Power Electronics & Simulation Lab

INDEX

Sl.No. EXPERIMENT NAME Page No.

1 Characteristics of SCR, BJT and MOSFET 2

2 GATE firing circuits of SCR 11

3 1-Φ AC voltage controller with R & RL loads 15

4 1- Φ Fully controlled bridge Converter with R & RL loads 18

5 DC JONES chopper 22

6 1- Φ Parallel inverter with R & RL loads 25

7 1- Φ Half-controlled Converter with R & RL loads 28

8 1- Φ Series inverter with R & RL loads 32

9PSPICE Simulation of 1- Φ Full converter using RLE loads and

1-Φ AC voltage controller using RLE loads35

10PSPICE Simulation of Resonant pulse commutation circuit and Buck

chopper41

Add-on Experiments

Sl.No. EXPERIMENT NAME Page No.

11 1-Φ Dual converter with R & RL loads. 48

12 PSPICE Simulation of single phase Inverter with PWM control 52

13 Forced commutation circuits of SCR 56

14 1- Φ Cyclo converter with R & RL loads 66

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1. CHARACTERISTICS OF SCR , BJT and MOSFET

(a) STATIC CHARACTERISTICS OF SCR

AIM :

To plot the characteristics of SCR and to find the forward resistance holding current and latching current.

APPARATUS :

1) CHARACTERISTICS STUDY UNIT2) 0-50V DC voltmeter3) 0-500mA DC ammeter4) 0-25mA DC Ammeter

CIRCUIT DIAGRAM:

PROCEDURE :

1) Make the connections as per the circuit diagram 2) Now switch on the supply and initially keep v1 and v2 at minimum3) Set load potentiometer R1 in the minimum position.4) Adjust Ig1 to some value say 10mA by varying V2 or gate current potentiometer

R2 vary V1 and note down VAK and IA READINGS.5) Further vary V1 till SCR conducts, this can be noted by sudden drop of VAK and IA

readings.6) Draw the graph VAK vs IA .

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7) Repeat the same procedure for different gate currents.

MODEL GRAPH:

TO FIND

1) Apply about 20V between anode and cathode by varying V1.2) Keep the load potentiometer R1 in minimum position.3) The device must be in the OFF state with gate open.4) Gradually increase gate voltage V2 till the device turns ON. This is the minimum gate

current required to turn ON the device.5) Adjust the gate voltage to a slightly higher value and set the load potentiometer at the

maximum resistance position, the device should come to OFF state.6) The gate voltage should be kept constant in this experiment.7) By varying R1, gradually increase anode current IA in steps.8) Open and close the gate voltage V2 switch after each switch.9) If the anode current is greater than the latching current of device, the device stays ON

even after the gate switch is opened otherwise the device goes in to blocking mode as soon as the gate switch is opened.

10) Note down the latching current.11) Increase the anode current from the latching current level by load pot R1 or V1.12) Open the gate switch permanently, the thyristors must be carefully ON.13) Now start reducing the anode current gradually by adjusting R1.

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14) If the thyristor does not turns OFF even after the R1 at maximum position then reduce V1.

15) Observe when the device goes to blocking mode.16) The anode current at this instant is holding current of the devices.17) Repeat the steps again to get the holding current IH.

TABULAR FORM :

S.NO. VBO1= , IG1= VBO2= , IG2=

VAK

(V)IA (A) VAK

(V)IA (A)

RESULT :

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(b) CHARACTERISTICS OF IGBT

AIM :To plot the input and transfer characteristics of an IGBT to find ON state resistance

and the transfer conductance.APPARATUS :

1) IGBT 1RG BC 20S2) Resistors 10KΩ/25w, 75Ω/25W3) DC voltmeter 0-50V4) DC voltmeter 0-15V5) DC ammeter 0-500mA

THEORY : It is a new development in the area of power MOSFET technology. This device combines in to advantages of both MOSFET and BJT. So an IGBT has high input impedance like as MOSFET and low ON state power like BJT. Further IGBT is free from second breakdown problem present in BJT. IGBT is also known as metal oxide insulated gate transistor.

It was also called as insulated gate transistor. The static characteristics or output characteristics of IGBT shows plot of collector current IC vs collector –emitter voltage VCE

for various values of gate emitter voltage. In the forward direction the shape of output characteristics is similar to that of BJT and have the controlling parameter is gate-emitter voltage VGE because IGBT is a voltage controlled device. The device developed by combining the areas of field effect concept and technology.

CIRCUIT DAIGARAM:

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PROCEDURE :TRANSFER CHARACTERISTICS :

1) Make connections as per the circuit diagram.2) Initially keep V1 and V2 minimum and set V1 ( VCE1) = 10V.3) Slowly vary V2 i.e., VGE and note down IC and VGE reading for every 1V.4) Repeat the same procedure for different values of VCE and draw the graph VGE vs. IC .

COLLECTOR CHARACTERISTICS :

1) Initially set V2 i.e., VGE to 5V. Slowly vary V1 and note down IC and VCE .2) For a particular value of VGE1 there is a pinch off voltage VP between collector and emitter.3) Repeat the same for different values of VGE and draw the graph between VCE vs IC

MODEL GRAPH:

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TABULAR FORM :

S.NO VCE1= VCE2=

VGE

(V)IC

(mA)VGE (V) IC

(mA)

S.NO VGE1= VGE2=

VCE

(V)IC

(mA)VCE (V) IC

(mA)

RESULT :

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(c) CHARACTERISTICS OF MOSFET

AIM:To plot the input and transfer characteristics of a MOSFET to find ON state

resistance and the transfer conductance.APPARATUS:

1) MOSFET IRF 7402) Resistors 10KΩ/25w, 75Ω/25W, 4KΩ/25W3) DC voltmeter 0-50V4) DC voltmeter 0-15V5) DC ammeter 0-500mA

THEORY: A metal oxide semiconductor field effect transistor is a recent device developed by

combining the areas of field effect concept and technology.The transfer characteristics of MOSFET shows the variation of drain current ID as a

fuction of gate source voltage VGS. The device is in OFF state upto some voltage called threshold device voltage. The output characteristics of Power MOSFET indicate the variation of Drain current ID as a function of Drain source voltage VDS as a parameter.

This device combines into advantages of IGBT and BJT. So this device has high impedence and low ON state power like BJT. It is a new development in the ared of power MOSFET technology. All the devices are mounted on proper heat sink. Each device is protected by snubber circuit.

PROCEDURE:TRANSFER CHARACTERISTICS:

1) Make connections as per the circuit diagram.2) Initially keep V1 and V2 minimum and set V1 ( VDS1)= 10V.3) Slowly vary V2 i.e., VGS and note down ID and VGS reading for every 1V.4) Repeat the same procedure for different values of VDS and draw the graph VGS vs

ID .

COLLECTOR CHARACTERISTICS:

1) Initially set V2 i.e., VGS to 3.5V.2) Slowly vary V1 and note down ID and VDS .3) For a particular value of VGS1 there is a pinch off voltage VP between drain and

source.4) Repeat the same for different values of VGS and draw the graph between VDS vs ID

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CIRCUIT DAIGRAM:

MODEL GRAPH:

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TABULARFORM:

TRANSFER CHARACTERISTICS :

S.NO VDS1= VDS2=

VGS (V) ID

(mA)VGS (V) ID

(mA)

DRAIN CHARACTERISTICS:

S.NO VGS1= VGS2=

VDS (V) ID

(mA)VDS (V) ID

(mA)

RESULT :

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2. GATE FIRING CIRCUITS OF SCR

AIM:

To study and compare resistance firing circuit with the resistance-capacitance firing circuits.

