for Ri (=(Vs-VZ)/IR), we obtain

Ri Vs

min

IL max /

Vz

Iz

min

Vs

max IL min /

Vz

Iz

max

. (Equation 8.48)

Once again using the rule of thumb approximation Izmin=0.1Izmax, and performing massive

quantities of algebra, we can get the expression for Izmax to be

I IL

min(Vz

Vs min) IL max(Vs

max

Vz ) , (Equation 8.50)

z max (Vs

min

0.9Vz 0.1Vs max)

or, by rearranging our rule of thumb to Izmax=10Izmin, the expression for Izmin is found to be

I IL

min(Vz

Vs min) IL max (Vs

max

Vz ) .

z min (10V

s

mi

n

9Vz Vs max)

Note that the above expressions for the current through the zener diode are the same as those in

Section B9, except that the zener diode current has been reduced by the of the transistor. The

design of the power supply is performed exactly as previously with the exception of the reduced

IZ.

p L

To estimate the capacitor CF, we determine the equivalent load seen by this capacitor. Assuming

the impedance of CL is very large for dc (ideally open), it may be neglected in the parallel

combination of ZCL||RL. This leaves us with the worst case situation of load resistance for the

equivalent load, or

RL (equivalent) RL (worst case) VS min

. (Equation 8.51) I L max

The expression of Equation 8.51 was defined as the worst case since it represents the smallest

load and, therefore, the largest load current. Substituting Equation 8.51 into Equation 3.62, we

get an expression for CF:

C Vs max

VZ

. (Equation 8.52)

F Vf R (equivalent)

Since the voltage gain of an EF (CC) amplifier may be approximated as unity, the output voltage

of the regulated power supply is

Vout

VE

VB VB

E

VZ VBE . (Equation 8.53, Modified)

If we assume that VBE remains constant, the percent regulation of this power supply is given by

%regulation VZ max

VZ

min * 100 RZ (IZ

max

IZ min) * 100 , (Equation 8.55)

VZ VZ

where VZ is the nominal zener voltage. The percent regulation has been significantly reduced by

using the BJT in the circuit since both IZmax and IZmin are divided by the of the transistor.

Power Supply Using IC Regulator (Three-Terminal Regulator)

The IC regulator further improves the performance of the zener diode regulator by incorporating

an operational amplifier. Using a single IC regulator and a few external components, excellent

regulation may be obtained, along with good stability and reliability and built-in overload

protection. In the following discussion, basic design considerations for IC regulators used in the

design of power supplies for low power applications will be presented.

A functional block diagram of a generic IC

regulator using series regulation is

presented to the right and in Figure 8.22 of

your text. The series regulator is based on

the use of one or more pass transistors that

possess a variable voltage that is in series

with the output voltage. The voltage across

the pass transistor(s) is varied by the

output of the error amplifier so as to keep

the output voltage constant. Specifically,

A reference voltage, VREF, is compared with the voltage divided output, vout. The resulting

error voltage, ve is given by

v V R1vou

t

. (Equation 8.55)

e REF

R1 R2

ve is amplified through a discrete amplifier (shown as an operational amplifier in the figure),

and is used to change the voltage drop across the pass transistor. This feedback system

generates a variable voltage across the pass transistor to force the error voltage to zero.

When the error voltage is zero, and since R1, R2 and VREF are constants,

we obtain an expression for vout that is independent of any variations in load current or input

voltage from Equation 8.55:

⎛ R2 ⎞

Vout ⎜1

R ⎟VREF . (Equation 8.56)

⎝ 1 ⎠

The thermal shutdown and current limit circuitry that exists between the error amplifier and the

pass transistor(s) protects the regulator in case the temperature becomes too high or the current

too large.

The maximum power dissipated in this type of series regulator is the power dissipated in the pass

transistor, which is approximately equal to

Pmax (Vin

max

Vout )IL max .

Therefore, as the load current increases, the power dissipated in the pass transistor increases. It is

extremely important that the manufacturer’s recommendations as to maximum current with

and/or without a heat sink be

followed. Your text uses the example of an ILmax 0f 0.75A without a heat sink that may be

increased to 1.5A when the IC package is correctly secured to a heat sink.

Power Supply Using 7800 Series IC Regulators

In the 7800 family of IC regulators, the last two digits

indicate the output voltage of the device. There are a number

of different voltages that may be obtained from the 7800 ICs.

The specific IC designations, as well as the output voltage

and maximum and minimum input voltages, are given to the

right in a modified version of Table

8.1 of your text. The specification sheets for the 7800 series

are given in the Appendix of your text (pages 949-954).

A typical circuit application

of the 7800 IC family is shown in

Figure 8.23b and is reproduced to

the right. The terminal designations

(shown in blue) are based on the

three-terminal package of Figure

8.23a. In order to design a regulator

using one of these ICs, and ensure that the

required minimum and maximum input voltages are maintained, we need to select an appropriate

transformer and filter capacitor.

