power electronics and simulation lab - mist.ac.in · 11. single phase series inverter with r and rl...

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MAHAVEER INSTITUTE OF SCIENCE & TECHNOLOGY (Approved by AICTE, Affiliated to JNTU, Hyd)

Vyasapur, Bandlaguda, Post : Keshavgiri, Hyderbad - 500 005. =============================================================

POWER ELECTRONICS AND

SIMULATION LAB

III-B.TECH

II SEMESTER

NAME OF THE STUDENT : REGISTERNUMBER : YEAR/ SEMESTER : STAFF INCHARGE : Ms .R.KALYANI Assis.Prof/EEE

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General Instructions to students for EEE Lab courses

Be punctual to the lab class.

Attend the laboratory classes wearing the prescribed uniform and shoes.

Avoid wearing any metallic rings, straps or bangles as they are likely to prove dangerous at times.

Girls should put their plait inside their overcoat

Boys students should tuck in their uniform to avoid the loose cloth getting into contact

with rotating machines.

Acquire a good knowledge of the surrounding of your worktable. Know where the various live points are situated in your table.

In case of any unwanted things happening, immediately switch off the mains in the

worktable.

This must be done when there is a power break during the experiment being carried out.

Before entering into the lab class, you must be well prepared for the experiment that you are going to do on that day.

Get the circuit diagram approved.

Prepare the list of equipment’s and components required for the experiment and get the

indent approved.

Make connections as per the approved circuit diagram and get the same verified. After getting the approval only supply must be switched on.

Get the reading verified. Then inform the technician so that supply to the worktable can

be switched off.

You must get the observation note corrected within two days from the date of completion of experiment. Write the answer for all the discussion questions in the observation note. If not, marks for concerned observation will be proportionately reduced.

Submit the record note book for the experiment completed in the next class.

If you miss any practical class due to unavoidable reasons, intimate the staff in charge

and do the missed experiment in the repetition class.

Such of those students who fail to put in a minimum of 75% attendance in the laboratory class will run the risk of not being allowed for the University Practical Examination.

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LIST OF EXPERIMENTS

Any eight of the experiments in power electronics lab

1. Study of characteristics of SCR, MOSFET, & IGBT.

2. Gate firing circuit for SCR’s.

3. Single phase AC voltage controller with R AND RL loads.

4. Single phase fully controlled bridge converter with R load and RL loads

5. Forced commutation circuits (Class A, Class B, Class C, Class D & Class E).

6. DC Jones chopper with R and RL loads.

7. Single phase parallel inverter with R and RL loads.

8. Single phase cycloconverter with R and RL loads.

9. Single phase half controlled converter with R loads.

10. Three phase half controlled bridge converter with R loads.

11. Single phase series inverter with R and RL loads.

12. Single phase bridge converter with R and RL loads.

13. Single phase dual converter with RL loads.

14. Operation of MOSFET based chopper.

Any two simulation experiments with PSPICE/PSIM

15. PSPICE simulation of single phase full converter using RLE loads and single

phase AC voltage controller using RLE loads.

16. PSPICE simulation of resonant pulse commutation circuit and buck chopper.

17. PSPICE simulation of single phase inverter with PWM control.

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LIST OF CYCLE-I

.

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LIST OF CYCLE-II

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TABLE OF CONTENTS

Sl.No Experiment Name Experiment Submission Marks Signature Date date

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

SUBJECT INCHARGE INTERNAL MARK---------

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Exp. No.:1 Date:

SINGLE PHASE AC VOLTAGE CONTROLLER WITH R AND RL LOADS

OBJECTIVE: To study the module and waveforms of a 1-Φ AC voltage controller with R and RL loads.

RESOURCES:

S. No. Name of the Apparatus Range Type Quantity

1 1 AC voltage regulator - - 1 power module

2 Loading Rheostat 50 , 2A - 1

3 Loading Inductor 150mH, 2A - 1

4 CRO & probe 20MHz Dual 1

5 Connecting wires - As required

SPECIFICATIONS:

1. Input 1 , 230V, 50Hz AC Supply

2. Load R and RL.

3. Thyristors 12A, 600V, type 25 RIA 120.

4. TRIACs 10A, 600V, BT136.

5. MCB Two pole 230V/16A.

6. Fuses 16A HRC.

7. Step down transformer 230V/24V-0-24V, 2A.

PRECAUTIONS:

1. Make sure all the connecting links are tightly fixed.

2. Ensure all the controlling knobs in fully counterclockwise position before starting experiment.

3. Handle everything with care.

4. Make sure the firing pulses are proper before connecting to the power circuit.

5. Make sure to connect firing pulses from the firing circuit to their corresponding

SCRs/TRIAC in the power circuit.

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PROCEDURE:

1. Switch ON the mains supply to the firing circuit. Observe the trigger outputs by varying

firing angle potentiometer and by operating On/OFF and SCR/TRIAC selector switch. Make

sure the firing pulses are proper before connecting to the power circuit.

2. Make the connections as per the circuit diagram.

3. Connect firing pulses from the firing circuit to the corresponding SCRs/TRIAC in the power

circuit.

4. Switch ON the step down transformer supply (MCB) and now switch ON the trigger pulses

by operating ON/OFF switch in the firing circuit.

5. Observe the output voltage waveform across load using oscilloscope.

6. Note down the input voltage, firing angle and output voltage readings in the TABULAR

FORMS.

7. Draw the waveforms in the graph at 0, 45, 90, 135 and 180 Deg. firing angles.

FORMULAE USED: OUT PUT VOLTAGE(Vo)=

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CIRCUIT DIAGRAM:

T1

K1

A1

P

G2

2 TYN616G1

A2

K2

230V 24V T2

1 , 230V, 0V

K1 G1 R

LOAD

50Hz, AC K2 G2 50 ,2A

0V 24V

N

TABULAR FORMS: 1. for R load

S.No. Input voltage (V)

Firing angle Output voltage Theoretical

( )

(V) output

voltage (V)

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MODEL GRAPH:

Vi (v) Input Waveform

0 π 2π 3π 4π

Output Waveform Across R and RL- Load =00

VL (v)

0

Output WaveformAcross R- Load =900

VL (v)

0

Output Waveform Across RL- Load =900

VL (v)

0

t (ms) t (ms)

t (ms) t (ms)

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MODEL CALCULATIONS:

RESULT:

Thus the single phase AC voltage controller with R & RL loads is studied and we plotted

the waveforms of different firing angle.

PRE LAB QUESTIONS:-

1. Why should the two trigger sources be isolated?

2. What are the advantages and the disadvantages of phase control?

3. What is phase control?

4. What are the advantages of bidirectional controllers?

5. What is meant by duty cycle in ON-OFF control method?

POST LAB QUESTIONS:- 1. What type of commutation is used in this circuit?

