1

POWER ELECTRONICS II

PRACTICAL FILE

SUBMITTED TO: SUBMITTED BY:

Prof. Dheeraj Joshi Group4 (E4)

Khushboo (108588) Pranav (108624) Ankur (108637) Bhavya (108638) Shweta (108640)

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INDEX

S.NO. EXPERIMENT PAGE NO. REMARKS

1 To simulate single phase series inverter.

5-7

2 To simulate single phase parallel inverter.

8-10

3 To study single phase series inverter.

11-13

4 To study single phase parallel inverter.

14-16

5 To study speed control of DC shunt motor using Dc john chopper.

17-19

6 To study 3 phase

inverter on psim.

20-23

7 To simulate single phase cyclo converter.

24-25

3

8 To study single phase step up chopper.

26-28

9 To study SCR based on cycloconverter.

29-30

10 To study single phase AC & DC power control.

31-33

List of Figures

Fig No. NAME OF FIGURE PAGE NO.1.1 Series Inverter 51.2 Waveforms for series inverter 61.3 Waveforms of series inverter

with sine output wave7

2.1 Parallel bridge inverter 82.2 Waveforms for parallel inverter 92.3 Output voltage waveform for

c=0.2uF9

2.4 Output voltage waveform for C=0.9uF

10

3.1 series inverter 12

3.2 Output voltage wave shape from CRO.

13

4.1 Single phase parallel bridge inverter with feedback diode

15

4.2 Waveform of output voltage obtained from CRO

16

5.1 D.C. Jones Chopper 19

4

6.1 Three phase inverter 206.2 Switching sequence for 180

mode20

6.3 Output waveform for 180° mode R load

21

6.4 Switching sequence for 120 mode

21

6.5 Output waveform for 120° mode R load

23

7.1 Single phase cycloconverter 247.2 Output waveform for step down

cycloconverter25

8.1 step up chopper 26

8.2 Output waveform across L 28

9.1 SCR based cycloconverter 29

9.2 Output waveform of step down cycloconverter

30

10.1 A.C. Power Control 31

10.2 D.C. Power Control 1 phase Half Wave Converter

31

10.3 D.C. Power Control 1 phase Full Wave Converter

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List of Observation Tables

Table no. Name of Observation Table Page no.5.1 Observation table for D.C. Jones

Chopper

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Experiment 1

1.1 A im : “To study 1 phase series inverter circuits through simulation”

1.2 A pparatus : Psim and Pc.

1.3 Ciruit diagram : Vt

6

Fig1.1 Series Inverter

1.4 T heory : Let f0 be the output frequency of the inverter, Toff be the duration for safe turn-off of each of the two thyristors and fr is the ringing frequency of the inverter circuit.

Then (1/fr) = [(a/f0)- 2Toff] = (1/f0)[1-2f0Toff]Or fr = fo/[1-2f0Toff]For f0 = 800 Hz and Toff = 125 us; fr = 800/0.8 = 1000 Hz

Design of Inductance L

A suitable value of L is chosen on the basis of the attenuation factor. When wct = π/2, the peak value of Imax [in the expression for current i(t) = Imax e(R/2L)tπsinwct] will be reduced. The optimum value of this attenuation factor(AF) is 0.5. Therefore, we can write,

R π/4wrL = ln2 or L = R/(8fcln2)R= 100 Ω and fr = 1000 Hz L = 100/(8000.ln2) = 18.034 mH

Design of Capacitance CCapacitance is obtained from the expression for fr, i.e. fr = 12π√(1LC-R24L2)C = 1.1757 uF

1.5 Discussion:

Vin

Vout

7

02 04 06 08 0

1 0 01 2 0

V i n

0

-5 0

-1 0 0

5 0

1 0 0

V o u t

0 0 . 0 1 0 . 0 2 0 . 0 3 0 . 0 4 0 . 0 5 0 . 0 6T i m e (s )

0

-4 0 0

4 0 0

8 0 0

V t

Fig 1.2 Waveforms for series inverter

Fig 1.2 shows the output waveform of series inverter for L=18.034mH, C=1.1757µF, R=100Ω. The output wave is not pure sinusoidal, for given values of L and C. This is because before T2 is switched on, after switching off T1, sometime has to be elapsed so that T1 regains its forward blocking capability. However, adjusting the values of L, C in the circuit sine wave output can be obtained as shown in fig 1.3. the values of inductance and capacitance are 48.963mH and 1.8uF, respectively.

