maharashtra state board of technical education (autonomous) (iso/iec...

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

(ISO/IEC - 27001 - 2005 Certified)

SUMMER-14 EXAMINATION

Subject Code: 12189 Model Answer Page No: ____/ N

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Important Instructions to examiners:

1) The answers should be examined by key words and not as word-to-word as given in the

model answer scheme.

2) The model answer and the answer written by candidate may vary but the examiner may try

to assess the understanding level of the candidate.

3) The language errors such as grammatical, spelling errors should not be given more

Importance (Not applicable for subject English and Communication Skills.

4) While assessing figures, examiner may give credit for principal components indicated in the

figure. The figures drawn by candidate and model answer may vary. The examiner may give credit for any

equivalent figure drawn.

5) Credits may be given step wise for numerical problems. In some cases, the assumed constant

values may vary and there may be some difference in the candidate’s answers and model answer.

6) In case of some questions credit may be given by judgement on part of examiner of relevant answer based on

candidate’s understanding. 7) For programming language papers, credit may be given to any other program based on equivalent concept.

Q.1 A) Attempt any three : 12 Marks

a) Draw the symbols of ( 1Mark Each)

i) IGBT ii) DIAC iii) GTO iv) SCS

Ans : i) ii)

iii) iv)





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b) Explain the need and use of the poly phase rectifiers

Ans:

Use of polyphase rectifier: (2 Marks)

1)Number of phases are more due to that average output can be more.

2)In polyphase rectifier have less ripple so small filter can be used.

3) More DC power can be generated in motor control operation.

Need of polyphase rectifier: (2 Marks)

The average o/p voltage that can be obtained from single phase full wave rectifier is about 0.636 Vm &

can support power of the order of 1.5Kw To obtain higher power three phase line can be further

increased by connecting transformer windings to give intermediate phase shift. This increase in number

of phases increases the smoothness of o/p Dc. When simplest single phase HWR is used efficiency is

about 41% while for 3phase HWR efficiently increases to 97%. It is observed that as large number of

phases are used.

c) Why phase control rectifier are called as ac to dc converters? List the application of controlled

rectifiers.

Ans : (1 Marks)

These controller convert fixed ac voltage to a variable dc voltage. These converter takes power from one

or more ac voltage source of single or multiple phases and deliver to a load. The output variable is low

ripple dc voltage. These controller circuits used line voltage for their commutation. Hence they are also

called as line commutated or naturally commutated ac to dc converter.

Application : (3 Marks Any Three)

1. High voltage dc transmission system

2. DC motor drives.

3. Regulated dc power supply

4. Static VAR compensator

5. Wind generator converter

6. Battery charger circuits.





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d) Draw the construction of GTO and draw its V-I characteristics.

Ans : Construction of GTO ) (2 Marks)

V-I characteristics of GTO : (2 Marks)





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Q. 1 B) Attemptany One (6 Marks)

a) draw the static V-I characteristics of SCR. Name the various region on it. Define latching current ,

holding current and break over voltage of SCR.

Ans : V-I characteristics of SCR (2 Marks Charact. And 1 Mark mention region )

Latching current (IL):- (1 Marks)

It is minimum current whish is required to latch the device from off state to on state. It is

also can be defined as the minimum value of current required to trigger device.

Holding current (IH) :- (1 Marks)

It is minimum value of current required to hold conduction of the device. It may be

defined as minimum current below which device get off or stop conduction.

Break over voltage: (1 Marks)

It is a forward break over voltage when diode is conducted without gate, because to make device

forward bias should overcome junction barrier which is at reverse (J2) and which added with voltage

across J1 & J3 may be (0.6 + 0.6 + VB J2) when is VBR is depend on temperature ratio of and due to

reversed thermal operation and ICEO, which may be smaller than VBO.





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b) What do you mean by chopper? Classify them. Explain any one type of chopper.

Ans :

A DC chopper is a static device (switch) used to obtain variable DC voltage from source of constant DC

voltage. (1 Marks)

Classification of Choppers: (2 Marks)

Choppers

Step- down chopper Step-up chopper (Type B- Chopper)

Type A Type C Type D Type E

Chopper chopper chopper chopper

Single quadrant Four quadrant chopper

chopper Two quadrant

chopper

Circuit Diagram (step down chopper) (1 Marks)

Explanation: (2 Marks)

When SCR S1 is closed for a time TON the input voltage V appear across the load. If the SCR remains are for a

period TOff the voltage across the load is zero. Practically any power device can the used as a switch this devices

have finite voltage drop. The average output voltage is depends on the value of Duty cycle if D has a value

between 0 & 1. Then VLdc will vary between 0 volts top V Volts. Thus the average output voltage of this chopper





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circuit is always less than or equal to the input voltage. Hence this type of chopper is known as the step down

chopper.