APPARATUS:

1) Control HWR and FWR using resistance circuits.2) Rheostat 50Ω/2A3) CRO4) connecting wires

THEORY:

The most common method for controlling the onset of conduction in an SCR is by means of gate voltage control. The gate control circuit is also called as firing or triggering circuits. These gating circuits are usually low power electronic circuits. The firing circuit should fulfill the functions. An SCR can be switched from OFF state to ON state in several ways. These are forward voltage triggering, dv/dt triggering, light triggering is used in some applications particularly in a series connected string gate triggering is the most common method of turning ON the SCR at desired instant of time.

PROCEDURE:

R-TRIGGERING:

1) Make connections as per the circuit diagram.2) Connect the rheostat of 50Ω/2A between the load points.3) Vary the control pot and observe the voltage waveform across the load, SCR and

different points of circuits.4) We can vary the firing angle from 0 to 90 degrees only in R-triggering. In this

synchronized firing angle can be obtained easily and economically in the half-cycle of the supply.

5) But there is a drawback that firing angle can controlled at the most at 900 since the gate current is in phase with the applied voltage.

6) A resistance is connected in series with the control pot so that the gate current does not cross the maximum possible value IG max.

7) Draw the waveform across the load and device for different values of firing angles.

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RC-TRIGGERING:

1) Connect a capacitor to R-triggering circuit to realize RC triggering.2) Repeat the above procedure and draw the waveforms across the load and device

for different values of firing angles..CIRCUIT DIAGRAMS:

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MODEL GRAPH:

CALCULATIONS:

Average output voltage:

VO = Vm2π

∫α

π

Vmsinωt dωt

= Vm2π

(1+COSα)

VRMS= [ 12 π

∫α

π

√Vm2sin2ωt dωt ]

TABULAR FORM:

R-FIRING:

S.NO Firing PRACTICAL THEORITICAL

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angle α VM( V) VAVG(V) VRMS(V) VM( V) VAVG(V) VRMS(V)

RC-FIRING:

S.NO Firing angle α

PRACTICAL THEORITICAL

VM( V) VAVG(V) VRMS(V) VM( V) VAVG(V) VRMS(V)

RESULT:

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3. SINGLE PHASE A.C. VOLTAGE CONTROLLER

AIM:To observe the output wave forms of 1-phase A.C. voltage controller with R and RL

loads using SCR’s.APPARATUS:

1) SCR(tyn612)-22) Rheostat 50Ω/2A3) Inductor 50Mh4) Cathode ray oscilloscope5) Connecting wires

THEORY:By connecting the reverse parallel pair of SCR’s or TRIAC between AC supply and

load the voltage applied to load can be controlled. This type of power controller or regulator. Some of the main applications of AC voltage regulator are for domestic and industrial heating.

Transformer tap changing, lightening, speed control of 1-phase and 3-phase AC drives and starting of induction motors earlier. The devices used for applications are Auto transformers, tap t/f, magnetic amps, saturable and triac based controller because of their high efficiency flexibility in control. Compact size and less maintaince. A.C.voltage controllers are also closed loop control systems.

PROCEDURE:1) Switch on the main supply of the firing circuit and observe trigger outputs by

varying the firing angle.2) Make sure that firing pulses are proper before connecting to the power circuit.3) Make the connections as connect the trigger pulses from firing circuit to

corresponding SCR TRIAC in power circuit 4) Connect the A.C. supply to power through step down transformer.5) Connect the load of 50Ω/2a rheostat.6) Switch on the step down t/f supply and trigger outputs and observe the

waveforms in the CRO.7) Note down the o/p voltage, firing angle, I/p voltage.

Tabular Forms:

S.no Firing angle Practical theoretical(α) Vavg(v)

R-loadVrms(V)R-load

Vavg(v)RL- load

Vrms(v)RL- load

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CIRCUIT DIAGRAM:

MODEL GRAPHS:

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Result:

VIVA QUESTIONS :1) What is the specialty of an AC voltage controller?2) What is the device which can replace the AC voltage controller?3) What type of commutation is employed in this circuit?

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4. 1-Φ FULLY CONTROLLED BRIDGE CONVERTER WITH R AND RL LOADS

AIM: To study single phase fully controlled bridge converter with R and RL loads.

APPARATUS : 1) 1-ph converter firing circuit2) 1-ph fully controlled power circuit3) Rheostat- 50Ω/2A4) Inductor-50mH5) Power scope6) CRO

THEORY: A 1-ph full bridge converter using four SCR’s is shown in figure. The load is assumed

to be R and RL. Thyristor pair T1 and T2 is simultaneously triggered and π radians after pair T3 and T4 is gated together.

During the positive half cycle SCR’s T1 and T1I are forward biased and when there

two thyristors are fired simultaneously at wt = α, the load is connected to the input through T 1

and T1I. In this case of inductive loads during the period π <wt < π+α the input voltage is

negative and freewheeling diode Dm is forward biased. Dm conducts to provide the conductivity of current in the inductive load. The load current is transferred from T1 and T1

I

to DM and thyristors T1 and T1I are turned off due to line or natural commutation.

During the negative half cycle of the input voltage thyristors T2 and TI2 are forward

biased. The firing of thyristors T2 and T2I simultaneously at wt = π+α will reverse bias DM. the

diode DM is turned off and the load is connected to the supply through T2 and T2I .

PROCEDURE: 1) Make the connections as shown in figure2) Apply triggering pulse to SCR T1 , T1

I , T2 , T2I

3) Now vary the firing angle α and note down the readings of the output voltage.4) Observe the fully controlled waveform in CRO5) Perform the same operations with RL loads and with freewheeling diode with RL load6) Avoid using two CRO probes as there is a problem of short circuit.

PRECAUTIONS :1) Avoid loose connections2) While turning on first turn on the converter and then turn on the firing circuit.3) While turning off first remove the firing the pulses and then turn off the converter.

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CIRCUIT DIAGRAMS

1Φ full converter with RL Load & free wheeling Diode.

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MODEL GRAPHS:

TABULAR FORM :R-LOAD

Sl No.Firing Angle (α)

Output VoltageR-LoadPractical

Output VoltageR-LoadTheoretical

Output CurrentIac(A)

VDC (V) VRMS (V) VDC VRMS

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(V) (V)

RL-LOAD

Sl.No.Firing Angle(α)

Output VoltageRL LoadPractical

Output VoltageRL LoadTheoretical

Output CurrentIac(A)VDC (V) VRMS (V) VDC

(V)VRMS

(V)

CALCULATIONS:AVERAGE OUTPUT VOLATGE

VO = 1π

∫∫∝

π

VmSinωt dωt

= Vmπ

(1+cosα )

VRMS =[ [ 1π ]∫

α

π

V m2sin2ω t dωt]2

Average load current

IDC = 2V mπ

(1+cosα)

RESULT:

VIVA QUESTIONS:

1) What type of commutation is employed in this circuit?2) Is second quadrant operation possible with R-load?3) For what firing angle in RL load the average output voltage is zero?4) The average output power is purely reactive from what firing angle?5) The output voltage waveform for R-load and RL-load with freewheeling diode are

similar. Why?

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5. DC JONES CHOPPERAIM :

To rig up DC Jones chopper and to control output average DC voltage both at constant frequency and variable frequency and at different firing cycles.