The information of Table 8.1 is used to select the turns ratio (n=N1/N2, where N1 is the number

of turns on the primary side and N2 is the number of turns on the secondary side). As a design

guide, and as a conservative method for selecting the transformer turns ratio, it is common

practice to proceed as follows:

Take the average of Vinmax and Vinmin of the particular IC regulator to calculate AVG. Your

text uses the example of the 7805 regulator, where Vinmax=25V and Vinmin=7V, for AVG=16.

Using the center tapped transformer provides a division by 2, so the peak voltage out of the

rectifier is

AVG (ac line voltage (rms)) * /(2n) . 2

Type vout vinmin vinmax

7805 5 7 25

7806 6 8 25

7808 8 10.5 25

7885 8.5 10.5 25

7810 10 12.5 28

7812 12 14.5 30

7815 15 17.5 30

7818 18 21 33

7824 24 27 38

2

p

Once again, for the example using the 7805, if the rms value of the ac

line voltage is 115V, we get 115 / 2n 16 , or, n=5.

The filter capacitor is chosen to maintain the input voltage to the regulator as specified in

Table 8.1 by calculating

C Vmax ,

F Vf R

where

Vmax is the average of Vinmax and Vinmin from Table 8.1, V = Vmax-Vinmin

fp=120 for a 60Hz input (full wave recitification), and

RL=Vinmin/ILmax (or the worst case value).

According to the manufacturer’s data sheet, the capacitor at the output (CL) is not needed for

stability; however, it does improve the transient response of the regulator. It is further noted that

capacitance values of less than 0.1 F could cause instability. Your author states that CL should

be a high quality tantalum capacitor with a capacitance of 1.0 F, and should be connected close

to the 78XX regulator using short leads to improve the stability performance. Other possible

applications for the 78XX regulator are illustrated on page 953 of your text.

Power Supply Using Three-Terminal Adjustable Regulator

The LM317 IC is an adjustable three- terminal

positive regulator that is capable of supplying

more than 1.5A over an output range of 1.2 to

37V. Figure 8.24, given to the right, shows a

connection diagram for the LM317 with the

terminal designations indicated in blue. Setting

the output voltage requires only two external

resistors,

denoted R1 and R2 in the figure. As for the 7800 family, the capacitor CL is optional. When it is

included, the transient response is improved through the rejection of transients that may appear

on the regulated supply line. Your author further states that the capacitor C1 is needed if the

device is physically located far from filter capacitors. However, an input bypass capacitor is

usually used to short out any high frequency variations that may occur in adjoining circuitry.

L

The voltage, VREF, maintains a nominal 1.25V that is developed from a precision internal voltage

reference. VREF appears between the output and adjustment terminals and across the program

resistor, R1. Since VREF and R1 are constants, there is a constant current through R1 of

I1=VREF/R1. The output voltage is then given by

⎛ R2 ⎞

Vout

VREF

(I1

I ADJ )R2

VREF ⎜1 R

⎟ IADJR2 . (Equation 8.58)

⎝ 1 ⎠

The LM317 is packaged in a standard transistor package and provides both current limiting and thermal

overload protection. Your author states that both line and load regulations are better than in standard fixed

voltage regulators.

Higher Current Regulator

The regulators that we’ve been talking about so far, in

common with most IC regulators, are limited to an

output current of about 1.5A due to the large amount

of power dissipated in the internal pass transistor(s).

The configuration presented in Figure 8.25, and

reproduced to the

right, allows the output current to increase to about 5A while still preserving the thermal shutdown and short

circuit protection of the IC.

The concept of this circuit is that the external power transistor Q1, which acts as a pass transistor for the

regulator, provides 80% of the load current, while the regulator carries only 0.2 iL. This current sharing is

accomplished by R1, R2 and D1. If the VBE of Q1 and the Von of D1 are made equal by design, and iB is

assumed negligible, the voltages across R1 and R2 are equal. If R2 is chosen to be 4R1, the current through

Q1 is four times the current through the LM317. The resistor R3 is included to provide a dc bias path for Q1

to ensure that the transistor is properly biased.

SCR and THYRISTOR

SILICON CONTROLLED RECTIFIERS (SCR)

A silicon controlled rectifier is a semiconductor device that acts as a true electronic

switch. it can change alternating current and at the same time can control the amount of power

fed to the load. SCR combines the features of a rectifier and a transistor.

CONSTRUCTION

When a pn junction is added to a junction transistor the resulting three pn junction device

is called a SCR. ordinary rectifier (pn) and a junction transistor (npn) combined in one unit to

form pnpn device. three terminals are taken : one from the outer p­ type material called anode a

second from the outer n­ type material called cathode K and the third from the base of transistor

called Gate. GSCR is a solid state equivalent of thyratron. the gate anode and cathode of SCR

correspond to the grid plate and cathode of thyratron SCR is called thyristor

WORKING

Load is connected in series with anode the anode is always kept at positive potential w.r.t

cathode.

WHEN GATE IS OPEN

No voltage applied to the gate, j2 is reverse biased while j1 and j3 are FB . J1 and J3 is

just in npn transistor with base open .no current flows through the load RL and SCR is cut off. if

the applied voltage is gradually increased a stage is reached when RB junction J2 breakdown .the

SCR now conducts heavily and is said to be ON state. the applied voltage at which SCR

conducts heavily without gate voltage is called Break over Voltage.