OR

Circuit diagram (step up chopper) (1 Marks)

Explanation (2 Marks)

When SCR S1 is closed for time =t1, the inductor current rises and energy stored in the inductor, L. the

voltage across the inductor is equal to the input dc voltage V. during this time the output voltage will be

zero.

If SCR S1 is open circuited, then in order to maintained the inductor current the same direction there will

be an induced voltage across L. this voltage comes in series with the input DC voltage V. thus the total

voltage across SCR S1 is (V+voltage across inductor). The diode D1 is forward biased and the stored

energy in the inductance is transferred to the load.

Q. 2 Attempt any Four (16 Marks)

A) Explain two transistor analogy of SCR with neat diagram

Ans : Two transistor analogy (2 Marks)





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The two transistor equivalent circuit shows that the collector current of the NPN transistor Q2 feeds

directly into the base of the PNP transistor Q1, while the collector current of TR1 feeds into the base of

Q2. These two inter-connected transistors rely upon each other for conduction as each transistor gets its

base-emitter current from the other‟s collector-emitter current. So until one of the transistors is given

some base current nothing can happen even if a Anode-to-Cathode voltage is present.

When the thyristors Anode terminal is negative with respect to the Cathode, the centre N-P junction is

forward biased, but the two outer P-N junctions are reversed biased and it behaves very much like an

ordinary diode. Therefore a thyristor blocks the flow of reverse current until at some high voltage level

the breakdown voltage point of the two outer junctions is exceeded and the thyristor conducts without

the application of a Gate signal. This is an important negative characteristic of the thyristor, as

Thyristors can be unintentionally triggered into conduction by an overvoltage as well as high

temperature or a rapidly rising dv/dt voltage such as a spike.

If the Anode terminal is positive with respect to the Cathode, the two outer P-N junctions are forward

biased but the center N-P junction is reverse biased. Therefore forward current is also blocked. If now a

positive current is injected into the base of the NPN transistor Q2, the resulting collector current flows in

the base of transistor Q1.This in turn causes a collector current to flow in the PNP transistor, Q1 which

increases the base current of Q2 and so on.

Very rapidly the two transistors force each other to conduct to saturation as they are connected in a

regenerative feedback loop. Once triggered into conduction, the current flowing through the device

between the Anode and the Cathode is limited only by the resistance of the external circuit. Then a

thyristor can be turned “ON” and made to act like a normal rectifying diode by the application of a

positive current to the base of transistor, TR2 which for a silicon controlled rectifier is called the “Gate”

terminal.

OR OR





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Once the thyristor has been turned “ON” and is conducting in the forward direction (anode positive), the

gate signal loses control due to the regenerative latching action of the two internal transistors. Then

applying a momentary Gate pulse to the device is enough to cause it to conduct and will remain

permanently “ON” even if the gate signal is completely removed. Then the thyristor can also be thought

of as a Bistable Switch having two states “ON” or “OFF”, as once it is triggered “ON” it cannot be

turned “OFF” again by its Gate

b) Draw the vertical structure of power transistor and explain it.

Ans : Vertical structure of power transistor :- (2 Marks)


Fig. shows the doping level in each layer. The thickness of the different layer will have a significant

effect on the characteristics of the device. The emitter layer is heavily doped . the base is moderately

doped. The n- region is known as the collector drift region and it is lightly doped. The n

+ that terminates.

The drift region has doping level similar to that of emitter. This n+ region serves as collector contact.

Due to the low doping level the n- drift layer will increase the voltage blocking capacity of the transistor.

The width of this layer decide the breakdown voltage of power transistor.

The current gain β of a transistor depends on the base thickness. As the base thickness reduces the gain

increases but the breakdown voltage of transistor will decrease. In power transistor high breakdown

voltage is more important than high current gain. Therefore the base thickness is much larger than that in

the logic level transistor.

c) List the various forced commutation methods. Explain self commutation by resonating load.