APPARATUS:

1) DC chopper power circuit2) DC chopper firing circuit3) DC regulated power supply 0-30V/2A4) Rheostat 50Ω/2A5) Cathode ray oscilloscope

THEORY:

In many industrial applications, it is required to connect a fixed voltage DC source into a variable voltage DC source. A DC chopper converts directly from fixed DC to variable DC and is also known as DC to DC converter. A chopper can be considered as DC equivalent to an AC transformer with a continuously variable turns ratio. Like a transformer, it could be used to step down or step up a DC voltage source. Choppers are widely used for traction motor control in electric automobiles, trolley cars, marine hoists, forklift trucks and mine haulers. They provide smooth acceleration control, high efficiency and fast dynamic response. Chopper can be used in regenerative braking of DC motors to return energy back to the supply and this feature results in energy savings for transportation systems with frequent stops. Chopper can also be in DC voltage regulators.

The Jones chopper is another example of class-D commutation in which a charged capacitor is switched by an auxiliary SCR to commutate the main SCR.

PROCEDURE:

1) Switch on the DC chopper firing circuit.2) Observe the point signals and trigger output signals by carrying duty cycle and

frequency potentiometer.3) Make sure that trigger outputs are proper before connecting to the power circuit.4) Now, make the interconnections as per the circuit diagram.5) Connect the DC supply from the variable DC source.6) Initially set input DC supply to 10V.7) Connect respective trigger outputs from the firing circuit to the respective SCR’s in

the power circuit.8) Initially keep firing circuit in OFF position.9) Switch on the DC supply and now supply main SCR trigger pulses.10) Observe the voltage waveform across load.11) If commutation fails we can see only the DC voltage, in that case switch off the DC

supply, switch off the pulses and check the connections and try again.

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12) Observe the voltage across the load, across capacitor, across main SCR and auxiliary SCR by varying duty-cycle and frequency.

13) Now vary the DC supply upto 30V.14) Draw the waveforms at different duty cycle and at different frequencies.15) Connect voltmeter and ammeter and note down the values.

CIRCUIT DIAGRAM:

MODELGRAPHS:

CALCULATIONS:

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DUTY CYCLE, α = TonT

Total time period T = TON + TOFF msOutput voltage VO = α VIN volts

TABULAR FORM:

s.no

Capacitor voltage VC (V)

VIN (V)Input voltage

TON

(ms)ON time

TOFF

(ms)OFF time

Duty cycle (α )

Output voltage (V) Frequency(KHz)Practical

VO (V)TheoreticalVO (V)

RESULT:

VIVA QUESTIONS:

1) What is chopper?2) What type of commutation is employed in Jones chopper?3) Is the commutation Voltage or current?4) Mention any applications of choppers.

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6. 1-Φ PARALLEL INVERTER

AIM: To study the performance of center tapped transformer type parallel inverter at

different frequencies.

APPARATUS:1) Parallel inverter2) DC regulated power supply3) Rheostat 50Ω/2A4) Cathode ray oscilloscope5) Connecting wires

THEORY:This circuit is a typical class-C parallel inverter. Assume TN to be in ON and TP to be

in OFF state. The lower end of commutating capacitor is charged to twice the supply voltage and remains at the value until TP is turned ON. When TP is turned ON, the current flows through the lower end of primary TP and commutating inductance L . Since the voltage across c can’t be instantaneous the common SCR cathode raises approximately to 2V DC and reverse bias TN. Thus TN turns OFF and C discharge through L and supply and then recharges in reverse direction. The auto transformation makes C to charge making now its upper point to reach +2V DC ready to commutate TP when TN is ON. The major purpose of commutation inductance L is to limit the commutation capacitance charging current during switching. Freewheeling diodes DP and DN assist the inverter in handling the wide range of loads and value of C may be reduced since the capacitor now does not have to carry the reactive current. To dampen the feedback diode currents with in the half period feedback diodes are connected to tapings of transformer of 25V tapping.

PROCEDURE:

1) Switch on the firing circuit and observe the trigger output TP and TN by varying potentiometer and by operating ON/OFF switch.

2) Connect DC power supply to power circuit.3) Connect trigger output to gate cathode by SCR TP and TN.4) Make the connections as in circuit diagram including Freewheeling diodes.5) Set input voltage to 15V and aplly trigger pulses to SCR and observe voltage

waveform across load.6) Remove freewheeling diode and observe the waveform. Vary the load and fo and

observe the waveform.7) To switch off the inverter switch off DC input supply first and then switch off

trigger pulses.

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CIRCUIT DIAGRAM:

MODEL GRAPHS:

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PRECAUTIONS:

1) Since the parallel inverter works on forced commutation there is a chance of commutation failure. If the commutation fails there is dead short circuit in the input DC supply which will lead to blowing of input fuse. Please check the fuse if the commutation fails. Preferably connect the input DC supply from 30V/2A Regulated power supply which has over current tripping facility thereby protecting the DC power supply unit.

2) If the commutation fails switch OFF the DC supply first and then trigger output. Check the connections again and restart.

TABULAR FORM :

S.NO VDC (V) TON (ms) Frequrncy (Hz) VLOAD (V)

CALCULATIONS :

Resonant frequency fR = 1

2π √LC

RESULT :

VIVA QUESTIONS :

1) To what voltage will the capacitor gets charged?2) What is the need of the transformer is the circuit?3) What type of commutation is employed in this circuit?

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7. 1-Φ HALF CONTROLLED CONVERTER WITH R AND RL LOADS

AIM: To study single phase half controlled bridge converter with R and RL loads.

APPARATUS:

1) 1-ph converter firing circuit2) 1-ph half controlled power circuit3) Rheostat- 50Ω/2A4) Inductor-50mH5) Power scope or CRO

THEORY:

The circuit arrangement of a 1-ph converter is shown in figure. In the positive half cycle thyristor T1 is forward biased. When SCR T1 is fired at ωt = α, the load is connected to the input supply through T1 and D2 during the period from α ≤ ωt ≤ π+α the input voltage is negative and freewheeling diode DM is forward biased. DM conducts to provide continuously current in case of inductive loads. In the negative half-cycle of input voltage T2 is forward biased and triggering of T2 at ωt = π +α will reverse bias DM and is turned OFF. Load is connected to supply through T2 and D1.

The converter has a better power factor due to the freewheeling diode and is commonly used in applications up to 15KW where one quadrant operation is acceptable.

The half controlled bridge has the inhering freewheeling action and analysis is more or less the same with or without a freewheeling diode is connected across the load. In practical it is always adjustable to provide a freewheeling diode in a half-controlled converter so that the commutation of SCR’s is assumed inductive loads.

PROCEDURE:

1) Connections are made as per the circuit diagram.2) Switch on the power circuit and next switch on trigger circuit.3) Now vary the firing angle α and note down the readings of the output voltage.4) Observe the half-controlled waveform in CRO5) Perform the same operations with RL loads.6) Avoid using two CRO probes as there is a problem of short circuit.

PRECAUTION:

1) Avoid loose connections

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2) While turning on first turn on the converter and then turn on the firing circuit.3) While turning off first remove the firing the pulses and then turn off the converter.

CIRCUIT DIAGRAMS:

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MODEL GRAPHS:

TABULAR FORM:

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R-LOAD

Serial No.

Firing Angle(α)

Output VoltageR-Load

Practical

Output VoltageR-Load

TheoreticalOutput CurrentIac(A)VDC (V) VRMS (V) VDC

(V)VRMS

(V)

RL-LOAD

Serial No.