WHEN GATE IS POSITIVE W.R.T CATHODE.

The SCR can be made to conduct heavily at smaller applied voltage by applying small

positive potential to the gate.J3 is FB and J2 is RB the electron from n type material start moving

across J3 towards left holes from p type toward right. electrons from j3 are attracted across

junction J2 and gate current starts flowing. as soon as gate current flows anode current increases.

the increased anode current in turn makes more electrons available at J2 breakdown and SCR

starts conducting heavily. the gate looses all control if the gate voltage is removed anode current

does not decrease at all. The only way to stop conduction is to reduce the applied voltage to zero.

BREAKOVER VOLTAGE

I t is the minimum forward voltage gate being open at which SCR starts conducting

heavily i.e turned on

PEAK REVERSE VOLTAGE( PRV)

It is the maximum reverse voltage applied to an SCR without conducting in the reverse

direction.

HOLDING CURRENT

It is the maximum anode current gate being open at which SCR is turned off from on

conditions.

FORWARD CURRENT RATING

It is the maximum anode current that an SCR is capable of passing without destruction

CIRCUIT FUSING RATING

It is the product of of square of forward surge current and the time of duration of the

surge

VI CHARACTERISTICS OF SCR

FORWARD CHARCTERISTICS

When anode is +ve w.r.t cathode the curve between V & I is called Forward

characteristics. OABC is the forward characteristics of the SCR at Ig =0. if the supplied voltage

is increased from zero point A is reached .SCR starts conducting voltage across SCR suddenly

drops (dotted curve AB) most of supply voltage appears across RL

REVERSE CHARCTERISTICS

When anode is –ve w.r.t.cathode the curve b/w V&I is known as reverse characteristics

reverse voltage come across SCR when it is operated with ac supply reverse voltage is increased

anode current remains small avalanche breakdown occurs and SCR starts conducting heavily is

known as reverse breakdown voltage

SCR as a switch

SCR Half and Full wave rectifier

Application

SCR as a static contactor

SCR for power control

SCR for speed control of d. c. shunt motor

Over light detector

Triggering (Turn on) Methods of Thyristor:

Triggering:­

The turning on Process of the SCR is known as Triggering. In other words, turning the SCR from

Forward­Blocking state to Forward­Conduction state is known as Triggering.The various

methods of SCR triggering are discussed here.

The various SCR triggering methods are

Forward Voltage Triggering

Thermal or Temperature Triggering

Radiation or Light triggering

dv/dt Triggering

Gate Triggering

(a) Forward Voltage Triggering:­

In this mode, an additional forward voltage is applied between anode and cathode. When the anode terminal is positive with respect to cathode(VAK) , Junction J1 and J3 is

forward biased and junction J2 is reverse biased.

No current flows due to depletion region in J2 is reverse biased (except leakage current).

As VAK is further increased, at a voltage VBO (Forward Break Over Voltage) the

junction J2 undergoes avalanche breakdown and so a current flows and the device tends to turn ON(even when gate is open)

(b) Thermal (or) Temperature Triggering:­

The width of depletion layer of SCR decreases with increase in junction temperature. Therefore in SCR when VAR is very near its breakdown voltage, the device is triggered

by increasing the junction temperature.

By increasing the junction temperature the reverse biased junction collapses thus the device starts to conduct.

(c) Radiation Triggering (or) Light Triggering:­

For light triggered SCRs a special terminal niche is made inside the inner P layer instead of gate terminal.

When light is allowed to strike this terminal, free charge carriers are generated.

When intensity of light becomes more than a normal value, the thyristor starts conducting.

This type of SCRs are called as LASCR

(d) dv/dt Triggering:­

When the device is forward biased, J1 and J3 are forward biased, J2 is reverse biased.

Junction J2 behaves as a capacitor, due to the charges existing across the junction.

If voltage across the device is V, the charge by Q and capacitance by C then, ic=dQ/dt

Q=CV ic=d(CV)/dt

C.dV/dt+V.dC/dt

as dC/dt = 0

ic = C.dV/dt

Therefore when the rate of change of voltage across the device becomes large, the device may turn ON, even if the voltage across the device is small.

(e) Gate Triggering:­

This is most widely used SCR triggering method.

Applying a positive voltage between gate and cathode can Turn ON a

forward biased thyristor.

When a positive voltage is applied at the gate terminal, charge carriers are

injected in the inner P­layer, thereby reducing the depletion layer thickness.

As the applied voltage increases, the carrier injection increases, therefore the

voltage at which forward break­ over occurs decreases.

CROWBAR CIRCUT

A crowbar circuit is an electrical circuit used for preventing an overvoltage condition of a

power supply unit from damaging the circuits attached to the power supply. It operates by putting

a short circuit or low resistance path across the voltage output (Vo).Crowbar circuits are

frequently implemented using a thyristor, TRIAC, trisil orthyratron as the shorting device.