2. What are the effects of load inductance on the performance of AC voltage controllers?

3. What is extinction angle?

4. What are the disadvantages of unidirectional controllers?

5. What are the advantages of ON-OFF control?

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Exp. No.:2 Date:

DC JONES CHOPPER WITH R AND RL LOADS

OBJECTIVE: To study the module and waveforms of a DC Jones chopper with R and RL loads.

RESOURCES:

S. No.

Name of the Range Type Quantity

Apparatus

Jones chopper firing

1 circuit module and - - 1

power circuit module





SPECIFICATIONS:

1. Input: 0 – 230V 1Φ AC supply.

2. Load R, & RL


4. Diodes: 25A, 1200V.

5. Commutating Capacitor 25µF, 440V


7. Fuses 16A HRC

PRECAUTIONS:





5. Make sure to connect firing pulses from the firing circuit to their respective SCRs in the

power circuit.

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MODEL GRAPH:

Voltage and Current waveforms in the Jones Chopper

Ig1

Ig2

ISCR2

VSCR2

Vc

Ic

ISCR1

`

VSCR1

VL

t

t

t

t

t

t

t

t

t

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PROCEDURE:

1. Switch On the mains supply to the firing circuit. Observe the trigger on by varying duty

cycle and frequency potentiometer by keeping the c switch in ‘INT’ position. Make sure

the firing pulses are proper be connecting to the power circuit.


3. Connect firing pulses from the firing circuit to the respective SCRs power circuit.

4. Initially set the input DC supply to 5V.

5. At the beginning, keep the ON/OFF switch in the firing circuit in OFF position.

6. Switch ON the DC supply and now ON the trigger pulses by open On/OFF switch in the

firing circuit.

7. Observe the DC chopped voltage waveform across load using oscilloscope.

8. If the commutation fails, pure DC voltage can be observed across the then switch OFF the

DC supply and trigger pulses. Check the connect and try again.

9. Observe the voltage waveform across load, capacitor, main SCR auxiliary SCR by varying

the duty-cycle potentiometer and frequency potentiometer, using oscilloscope.

10. Now, vary the DC supply up to the rated voltage, 30VDC.

11. Note down the readings in the TABULAR FORMS.

12. Draw the waveforms in the graph at different duty cycles and at different

Formula used:

Theoretical value = Ton/T 100

T= Ton +Toff

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CIRCUIT DIAGRAM: Circuit Diagram of DC Jones Chopper with RL Load

TABULAR FORMS:

At F1 (middle)

S. Input Time in milli sec. Duty Output Theoretical

No. voltage cycle voltage value =

(Vin) Ton(ms) Toff(ms) (%) (V0) Ton/T 100

\

RESULT: Thus the module and waveforms of a DC Jones chopper with R and RL loads was studied.

PRE LAB QUESTIONS:-

1. What is a chopper? Where it is normally employed? 2. Explain the principle of operation of a chopper. 3. What are the control strategies used for a chopper? 4. What is time ratio control (TRC) of a chopper? How is it classified?

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2.11 POSTLAB QUESTIONS:-

1. Why is forced commutation required in dc choppers?

2. What are the effects of turn on and turn off times of thyristor on the performance of the chopper.

3. What are the merits and demerits of this circuit? 4. What is current limit control of a chopper?

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Exp. No.: 3 Date:

SINGLE PHASE PARALLEL INVERTER WITH R AND RL LOADS

OBJECTIVE: To study the module and waveforms of a 1-Φ Parallel inverter with R and RL loads.

RESOURCES:

S. No.


Apparatus

1-Φ parallel inverter

1 firing module and - - 1

power circuit module





SPECIFICATION:

1. Input 230V, 50Hz, 1-Φ AC supply.

2. Load R and RL.

3. Thyristors 10A, 600V.

4. Diodes 10A, 600V.

5. Capacitors 6.8µF, 100V.

7. Inductor 300mH, 2A.

9. Fuses 2A Glass fuse.

PRECAUTIONS:





5. Make sure to connect firing pulses form the firing circuit to their respective SCRs in the

power circuit.

6. Ensure to switch OFF the input supply first and then trigger pulses to short circuit.

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CIRCUIT DIAGRAM:

MODEL GRAPH:

TRIGGER OUTPUTS

T1

T2

+VDC

--VDC

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PROCEDURE:

1. Switch ON the mains supply to the firing circuit. Observe the trigger output in the firing

circuit by varying frequency potentiometer and by operating OFF switch. Make sure the

firing pulses are proper before connecting to the power circuit.


3. Connect firing pulses form the firing circuit to the respective SCRs in power circuit.

4. Connect the DC input from a 30V, 2A regulated power supply.

5. Switch ON the DC supply, set input voltage to 15V and switch ON the trigger pulses


6. Observe the voltage waveform across load using oscilloscope.

7. Vary the frequency and observe the voltage waveforms across load with without

freewheeling diode.

8. Draw the waveforms in the graph at different frequencies.

9. To switch off the inverter, switch OFF the input supply first and then trigger pulses.

10. Since the parallel inverter works on forced commutation, there is a chopper commutation

failure. If the commutation fails, switch off the DC supply and then trigger outputs. Check

the connections and try again.

TABULAR FORMS: 1For R Load:

S. Input Time in milli sec. Output Theoretical

No. voltage

TOTAL TIME PERIOUD average output

(Vin) Ton(ms) Toff(ms) voltage frequency

(V0) in Hz

RESULT: Thus the module and waveforms of a 1-Φ Parallel inverter with R and RL loads was studied.

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PRE LAB QUESTIONS:-

1. What is parallel inverter? Why is it called so? 2. What is the purpose of capacitor in the parallel inverter? 3. What is the purpose of transformer in the parallel inverter? 4. IS the parallel inverter naturally commutated or force commutated? 5. What are the advantages of parallel resonant inverters?

3.11 POST LAB QUESTIONS:-

1. What is the purpose of the inductor in the parallel inverter? 2. During its operation, capacitor voltage reaches 2Vs. How? 3. What is the significance of the split phase transformer? 4. During operation, what is the voltage across primary winding of the transformer? 5. Capacitor current flows in how many modes of the operation of parallel inverter?

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Exp. No.:4 Date:

SINGLE PHASE HALF CONTROLLED CONVERTER WITH R LOAD

OBJECTIVE: To study the module and waveforms of a 1-Φ Half controlled converter with R load at different

firing angles.

RESOURCES:


1 Half controlled

1 converter power and - - 1

firing module





SPECIFICATIONS:

1. Input : 1 , 230V , 50Hz AC supply.

2. Load : R, RL

3. Thyristors : 25A, 1200V, type 25 RIA 120/TYN616.

4. Diode : 25A, 1200V, BY126/BY127.

5. MCB : Two pole 230V/16A

6. Fuses : 16A HRC.

7. Field Supply bridge rectifier : 10A, 600V.

8. Field Supply : 220V + 10%.

PRECAUTIONS:





5. If the output is zero even after all power connections, switch OFF the MCB and just

interchange AC input connections to the power circuit. This is to make the firing circuit

and power circuit to synchronize.

