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CENTRAL VALUATION – NOVEMBER 2019

Answer Key & Scheme of Valuation

Branch : 1042- Instrumentation & Control Engg.

Subject : 34273- Industrial Power Electronics

Q.P Code : 374

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Part – A (Answer any 5 questions & Qn. No. 8 is compulsory)

1. Give the example for thyristor family. (any two, 2x1=2 marks)

SCR, IGBT, GTO, MOSFET

2. Draw the UJT symbol. (2 marks)

3. What is converter? (2 marks)

Converter is a circuit that converts fixed AC voltage into variable DC voltage.

4. Mention the types of forced commutation. (any two, 2x1=2 marks)

(i) Class-A (Self commutation by resonating the load)

(ii) Class-B (Self commutation by LC circuit)

(iii) Class-C (Complementary commutation)

(iv) Class-D (Auxilary commutation)

(v) Class-E (External pulse commutation)

5. What do you mean by control strategy? (2 marks)

Control strategy is the technique adopted to control the average output voltage of

choppers. There are two kinds of control strategies used in DC choppers namely,

Time Ratio Control and Current Limit control.

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6. Write about the Morgan chopper. (Diagram-2 marks, Explanation-1 mark)

Morgan chopper is a class-B commutated chopper. Here, saturable reactor is used as a

commutating inductor. Saturable reactor offers high impedance on before saturation and

low impedance on after saturation. So the ON Time of chopper is increased.

7. What are the requirements of an inverter? (any two, 2x1=2 marks)

(i) Output voltage should be controllable.

(ii) Output waveform should be sinusoidal.

(iii) It should be able to operate with inductive load.

(iv) It should be able to operate even without load.

8. State the basic concept of numerical control. (3 marks)

Controlling a machine tool by a prepared program which consists of series of numbers is

known as numerical control. The numerical data required for producing a part is

maintained on a punched tape. Each instruction is represented by a sequence of holes on

the tape. The dimensional informations like length, breadth etc are taken from the

drawing. Dimensions are given separately for each axis of motion. Cutting speed, feed

rate, coolant ON/OFF, gear changes, spindle direction etc are programmed according to

the surface finish and tolerance requirements. The programmer writes the program in

manual or in a computer language called APT. The program is punched on a tape using a

flexo-writer.

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Part –B (Answer any 5 questions & Qn. No. 16 is compulsory)

9. Draw the VI characteristics of SCR. (3 marks)

10. Brief about the importance of flywheel diode. (any three, 3x1=3 marks)

(i) It transfers the inductive current away from the main rectifier.

(ii) It prevents the reversal of load voltage.

(iii) It improves the commutation.

(iv) It provides the higher output voltage.

11. Write the principle of chopper. (Diagram- 2 marks, Explanation- 1 mark)

Chopper is a circuit that converts fixed DC voltage into variable DC voltage.

Fig above shows the basic circuit of DC chopper. Here, SCR is used as a

chopping element .

During the ON time of SCR the load is connected with the supply , now

Vo=+Vs.

During the OFF time of SCR, the load is disconnected from the supply, now

Vo=0. Thus the average output voltage can be controlled from 0% to 100% by

varying the duty cycle of chopper.

(ie) Vo = α Vs Where, α = Ton/(Ton+Toff)

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12. Draw the circuit diagram of single phase inverter with RL load. ( 3 marks)

13. Compare online UPS with offline UPS. (any three , 3x1=3 marks)

ON Line UPS OFF Line UPS

1. No break UPS

1. Short break UPS

2. Sinusoidal output

2. Quasi square output

3. Suitable for critical load 3. Suitable for general purpose

load

4. Power rating of charger is

high

4. Power rating of charger is low

5. High cost

5. Low cost

14. Draw the block diagram of ACC. (3 marks)

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15. State the advantages of CNC system. (any three, 3x1= 3 marks)

(i) Increased productivity

(ii) Increased flexibility

(iii) Compatible with DNC

(iv) It is software based system, so making changes are ceasy

(v) Reduced the usage of tape reader

(vi) Information about performance is displayed in the monitor.