Fig1.1Circuitdiagramofcrowbarprotectioncircuit

Operation of the circuit

During normal operating conditions, the voltage across R2 is slightly lower than VREF of the

LM431.Since this voltage is below the minimum reference voltage of the LM431, it

remainsoffandverylittlecurrentisconductedthroughtheLM431.Ifthesupplyvoltage

increases,thevoltageacrossR2willexceedVREFandtheLM431cathodewillbeginto draw current.

The voltage at the gate terminal will be pulled down, exceeding the gate trigger voltage of the

TRIAC and latching.

Applications of crowbar circuit

High voltage crowbars are used for HV tube (Klystron and IOT) protection.

Many bench top power supplies have a crowbar circuit to protect the connected

equipment.

Microwave ovens often use a micro switch that acts as a crowbar circuit in the door latch

assembly. This will absolutely prevent the magnetron from being energized with the door

open. Activation will blow the main fuse and ruin the micro switch.

FOLDBACKPOWERSUPPLYDESIGN

The disadvantage of constant current limit is relatively large power dissipation in the pass

transistor when the load terminals are shorted. Thus, large power rating transistor is required. The

fold back limiting technique allows us to provide the necessary load current at rated voltage but

reducing the short circuit current. Thus the series pass transistor gets utilized efficiently. The

basic circuit for fold back limiting is shown in the figure below.

Circuit analysis

All the voltages at point AbeVAandthecurrentflowingthroungR4isalmostIL

VA=ILR4+Vo

NeglectingthebasecurrentflowingthroughR5andR6isthesameasI I=

VA

(R5+R6)

HencethevoltageatthebaseofQ3isthevoltageacrossR6

VB3= VA

.R6 R5+R6

R5+R6}

k= R6

R5+R6

VB3=k(ILR4+Vo)

ThevoltageacrosstheemitterofQ3is

VB=Vo

VBE3=VB-VE3

=kILR4+(k-1)Vo

IL=VBE3+ (1-k)Vo

kR4

Thusiftheoutputterminalsareshorted,theoutputvoltageVoreducestozero

Hence we get from the equation(6)

Isc=IL=VBE3/kR4

The rated current can be written as,

IL=Isc+ (1-k)Vo

{kR4}

IL =Rated load current

Isc=Short circuit current

The rated load current IL is also called Iknee known as knee current

Switched-mode power supply

A switched-mode power supply (switching-mode power supply, switch-mode power

supply, switched power supply, SMPS, or switcher) is an electronic power supply that

incorporates a switching regulator to convert electrical power efficiently. Like other power

supplies, an SMPS transfers power from a DC or AC source (often mains power) to DC loads, such

as a personal computer, while converting voltage and current characteristics. Unlike a linear power

supply, the pass transistor of a switching-mode supply continually switches between low-

dissipation, full-on and full-off states, and spends very little time in the high dissipation transitions,

which minimizes wasted energy. Ideally, a switched-mode power supply dissipates no

power. Voltage regulation is achieved by varying the ratio of on-to-off time. In contrast, a linear

power supply regulates the output voltage by continually dissipating power in the pass transistor.

This higher power conversion efficiency is an important advantage of a switched-mode power

supply. Switched-mode power supplies may also be substantially smaller and lighter than a linear

supply due to the smaller transformer size and weight.

Need of switched mode power supply

A linear regulator power supply has following limitations.

1. There quire input step down transformer is bulky and expensive.

2. Due to low line frequencies (50Hz),large value so filter capacitors are required.

3. The efficiencyisverylow

4. Inputmustbegreaterthantheoutputvoltage

5. As large is the difference between input and output voltage, more is the power dissipation

intheseriespasstransistor.

Switching regulators are used as replacements for linear regulators when higher efficiency, smaller

size or lighter weights are required. They are, however, more complicated; their switching currents

can cause electrical noise problems if not carefully suppressed, and simple designs may have a

poor power factor.

A linear regulator provides the desired output voltage by dissipating excess power in ohmic

losses (e.g., in a resistor or in the collector–emitter region of a pass transistor in its active mode). A

linear regulator regulates either output voltage or current by dissipating the excess electric power in

the form of heat, and hence its maximum power efficiency is voltage-out/voltage-in since the volt

difference is wasted.

In contrast, a switched-mode power supply changes output voltage and current by switching ideally

lossless storage elements, such as inductors and capacitors, between different electrical

configurations. Ideal switching elements (approximated by transistors operated outside of their

active mode) have no resistance when "on" and carry no current when "off", and so converters with

ideal components would operate with 100% efficiency

For example, if a DC source, an inductor, a switch, and the corresponding electrical ground are

placed in series and the switch is driven by a square wave, the peak-to-peak voltage of the

waveform measured across the switch can exceed the input voltage from the DC source. This is

because the inductor responds to changes in current by inducing its own voltage to counter the

change in current, and this voltage adds to the source voltage while the switch is open. If a diode-

and-capacitor combination is placed in parallel to the switch, the peak voltage can be stored in the

capacitor, and the capacitor can be used as a DC source with an output voltage greater than the DC

voltage driving the circuit. This boost converter acts like a step-up transformer for DC signals.

A buck–boost converter works in a similar manner, but yields an output voltage which is opposite

in polarity to the input voltage. Other buck circuits exist to boost the average output current with a

reduction of voltage.