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MODEL GRAPH:

Input Waveform

Vi (v)

t (ms)

0 π 2π 3π 4π

Output Waveform R and RL - Load at =00

VL (v)

t (ms) 0

VL (v) Output Waveform R and RL -Load at=45

0

t (ms) 0

VL (v)

0

0


t (ms)


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PROCEDURE:

1. Switch ON the main supply to the firing circuit. Observe the trigger output by varying firing

angle potentiometer and by operating ON/OFF switch and their phase sequence. Make sure

the firing pulses are proper before connecting to the power circuit.


3. Connect 30V tapping of the transformer secondary to the power circuit.

4. Connect firing pulses 0from the firing circuit to their respective SCRs in power circuit.

5. Switch ON the MCB and now switch ON the trigger pulses by operate ON/OFF switch in

the firing circuit.

6. Observe the output voltage waveforms across load and devices us oscilloscope.

7. Note down the input voltage, firing angle, Output voltage and output circuit reading in the

TABULAR FORMS.

8. Repeat the same for different input voltage up to max. voltage as provided in the

isolation transformer.

9. Repeat the same for R-L and RLE loads with and without freewheeling diode.

10. Draw the waveforms in the graph at firing angles 00, 45

0, 90

0, 135

0 and 180

0.

FORMULAE USED:

Output voltage V0 = Vdc = Vm/ (1 + cos )

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CIRCUIT DIAGRAM: SINGLE PHASE HALF CONTROLLED CONVERTER WITH R LOAD 2 TYN616

P

K1G1 K2 G2

230V 30V T1 T2 R LOAD

A1

A2 1 , 230V,

150 , 5A

50Hz, AC K1

K2

0V

0V

D1 D2

A1 A2 N

2 1N4007 Tabular form with R load

S. No. Input Firing Output voltage Output

voltage angle (a) (V) Theoretical

(V) voltage

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SINGLE PHASE HALF CONTROLLED CONVERTER WITH RL LOAD

K1

K2 G2

p

230V 30V T1 T2

R LOAD A1 A2

1 , 230V,

150 , 5A

0-150mH, 5A 50Hz, AC K1

K2 L

0V 0V

D1

D2

N A1 A2

2 1N4007



(V) voltage

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MODEL CALCULATIONS:

RESULT:

Thus the single phase half controlled converter with R and RL load is studied and also

we plotted the waveforms of different firing angles. PRE LAB QUESTIONS:-

1. What is the delay angle control of converters? 2. What is natural or line commutation? 3. What is the principle of phase control? 4. What is extinction angle? 5. Can a freewheeling diode be used in this circuit and justify the reason?

POSTLAB QUESTIONS:- 1. What is conduction angle? 2. What are the effects of adding freewheeling diode in this circuit? 3. What are the effects of removing the freewheeling diode in single phase semi converter? 4. Why is the power factor of semi converters better than that of full converters? 5. What is the inversion mode of converters?

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Exp. No.: 5 Date:

PSPICE SIMULATION OF SINGLE PHASE FULL CONVERTER AND

SINGLE PHASE AC VOLTAGE CONTROLLER USING RLE LOADS

OBJECTIVE: To study the output waveforms of single-phase full converter using RLE loads and single-phase

AC voltage controller using RLE loads using PSPICE simulation.

RESOURCES:PSPICE Software AC Model of SCR:

F1= P1Ig + P2Ia

= 50Ig + 11Ia

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Circuit diagram of single phase full converter:

Circuit file for Single phase full converter:

VS 10 0 SIN (0 169.7V 60HZ) VG1 6 2 PULSE (0V 10V 2777.8US 1NS 1NS 100US 16666.7US) VG2 7 0 PULSE (0V 10V 2777.8US 1NS 1NS 100US 16666.7US) VG3 8 2 PULSE (0V 10V 11111.1US 1NS 1NS 100US 16666.7US) VG4 9 1 PULSE (0V 10V 11111.1US 1NS 1NS 100US 16666.7US) R 2 4 10 L 4 5 20MH C 2 11 793UF RX 11 3 0.1 VX 5 3 DC 10V VY 10 1 DC 0V * SUBCIRCUIT CALLS FOR THYRISTOR

MODEL XT1 1 6 2 SCR XT2 0 8 2 SCR

XT3 3 7 0 SCR

XT4 3 9 1 SCR . SUBCKT SCR 1 3

2 S1 1 5 6 2 SMOD

RG 3 4 50 VX 4 2 DC 0V

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VY 5 2 DC 0V RT 2 6 1 CT 6 2 10UF F1 2 6 POLY (2) VX VY 0 50 11 .MODEL SMOD VSWITCH (RON=0.01 ROFF=10E+5 VON=0.1V VOFF=0V) .ENDS SCR .TRAN 10US 35MS 16.67MS .PROBE .OPTIONS ABSTOL=1.00U RELTOL=1.0M VNTOL=0.1 ITL5=10000 .FOUR 120HZ I (VX)

Circuit diagram of single phase Ac Voltage Controller:

5.6 Circuit file for Single phase ac voltage controller:

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VS 10 0 SIN (0 169.7V 60HZ) VG1 2 4 PULSE (0V 10V 4166.7US 1NS 1NS 100US 16666.7US) VG2 3 1 PULSE (0V 10V 12500.0US 1NS 1NS 100US 16666.7US) R 4 5 2.5 L 5 6 6.5MH VX 6 0 DC 0V CS 1 7 0.1UF RS 7 4 750 * SUBCIRCUIT CALLS FOR THYRISTOR

MODEL XT1 1 2 4 SCR XT2 4 3 1 SCR . SUBCKT SCR 1 3

2 S1 1 5 6 2 SMOD

RG 3 4 50 VX 4 2 DC 0V

VY 5 2 DC 0V RT 2 6 1 CT 6 2 10UF F1 2 6 POLY (2) VX VY 0 50 11 .MODEL SMOD VSWITCH (RON=0.01 ROFF=10E+5 VON=0.1V VOFF=0V) .ENDS SCR .TRAN 10US 33.33MS .PROBE .OPTIONS ABSTOL= 1.00N RELTOL = 1.0M VNTOL=1.0M ITL5=10000 .FOUR 60HZ V (4) .END

5.7RESULT : The output waveforms of single-phase full converter using RLE loads and single-phase AC

voltage controller using RLE loads using PSPICE simulation are studied.

5.8 PRE LAB QUESTIONS:- 1. What is the difference between a diode rectifier and a thyristor rectifier? 2. What is controlled rectification? 3. What is meant by firing angle of a converter? 4. What is an ac voltage controller? 5. How does the load inductance effect the conduction angle of a controller?

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5.9 POST LAB QUESTIONS:- 1. What is an integral cycle control? 2. What is phase control? 3. What is discontinuous in thyristor power converters? 4. How is it achieved in thyristor power converters? 5. What are the effects of load inductance on the performance of a power converter

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Exp. No.:6 Date:

GATE FIRING CIRCUITS FOR SCRs

OBJECTIVE: To study the following various firing schemes for triggering SCRs when they are different

converter topologies employing line commutation.