Firing Angle(α)

Output VoltageRL LoadPractical

Output VoltageRL LoadTheoretical

Output CurrentIac(A)

VDC (V) VRMS (V)VDC

(V)VRMS

(V)

CALCULATIONS:

AVERAGE OUTPUT VOLATGE

VO = 1π

∫∫∝

π

VmSinωt dωt

= Vmπ

(1+cosα )

VRMS =[ [ 1π ]∫

α

π

V m2sin2ω t dωt]2

Average load current

IDC = 2V mπ

(1+cosα)

RESULT:

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8. 1-Φ SERIES INVERTER

AIM: To study the behavior of modified series inverter by varying load resistance center tap inductance of combination of capacitance at different inverter frequencies

APPARATUS :1) Modified series inverter2) D.C.regulated power supply 3) Rheostat 50Ω/2A4) Cathode ray oscilloscope5) Connecting wires

THEORY : Inverters in which commutating components are permanently connected in series with the load are called series inverter .The series circuit so formed must be under damped. As the current attains zero value due to the nature of series circuit, series inverters are classified as self-commutated (or) load commutation inverter. These inverter operate at high frequencies (200hzto100khz) the size of commutation components is small these inverters are used in induction heating fluorescent lighting etc. Two SCR’s T1 and T2 are turned on app. so that o/p voltage of desired frequency can be obtained when T1 ON T2 OFF. Current P starts building up, after reaching some peak value decays to zero at one point. so that the load current tends to reverse SCRT 1 is OFF after T2 turns ON at point B capacitor begins to discharge and load current in reversed direction builds up to some –ve peak value and the process repeats at point c.T1 again turned and the process repeats. In this manner DC is converted into AC with help of series inverter.

PROCEDURE:1) Connections are made as per the circuit diagram 2) Now connect trigger o/p from the firing circuit to gate and cathode at SCR’sT1&T2

3) Connect DC input from a 30V/2A regulated power supply4) Switch ON the input DC supply by means of applying triggering pulses to the sets5) Repeat the experiment for different frequencies and observe the voltages wave forms

across the load.

TABULAR FORM :

S.NO VDC (V) TON (ms) Frequency (Hz) VLOAD (V)

CALCULATIONS :

Resonant frequency fr = 1

2π √LC

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CIRCUIT DIAGRAM :

MODEL GRAPHS:

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RESULT:

VIVA QUESTIONS:

1) What is an inverter?2) What should be the condition between R-L-C components?3) What type of commutation is employed to turn off SCR?4) Can two thyristors be in ON condition at the same time? 5) While a capacitor is charging the other capacitor discharges. YES/NO?

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9. PSPICE ANALYSIS OF SINGLE PHASE FULL CONVERTER & AC VOLTAGE CONTROLLER

PSPICE ANALYSIS OF SINGLE PHASE FULL CONVERTER WITH RLE LOADAIM: To analyze the single phase full converter with RLE Load.

SIMULATION TOOLS REQUIRED:

PC with PSPICE Software.

CIRCUIT DIAGRAM:

5

8

R1 0 O H M S

3 4

7

V s

L1 0 0 M H

E1 0 0 V

9

1

2

0

XT1

XT2

XT3

XT4

6

SPECIFICATIONS:

Sinusoidal input: VOFF = 0V, VAMPL = 169.7V, FREQ = 50 Hz.T1 and T2: V1 = 0V, V2 = 100V, TD = 3333.34us, TR = TF = 1ns, PW = 100us, PER = 20ms. T3 and T4: V1 = 0V, V2 = 100V, TD = 13333.34us, TR = TF = 1ns, PW = 100us, PER = 20ms. Firing circuit: RG = 50 Ώ, VX, VY = 0V, RT = 1 Ώ, CT = 10uf, RON = 0.0125, ROFF = 10E+5, VON = 0.5V, VOFF = 0V, IS = 2.2E-15, BV = 1800V, TT = 0 sec.

THEORY:Single phase full controlled converters are thyristors based circuits which

convert fixed alternating voltage into variable direct voltage change in magnitude. The circuit consists to four thyristors connected in in bridge manner.

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Power Electronics & Simulation Lab

PROCEDURE:FOR ANALYSIS USING PROGRAM:

1. Write the program in a new text file in PSpice AD.2. Save the file using the notation file name .cir.3. Activate the file by opening it.4. Run the simulation process using blue button.5. By clicking Add Trace icon, get the required waveform.

PROGRAM:With RLE LoadSIGLE-PHASE FULL CONVERTER CIRCUIT WITH RLE LOADVS1 1 2 SIN(0 169.7V 50HZ)R1 7 8 10OHML1 8 9 100MHVDC 9 0 DC 100VVG1 3 7 PULSE(0 100V 3333.34US 1NS 1NS 100US 20000US)VG3 4 7 PULSE(0 100V 13333.34US 1NS 1NS 100US 20000US)VG2 5 2 PULSE(0 100V 3333.34US 1NS 1NS 100US 20000US)VG4 6 1 PULSE(0 100V 13333.34US 1NS 1NS 100US 20000US)XT1 1 7 3 7 SCRXT2 0 2 5 2 SCRXT3 2 7 4 7 SCRXT4 0 1 6 1 SCR.SUBCKT SCR 1 2 3 2S1 1 5 6 2 SMODRG 3 4 50OHMSVX 4 2 DC 0VVY 5 7 DC 0VDT 7 2 DMODRT 6 2 1OHMCT 6 2 10UFF1 2 6 POLY(2) VX VY 0 50 11.MODEL SMOD VSWITCH(RON=0.0125 ROFF=10E+5 VON=0.5V VOFF=0V).MODEL DMOD D(IS=2.2E-15 BV=1800 TT=0).ENDS SCR.TRAN 1US 60MS.PROBE.END

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Power Electronics & Simulation Lab

Model Graphs for Full Converter with RLE Load:

INPUT WAVEFORM

Time

0s 10ms 20ms 30ms 40ms 50ms 60msV(1,2)

-200V

-100V

0V

100V

200V

OUTPUT WAVEFORM

Time

0s 10ms 20ms 30ms 40ms 50ms 60msV(7)

-100V

0V

100V

200V

300V

RESULT:

VIVA QUESTIONS:

1. What is your conclusion about this experiment?2. Briefly explain the operation of single phase full converter.3. Give the syntaxes for defining the following elements:

i. Pulse voltageii. Sinusoidal voltage iii. Thyristoriv. Switch

4. Explain about the command ‘SUBCKT’.5. How will you give specifications for diode?

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Power Electronics & Simulation Lab

PSPICE ANALYSIS OF SINGLE PHASE AC VOLTAGE CONTROLLER WITH RLE LOAD

AIM: To analyze the single phase full converter with RL and RLE Loads.

SIMULATION TOOLS REQUIRED:

PC with PSPICE Software.

CIRCUIT DIAGRAM:

XT1

XT2

V S

V D C

L 1

R 1

SPECIFICATIONS:Sinusoidal input: VOFF = 0V, VAMPL = 169.7V, FREQ = 50 Hz.T1: V1 = 0V, V2 = 100V, TD = 3333.34us, TR = TF = 1ns, PW = 100us, PER = 20ms.T2: V1 = 0V, V2 = 100V, TD = 13333.34us, TR = TF = 1ns, PW = 100us, PER = 20ms.Firing circuit: RG = 50 Ώ, VX, VY = 0V, RT = 1 Ώ, CT = 10uf, RON = 0.0125, ROFF = 10E+5, VON = 0.5V, VOFF = 0V, IS = 2.2E-15, BV = 1800V, TT = 0 sec.