In an SMPS, the output current flow depends on the input power signal, the storage elements and

circuit topologies used, and also on the pattern used (e.g., pulse-width modulation with an

adjustable duty cycle) to drive the switching elements. The spectral density of these switching

waveforms has energy concentrated at relatively high frequencies. As such, switching transients

and ripple introduced onto the output waveforms can be filtered with a small LC filter.

Block diagram of SMPS

The figure below shows the functional block diagram of basic switching voltage

regulatorwhichusestransistorQ1asaswitch.

Operation

ThepartR2/R1+R2oftheoutputisfedbacktotheinvertinginputoftheerroramplifier. It is compared

with the reference voltage. The differnce is amplified and given to the

comparatorinvertingterminal.

The oscillator generates a triangular waveform at a fixed frequency. It is applied to

thenon-invertingterminalofthecomparator.Theoutputofthecomparatorishighwhen the

triangular voltage waveform is above the level of the error amplifier output. Due to

thisthetransistorQ1remainsinthecut-offstate.Thustheoutputofthecomparatoris

nothingbutareguiredpulsewaveform.

The period of this pulse waveform is same as that of oscillator output say T. The duty

cycleisdenotedas¤=ton/Tortonf.Thisdutycycleiscontrolledbythedifferencbetween

thefeedbackvoltageandthereferencevoltage.

When Q1 is on in saturation state, Vce for Q1 is zero. Hence entire input voltage Vin

appearsatpointA.ThusthecurrentflowsthroughinductorL1.

WhenQ1 isoff,L1 stillcontinuetosupplycurrentthroughitselftotheload.Thediode

isprovidesthereturnpathforthecurrent.

The capacitor C1 acts to smooth out the voltage and the voltage at the output is

almost d.c in nature. The output voltage Vo of the switching regulator is a function of

dutycycleandtheinputvoltageVin.Mathematicallyitisexpressedas

Vo=to

n/T×

Vin

App

licat

ions

1.Indomesticproductslikecomput

ers

2.Inswitchedmodemobilephonecharg

er

3.UsedforDCtoDCconversion(inoutomobiles

)

4.Usedasextralow-

voltagesourceforlighting

Advantages and disadvantages

The main advantage of the switching power supply is greater efficiency than linear

regulators because the switching transistor dissipates little power when acting as a switch.

Other advantages include smaller size, low-noise, and lighter weight from the elimination

of heavy line-frequency transformers, and comparable heat generation. Standby power

loss is often much less than transformers.

Disadvantages include greater complexity, the generation of high-amplitude, high-

frequency energy that the low-pass filter must block to avoid electromagnetic

interference (EMI), a ripple voltage at the switching frequency and the harmonic

frequencies thereof.

Very low cost SMPSs may couple electrical switching noise back onto the mains power

line, causing interference with A/V equipment connected to the same phase. Non-power-

factor-corrected SMPSs also cause harmonic distortion.

FORWARD CONVERTER

The forward converter is a DC/DC converter that uses a transformer to increase or decrease the output

voltage (depending on the transformer ratio) and provide galvanic isolation for the load. With multiple

output windings, it is possible to provide both higher and lower voltage outputs simultaneously.

While it looks superficially like a fly back converter, it operates in a fundamentally different way, and is

generally more energy efficient. A fly back converter stores energy in the magnetic field in the inductor air

gap during the time the converter switching element (transistor) is conducting. When the switch turns off,

the stored magnetic field collapses and the energy is transferred to the output of the flyback converter as

electric current. The fly back converter can be viewed as two inductors sharing a common core with

opposite polarity windings

.

In contrast, the forward converter (which is based on a transformer with same-polarity windings, higher

magnetizing inductance, and no air gap) does not store energy during the conduction time of the switching

element — transformers cannot store a significant amount of energy, unlike inductors.[1] Instead, energy is

passed directly to the output of the forward converter by transformer action during the switch conduction

phase.

While the output voltage of a flyback converter is theoretically infinite, the maximum output voltage of the

forward converter is constrained by the transformer turns ratio.

FLYBACK CONVERTER

The flyback converter is used in both AC/DC and DC/DC conversion with galvanic Isolation

between the input and any outputs. The flyback converters use galvanic Isolation because the

switch and the load are at different ground potentials. It is an effective way of breaking the

ground loops by preventing unwanted current from flowing between two units sharing a

ground conductor. It also prevents accidental current from reaching ground through a person’s

body. The galvanic Isolation can be done in various ways such as applying the use of

Transformer, Opto-isolator, Capacitor, Hall Effect, Magneto resistance and Relay. The

flyback converter is a buck-boost converter with the Inductor split to form a Transformer.

The flyback converter is an isolated power converter. The two prevailing control schemes are

voltage mode control and current mode control (in the majority of cases current mode control

needs to be dominant for stability during operation). Both require a signal related to the output

voltage. There are three common ways to generate this voltage. The first is to use an Opto-

coupler on the secondary circuitry to send a signal to the controller. The second is to wind a

separate winding on the coil and rely on the cross regulation of the design. The third consists

of sampling the voltage amplitude on the primary side, during the discharge, referenced to the

standing primary DC voltage.