1. Resistance firing circuit.

2. Resistance capacitance (RC) firing circuit.

3. UJT firing scheme.

RESOURCES:

S. No.


Apparatus

1 R & RC power module - - 1

2 UJT power module - - 1

3 R load 50 , 2A - 1

4 CRO & probe Dual - 1


SPECIFICATIONS:

1. SCRs : 400V, 4A, type 106 D

2. Diodes : 1N4007

3. Diacs : D3202U

4. Zeners : 20V, 1W

5. UJTs : 2N2646

6. Pulse transformer : 1:1:1

PRECAUTIONS:

1. Make sure all the connections are tight.

2. Ensure all the controlling knobs are kept in fully counterclockwise position before

starting experiment.


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MODEL GRAPH:


0

π 2π 3π 4π t (ms)

Output Waveform=00

VL (v)

t (ms)

VSCR (v)0

0 t (ms)

Output Waveform =90 0

VL (v)

0 t (ms)

VSCR (v)

0 t (ms)

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PROCEDURE: (a). R firing circuit:


2. Connect a load rheostat of 50Ω, 2 A between the load points

3. Switch ON the power supply

4. Vary the control pot and observe the voltage waveforms across load and at different points

in the circuit using oscilloscope.


0, 90

0, 135

0.

6. Bring the pot to the original position.

7. Switch OFF the power Supply.

(b). RC firing circuit:




4. Vary the control pot and observe the voltage waveforms across load and at different

points in the circuit using oscilloscope.


0, 90

0, 135

0 and 180

0.



(c). UJT firing circuit:




4. Vary the control pot and observe the voltage waveforms across load and at different

points in the circuit using oscilloscope.


0, 90

0, 135

0 and 180

0.



FORMULAE USED:

Firing Angle

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GATE FIRING CIRCUIT FOR SCR’S-RESISTANCE FIRING CIRUCIT

R LOAD

50 , 2A

20V, 2A R

AC T1

RC TYN616

D

K1

K

Rg A

1N4007

TABULAR FORMS:

Resistance Firing Circuit: -

s.no T on T off firing angle(α)

output voltage

36

GATE FIRING CIRCUIT FOR SCR’S-RESISTANCE CAPACITANCE (RC) FIRING CIRUCIT

R LOAD

50 , 2A

20V, 2A

R A1

AC T1

RC D1 TYN616

D2

G1 K1 A K Rg

C 4.7 F 1N4007

B. i. Resistance Capacitance Firing Circuit: -

s.no T on T off firing angle(α) output voltage

37

C. i. UJT Firing Circuit: -

GATE FIRING CIRCUIT FOR SCR’S-UJT FIRING CIRUCIT R

P

R

D1 D3 RC

1 , 230V, K 50Hz AC 20V

C UJT

, 2A 2N2646

1000 F ZD

A

D4 D2 C

4.7 F

N

Pulse TFR

1:1:1

R 50 L

, 2A O

A A

T1

G K

AC

20V, 2A

Sl. Ton Toff Firing angle

OUTPUT VOLTAGE

No.

( ) in

1.

2.

3.

4.

5.

38

RECTIFICATION CHARACTERISTICS OF SCR BY AC/DC GATE CONTROL Aim This unit consists of a transformer, a DC supply with potentiometer for DC gate control and AC supply with potentiometer for AC gate control, an SCR to study the rectification characteristics of SCR by AC/DC gate control. Apparatus

1) Mains :- Mains ON/OFF for the unit with built in indicator.

2) Fuse :- 500mA to protect the SCR against short circuit.

3) VDC :- 0 to 9 Volts unregulated for DC gate control.

4) SCR :- TYN 612 G- Gate,

K- Cathode, A- Anode.

5) Transformer :- 0-10 V/ 500mA.

Theory One important application of an SCR is the controlled half-wave rectification.The A.C supply to be rectified is supplied through thetransformer. The load resistance RL is connected in series with the anode. A variable resistance isinserted in the gate circuit to control the gate current.Suppose the peak reverse voltage appearing across secondary is less than the reverse breakdown voltage of the SCR. This condition ensures that SCR will not break down during negative half-cycles of A.C supply. During the negative half-cycles of A.C voltage appearing across secondary, the SCR does notconduct regardless of the gate voltage. It is because in this condition, anode is negative w.r.t. cathodeand also PRV is less than the reverse breakdown voltage. The SCR will conduct during the positive half-cycles provided proper gate current is madeto flow. The greater the gate current, the lesser the supply voltage at which SCR is turned ON. The gate current can be changed by the variable resistance. Procedure for Rectification characteristics of SCR using AC control

1) Connections are made as shown in connection diagram.

2) Keep the potentiometer 2K at minimum so that Ig = 0.

3) Connect load of 50 Ohms/25 Watt fixed Resistor across Load terminals.

4) Vary potentiometer and observe the waveform across Load and SCR

5) To calculate firing angle note down the values of X,Y using CRO

39

Connection diagram for Rectification characteristics of SCR by AC gate control

Rectification characteristics of SCR by AC gate control circuit diagram

Tabular column for AC triggering circuit

Procedure for Rectification characteristics of SCR using DC control 1) Connections are made as shown in connection diagram.

2) Keep the potentiometer 2K at minimum so that Ig = 0.

3) Connect load of 50 Ohms/25 Watt fixed Resistor across Load terminals.

4) Vary potentiometer and observe thatSCR conducts after Ig min require to SCR conducts.

This is one of basic triggering of SCR by AC and DC gate control. 5) To calculate firing angle note down the values of X,Y using CRO

s.no Firing circuit Vo(volts) X Y

40

Rectification characteristics of SCR by DC gate control circuit diagram

Connection diagram for Rectification characteristics of SCR by DC gate control

Tabular form for D.C triggering circuit

s.no Firing circuit Vo(volts) X Y

41

MODEL CALCULATIONS:

42

Model waveforms for AC and DC triggering circuit

RESULT: Gate firing circuits using AC and DC triggering circuit is verified and wave forms are observed

RESULT: Thus the different types of gate firing circuits of SCR’s i. R Firing circuit, ii.RC Firing circuit and iii. UJT firing circuit is studied and also plotted its waveforms.

PRE LAB QUESTIONS:-

1. UJT triggering circuit is also known as?

2. Types of triggering circuit?

3. What is the purpose of series resistor?

4. What is the condition for triggering the circuit?

5. What is the function of pulse transformer in firing circuit?

POST LAB QUESTIONS:-

1. Explain how synchronization of the triggering circuit with the supply voltage across SCR

is achieved?