16. Write about the DC transmission. (Diagram – 2marks, Explanation – 1 mark)

In DC transmission , the Ac power of two power systems are linked through DC. Here, two

bridges are used. The firing angle of two bridges are adjusted, to make one bridge function

as rectifier and the other bridge as inverter. It has two types namely unipolar system and

bipolar system. The advantages of DC transmission are,

(i) Transmission cost is less

(ii) High stability

(iii) No skin effect

(iv) No synchronizing problems

(v) Less corona loss

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Part –C (Answer either (a) or (b) of each questions)

17. (a) Explain the working principle and VI characteristics of IGBT.

(Diagrams-5 marks , explanation-5 marks)

IGBT means Insulated Gate Bipolar transistor. It is a voltage controlled power

transistor. It has the following features,

(i) Low on-state voltage drop as like BJT

(ii) High input impedance as like MOSFET

(iii) Reverse blocking capability as like GTO

It is a four layers (PNPN) and three junctions device. It has three terminals namely

emitter, collector and gate. It’s structure is similar to that of vertically diffused

MOSFET except for the presence of P+ layer forms the drain of IGBT. This layer

forms a PN junction J3 which inject minority carriers into the drain drift region.

The N+

buffr layer between the P+ drain and N

- drift layer with proper doping density

and thickness can significantly improve the operation of IGBT in two aspects,

(i) It lowers the on state voltage drop and shortens the turn off time.

(ii) It improves the capacity of breakdown voltage.

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OPEARTION

When a positive voltage is applied to the collector w.r.t emitter , the junction J2

becomes reverse biased and the device operates in its forward blocking mode.

However, if a positive gate voltage w.r.t emitter is applied, the device switches to its

forward conducting state because the current can now flow from the emitter N+

region into the N- base region.

In this forward conducting state, J3 is forward biased and the substrate P+ region

injects holes into the N- base region. When the forward bias is increased, the injected

hole concentration is also increased.

b).Explain the synchronized UJT triggering with a circuit diagram.

(Diagram-5 marks , explanation-5 marks)

Synchronized UJT triggering is also called as ramp triggering. Here, the delay angle

of firing pulse depends upon the ramp voltage.

Fig above shows the ramp triggering circuit.

The diodes D1 to D4 act as bridge rectifier and it converts AC into DC.

The resistor Rs and Zener diode arrangement act as voltage regulator and provides

flat-top DC voltage.

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The capacitor C is charged through R exponentially. When the capacitor voltage

reaches the peak point voltage of UJT , the UJT fires ON.

Now the capacitor is discharged through UJT and pulse transformer. Thus it produces

the firing pulse. This firing pulse is used to trigger ON the SCR in the power circuit.

The delay angle of the firing pulse can be varied by the variable resistor R.

In this method, the firing angle can be controlled from 00 to 180

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18. (a) Draw and explain single phase fully controlled bridge converter with RL load.

(Diagram-5 marks , Explanation-5 marks)

Converter is a circuit that converts fixed AC voltage into variable DC voltage.

When a bridge type converter is built with 4 SCRs, it become a “fully controlled

bridge”. It is also called as Full Converter.

Fig above shows the single phase Full converter with RL load.

When terminal A is positive, T1 & T4 are turned ON and T2 & T3 are turned OFF.

Now the current flows from A to B through T1, load(x to y) and T4.

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When terminal B is positive, T2 & T3 are turned ON and T1 & T4 are turned OFF.

Now the current flows from B to A through T2, load(x to y) and T3.

Thus the direction of load current in both the half cycles is same. Hence we get DC

output across the load. Also, variable DC voltage is obtained by varying the firing

angle of SCRs in the bridge.

Suppose the load is highly inductive, the current does not become zero, when its load

voltage reaches zero. It extends till the next pair SCRs are triggered into conduction.

When the firing angle is more than 900, the output DC is negative. Now the power is

pumped from load to source. Thus the bridge act as an inverter.

The average output DC voltage is given by,

VDC = 2VM COS α / π

Where, VM is the maximum value of AC input

α is the delay angle

(b). Discuss dual converter. (Diagram-5 marks ,Explanation-5 marks)

Dual converter is a four quadrant converter. Here, two full converters are connected

together back to back.

It operates in two modes namely non-circulating current mode and circulating current

mode.

Non-circulating current mode

In this mode, one bridge operates at a time while the other is in blocked state. No

circulating current can flow between two converters and hence this mode is called as

non-circulating current mode.

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When we want the reverse output voltage, to stop the first bridge and also trigger the

second bridge. A small transition time of 10 to 15 ms is necessary, otherwise, the AC

input will be shorted through the two bridges.