Ans:

Forced commutation methods: (2 Marks)

1. Class A Self commutated by resonating the load.

2. Class B Self commutated by a LC circuit.

3. Class C Complementary commutation





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4. Class D Auxillary commutation

5. Class E an External pulse source for commutation

6. Class F AC line commutation

Circuit Diagram : (1 Marks)

Class A Self commutated by resonating the load.


Class A turn off method:- for low values of R L& C are connected in series with R and for high

value of R Load R is connected across C, the essential required for this commutation circuit the overall

circuit must be under damped. The nature of the circuit should be such that when it is energised from

DC source current must have a natural tendency to zero for the load commutation to occur in thyristor

circuit. This class A commutation is also called as resonant commutation or self-commutation. When the

circuit is energized with DC source it is seen that current i is first rises to maximum value and then begin

to fall. When current decays to zero and tends to reverse thyristor turns off on its

d) Compare SCR and TRIAC ( any 4points)

Ans : (1 Marks Each point)

SCR TRIAC

1. Unidirectional 1. bidirectional

2. Terminal: gate, anode, cathode 2. Gate, mt1, mt2

OR





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3. Symbol:

3.

4. Application : specially converter

,inverter

4. Light dimmer,, ac speed controlled,

ac motor controlled

5. characteristics

5. characteristic

e) Draw the output waveform of 1 ¢ full controlled bridge converter with R-load for output voltage and

output current for firing angle. I) α =00 ii) α =90

0

Ans:- For α =00 ( 2 Marks)

α =900 ( 2Marks)





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f) With neat circuit diagram and waveforms explain half wave control rectifier with R-Load.

Ans:-

Circuit diagram :- ( 1 Marks)

Waveforms : ( 1 Marks)

Explanation: ( 2 Marks)

During positive half cycle, of the supply voltage anode is positive with respect to its cathode and until

the thyristor is triggered by a proper gate pulse, it block the load current in the forward direction. When

the thyristor is fired at an angle α. Hence the load is directly connected to the ac supply with a zero

resistance and purely resistive load. The current wave forms after the thyristor trigger will be identical to

the applied voltage and of the magnitude depends on the amplitude of the voltage and the value of load

resistance R. the load current will flow until it is commutated by reversal of supplied voltage at Ѡ t= π .

During negative half cycle of the supply voltage the thyristor block the flow of load current and no

voltage is appear to the load R.





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Q 3 Attempt any Four : (16 Marks)

a) Describe Class-C commutation with the help of neat diagram.

Ans : Circuit diagram (2Marks )

Explanation (2Marks)

Here complementary SCR T2 is connected in parallel with the main SCR. Initially both the SCR are

OFF.

When a triggering pulse is applied to the gate of T1 the SCR T1 is triggered. Therefore current starts

flowing through the load as well as R2 & C. Capacitor C will get charged by the supply voltage Edc as

shown in figure.

When a triggering pulse is applied to the gate of T2 the SCR T2 is on, the negative polarity of the

capacitor C is applied to the anode of T1 & positive to the cathode. This causes the reverse voltage

across the main SCR T1 and immediately turns it off.

b) Explain 1 ф full converter with L-load with neat circuit diagram and necessary waveforms.

Ans: Circuit diagram ( 1 Marks)





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Explanation (2 Marks)

Here the conduction does not takes place until the SCRs are fired and in order for current to flow, SCRs

T1 & T2 must be fired together and SCRs T3 & T4 in the next half cycle. To ensuring simultaneous firing

both SCRs T1 & T2 are fired from the same firing circuit.

At firing angle α = 600 SCRs T1 & T2 are triggered. Current flows through the path L – T1 – A – L – R

– B – T2 - N. Supply voltage from this instant appears across output terminal & forces the current

through load. This load current Id is assumed to be constant. This current also flows through the supply

& the direction is from line to neutral, which is taken positive. At instant π voltage reverses. However

because of very large inductance L. the current is maintained in the same direction at instant constant

magnitude Id which keeps the SCRs T1 & T2 in conducting state and hence, the negative supply voltage

appears across output terminals.

At an angle Ω + α SCRs T3 & T4 are fired. With this the negative line voltage reverse biases SCRs T1 &

T2. The current flows through the path N – T3 – A- L – R – B – T4 – L. This continues in every halt

cycle and we get the output voltage as shown in the fig.

Waveforms ( 1 Marks)

c) List the various gate triggering methods of SCR. What do you mean by synchronization in triggering?

Draw the circuit of synchronized SCR gate triggering using UJT.