THEORY:Ac voltage controllers are thyristors based devices which convert fixed

alternating voltage directly to variable alternating voltage without a change in frequency. The circuit consists to two thyristors connected in anti-parallel.

PROCEDURE:FOR ANALYSIS USING PROGRAM:

1. Write the program in a new text file in PSpice AD.2. Save the file using the notation file name.cir.3. Activate the file by opening it.4. Run the simulation process using blue button.5. By clicking Add Trace icon, get the required waveform.

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Power Electronics & Simulation Lab

PROGRAM:

SIGLE-PHASE AC VOLTAGE CONTROLLER CIRCUIT WITH RL LOADVS1 1 0 SIN(0 169.7V 50HZ)R1 2 3 10OHML1 3 8 10MHVDC 8 0 100VVG1 4 2 PULSE(0 100V 3333.34US 1NS 1NS 100US 20000US)VG2 5 1 PULSE(0 100V 13333.34US 1NS 1NS 100US 20000US)XT1 1 2 4 2 SCRXT2 2 1 5 1 SCR.SUBCKT SCR 1 2 3 2S1 1 5 6 2 SMODRG 3 4 50OHMSVX 4 2 DC 0VVY 5 7 DC 0VDT 7 2 DMODRT 6 2 1OHMCT 6 2 10UFF1 2 6 POLY(2) VX VY 0 50 11.MODEL SMOD VSWITCH(RON=0.0125 ROFF=10E+5 VON=0.5VVOFF=0V).MODEL DMOD D(IS=2.2E-15 BV=1800 TT=0).ENDS SCR.TRAN 1US 60MS.PROBE.END

MODEL WAVEFORMS:INPUT WAVEFORM

Time

0s 10ms 20ms 30ms 40ms 50ms 60msV(1)

-200V

-100V

0V

100V

200V

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Power Electronics & Simulation Lab

OUTPUT WAVEFORM

Time

0s 10ms 20ms 30ms 40ms 50ms 60msV(2)

-200V

-100V

0V

100V

200V

RESULT:

REVIEW QUESTIONS:

1. What is Full Converter?2. Briefly explain about single phase AC voltage controller?3. What is the command for calling a subprogram in Pspice?4. Explain about the command ‘SUBCKT’?5. What is your conclusion after doing the analysis?6.

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Power Electronics & Simulation Lab

10. PSPICE ANALYSIS OF RESONANT PULSE COMMUTATION CIRCUIT AND BUCK CHOPPER

PSPICE ANALYSIS OF RESONANT PULSE COMMUTATION CIRCUIT

AIM: Pspice analysis of resonant pulse commutation circuit.

SIMULATION TOOLS REQUIRED:

PC with PSPICE Software.

CIRCUIT DIAGRAMS:

Resonant Pulse Commutation Circuit

0

3

R 11 0 M E G

L m5 M H

T1

7

T3

11

V g 3

C

3 1 . 2 U F

T2

4

8

V g 1R 2

1 0 M E GV g 2

1

V x0 V9

R 31 0 M E G

2

D m

R s

7 5 0 o h m s

D 1

R m0 . 5

10

C s

0 . 1 U F

V s

L

6 . 4 U H

5

6

V y

0 V

SPECIFICATIONS:

VS = 200V, for diode DM : IS=1E-25 BV=1800, Vg1: V1 = 0, V2 = 100V, TD = 0, TR = TF = 1us, PW = 0.4ms, PER = 1ms.Vg2: V1 = 0, V2 = 100V, TD = 0.4ms, TR = TF = 1us, PW = 0.6ms, PER = 1ms.Vg3: V1 = 0, V2 = 100V, TD = 0, TR = TF = 1us, PW = 0.2ms, PER = 1ms.Internal thyristor circuit: switch: RON=0.1, ROFF=10E+6, VON=10V, VOFF=5V, diode: IS=1E-25, BV=10000. Transient response specifications: 0.5US 3MS 1.5MS 0.5US. Conditions: abstol=1.000u, reltol=0.01, vntol=0.1, ITL5=20000

EEE Department, SVEC, Tadepalligudem Page 41

10 V

10 V

Power Electronics & Simulation Lab

PROGRAM:

RESONANT PULSE COMMUTATION CIRCUITVS1 1 0 DC 200VVG1 7 0 PULSE(0 100V 0 1US 1US 0.4MS 1MS)VG2 8 0 PULSE(0 100V 0.4MS 1US 1US 0.6MS 1MS)VG3 9 0 PULSE(0 100V 0 1US 1US 0.2MS 1MS)RG1 7 0 10MEGRG2 8 0 10MEGRG3 9 0 10MEGCS 10 11 0.1UFRS 11 4 750OHMSC 1 2 31.2UF IC=200VL 2 3 6.4UHD1 4 1 DMODDM 4 0 DMOD.MODEL DMOD D(IS=1E-25 BV=1800)RM 4 5 0.5OHMLM 5 6 5MHVX 6 0 DC 10VVY 1 10 DC 10VXT1 10 4 7 0 DCSCRXT2 3 4 8 0 DCSCRXT3 1 3 9 0 DCSCR.SUBCKT DCSCR 1 2 3 4DT 5 2 DMODST 1 5 3 4 SMOD.MODEL SMOD VSWITCH(RON=0.1 ROFF=10E+6 VON=10VVOFF=5V).MODEL DMOD D(IS=1E-25 BV=10000).ENDS DCSCR.TRAN 0.5US 3MS 1.5MS 0.5US.PROBE.options abstol=1.000u reltol=0.01 vntol=0.1 ITL5=20000.END

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Power Electronics & Simulation Lab

PROCEDURE:

FOR ANALYSIS USING PROGRAM:

1. Write the program in a new text file in PSpice AD.2. Save the file using the notation filename.cir.3. Activate the file by opening it.4. Run the simulation process using blue button.5. By clicking Add Trace icon, get the required waveform.

MODEL WAVEFORMS:

V (1, 2)

Time

1.4ms 1.6ms 1.8ms 2.0ms 2.2ms 2.4ms 2.6ms 2.8msV(1,2)

-1.0KV

-0.5KV

0V

0.5KV

1.0KV

I(C)

Time

1.4ms 1.6ms 1.8ms 2.0ms 2.2ms 2.4ms 2.6ms 2.8msI(C)

-2.0KA

-1.0KA

0A

1.0KA

2.0KA

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Power Electronics & Simulation Lab

I(VX)

Time

1.4ms 1.6ms 1.8ms 2.0ms 2.2ms 2.4ms 2.6ms 2.8msI(VX)

-200mA

-100mA

0A

100mA

200mA

RESULT:

REVIEW QUESTIONS:

1. What is resonant pulse commutation?2. What is carrier signal?3. What is the procedure to define a switch and diode?4. What is sub circuit?5. What is your conclusion after doing the analysis?

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Power Electronics & Simulation Lab

PSPICE ANALYSIS OF BUCK CHOPPER

AIM: To analyze Buck chopper using PSPICE.

SIMULATION TOOLS REQUIRED:

PC with PSPICE Software.CIRCUIT DIAGRAM:

R B2 5 0 O H M S

0

C e8 . 3 3 U F

T1

V x 0 VV g

L

4 0 . 9 1 U H

D m

V y

0 V

R3 O H M S

V s1 1 0 V

L e

6 8 1 . 8 2 U H

SPECIFICATIONS:

VG: V1 = 0V, V2 = 20V, TD = 0, TR = TF = 0.1ns, PW = 27.28us, PER = 50us. For Ce, IC = 60V. Firing circuit: RON = 0.1, ROFF = 10E+8, VON = 10V, VOFF = 5V, For DT: IS = e-25, BV = 1000V. For DM, IS = 2.2e-15

, BV = 1000V, TT = 0.