Circulating current mode

In this mode, both bridges are triggered simultaneously, one operating in the

rectifying mode and the other in the inverting mode to avoid short circuit.

The condition for such control is that α1+α2 = 1800

Where α1 is the firing angle of bridge 1 and α2 is the firing angle of bridge 2

19. (a) Brief about any two types of chopper with a circuit diagram.

( any two types, Diagrams 3+3 = 6 marks, Explanation 2+2 = 4 marks)

TYPE – A CHOPPER

This is a single quadrant ( I quadrant ) chopper. Fig above shows the basic circuit of

Type– A chopper. Here, SCR T is a chopping element and diode D is a freewheeling

diode.

When T is ON, Vo = +Vs & Io is positive. When T is OFF, Vo = 0 but Io free wheels

through D in the same direction.

Thus the average load voltage and load current are positive. It means that power

flows from source to load and is said to be operated as a rectifier.

TYPE – B CHOPPER

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This is a single quadrant ( II quadrant ) chopper. Fig above shows the basic circuit of

Type- B chopper. Here, SCR T is a chopping element and diode D is a freewheeling

diode.

When T is ON, Vo = 0 and When T is OFF & D conducts , Vo = +Vs and the load

current has the direction opposite to that shown in fig.

Thus the average load voltage is positive and load current is negative. It means that

power flows from load to source and is said to be operated as an inverter.

TYPE – C CHOPPER

This is a two quadrant ( I & II quadrants ) chopper. Fig above shows the basic circuit

of Type – C chopper. Here, SCRs T1 & T2 are the chopping elements and diodes D1

&D2 are the freewheeling diodes.

When T2 is ON or D1 conducts, Vo = 0 and when T1 is ON or D2 conducts,

Vo = +Vs

When T1 is ON or D1 conducts, Io is positive and when T2 is ON or D2 conducts, Io

is negative.

Thus the average load voltage is positive but load current may be positive or

negative. It means that power flows from source to load or from load to source.

TYPE – D CHOPPER

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This is a two quadrant ( I & IV quadrants ) chopper. Fig above shows the basic circuit

of Type – D chopper. Here, SCRs T1 & T2 are the chopping elements and diodes D1

& D2 are the freewheeling diodes.

When T1 & T2 are turned ON and D1 & D2 are turned OFF, Vo = +Vs & Io is

positive. Now the power flows from source to load.

When T1 & T2 are turned OFF and D1 & D2 are conducts, average Vo is negative

and Io is positive. Now the power flows from load to source.

Thus the load current is positive but the average load voltage may be positive or

negative. It means that power flows from source to load or from load to source.

TYPE – E CHOPPER

This is a four quadrant chopper. For I & III quadrants it is operated as a step down

chopper and for II & IV quadrants it is operated as a step up chopper. Fig above

shows the basic circuit of Type – E chopper. Here, SCRs T1,T2,T3 & T4 are the

chopping elements and diodes D1,D2,D3 & D4 are the freewheeling diodes.

When T1 & T2 are turned ON, Vo is positive & Io is positive. Now the power flows

from source to load. Thus we get I quadrant operation.

When T1 & T2 are turned OFF and D3 & D4 are conducts, Vo is negative & Io is

positive. Now the power flows from load to source. Thus we get IV quadrant

operation.

When T3 & T4 are turned ON, Vo is negative & Io is negative. Now the power flows

from source to load. Thus we get III quadrant operation.

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When T3 & T4 are turned OFF and D1 & D2 are conducts, Vo is positive & Io is

negative. Now the power flows from load to source. Thus we get II quadrant

operation.

b). Explain Jones chopper. (Diagram-5 marks , Explanation-5 marks)

Jones chopper is an auxiliary commutated chopper. The circuit diagram of jones

chopper is shown in the fig above.

Here, T1 is the main SCR, Whereas, SCR T2, capacitor C, diode D and auto

transformer forms the commutating circuit for T1.

Because of the auto transformer, the capacitor always gets sufficient energy to turn

OFF the main SCR T1.

OPERATION

We assume that, initially the capacitor is charged to Vs with the polarity as shown in

fig. When T1 is turned ON, the capacitor discharges from top plate to bottom plate

through T1,L2 & D and C charges to opposite polarity with 2Vs level. This charge is

maintained in the capacitor by the diode. This is ON time of chopper.