A variation in primary-side sensing technology is where the output voltage and current are

regulated by monitoring the waveforms in the auxiliary winding used to power the Control IC

itself, which have improved the accuracy of both voltage and current regulation. The auxiliary

primary winding is used in the same discharge phase as the remaining secondary, but it builds

a rectified voltage referenced commonly with the primary DC, hence considered on the

primary side.

Previously, a measurement was taken across the whole of the flyback waveform which led to

error, but it was realized that measurements at the so-called knee point (when the secondary

current is zero, see Fig. 1) allow for a much more accurate measurement of what is happening

on the secondary side. This topology is now replacing ringing choke converters (RCCs) in

applications such as mobile phone chargers.

Fig 1.1Wave form-using primary side sensing techniques- showing the knee

Fig1.2 ON state of the flyback converter

Fig 1.3 OFF state of the flyback converter.

OPERATION When the switch is ON

When the switch is closed (see Fig. 2), the primary of the transformer is directly connected to the

input voltage source. The primary current and magnetic flux in the transformer increases storing

energy in the transformer. The voltage induced in the secondary winding is negative, so the diode is

reverse-biased (i.e., blocked). The output capacitor supplies energy to the output load.

When the switch is OFF

When the switch is opened (see Fig. 3), the primary current and magnetic flux drops. The secondary

voltage is positive, forward-biasing the diode, allowing current to flow from the transformer. The

energy from the transformer core recharges the capacitor and supplies the load.

BUCK CONVERTER

This is the DC-DC converter which reduces the input voltage to the output voltage (load). Its

Objective is to efficiently reduce DC voltage.

Lossless objective: 𝑃𝑖 𝑛=𝑃𝑜𝑢𝑡 , This means

Capacitors and Inductors

With infinite capacitor acts as a constant voltage source. Capacitors tend to keep the voltage

constant. Thus a capacitor cannot be connected in parallel with a voltage source or a switch

otherwise KVL would be violated thus there will be no short circuit.

In inductors

Inductors tend to keep the current constant. An ideal inductor with infinite inductance acts as a

constant current source thus inductor cannot be connected in series with a current source.

𝑉𝑖𝑛 𝑉𝑜𝑢𝑡

𝐼𝑖𝑛 𝐼𝑜𝑢𝑡

+ +

− −

𝑉𝑖𝑛

𝐼𝑖𝑛

𝐼𝑖𝑛=

𝑉𝑜𝑢𝑡 𝐼𝑜𝑢𝑡

4

The input/output equation for DC-DC converters usually comes by examining

inductor voltages

Vin

+Vout

–

L

C

Ioutiin

+ (Vin – Vout) –

iL

(iL – Iout)

Reverse biased, thus the diode is

open

,dt

diLv L

L L

VV

dt

di outinL ,

dt

diLVV L

outin ,outinL VVv

for DT seconds

Note – if the switch stays closed, then Vout = Vin

Switch closed for DT seconds

5

Vin

+Vout

–

L

C

Iout

– Vout +

iL

(iL – Iout)

Switch open for (1 − D)T seconds

iL continues to flow, thus the diode is closed. This is

the assumption of “continuous conduction” in the

inductor which is the normal operating condition.

,dt

diLv L

L L

V

dt

di outL ,

dt

diLV L

out ,outL Vv

for (1−D)T seconds

6

Since the average voltage across L is zero

01 outoutinLavg VDVVDV

outoutoutin VDVVDDV

inout DVV

From power balance, outoutinin IVIV

D

II inout

, so

The input/output equation becomes

Note – even though iin is not constant (i.e., iinhas harmonics), the input power is still simply

Vin • Iin because Vin has no harmonics.

!

The input/ output equation can be found by the above equation formula which shows the input

voltage lower than the output voltage.

The buck converter operates in two states the ON and OFF state where the diode will be in

reverse bias and in forward bias depending on the condition of the transistor.

ON state

The voltage across the inductor is 𝑉𝐿=𝑉𝑖𝑛-𝑉𝑜𝑢𝑡, the current through the inductor rises linearly

thus when the voltage is almost constant the diode will be reversed biased by voltage source 𝑉𝑖𝑛

the inductor will be charging.

OFF state

The diode now will be in forward bias as the transistor as a switch will be off thus the current

across the inductor will start to decrease (inductor will be discharging) thus the voltage across

the inductor will be equal to the negative polarity of the output voltage. Thus we will have the

wave forms like these below

BOOST CONVERTER

A boost converter (step-up converter) is a DC-to-DC power converter that steps up voltage

(while stepping down current) from its input (supply) to its output (load). It is a class of

switched-mode power supply (SMPS) containing at least two semiconductors (a diode and a

transistor) and at least one energy storage element: a capacitor, inductor, or the two in

combination. To reduce voltage ripple, filters made of capacitors (sometimes in combination

with inductors) are normally added to such a converter's output (load-side filter) and input

(supply-side filter).

Fig1.1 The basic schematic of a boost converter.

OPERATION

The key principle that drives the boost converter is the tendency of an inductor to resist

changes in current by creating and destroying a magnetic field. In a boost converter, the output

voltage is always higher than the input voltage. A schematic of a boost power stage is shown in

Figure 1.