THEORY:

A chopper is a high speed on (or) off switch. It connects source to load and

disconnects the load from source at a fast speed. If the average output voltage V 0 is less

than the input voltage VS i.e, V0 < VS, then the corresponding chopper circuit is called

step-down chopper.

The principle of operation is as follows during the period in which

transistor (Tr ) is ON, the load voltage is equal to source voltage Vs. During the interval

in which Tr is off, the load current flows through the free-wheeling diode FD and load

voltage is therefore zero during T off. In this manner, a chopper DC voltage is produced

at the load terminals.

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10 V

Power Electronics & Simulation Lab

Where

= duty cycle = ON time of transistor(TON) / Total Time(T)

Hence, T = constant and Vs = constant

Therefore, The load voltage V0 can be controlled by varying duty cycle , the load

voltage is independent of load current.

PROGRAM:

BUCK CHOPPERVS 1 0 DC 110VVY 1 2 DC 0VVG 7 3 PULSE(0 20 0 0.1NS 0.1NS 27.28US 50US)RB 7 6 250OHMSLE 3 4 681.28UHCE 4 0 8.33UF IC=60VL 4 8 40.91UHR 8 5 3OHMSVX 5 0 DC 0VDM 0 3 DMOD.MODEL DMOD D(IS=2.2E-15 BV=1000V TT=0)XT1 2 3 6 3 DCSCR.SUBCKT DCSCR 1 2 3 4DT 5 2 DMODST 1 5 3 4 SMOD.MODEL DMOD D(IS=1E-25 BV=1000V).MODEL SMOD VSWITCH(RON=0.1 ROFF=10E+8 VON=10VVOFF=5V).ENDS DCSCR.TRAN 1US 1.6MS 1.5MS 1US.PROBE.END

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Power Electronics & Simulation Lab

PROCEDURE:

FOR ANALYSIS USING PROGRAM:

1. Write the program in a new text file in PSpice AD.2. Save the file using the notation filename.cir.3. Activate the file by opening it.4. Run the simulation process using blue button.5. By clicking Add Trace icon, get the required waveform.

MODEL WAVEFORMS:

INPUT & OUTPUT WAVEFORMS

Time

1.50ms 1.51ms 1.52ms 1.53ms 1.54ms 1.55ms 1.56ms 1.57ms 1.58ms 1.59msV(3) V(1)

-40V

0V

40V

80V

120V

RESULT:

REVIEW QUESTIONS:

1. What is a chopper?2. What is the syntax for defining an element?3. What is the procedure to define a switch and diode?4. What is sub circuit?5. What is your conclusion after doing the analysis?

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Power Electronics & Simulation Lab

11.SINGLE-PHASE DUAL CONVERTER AIM:

To construct a single phase dual converter and to apply reversible voltage to load.

APPARATUS:

1) Auto transformer2) Isolation transformer 3) Dual converter power and firing module4) Loading rheostat-50Ω/8A5) Digital millimeters6) Cathode ray oscilloscope7) Patch cards

THEORY:

Dual converter consists of two converters both are connected to the same load. The purpose of a dual converter is to provide a reversible DC voltage to the load. It is needed for DC motor drives where reversal is required. Dual converter provides four quadrant operations hence the name dual. The two modes of operations are the non-circulating current mode and circulating current mode. In the former only one bridge is triggered. When reversal of output voltage is required, the firing pulses for concreting bridge are stopped and second bridge is gated. Since the conducting SCR’s in the first bridge will turn off only when the current goes to zero, a small dead time must be allowed before the second bridge is gated otherwise: the AC input will be shorted through the two bridges.

In the circulating current mode, both bridge are gated simultaneously, one operating in the rectifying mode and the other in the inverting mode to avoid short circuits. This scheme requires fully controlled bridges. The internal voltage of rectifier is higher and that of inverter is lower than the output voltage. This can be done by two ways 1) by keeping supply voltage V constant and firing bridge 1 (P- converter) at α and bridge 2 (N-converter) at (π-α). By keeping firing angle constant and maintaining supply voltage at rectifier bridge greater than supply voltage at inverter bridge.

The dual converters can be operated with or without a circulating current. In this case of operation without circulating current, only one converter operates at a time and carries the load current and the other converter is completely blocked by inhibiting gate pulses. However, the operation with circulating current has the following advantages.

1) The circulating current maintains continuous conduction of both converters over the whole control range, independent of the load.

2) Since one converter always operates as a rectifier and the other converter operates as an inverter, the power flow in either direction at any time is possible.

3) Since both converters are in continuous conduction the time response for changing from quadrant to another is faster.

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Power Electronics & Simulation Lab

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Power Electronics & Simulation Lab

PROCEDURE:

DUAL CONVERTER IN NON-CIRCULATORY CURRENT MODE:

I) P-Converter is ON & N-Converter is OFF:

1) Connections are made as per the circuit diagram.2) Connect rheostat at 50Ω/8A.3) Connect CRO across load.4) Apply AC input voltage using isolation transformer.5) Made P-converter ON & OFF the N-converter.6) Vary firing angle observe load voltage waveforms on CRO.7) Note down load voltage in steps by varying firing angle α using multimeter.

Sl.NOFiring angle in degrees

(N-converter) π-αLoad voltage

VL(DC)In volts

1 00

2 300

II) N-Converter is ON & P-Converter is OFF:

1) Connections are made as per the circuit diagram.2) Connect rheostat at 50Ω/8A.3) Connect CRO across load.4) Apply AC input voltage using isolation transformer.5) Made N-converter ON & P-converter OFF.6) Vary firing angle observe load voltage waveforms on CRO.7) Note down load voltage in steps by varying firing angle α using multimeter.

Sl.NOFiring angle in degrees

(N-converter) π-αLoad voltage

VL(DC)In volts

1 00

2 300

Firing angle of N-converter = П – firing angle of P-converter

= П – α

III) DUAL CONVERTER IN CIRCULATORY CURRENT MODE:

1) Connections are made as per the circuit diagram.2) Connect rheostat at 2Ω/1A.3) Connect CRO across load.

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Power Electronics & Simulation Lab

4) Apply AC input voltage using isolation transformer. Say 30V range.5) Made N-converter ON & P-converter ON.6) Vary firing angle, observe load voltage waveforms on CRO.7) Note down load voltage in steps by varying firing angle α using multimeter.

Sl.No.firing angle α in

degrees ( P-Converter)

Firing angle П-α in degrees (N-

converter)

Load voltage VL

(DC) in volts

1 1800 00

2 1500 300

3 1200 600

4 900 900

5 600 1200

6 300 1500

7 00 1800

RESULT:

Single phase dual converter is constructed and its performance is studied.

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Power Electronics & Simulation Lab

12.SINGLE PHASE INVERTER WITH PWM CONTROL

AIM: PSpice analysis of single phase inverter with PWM control.

SIMULATION TOOLS REQUIRED:

PC with PSPICE Software.