Ans:

Gate triggering methods of SCR:- ( 1 Mark)

1. D.C. gate triggering :

2. A.C. gate triggering

3. Pulse gate triggering





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Synchronization:- ( 1 Marks)

As the zener diode voltage Vz goes to zero at the end of each half cycle, the synchronization of the

trigger circuit with the supply voltage across SCR is achieved. Thus the time „t‟ when the pulse is

applied to SCR for the first time in each half cycle will remain constant for the same value of R

The circuit of synchronized SCR gate triggering using UJT:- ( 2 Marks)

d) State the function of free wheel diode in controlled rectifier with neat sketch.

Ans: Diagram ( 2 Marks)

Functions of free wheel diode: ( 2 Marks)

i) It prevents reversal of load voltage except for small diode voltage – drop.

ii) It transfers the load current away from the main rectifier, thereby allowing all of its thyristors to

regain their blocking states.





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e) Draw the symbol and construction of LASCR and describe it in brief.

Ans: Symbol Construction (1Marks Each)

Explanation ( 2marks)

It is SCR which will be turned on only when light falls on it because it is made from a material which is

light sensitive. Its symbol is shown in the diagram. It is having three junction and three terminals Anode,

Cathode and Gate.

As the light falls on LASCR this light energy generate large number of electrons and holes in the

junction J2 and current starts conducting because when LASCR is in the forward bias junction J1 and J3

are already forward biased so the device start conducting and the current increases rapidly which latch

the LASCR in the On state

f) Describe with neat ckt. Diagram and waveforms of the inverting mode operation of 1 ϕ full wave

controlled bridge rectifier.

Ans: Circuit diagram ( 1 mark)





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Waveforms ( 1 marks)

Explanation ( 2 marks)

Here firing pulses are retarded by an angle of 135 . The D.C. terminal voltage waveform now contains a

mean negative component & the fundamental component of the A.C. line current waveform lags the

voltage by an angle 135 . Since the mean D.C. terminal voltage is negative (α>90 the D.C. power and

hence also the mean A.C. power must also be negative. In other words power must also be negative. In

other words power is now being delivered from the D.C. side of the convertor to the A.C. side, and the

convertor is operating as a “line commutated invertor.”





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Q. 4 A) Attempt any three

a) Explain the necessity of specifying SOA by manufacturer. Also explain the second breakdown.

Ans:

Necessity of specifying SOA:- ( 2 Marks)

FBSOA:- During turn on and on state conditions, the average junction temperature and second

breakdown limit the power handling capacity of a transistor. The manufacturers uasualy provide the

FBSOA curves under specified test conditions. FBSOA indicates the IC - VCE limit of the transistor

and for reliable operation the transistor must not be subjected to greater power dissipation than that

shown by the FBSOA curve.

RBSOA:- During turn off a high current and high voltage must be sustained by the transistor, in most

cases with the base to emitter junction reverse bias. The collector emitter voltage must be held to a

safe level at or below a specified value of collector current. Tme manufacturers provide the IC - VCE

limits during reverse bias turn off as RBSOA..

Second Breakdown:- ( 2 Marks)

It is a destructive phenomenon, result from the current flow to a small portion of the base producing

localized hot spots. If the energy in these hot spots is sufficient the excessive localized heating may

damage the transistor. The secondary breakdown is caused by a localized thermal runway, resulting

from high current concentration. The current concentration may be caused by the defect in the transistor

structure.

b) Explain class A commutation with neat circuit diagram and waveforms.

Ans: Circuit diagram ( 1 mark)

OR





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Waveforms ( 1 Mark)

Explanation ( 2 marks)

In this process of commutation the forward current passing through the device is reduced to less than the

level of holding current of the device.

When the SCR is turned on by a gate pulse, an oscillatory current i flows in the circuit and charges up

the capacitor C. Capacitor C is charged up to the supply voltage Edc as the oscillatory current reaches its

peak value. Further capacitor charges from Edc value to 2Edc because inductor induces Edc voltage in the

capacitor. When the capacitor is completely charged up to 2 Edc, a differential voltage of ( 2Edc – Edc) =

Edc appear across the capacitor reflects at the cathode of the SCR and turns it off.





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c) Explain SUS with the construction, symbol and V-I characteristics.

Ans: Construction Symbol ( 1 Mark Each)

Characteristics ( 1Mark)

Explanation : ( 1 Mark)

It is a four layer, three junction PNPN type device. It has three terminals anode, cathode and gate. The

gate terminal is towards anode i. e. anode gate.