When T2 is turned ON, the capacitor discharges from bottom plate to top plate

through T2 & T1. The discharge of capacitor reverse biases the T1 and turned it OFF.

The capacitor again charged up with its original polarity and hence T2 is also turned

OFF.

This cycle repeats when T1 is again turned ON.

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20.(a) Explain the operation of parallel inverter using IGBT.

(Diagram-5 marks , Explanation-5 marks)

The basic parallel inverter using IGBT is shown in the fig above.

Here, the commutating capacitor is connected across the load, hence it is called as

parallel inverter.

IGBTsT1 & T2 are used as switching elements. The function of L is to make the

source current constant.

OPERATION

When T1 is conducting, the current flows in the upper half of primary winding. Now

T2 is OFF. As a result an emf 2Vs is induced across the primary winding. Hence the

capacitor charges to a voltage of 2Vs with top plate as positive. Now VL = +Vs.

When T2 is turned ON, the capacitor voltage 2Vs appears as a reverse bias across T1,

it is therefore turned OFF. Now the current begins to flow through T2 and lower half

of primary winding. Hence the capacitor has charged from +2Vs to -2Vs.

Now VL = -Vs.

This cycle repeats when T1 is again turned ON and hence we get AC output across

the load.

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(b) Draw the block diagram of online UPS and explain.

(Diagram – 5 marks, Explanation – 5 marks)

ON-Line UPS is the No-Break UPS. This system operates continuously and its output

is connected to the load.

The block diagram of online UPS is shown in the fig above.

In this system, AC main supply is rectified and the rectifier delivers power to

maintain required charge on the batteries.

Rectifier also supplies power to inverter continuously which is then given to AC load

through filter and normally ON switch.

In case of main supply failure, batteries at once take over with no break of supply to

the critical load. No dip or discontinuity in the illumination is observed in case of

short break UPS.

In case inverter failure is detected, the load is switched ON to the main AC supply

directly by turning ON the normally OFF switch and opening the normally ON

switch.

After inverter failure is cleared, UPS is again restored to the load through the

normally ON switch.

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21. (a) With block diagram explain numerical control system.

(Diagram-5 marks , Explanation-5 marks)

A N/C machine tool system consists of the Machine Control Unit (MCU) and the

machine tool.

The MCU consists of two main units. The Data Processing Unit (DPU) and the

Control Logic Unit (CLU).

The DPU consists of the data input device, data reading circuits, the parity checking

logic, decoding circuits and an interpolator.

The DPU reads the punched tape using a tape reader. The coded information from the

tape reader passes to the decoding circuits.

The data contains the required new position of each axis, its direction of motion, feed

and auxiliary function control signals .

The decoding circuits distribute the data among the controlled axes. The interpolator

supplies current velocity commands between two different points from the

drawing .This data is then sent to the control loops unit.

The CLU operates the driving devices of the machine lead screws and receives the

feedback signals about the position and velocity of each of the axes. Each lead screw

under each axis of control has a separate driving device and a separate feedback.

The driving device can be a dc motor, a stepper motor or a hydraulic system.

The feedback devices are the measuring devices such as the encoders, digitizers,

resolvers, inductosyn, tachometers and digital to analog converters.

The CLU has position control loops, velocity control loops, decelerating and

backlash take-up circuits and auxiliary function control.

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(b) (i) Write short notes on post processor.(Diagram – 3 marks, explanation – 2 marks)

Post processor is a device that converts APT output into coded information.

The post processor elements are the input, motion, auxiliary, output and control.

1 .The input element reads the information which is the APT output. Reading may be

performed directly with punched cards, punched tape or the magnetic tape.

2. The motion element performs all the instructions concerned with the tool movement. The

motion element includes two functions denoted as the dynamic and geometric portion of the

package.

3. The auxiliary element is fed by the preparatory and miscellaneous functions that can be

performed by the machine tool /control unit combination and accepts from the input

element all data concerning these functions.

4. The output element receives the output data from the motion and auxiliary elements. The

data is converted into the format acceptable by the MCU.

5. The control element generates the timing of the post processor and adapts all elements

and permits program flow. It also controls the flow of data to the external output.

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(ii) Block diagram of ACO. (Diagram – 5 marks )

Prepared by,

I.KUMARAN,

LECTURER/INSTRUMENTATION

224, ANNAMALAI POLYTECHNIC COLLEGE

CHETTINAD – 630102.