CIRCUIT DIAGRAM:

D 4

5

12D 1

T2

V y 0 V

9

T3R g 1

R

2 . 5 O H M S

1

7V x

0 V

T4

13

R g 4

2

4

6V s1 0 0 V

8T1

D 2

R g 3

11

L

1 0 M H

0

14 10R g 2

D 3

3

PWM Generator

R in2 M E G

R 4

1 0 0 K O H M S

4

2

V c

1

C o1 0 p f

R 2

1 k

6R o

7 5 O H M S

0

V r

3

R 1

1 k5

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Power Electronics & Simulation Lab

Carrier and Reference signals

0

R c 12 M E G

1715

V rV c 1 V c 3R c 32 M E G

16

R r2 M E G

SPECIFICATIONS:

VR: V1 = 50V, V2 = 0V, TD = 0, TR = TF = 8333.33us, PW = 1ns, PER = 16666.67us.VC1: V1 = 0V, V2 = -30V, TD = 0, TR = TF = 1ns, PW = 8333.33us, PER = 16666.67us. VC3: V1 = 0V, V2 = -30V, TD = 8333.33us, TR = TF = 1ns, PW = 8333.33us, PER = 16666.67us.Firing circuit: RG = 50 Ώ, VX, VY = 0V, RT = 1 Ώ, CT = 10uf, RON = 0.0125, ROFF = 10E+5, VON = 0.5V, VOFF = 0V, IS = 2.2E-15, BV = 1800V, TT = 0 sec. For Dt, IS = e-15, BV = 1000V.

THEORY:

Dc to ac converters are known as inverters. The function of an inverter is to change a dc input voltage to a symmetrical ac output voltage of desired magnitude and frequency. The output voltage could be fixed or variable at a fixed or variable frequency. A variable voltage can be obtained by varying the input DC voltage and maintaining the gain of inverter constant. On the other hand, if the dc input voltage is fixed and it is not controllable, a variable output voltage can be obtained by varying the gain of the inverter, which is normally accomplished by pulse width modulation (PWM) control within inverter. The inverter gain may be modified as the ratio of the ac output voltage to dc input voltage.

The output voltage waveforms of ideal inverters should be sinusoidal. However, the waveforms of practical inverters are non-sinusoidal and contain certain harmonics. For low and medium power applications, square wave or quassi – square wave voltages may be acceptable; and for high-power applications, low distorted sinusoidal waveforms are required. With the availability of high-speed power semiconductor devices, the harmonic contents of output voltage can be minimized or reduced significantly by switching techniques.

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Power Electronics & Simulation Lab

PROGRAM:PULSE WIDTH MODULATIONVS 1 0 DC 100VVR 17 0 PULSE(50V 0V 0 833.33US 833.33US 1NS 16666.67US)RR 17 0 2MEGVC1 15 0 PULSE(0 -30V 0 1NS 1NS 8333.33US 16666.67US)RC1 15 0 2MEGVC3 16 0 PULSE(0 -30V 8333.33US 1NS 1NS 8333.33US 16666.67US)RC3 16 0 2MEGR1 4 5 2.5L 5 6 10MHVX 3 4 DC 0VVY 1 2 DC 0VD1 3 2 DMODD2 0 6 DMODD3 6 2 DMODD4 0 3 DMOD.MODEL DMOD D(IS=2.2E-15 BV=1800V TT=0)XT1 2 3 7 3 DCSCRXT2 6 0 9 0 DCSCRXT3 2 6 11 6 DCSCRXT4 3 0 13 0 DCSCR.SUBCKT DCSCR 1 2 3 4DT 5 2 DMODST 1 5 3 4 SMOD.MODEL DMOD D(IS=1E-15 BV=1000V).MODEL SMOD VSWITCH(RON=0.1 ROFF=10E+8 VON=10VVOFF=5V).ENDS DCSCRRG1 8 7 100RG2 10 9 100RG3 12 11 100RG4 14 13 100XPW1 17 15 8 3 PWMXPW2 17 15 10 0 PWMXPW3 17 16 12 6 PWMXPW4 17 16 14 0 PWM.SUBCKT PWM 1 2 3 4R1 1 5 1KR2 2 5 1K

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Power Electronics & Simulation Lab

RIN 5 0 2MEGRF 5 3 100KRO 6 3 75CO 3 4 10PFE1 6 4 0 5 2E+5.ENDS PWM.TRAN 10US 3MS 0 10US.PROBE.END

PROCEDURE:

1. Write the program in a new text file in PSpice AD.2. Save the file using the notation filename.cir.3. Activate the file by opening it.4. Run the simulation process using blue button.5. By clicking Add Trace icon, get the required waveform.

MODEL WAVEFORM:

OUTPUT WAVEFORM

Time

0s 0.5ms 1.0ms 1.5ms 2.0ms 2.5ms 3.0msV(3,6)

-120V

-80V

-40V

-0V

40V

80V

120V

RESULT:

REVIEW QUESTIONS:

1. What is an inverter?2. What is the procedure to trace a waveform?3. What is the procedure to define a switch and diode?4. What is subckt?5. What is your conclusion after doing the analysis?

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Power Electronics & Simulation Lab

13.FORCED COMMUTATION CIRCUITS

CLASS –A COMMUTATION

AIM:To rig up turn off circuit for SCR by CLASS-A commutation.

APPARATUS:1) Forced commutation study kit2) DC regulated power supply 0-30V/2A3) Connecting wires4) Cathode ray oscilloscope

THEORY:The current reversing property of load will force the device commutation. L,C,R

values are chosen such that the circuit is under damped. Since the commutation elements carry load current on a continuous basis their ratings are generally high. For low frequency operation, large values for L and C are required.

CIRCUIT DIAGRAM & MODEL GRAPHS:

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Power Electronics & Simulation Lab

PROCEDURE :1) Make the connections as per the circuit diagram.2) Connect the trigger output T1 to gate and cathode of SCR T1.3) Switch ON the DC supply to power circuit and observe the voltage waveform

across the load by varying the frequency potentiometer.4) Duty cycle potentiometer is of no use in this experiment.5) Repeat the same for different values of L, C , R.

TABULAR FORM:

RESULT :

VIVA QUESTIONS:1) What is the other name for this commutation?2) What condition should be satisfied for R-L-C elements?3) Mention any application of class-A commutation.

EEE Department, SVEC, Tadepalligudem Page 57

S.NO.

TON (ms) TOFF (ms)DUTY CYCLE

TON/TT=TON +TOFF

FREQUENCY(Hz)OUTPUT

VOLTAGE VO (V)

Power Electronics & Simulation Lab

CLASS-B COMMUTATION

AIM : To study CLASS-B commutation circuit.

APPARATUS :1) Forced commutation study kit2) DC regulated power supply 0-24V/2A3) Connecting wires4) Cathode ray oscilloscope5) Rheostat 50 Ω/2A

THEORY :

In this type of commutation reverse voltage is applied to the SCR by the over swinging of under damped circuit connected across the SCR.

A capacitor charge up to the supply voltage before the trigger pulse is applied to the gate. When the SCR is triggered, a load current flows through the external circuit and a pulse of current through LC circuit and SCR is in opposite direction. This rheostat current tends to turn OFF SCR.

CIRCUIT DIAGRAM & MODEL GRAPH:

PROCEDURE :1) Make the connections as per the circuit diagram.2) Connect the trigger pulses to SCR1

3) Switch ON the DC supply and observe the voltage waveform across the load by varying the frequency potentiometer.

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Power Electronics & Simulation Lab

4) Repeat the same for different values of R, L and C. Model waveforms are drawn by observing the load voltage using CRO.

TABULAR FORM:

S.NO. TON (ms) TOFF (ms)

DUTY CYCLE

TON/TT=TON +TOFF

FREQUENCY(Hz)OUTPUT

VOLTAGE VO (V)

RESULT :

VIVA QUESTIONS :

1) What is the other name of this commutation?2) Is it current commutation or voltage commutation?3) What should be the condition for R-L-C components?