It has one inbuilt low voltage avalanche diode between the gate and the cathode.





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d) Compare R and RC triggering methods of thyristors (any 4 points)

Ans: ( Each Point 1 Mark)

Parameter R triggering RC triggering

Components used Resistor Resistor and Capacitor

Maximum firing angle Cannot be greater than 900 Can be changed from 0

0 to 180

0

Isolation of control circuit and

power circuit

Not possible Not possible

Effect of supply fluctuations α changes α changes

Type of triggering AC gate triggering AC gate triggering

Circuit Diagram

Waveforms





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B) Attempt any one : ( 6 Mark)

a) Draw M2 ( mid – point) converter with L-Load. Explain with waveforms for the firing pulses. Load

voltage, load current and voltage across thyristors.

Answer: Circuit Diagram. ( 2 Marks)

Wave forms: ( 2 Marks)





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With reference to above fig SCR T1 can be fired into the on state at any time after e1 goes positive. Once

SCR T1 is turned on current builds up in the inductive load, maintaining SCR T1 in the on state upto the

period when e1 goes negative. Since the load consists of inductance & resistance the w/f will be different

from that obtained with a pure resistive load. In this case as the AC voltage passes through natural zero

after negative half cycle, At negative half cycle SCR T2 will on after gate is trigger.

b) Draw the construction and V-I characteristics of TRIAC. Describe mode – III operation of TRIAC.

Ans : Construction (2 Marks)

Characteristics: ( 2Marks)





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When terminal MT2 is negative and terminal MT1 is positive the device can be turned on by applying a

positive voltage between the gate and terminal MT1.

In this mode the device operates in the third quadrant when it is triggered into conduction. The turn on is

initiated by remote gate control. The main structure that leads to turn on is P2N1P1N4 with N2 acting as a

remote gate as shown in fig.

The external gate current IG forward biases P2N2 junction. Layer N2 injects electrons into P2 layer as

shown by dotted arrows and are collected by the junction P2N1. The electrons from N2 collected by P2N1

junction cause an increase of current through the junction P2 N1. The holes injected from P2, diffused

through N1 and arrived in P1. Hence a positive space charge builds a in the P1 region. More electrons

from N4 diffuse into P1 to neutralize the positive space charge. These electrons arrived at the junction J2.

They produced a negative space charge in the N1 region. Which results in more holes being injected

from P2 into N1.

This regenerative process continues until the structure P2N1P1N4 completely turns on and conducts the

current which is limited by the external load.

Q.5 Attempt any four (4x4=16)

a) Draw neat sketch of construction of IGBT and draw its V-I characteristics

Ans : Construction ( 2 Marks),

V-I characteristics ( 2 Marks)





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b) With neat circuit diagram and waveform explain symmetrical configuration of 1ф half controlled

bridge rectifier with R-load.

Ans: Circuit diagram (1 Marks)

Wave Form (2 marks)

Explanation (1 Mark)

Working –During +ve half cycle of the ac supply thyristor T1, and Diode D1, are forward biased and are

in forward Blocking mode. When the SCR T1 is triggered, at firing angle α , current flow through the

path L-T1-R-D1- N as shown in diagram, load current will flow until it is commutated by reversal of

supply voltage at Ѡt = π

During –ve half cycle of the ac supply T2 and D2 are forward biased When SCR T2 is triggered at an

angle (π+α), current would flow through the path N-T2 – R-D2-L load current is continuous till angle

2π,when SCR T2 is turned off.





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c) Compare natural commutation and forced commutation

Ans: [Any 4 points 4 Marks]

Sr.No. Parameter Natural Commutation Forced Commutation

1) Need of external

commutating

components

Not Necessary Necessary

2) Type of Supply AC DC

3) Power loss in

cumulating

components

NIL Some power loss takes

place

4) Type of Commutation SCR turns off due to anode

current going below IH or

due to the application of

reverse line voltage

Voltage commutation or

current commutation

5) Size of circuit Small due to absence of

commutating components

Big due to large

commutating components

6) Application In line commutated

converters and inverters

In series, parallel inverter,

chopper

d) Differentiate between 3ф half wave and 3ф full wave rectifiers with R-load with respect to

a) No. of diodes

b) PIV

c) Vdc

d) Ripple factor

Ans: (Each point 1 Mark)