2. How can the capacitor charging be controlled?

3. What is the maximum value of firing angle which can be obtained from the circuit?

4. How is the output power to the triggering circuit controlled?

5. Compare UJT triggering circuit with RC firing circuit?

43

Exp. No.:7 Date: SINGLE PHASE FULLY CONTROLLED BRIDGE CONVERTER WITH R AND RL LOADS

OBJECTIVE: To study the module and waveforms of a 1-Φ Full Bridge Converter with RL and RL loads.

RESOURCES:


Full bridge

1

controlled converter -

- 1

power and firing

module





SPECIFICATIONS:

1. Input : 1 , 230V 50Hz, AC supply.

2. Load : R and RL loads.

3. Thyristors : 16A, 1200V, type 16 TTS/TYN616

4. Diode : 25A, 1200V, BY126/BY127


6. Fuses : 16A HRC.

7. Field Supply bridge rectifier: 10A, 600V.


PRECAUTIONS:








44

MODEL GRAPH:

Input Waveform

Vi (v)

t (ms)

0 π 2π 3π 4π

Output Waveform R- Load at =00

VL (v)

t (ms) 0

Output Waveform R- Load at=450

VL (v)

t (ms) 0

RL- Load at =450

VL (v)

t (ms) 0

VL (v) R- Load at =90

0

0 t (ms)

VL (v) RL- Load at =90

0

t (ms) 0


0

t (ms)

0

45

PROCEDURE:

1. Switch ON the main supply to the firing circuit. Observe the trigger output by varying

firing angle potentiometer and by operating ON/OFF switch their phase sequence. Make




4. Connect firing pulses from the firing circuit to their respective SCRs in power circuit.


the firing circuit.



TABULAR FORMS.





0, 90

0, 135

0 and 180

0.

FORMULAE USED:

Average output voltage – R load, VAvg= Average output voltage – RL load, VAvg=

46

CIRCUIT DIAGRAM :

SINGLE PHASE FULL CONTROLLED BRIDGE CONVERTER WITH R LOAD

4 TYN616

P

K1

G1 K3 G3

230V 30V T1 T3

1 , 230V, A1 A3

R

LOAD

50Hz, AC 150 , 5A

0V 0V K4

G4 K2 G2

T4 T2

N

A4

A2

TABULAR FORMS:

a. For R load

Firing

Output voltage

Theoretical

S.No.

Input voltage (V)

Output

angle ( )

(V)

voltage (V)

47

SINGLE PHASE FULL CONTROLLED BRIDGE CONVERTER WITH RL LOAD

4 TYN616

P K1 G1 K3 G3

230V 30V T1 T3

R LOAD A1

A3

1 , 230V, 150 , 5A

0-150mH, 5A 50Hz, AC L

G2

0V 0V K4 G4 K2

T4 T2

N A4

A2

b. For RL load without freewheeling diode:

Firing Output voltage

Theoretical

S.No. Input voltage (V) Output angle ( ) (V)

voltage (V)

48

SINGLE PHASE FULL CONTROLLED BRIDGE CONVERTER WITH FREEWHEELING DIODE FED RL LOAD

4 TYN616

P K1 G1 K3 G3

230V 30V T1 T3

K R LOAD A1

A3

1 , 230V, 150 , 5A

FD 0-150mH, 5A 50Hz, AC

G2 A L

0V 0V K4

G4 K2

T4 T2

N A4

A2

c. For RL load with freewheeling diode:


Theoretical


voltage (V)

49

MODEL CALCULATIONS:

50

RESULT:

Thus the single phase Full controlled bridge converter with R and RL load is studied and

also plotted the waveforms of different firing angles.

PRE LAB QUESTIONS:-

1. State the type of commutation used in this circuit?

2. What will happen if the firing angle is greater than 90 degrees?

3. What are the performance parameters of rectifier?

4. What are the advantages of three phase rectifier over a single phase rectifier?

5. What is the difference between half wave and full wave rectifier?


1. If firing angle is greater than 90 degrees, the inverter circuit formed is called as?

2. What is Dc output voltage of single phase full wave controller?

3. What are the effects of source inductance on the output voltage of a rectifier?

4. What is commutation angle of a rectifier?

5. What are the advantages of three phase rectifier over a single phase rectifier

51

Exp. No.:8 Date:

SINGLE PHASE CYCLO CONVERTER WITH R & RL LOADS

OBJECTIVE:

To verify the operation of single phase Cyclo Converter with R and RL Loads and to

observe the output and input waveforms

RESOURCES:

S.No EQUIPMENT Qty

1. I-φ Center tapped

1 Transformer

2. I-φ Cyclo Converter power

1 circuit with firing unit

3. Rheostat 1

4. Inductive load 1

5. Voltmeter(MI) 1

6. CRO with (1:10) Probe 1

7. Patch cards 1 set

PRECAUTIONS:








52

MODEL GRAPHS:

1/2f cycloconverter waveforms

1/3f cyclo converter waveforms

1/4f cycloconverter waveforms

54

CIRCUIT DIAGRAMS:

T1

Ph

T31-Φ, 230V 50Hz

AC Supply

T2

N

center tapped transformer

T4

Fig1-Single phase cyclo converter with R-load

T1

Ph

T3 1-Φ, 230V 50Hz AC Supply

T2

N center tapped transformer

T4

Fig2-Single phase cyclo converter with RL-loadFig-2

To CRO

R

R

To CRO

L

55

PROCEDURE:

A) For R-Load:

1. Connect the circuit as shown in figure.

2. Verify the connections from the lab instructor before switch on the supply.

3. Keep the rheostat position value given by the lab instructor

4. Switch ON the supply and note down the frequency of input voltage from the CRO.

5. Set the frequency division switch at 2 and note the readings of time period of

output voltage waveform for different set of firing angles

6. Calculate the practical value of output frequency by reciprocating the value of time period

and theoretical value of frequency will be found from frequency division setting

7. Repeat the above process from step 5 to 6 for frequency division of 3 and 4.

B). For RL-Load:

1. Connect the circuit as shown in figure.

2. Connect an inductance of given value in series with the load resistance.

3. Verify the connections from the lab instructor before switch on the supply.

4. Keep the rheostat position value given by the lab instructor

5. Switch ON the supply and note down the frequency of input voltage from the CRO.

6. Set the frequency division switch at 2 and note the readings of time period of

output voltage waveform for different set of firing angles

7. Calculate the practical value of output frequency by reciprocating the value of

time period and theoretical value of frequency will be found from frequency

division setting

8. Repeat the above process from step 5 to 6 for frequency division of 3 and 4

56

TABULAR FORMS:

A) For R-Load:

The input voltage Vph = V (As given by the instructor)

Value of load resistance RL= Ω(As given by the instructor) Input frequency = Hz

S.NO. Frequency Firing angle(α) Time period in Frequency Frequency division in degrees msec (practical) (theoretical)

B) For RL-Load:

The input voltage Vph =

Value of load resistance RL=

Value of Load inductance L=

V (As given by the instructor) Ω(As given by the instructor) mH(As given by the instructor)

S.NO. Frequency Firing angle(α) Time period in Frequency Frequency

division in degrees msec (practical) (theoretical)

57

RESULT: The operation of I-φcyclo converter is verified and the theoretical and practical

valuesof output frequencies at different frequency divisions are found both for R & RL loads

PRE LAB QUESTIONS:- 1. On what principle does cycloconverter works? 2. What is the major difference between AC voltage controller and cycloconverter? 3. What type of commutation is employed in cycloconverter?