CLASS-C COMMUTATION

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Power Electronics & Simulation Lab

AIM :To rig up the TURN OFF circuit for SCR by CLASS-C commutation.

APPARATUS :1) Forced commutation study kit2) DC regulated power supply 0-24V/2A3) Connecting wires4) Cathode ray oscilloscope5) Rheostat 50 Ω/2A- 2

THEORY: This commutation is used to transfer current between loads. The firing of one SCR

commutates the other one. Both the SCR’s are conducting through the load current. However in some cases the SCR used for turn OFF may cause very small amount of current required for charging. In such cases SCR is called as “AUXILIARY SCR “

CIRCUIT DIAGRAM & MODEL GRAPH:

PROCEDURE:

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Power Electronics & Simulation Lab

1) Make the connections as per the circuit diagram.2) Connect T1 and T2 from firing circuit to gate and cathode of SCR T1 and T2.3) Observe the waveforms across R1, R2, and C by varying frequency and also duty

cycle potentiometer.4) Repeat the same for different values of C and R.

TABULAR FORM:

S.NO.

TON (ms) TOFF (ms)

DUTY CYCLE

TON/TT=TON +TOFF

FREQUENCY(Hz)OUTPUT

VOLTAGE VO (V)

RESULT:

VIVA QUESTIONS:

1) What is the other name of this commutation? Why? 2) Is it current commutation or voltage commutation?3) Is there a chance for both the thyristors to be in ON condition?

CLASS-D COMMUTATION

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Power Electronics & Simulation Lab

AIM:To rig up the turn OFF for SCR by using Class-D commutation.

APPARATUS:1) Forced commutation study kit2) DC regulated power supply 0-24V/2A3) Connecting wires4) Cathode ray oscilloscope5) Rheostat 50 Ω/2A

THEORY:This type of commutation is popular due to the design flexibility. These are may be

choppers and inverters under this class.T2 must be triggered first in order to change capacitor C. T2 is commutated OFF

owing to lack of current. When T1 is triggered current flows in two paths, load current through R1 and commutating current through C, T1, L nad D. the charge on the capacitor is reversed and held with the hold OFF diode D1. At any desired instant T2 may be triggered which then places C across T1 via T2 and T1 is turned OFF.

CIRCUIT DIAGRAM & MODEL GRAPH:

PROCEDURE:

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Power Electronics & Simulation Lab

1) Make the connections as per the circuit diagram 2) Connect T1 and T2 gate pulses from firing circuit to the corresponding SCR’s in

the power circuit.3) Initially keep the trigger to OFF position to initially charge capacitor.4) Now switch ON the trigger output switch and observe the voltage waveform at

different frequencies of chopping and also at different duty cycles.5) Repeat the experiment for different values of load resistance, L and C.

TABULARFORM:

S.NO. TON (ms) TOFF (ms)

DUTY CYCLE

TON/TT=TON +TOFF

FREQUENCY(Hz)

OUTPUT VOLTAGE

VO (V)

RESULT:

VIVA QUESTIONS:

1) What is other name of this commutation?2) Is it current commutation or voltage commutation?3) What is current commutation and voltage commutation?4) Mention any application of class-D commutation.

CLASS –E COMMUTATION

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Power Electronics & Simulation Lab

AIM: To rig up turn OFF circuit for SCR by class-E commutation with an external source of pulse for commutation.

APPARATUS:

1) Forced commutation study kit with transistor2) DC regulated power supply 0-24V/2A two numbers3) Connecting wires4) Cathode ray oscilloscope5) Rheostat 50 Ω/2A

THEORY: In class-E for turn OFF, reverse voltage is applied to load carrying thyristor from an

external source across as in series with the conducting SCR. The turn OFF time of SCR is smaller than the width of the pulse. The conducting period of SCR is from the instant of application of triggering pulses till the external turn OFF voltage is applied. When SCR is triggered load current flows, this connects the negative auxiliary voltage to the SCR to turn it OFF.

CIRCUIT DIAGRAM & MODEL GRAPH:

PROCEDURE:1) Make the connections as per the circuit diagram.

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Power Electronics & Simulation Lab

2) Connect the trigger output T1 from the firing circuit to the SCR.3) Connect T2 to the SCR’s base and emitter points.4) Switch ON the DC supply and external DC supply.5) Switch ON the trigger output and observe the waveforms across the load.6) Repeat the same by varying frequency and duty cycle.

TABULAR FORM:

S.NO. TON (ms)TOFF

(ms)

DUTY CYCLE

TON/TT=TON

+TOFF

FREQUENCY(Hz)OUTPUT

VOLTAGE VO (V)

RESULT:

VIVA QUESTIONS :

1) What is other name for this commutation? Why?2) Explain the use of a transistor in this commutation.

14.SINGLE PHASE CYCLOCONVERTER

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Power Electronics & Simulation Lab

AIM:To study the performance of single phase Cyclo converter.

APPARATUS:1) Cyclo converter power circuit2) Microcontroller based firing unit3) Rheostat 50Ω/2A4) Inductor 50Mh5) Cathode ray oscilloscope6) Connecting wires

THEORY :

A Cycloconverter is a type of power controller in which an AC voltage at supply frequency is converted directly to an AC voltage at load frequency.

There are mainly two configurations for this type of Cycloconverter. They are 1) Centre tapped T/F configuration2) Bridge configurationIn Centre tapped transformer configuration, during the first half cycle when point P is

positive and point Q is negative SCR P1 being in conducting mode is gated. The current flows through +ve point P, SCR P1, load and the –ve point Q. In negative half cycle, when Q is positive and p is negative .SCR p1 is automatically turned off and SCR p2 is triggered simultaneously path for current flow in this configuration or condition will be from positive point Q.SCR p2,load and negative point ‘o’. Direction of flow of current through the load remains the same as in the positive half cycle .Next moment again p becomes positive and Q becomes negative, thus SCR p2 is automatically line commuted SCR p1 is gated simultaneously. The current path gain becomes again the previous case when SCR p1 was conducting .Thus it is seen that the direction of current through the load remains in all three phase circuits

PROCEDURE:1) Switch on the main supply of the firing circuit and observe test points and trigger

outputs by changing frequency division and by varying the firing angle.2) Make sure that firing pulses are proper before connecting to the power circuit.3) Make the connections as per the circuit diagram.4) Also connect voltmeter and ammeter in the power circuit.5) Connect the firing pulses from the firing circuit to respective SCR’s in the power

circuit.6) Observe the waveforms in the CRO.7) Before turning off first remove the firing pulses and then switch off the circuit.

CIRCUIT DIAGRAM WITH R & RL LOADS:

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Power Electronics & Simulation Lab

MODEL GRAPH:

TABULAR FORM:

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Power Electronics & Simulation Lab

R-LOAD:

S.NOFiring Angle

(α)FREQUENCY 1 FREQUENCY 4 FREQUNCY 6

VDC(V) VRMS(V) VDC(V) VRMS(V) VDC(V) VRMS(V)

RL-LOAD:

S.NOFiring

Angle (α)FREQUENCY(Hz) FREQUENCY 4 FREQUNCY 6

VDC(V) VRMS(V) VDC(V) VRMS(V) VDC(V) VRMS(V)

RESULT :

VIVA QUESTIONS:

1) How many AC-voltage controllers are required in a Cycloconverter .2) What is the difference between a Cycloconverter and an AC voltage controller?3) What type of commutation is employed in cycloconverter?

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