(a)When the switch is closed, current flows through the inductor in clockwise direction and the

inductor stores some energy by generating a magnetic field. Polarity of the left side of the

inductor is positive.

(b)When the switch is opened, current will be reduced as the impedance is higher. The magnetic

field previously created will be destroyed to maintain the current towards the load. Thus the

polarity will be reversed (meaning the left side of the inductor will become negative). As a

result, two sources will be in series causing a higher voltage to charge the capacitor through the

diode D.

If the switch is cycled fast enough, the inductor will not discharge fully in between charging

stages, and the load will always see a voltage greater than that of the input source alone when

the switch is opened. Also while the switch is opened, the capacitor in parallel with the load is

charged to this combined voltage. When the switch is then closed and the right hand side is

shorted out from the left hand side, the capacitor is therefore able to provide the voltage and

energy to the load. During this time, the blocking diode prevents the capacitor from discharging

through the switch. The switch must of course be opened again fast enough to prevent the

capacitor from discharging too much.

Fig 1.2 Operation of the boost converter

During the On-state, the switch S is closed, which makes the input voltage ( ) appear across

the inductor, which causes a change in current ( ) flowing through the inductor during a time

period (t) by the formula:

Where L is the inductor value.

At the end of the On-state, the increase of IL is therefore:

D is the duty cycle. It represents the fraction of the commutation period T during which the

switch is On. Therefore, D ranges between 0 (S is never on) and 1 (S is always on).

During the Off-state, the switch S is open, so the inductor current flows through the load. If we

consider zero voltage drop in the diode, and a capacitor large enough for its voltage to remain

constant, the evolution of IL is:

Therefore, the variation of IL during the Off-period is:

As we consider that the converter operates in steady-state conditions, the amount of energy

stored in each of its components has to be the same at the beginning and at the end of a

commutation cycle. In particular, the energy stored in the inductor is given by:

So, the inductor current has to be the same at the start and end of the commutation cycle. This

means the overall change in the current (the sum of the changes) is zero:

Substituting and by their expressions yields:

This can be written as:

The above equation shows that the output voltage is always higher than the input voltage (as the

duty cycle goes from 0 to 1), and that it increases with D, theoretically to infinity as D

approaches

1. This is why this converter is sometimes referred to as a step-up converter.

Rearranging the equation reveals the duty cycle to be:

Discontinuous mode

Fig. 4: Waveforms of current and voltage in a boost converter operating in discontinuous mode.

If the ripple amplitude of the current is too high, the inductor may be completely discharged

before the end of a whole commutation cycle. This commonly occurs under light loads. In this

case, the current through the inductor falls to zero during part of the period (see waveforms in

figure 4). Although the difference is slight, it has a strong effect on the output voltage equation.

The voltage gain can be calculated as follows:

As the inductor current at the beginning of the cycle is zero, its maximum value

(at ) is

During the off-period, IL falls to zero after

:

Using the two previous equations, δ is:

The load current Io is equal to the average diode current (ID). As can be seen on figure 4, the

diode current is equal to the inductor current during the off-state. Therefore, the output current

can be written as:

Replacing ILmax and δ by their respective expressions yields:

Therefore, the output voltage gain can be written as follows:

Compared to the expression of the output voltage gain for continuous mode, this expression is

much more complicated. Furthermore, in discontinuous operation, the output voltage gain not

only depends on the duty cycle (D), but also on the inductor value (L), the input voltage (Vi),

the commutation period (T) and the output current (Io).

Substituting Io=Vo/R into the equation (R is the load), the output voltage gain can be rewritten

as:

Where

THE DESIGN OF TRANSFORMERS AND THE CONTROL CIRCUIT FOR

SMPS Transformers are required for galvanic isolation between input and output voltages and for voltage

and current scaling. It also helps in optimizing the device voltage and current ratings. The switches,

diodes and other circuit elements on the high voltage side of the transformer are subjected to higher

voltages but only lower currents.

Case A

The circuit in Fig(a) uses a step down transformer with proper turns ratio

Case B

the switch and diode and the filter inductor in Fig. 25.1(b) need to withstand both input side

voltage and output side current. Also, the switch in case (b) will be constrained to operate in a

narrow range, which may cause lesser accuracy in output voltage control.

(a) (b)

Fig. 25.1: DC to DC buck converters: (a) Isolated type (b) Non-isolated type Transformers used in switched mode power supply circuits are significantly different from the

power transformers that are used in utility ac supply system.

(i) The input and output voltages and currents of a SMPS transformer are mostly non-

sinusoidal, whereas the transformers connected to utility ac supply are almost

always subjected to sinusoidal voltages and currents.

(ii) The currents and voltages of SMPS transformer are of very high frequency

whereas utility type transformers are subjected to low frequency supply voltages.

(iii) SMPS transformers generally handle much smaller power than the utility

transformer.