1. What is the purpose of reactor connected in cycloconverter? 2. What happens to the output if the frequency of operation is beyond suggested limit? 3. What are the applications of cycloconverter?

58

Exp. No.:9 Date:

PSPICE SIMULATION OF SINGLE PHASE INVERTER WITH PWM CONTROL

OBJECTIVE: To study the output of single phase Inverter with PWM control using

PSPICEsimulation.

Resources: PSPICE Software

CIRCUIT DIAGRAMS OF SINGLE PHASE INVERTER WITH PWM CONTROL

(a) Circuit

(b) PWM generator

(c) carrier and reference signals

10.4 CIRCUIT MODEL FOR SINGLE PHASE INVERTER WITH PWM

CONTROL VS 1 0 DC 100V VR 17 0 PULSE (50V 0V 0 833.33US 833.33US 1NS 16666.67US) RR 17 0 2MEG VC1 15 0 PULSE (0 -30V 0 1NS 1NS 8333.33US 16666.67US) RC1 15 0 2MEG VC3 16 0 PULSE (0 -30V 8333.33US 1NS 1NS 8333.33US 16666.67US) RC3 16 0 2MEG R 4 5 2.5 L 5 6 10MH VX 3 4 DC 0V VY 1 2 DC 0V D1 3 2 DMOD D2 0 6 DMOD D3 6 2 DMOD D4 0 3 DMOD .MODEL DMOD D (IS=2.2E-15 BV=1800V TT=0) Q1 2 7 3 QMOD Q2 6 9 0 QMOD Q3 2 11 6 QMOD Q4 3 13 0 QMOD .MODEL QMOD NPN(IS=6.734F BF=416.4 CJC=3.638P CJE=4.493P) RG1 8 7 100 RG2 10 9 100 RG3 12 11 100 RG4 14 13 100 * SUBCIRCUIT CALL FOR PWM

CONTROL XPW1 17 15 8 3 PWM

59

10.4 CIRCUIT MODEL FOR SINGLE PHASE INVERTER WITH PWM

VR 17 0 PULSE (50V 0V 0 833.33US 833.33US 1NS 16666.67US)

30V 0 1NS 1NS 8333.33US 16666.67US)

30V 8333.33US 1NS 1NS 8333.33US 16666.67US)

15 BV=1800V TT=0)

.MODEL QMOD NPN(IS=6.734F BF=416.4 CJC=3.638P CJE=4.493P)

SUBCIRCUIT CALL FOR PWM

CONTROL XPW1 17 15 8 3 PWM

60

XPW2 17 15 10 0 PWM XP3 17 16 12 6 PWM XP4 17 16 14 0 PWM * SUBCIRCUIT FOR PWM CONTROL .SUBCKT PWM 1 2 3

4 R1 1 5 1K R2 2 5 1K RIN

5 0 2MEG RF

5 3 100K RO 6

3 75 CO 3 4

10PF E1 6 4 0

5 2E+5 .ENDS PWM .TRAN 10US 16.67MS 0 10US .PROBE .OPTIONS ABSTOL 1.00N RELTOL=0.01 VNTOL=0.1 ITL5=20000 .FOUR 60HZ V (3, 6) .END

RESULT:

PSPICE simulation of single phase Inverter with PWM control is studied and

output waveforms are observed.

PRE LAB QUESTIONS:-

1. What are the disadvantages of PWM control? 2. What are the methods of reduction of harmonic content? 3. What is meant by PWM control? 4. What are the main classifications of inverter? POST LAB QUESTIONS:-

1. What is meant by inverter? 2. What is McMurray Inverter? 3. How is the inverter circuit classified based on commutation circuitry? 4. What are the applications of an inverter?

61

Exp. No.:10 Date:

PSPICE SIMULATION OF BUCK CHOPPER AND RESONANT PULSE COMMUTATION

OBJECTIVE: Study of resonant pulse commutation circuit and Buck chopper with PSPICE simulation RESOURCES: PSPICE Software CIRCUIT DIAGRAM OF RESONANT PULSE COMMUTATION

CIRCUIT FILE FOR RESONANT PULSE COMMUTATION VS 1 0 DC 12V VY 1 2 DC 0V VG 8 0 PULSE(0V 20V 0 1NS 1NS 12.24US 40US) RB 8 7 250 R 6 0 10 LE 2 3 25.47UH CE 3 0 1.38UF C 3 4 0.0958UF L 5 6 445.63UH VX 4 5 DC 0V Q1 3 7 0 MODQ1 .MODEL MODQ1NPN (IS=6.734F BF=416.4 ISE=6.734F BR=.7371 + CJE=3.637P MJC=0.3085 VJC=.75 CJE=4.493P MJE=.2593 VJE=.75 + TR=239.5N TF=301.2P) .TRAN 2US 300US 180US 1US UIC

62

.PROBE .OPTIONS ABSTOL=1.00N VNTOL=0.1 ITL5=20000 .END

11.5 Circuit diagram of buck converter

11.6 CIRCUIT MODEL FOR BUCK CHOPPER VS 1 0 DC 110V VY 1 2 DC 0V VG 7 3 PULSE (0V 20V 0 0.1NS 0.1 NS 27.28US

50US RB 7 6 250 LE 3 4 681.82UHCE 4 0 8.33UF IC=60V L 4 8 40.91UH R 8 5 3 VX 5 0 DC 0V DM 0 3 DMOD .MODEL DMOD D (IS=2.2E-15 BV=1800V TT=0) Q1 2 6 3 QMOD .MODEL QMOD NPN (IS=6.734F BF=416.4 BR=.7371 CJC=3.638P + CJE=4.493P TR=239.5N TF=301.2P) .TRAN 1US 1.6MS 1US UIC .PROBE .OPTIONS ABSTOL=1.00N RELTOL=0.01 VNTOL=0.1 ITL5=50000 .FOUR 20KHZ I(VY) .END

RESULT: PSPICE simulation of resonant pulse commutation circuit and Buck chopper

isstudied and output waveforms are observed.

63

11.8 PRE LAB QUESTIONS:-

1. What is PSPICE?

2. What is the principle of Buck Chopper?

3. What are the different types of chopper with respect to commutation process?

11.9 POST LAB QUESTIONS:-

1. What are the applications of dc chopper?

2. What is commutation angle or overlap?

3. What are the advantages of current commutated chopper?

64

Exp. No.:11 Date:

SINGLE PHASE SERIES INVERTER WITH R AND RL LOADS

OBJECTIVE:

To study the operation of Single-phase series inverter with R and RL loads and plot its output

waveform.