Parameter 3 ф HWR 3 ф full wave rectifiers

a) No. of diodes 3 6

b) PIV Low High

c) Vdc Low High

d) Ripple factor High ripple factor Low ripple factor





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e) Compare 1ф half wave controlled rectifier and 1ф half controlled bridge rectifier

Ans: (Any 4 points)

Parameter 1ϕ half wave

controlled rectifier

1ϕ half controlled

bridge rectifier

No. of SCR One Two

No. of diodes No (zero) Two

Vdc Vdc = Vm/2π ( 1+cos) Vdc = Vm/π (1+cos)

Region of working In first quadrant In first quadrant

Ripple factor Large Less

Input power factor Very poor Good

Q.6 Attempt any Four : (16 Marks)

a) Draw neat diagram of 6 ф star half wave rectifier and explain it with neat waveforms.

Ans: Circuit diagram ( 1 Mark)

Wave forms (2 Mark)





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Working (1 Mark)

Phase voltages VRN VYN VBN are 120° out of phase with each other and VRN VYN VBN are 180°

phase shifted with respect to VRN VYN VBN respectively.

Diode which corresponds to maximum positive phase voltage would conduct, at any given

instant. At any given instant only one diode will be conduct and the load voltage will be equal to

the corresponding phase voltage with respect to N.

From waveform phase voltages cross each other at point 1, 2, 3…6 at these phase crossover

points incoming diodes will come into conduction and outgoing diode will be turn off.

b) Compare class – A and class – B commutation circuits.

Ans : ( Any 4 Points Each Point 1 Mark)

Parameter Class –A commutation Class –B commutation

Diagram

Position of the

commutating

components.

Commutating components are

connected in series with SCR

Commutating components are

connected in parallel with SCR

Load current Commutating components carry

load currents

Commutating components does

not carry load currents

Size and cost Due to commutating component

carry load current their size is

big, so circuit becomes bulky

and costly.

Size of circuit is small and cost

is less.

Application It is used in series inverter. It is used in DC chopper.

OR





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c) Draw the neat circuit diagram of parallel inverter and write its two applications.

Ans : Circuit diagram ( 5Mark)

Application ( Each ½ Mark)

1. Emergency system

2. In industrial application

3. High voltage dc transmission system

d) With the help of neat circuit diagram explain the resistance firing circuit and state its

limitations.

Ans : Circuit diagram ( 1 Mark)

Working :- ( 2 marks)

As es goes positive the SCR becomes forward biased from anode to cathode. It also forward

biased diode; this causes flow of a gate current Ig.





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Gate current will increase as es increase towards its peak value. When Ig reaches a value equal to

Ig[min] SCR turns on and eL = es.

SCR remains ON and eL = es until es decreases to the point where load current is below the SCR

holding current.

SCR turns off when es become negative. Since SCR is in reverse biased mode. It will act Ѡ open

switch and now load voltage is zero.

Same sequence is repeated when es again goes positive.

Limitations :- ( ½ Mark each)

1. Trigger angle α is greatly dependent on the SCR Ig(min)

2. Load voltage wave forms can only be varied from

e) Explain with neat circuit diagram and waveforms 1 ф controlled bridge rectifier with R-load.

Ans : Circuit Diagram ( 1 Mark)

Wave form ( 1 mark)

Working : ( 2 Mark)

During positive half cycle SCR1, SCR3 are forward biased and they are triggered

simultaneously, then current flows through the path L – SCR1 – RL – SCR3 – N.

During negative half cycle SCR2, SCR4 are forward biased and they are triggered

simultaneously, then current flows through the path N-SCR2 – RL –SCR4 –L.

When supply voltage falls to zero, SCR1, SCR3 in positive half cycle and SCR 2,SCR 4 in

negative half cycle turns off by natural commutation.





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f) Write the different turn ON methods of SCR. Explain thermal triggering of SCR.

Ans:

Turn on methods of SCR ( Any Four ½ mark Each)

1) Forward voltage triggering

2) Thermal Triggering

3) Radiation Triggering

4) Dv/dt triggering

5) Gate triggering

Thermal triggering ( 2marks)

Like any other semi-conductor, the width of the depletion layer of a thyristor decreases on increasing the

junction temperature. When the voltage applied between the anode and cathode is very near to its break

down voltage, the device can be triggered by increasing its junction temperature. This method of

triggering the device by heating is known as the thermal triggering process.

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