RECAPITULATION OF GOVERNING EQUATIONS FOR UTILITY

TRANSFORMER In case of sinusoidal flux of peak magnitude ‘φm ’ and frequency ‘f’ linking the transformer

windings, the emf generated per turn of the winding will have a rms magnitude ‘Et’ givenby:

Et = 4.44 f φm………………………………...(i)

The peak flux through the core is the product of peak flux density (Bm) and the core area (Ac),

i.e.,

φm = Bm Ac …………………………………(ii)

The windings are placed around the core and are accommodated in the window of the

transformer. The transformer window area (Aw) is related with the winding’s current rating and

the number of turns. For a single-phase transformer the relation between them is given by:

Aw kwδ = 2 NI ………………………………. (iii)

, where kw is the window utilization factor and δ is the current density through the crosssectional

area of the transformer windings. Window utilization factor, roughly varies between 0.35 to 0.6

and is dependent on the insulation requirements of the windings. A typical figure for the current

density through copper conductors of naturally cooled transformers is 3X106 amps per square

meter. The VA rating of a single phase transformer (= N Et I) can now be found from the above

equations as:

VA rating = 2.22 f Bm δ kw Ac Aw ----------------------------(iv)

For the given operating frequency (f) the product ‘Ac Aw’, known as area product is roughly

proportional to the VA rating of the transformer as other parameters have nearly fixed

magnitudes.

TRANSFORMER WITH SQUARE-WAVE VOLTAGE AND BIPOLAR FLUX

Determination of primary to secondary turns ratio (NP/ NS)

Let the input voltage vary from Vmin to Vmax. With minimum input voltage ‘Vmin’ and duty ratio

‘D’ = 0.5, the magnitude of square-shaped secondary side voltage should equal (Vo + VR), where

VR is the estimated voltage drop in the transformer winding, output rectifier and filter circuit

under maximum load condition. The transformer turns ratio can thus be estimated to

NP/ NS = Vmin/(Vo + VR)

Determination of peak magnitude of flux in the transformer core

The maximum flux in the core will correspond to a square wave voltage of magnitude Vmax

across the primary winding (refer to Figure with D=0.5). The frequency of voltage waveform

‘f’(=1/T) is same as the frequency at which the converter switches are turned on and is fixed

beforehand. Now by simple integration of the square wave voltage waveform, the peak flux ‘φm ’

is related to the input voltage as, Vmax = 4.0 f φm NP

Vmax = 4.0 f Bm Ac NP

Determination of winding current rating and requirement of window area

: Let ‘Iom’ be the peak expected load current

above and they carry the load (dc) current only in alternate half cycles. Thus the rms current

rating of each half equals Iom

√2 andthe net copper cross-sectional area required for the secondary

winding √2N IS om

δ.If the secondary was not center-tapped, the rectifier used would be bridge type

and the copper area for the secondary would have been just NsIom

δ .The primary side carries the

reflected secondary current.

The total window area requirement for the transformer can now be given as:

Aw kw= NsIom

δ(1+ √2)

where Aw is the window area and kw is the window utilization factor

Expression for VA rating of the transformer

VmaxNs

NPIom (1+ √2) = 4fBmδkw AcAw

Using relations derived above, Eqn. may be rewritten as:

V IomK1 K2(1+√2) = 4fBmδk Ac Aw

where K1 = Vmax

Vmin a factor allowing for input voltage variation and K2 =

(Vo + VR)

Vo

a factor coming due to voltage drop in rectifier diode, filter inductor etc. Vo Iom is the peak

output power from the SMPS.

Selection of transformer core and determination of number of turns in the windings:

Knowing the area product ‘Ac A w’, as given by Eqn.25.8, the appropriate transformer core is to be

selected from the core-manufacturer’s catalog. Once the area product matches, the details of other

dimensions of the transformer core are found from the catalog. Knowing window area (Aw) and

core area (Ac), the number of turns in the windings can be decided using Eqns. Like (25.5) or (25.6).

TRANSFORMER WITH UNIPOLAR FLUX

Many switched mode power supply circuits use only one controlled switch. The winding current and

core-flux for most of these transformers are unidirectional. when the forward converter switch is

turned on the primary winding is subjected to input dc voltage. As soon as the primary winding is

turned-off, the tertiary winding starts conducting and the voltage across primary goes negative with

a magnitude that equals the product of input voltage and the turns ratio between the primary and

tertiary windings. The maximum duty ratio (Dmax) of the switch is also limited by the turns ratio

between the primary and tertiary winding to allow resetting of the transformer flux. The maximum

input voltage (Vmax), switching frequency ‘f’(=1/T) and the maximum duty ratio (Dmax) are

related with the peak magnitude of core-flux is calculated as

Vmax Dmax = fφmNP= f Bm Ac NP

Because of unipolar nature of flux the utilization of core (in terms of emf generation) is poorer here.

The primary to secondary turns ratio (NP/ NS ) for the forward converter can be estimated as done

previously for the H-bridge converter. Accordingly, NP/ NS = Vmin Dmax /(Vo + VR).

The maximum rms current through the secondary winding can be equated to om max Iom√ Dmax and

the window area (Aw) requirement is given by

Aw kw = 2 Ns Iom√ Dmax

δ

Hence the VA ratings can be given as

V IomK1 K2(√Dmax) = 4fBmδk Ac Aw