RESOURCES:

S.No. ITEM RANGE TYPE QUANTITY

1 Series inverter power circuit kits 1 Ф ,230 V , 2 A 1

2 Series inverter firing circuit kit 1 Ф ,230 V , 2 A 1

3 Loading rheostat 100 / 2A 1


5 Regulated power supply (0 – 30 V) / 2 A 1

6 CRO 20 MHZ 1

7 Patch chords 15

MODEL GRAPH (SERIES INVERTER):

T1

T2 t

ec1 t

ec2 t

t

eo

t

65

PROCEDURE (SERIES INVERTER) :


2. Switch on the thyristor firing circuit

3. Keep the frequency knob of the firing circuit kit below the resonance Frequency of power

circuit kit

4. Switch on the DC power supply connected to the power circuit kit and Switch on the firing

circuit kit

5. Vary the frequency knob of the firing circuit kit

6. Observe the waveform from the CRO.

7. Repeat the same procedure for different values of L,C and load resistance.

8. Switch of the power supply and disconnect the connection

9. Calculate the frequency of the output waveform.

CIRCUIT DIAGRAM -SERIES INVERTER

T1 (0-30V), M.I C1

D1

FUSE2A

L1

V

(0-30)V L2

LOAD

RPS

CRO

T2 C2

D2

66

TABULAR FORMS (SERIES INVERTER):

RESONANCE FREQUENCY = ____________: FIRING ANGLE = __________

S.No. Input Voltage Frequency Of Firing Output Voltage

(Vi) Volts Circuit (Hz) Vo (Volts)

CALCULATIONS

67

RESULT:

Thus a single-phase series inverter operation was studied and its output waveform was plotted.

PRE LAB QUESTIONS:-

1. What is series inverter?

2. What are the advantages of basic series inverter?

3. Compare basic &modified series inverter?


1. What is the condition for resonant circuit behave like a capacitive load and inductive load

in series resonant inverter

2. What are the drawbacks of a basic series inverter?

3. What are the applications of series inverters?

4. Why are the inductors L1, L2 and why are two capacitors needed?

68

THREE PHASE Exp. No.:12 Date:

HALF CONTROLLED BRIDGE CONVERTER WITH R LOADS.

OBJECTIVE The objective of Experiment is to analyze the operation (Switching) of three phase

half controlled rectifiers with resistive load.

RESOURCES:


3 Half controlled


firing module




SPECIFICATIONS:

Input Supply : 415V / 3ph. Supply for phase synchronization and 230V, 50Hz Single phase supply for the power supply Output : Six pairs of pulse transformer isolated trigger pulses. Gate drive current : 230mA. Gate Voltage : Open circuit- 5.1V, SCR LOAD-1.2V. Gate pulse width : Fixed 6.3 msec. Firing angle control : Internal 180° to 0° phase control by Potentiometer External 180° to 0° phase control obtained by external control voltage between Vc and GND. Test points : R, Y, B isolated signals for monitoring with respect to GND 1 to 8 – provide the test signals at various points of the trigger circuit.

PRECAUTIONS:








69

Waveform: Circuit diagram

70

PROCEDURE : 1. Connect the three-phase half wave controlled rectifier circuit shown in Fig.(1) on the power electronic trainer. 2. Turn on the power. 3. By use oscilloscope, plot the input and output waveforms on the same graph paper" same axis". 4. Measure the average and RMS output voltage by connect the AVO meter across load resistance. 5. Turn off the power 6. Use an inductive load. With L=10mH measure the output voltage and plot the output waveform. 7. Repeat step 6 with L=100mH measure the output voltage and plot the output waveforms. 8. Repeat step 6 & 7 with connect the freewheeling diode across the load




(V) voltage

Result: Thus the operation (Switching) of three phase half controlled rectifiers with resistive load was studied.

PRE LAB QUESTIONS:-

1. What is the delay angle control of converters?

2. What is natural or line commutation?


4. Can a freewheeling diode be used in this circuit and justify the reason? POSTLAB QUESTIONS:-

1. What is conduction angle?

2. What are the effects of adding freewheeling diode in this circuit?

3. What are the effects of removing the freewheeling diode in three phase semi converter?

71

Exp. No.:13 Date:

SINGLE PHASE DUAL CONVERTER WITH RL LOAD

OBJECTIVE:

To study and observer the operation of single phase dual converter with RL loads.

RESOURCES:


1 dual converter

1 power and firing - - 1

module





SPECIFICATIONS:

1. Input : 1 , 230V 50Hz, AC supply.

2. Load : R and RL loads.

3. Thyristors : 16A, 1200V, type 16 TTS/TYN616

4. Diode : 25A, 1200V, BY126/BY127


6. Fuses : 16A HRC.

PRECAUTIONS:








72

MODEL GRAPH:

NON CIRCULATING CURRENT MODE: P-TYPE CONVERTER AND N TYPE CONVERTER:

CIRCULATING CURRENT MODE: N TYPE CONVERTER P TYPE CONVERTER

PROCEDURE:

1. Switch ON the single phase dual converter firing circuit. Make sure all the pulses are

proper before connecting to the power circuit.

2. Make the connections as per the circuit diagram for non circulating current mode.




the firing circuit.


7. Note down the input voltage, firing angle, Output voltage and output circuit reading in

the tabular forms.

8. Repeat the same for different input voltage up to max. Voltage as provided in the



0, 90

0, 135

0 and 180

0.

10. Repeat the same step for circulating current mode also.

73

17.7 Circuit diagram: Non circulating current mode:

Circulating current mode:

TABULAR FORMS: For non circulating current mode

S.No. Input voltage (V) Firing Output voltage Ton +Toff

angle ( ) (V) (sec)

74

For circulating current mode:

S.No. Input voltage (V) Firing Output voltage Ton +Toff

angle ( ) (V) (sec)

RESULT: Thus the single phase dual converter with RL for circulating and non circulating mode of current was studied and also plotted the waveforms of different firing angles

75

.

PRE LAB QUESTIONS:-

1. What is the four quadrant operation?


3. What are modes of operation carried out in dual converter?

4. What are the advantages of dual converter?



2. How to change the circulating to non circulating current mode?

3. Write advantages of dual converter?

4. What are the drawbacks of the dual converter?

76

Exp. No.:14 Date:

SINGLE PHASE AC VOLTAGE CONTROLLER WITH MOTOR LOADS

OBJECTIVE: To study the module and waveforms of a 1-Φ AC voltage controller with motor loads.

RESOURCES:


1 1 AC voltage regulator - - 1 power module





SPECIFICATIONS:

1. Input 1 , 230V, 50Hz AC Supply

2. Load R and RL.


4. TRIACs 10A, 600V, BT136.


6. Fuses 16A HRC.

7. Step down transformer 230V/24V-0-24V, 2A.

PRECAUTIONS:





10. Make sure to connect firing pulses from the firing circuit to their corresponding

SCRs/TRIAC in the power circuit.

1.6 PROCEDURE:

77

8. Switch ON the mains supply to the firing circuit. Observe the trigger outputs by varying

firing angle potentiometer and by operating On/OFF and SCR/TRIAC selector switch. Make



10. Connect firing pulses from the firing circuit to the corresponding SCRs/TRIAC in the power

circuit.

11. Switch ON the step down transformer supply (MCB) and now switch ON the trigger pulses


12. Observe the output voltage waveform across load using oscilloscope.

13. Note down the input voltage, firing angle and output voltage readings in the TABULAR

FORMS.

14. Draw the waveforms in the graph at 0, 45, 90, 135 and 180 Deg. firing angles.

FORMULAE USED: OUT PUT VOLTAGE(Vo)=

78

T1

K1

A1

P

G2

2 TYN616G1

A2

K2

230V 24V T2 MOTOR

1 , 230V, 0V

K1 G1 R

LOAD

50Hz, AC K2 G2

0V 24V

N


S.No. Input voltage (V)

Firing angle Output voltage Theoretical

( )

(V) output

voltage (V)

79

MODEL GRAPH:


0 π 2π 3π 4π

Output Waveform Across R and RL- Load =00

VL (v)

0

Output WaveformAcross R- Load =900

VL (v)

0

Output Waveform Across RL- Load =900

VL (v)

0

u (ms) t (ms)

u (ms) t (ms)

80

MODEL CALCULATIONS:

RESULT:

Thus the single phase AC voltage controller with R & RL loads is studied and we plotted

the waveforms of different firing angle.

PRE LAB QUESTIONS:-

6. Why should the two trigger sources be isolated?

7. What are the advantages and the disadvantages of phase control?

8. What is phase control?

9. What are the advantages of bidirectional controllers?

10. What is meant by duty cycle in ON-OFF control method?

POST LAB QUESTIONS:- 2. What type of commutation is used in this circuit?

6. What are the effects of load inductance on the performance of AC voltage controllers?


8. What are the disadvantages of unidirectional controllers?

9. What are the advantages of ON-OFF control?

81

Exp. No.:15 Date:

SINGLE PHASE HALF AND FULL CONTROLLED BRIDGE CONVERTER WITH MOTOR LOAD

OBJECTIVE: To study the module and waveforms of a 1-Φ Half and full controlled converter with motor load at

different firing angles. RESOURCES:


1 Half controlled


firing module





SPECIFICATIONS:

1. Input : 1 , 230V , 50Hz AC supply.

2. Load : R, RL

3. Thyristors : 25A, 1200V, type 25 RIA 120/TYN616.

4. Diode : 25A, 1200V, BY126/BY127.


6. Fuses : 16A HRC.

7. Field Supply bridge rectifier : 10A, 600V.


PRECAUTIONS:








82

MODEL GRAPH:

Input Waveform

Vi (v)

t (ms)

0 π 2π 3π 4π


VL (v)

t (ms) 0

VL (v) Output Waveform R and RL -Load at=45

0

t (ms) 0

VL (v)

0

0


t (ms)


83

PROCEDURE:

11. Switch ON the main supply to the firing circuit. Observe the trigger output by varying firing

angle potentiometer and by operating ON/OFF switch and their phase sequence. Make sure

the firing pulses are proper before connecting to the power circuit.



14. Connect firing pulses 0from the firing circuit to their respective SCRs in power circuit.


the firing circuit.



tabular forms.





0, 90

0, 135

0 and 180

0.

FORMULAE USED:

Output voltage V0 = Vdc = Vm/ (1 + cos )

84

CIRCUIT DIAGRAM:

4.7.1SINGLE PHASE HALF CONTROLLED CONVERTER WITH MOTOR LOAD 2 TYN616

P

K1G1 K2 G2

230V 30V T1 T2

R LOAD

A1

A2 1 , 230V,

150 , 5A

50Hz, AC K1

K2

0V

0V

D1 D2

A1 A2 N

2 1N4007 Tabular form with R load



(V) voltage

85

MODEL GRAPH:

Input Waveform

Vi (v)

t (ms)

0 π 2π 3π 4π

Output Waveform R- Load at =00

VL (v)

t (ms) 0

Output Waveform R- Load at=450

VL (v)

t (ms) 0

RL- Load at =450

VL (v)

t (ms) 0


0

0 t (ms)

VL (v) RL- Load at =90

0

t (ms) 0


0

t (ms)

0

86

PROCEDURE:

11. Switch ON the main supply to the firing circuit. Observe the trigger output by varying firing angle

potentiometer and by operating ON/OFF switch their phase sequence. Make sure the firing pulses

are proper before connecting to the power circuit.




15. Switch ON the MCB and now switch ON the trigger pulses by operate ON/OFF switch in the firing

circuit.



TABULAR FORMS.

18. Repeat the same for different input voltage up to max. voltage as provided in the isolation

transformer.



0, 90

0, 135

0 and 180

0.

FORMULAE USED:

Average output voltage – R load, VAvg= Average

87

SINGLE PHASE FULL CONTROLLED BRIDGE CONVERTER WITH FREEWHEELING DIODE FED MOTOR LOAD

4 TYN616

P K1 G1

K3 G3

230V 30V T1 T3

K R MOTOR A1

A3

1 , 230V,

FD MM 50Hz, AC

G2 A L

0V 0V K4

G4 K2

T4 T2

N A4

A2

c. For RL load with freewheeling diode:


Theoretical


voltage (V)

88

MODEL CALCULATIONS:

RESULT:

Thus the single phase Full controlled bridge converter with motor load is studied and also

plotted the waveforms of different firing angles.

PRE LAB QUESTIONS:-

2. State the type of commutation used in this circuit?


6. What are the performance parameters of rectifier?

7. What are the advantages of three phase rectifier over a single phase rectifier?

6. What is the difference between half wave and full wave rectifier?



7. What is Dc output voltage of single phase full wave controller?

8. What are the effects of source inductance on the output voltage of a rectifier?

9. What is commutation angle of a rectifier?

10. What are the advantages of three phase rectifier over a single phase rectifier

89

CALCULATIONS:

RESULT:

Thus the single phase half controlled and full controlled converter with motor load is

studied and also we plotted the waveforms of different firing angles. PRE LAB QUESTIONS:-

90

6. What is the delay angle control of converters? 7. What is natural or line commutation? 8. What is the principle of phase control? 9. What is extinction angle? 10. Can a freewheeling diode be used in this circuit and justify the reason?

POSTLAB QUESTIONS:- 6. What is conduction angle? 7. What are the effects of adding freewheeling diode in this circuit? 8. What are the effects of removing the freewheeling diode in single phase semi converter? 9. Why is the power factor of semi converters better than that of full converters? 10. What is the inversion mode of converters?

power electronics and simulation lab - mist.ac.in · 11. single phase series inverter with r and rl...

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