8

0

20

40

60

80

10 0

12 0V in

0

-5 0

-1 00

50

10 0

V out

0 0 .0 1 0 .02 0 .03 0 .04 0 .05 0 .06T im e (s)

0

-5 0

-1 00

50

10 0

15 0

20 0

V t

Fig 1.3 waveforms of series inverter with sine output wave.

1.6 Result : Circuit of single phase series inverter has been designed on Psim and various plots are taken as shown and results obtained are same as theoretically predicted.

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Experiment 2

2.1 Aim : “To study 1-Φ parallel inverter circuit through simulation.”

2.2 Apparatus : Psim and Pc.

2.3 Circuit Diagram:

Fig 2.1 parallel bridge inverter.

2.4Theory: Parallel inverter inthe given 1- Φ capacitor commutated parallel inverter using two Thyristors and a centre-tapped transformer the source voltage is 200v dc. The centre-tapped transformer has a turn ratio from each half primary winding to secondary winding Of 3:1.

For tc= 40 μ s and R = 20 ΩCommutating capacitor:-c=n2tc4R ln2=0.0815 μ F

Taking f0 = 1Khz and L = 100 m h . The wave form for parallel inverter ckt are shown in psim simulation.

2.5 Discussion: Waveforms for parallel bridge inverter are shown in fig 2.2, for given values of L=100mH, C= 0.0185uF, R-20Ω. Output voltage waveform is like a square wave, not sine wave as required, indicating the presence of harmonics. Waveform can made better by adjusting the value of capacitor. Output voltage waveforms for different values of capacitor are shown in fig2.3 and 2.4, with C=0.2uF and C=0.9uF respectively.

Vin Vo

Vt

VR=20

L=100

10

0

4 0

8 0

V i n

0-2 0-4 0

2 04 0

V o u t

0

-2 0 0

2 0 0

V c

0 0 . 0 0 2 0 . 0 0 4 0 . 0 0 6 0 . 0 0 8 0 . 0 1T i m e ( s)

0

1 0 0

2 0 0

V t 1

Fig 2.2 Waveforms for parallel inverter

0

20

40

60

80

100

Vin

0

-20

-40

-60

20

40

60

Vout

0

-200

-400

200

400

Vc

0 0.002 0.004 0.006 0.008 0.01

T ime (s)

0

-200

-400

200

400

Vt1

Fig 2.3 Output voltage waveform for c=0.2uF

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0

20

40

60

80

100

Vin

0

-100

-200

100

200

Vout

0K

-0.5K

-1K

-1.5K

0.5K

1K

1.5K

Vc

0 0.002 0.004 0.006 0.008 0.01Time (s)

0K

-0.5K-1K

-1.5K

0.5K

1K1.5K

Vt1

Fig 2.4 Output voltage waveform for C=0.9uF

2.6 Result: Circuit of single phase parallel bridge inverter has been designed on Psim and various plots are taken as shown and results obtained are same as theoretically predicted.

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E xperiment 3

3.1 Aim: “To study single phase series inverter.”

3.2 Apparatus: 1. Inverter Circuit

2. DC Power Supply

3. Firing Circuit

4. CRO

5. Load

3.3 Circuit Diagram:

Fig 3.1 series inverter

3.4 Theory: Basic series inverter circuit shows the circuit of a series inverter. The commutating elements L and C are such that R1L and C for an under damped circuit. The capacitor has an initial voltage Ec. Thyristor TH1 is turned on first. Since their already forward biased TH1 starts conducting and a current I flows in the circuit through TH1 and C1, L and load. Because of under damped nature of the circuit the current is not constant. It rises to a maximum value and then decreases to zero. When the current is at its peak the voltage across the capacitor is equal to the source voltage V. After this the current starts decreasing but voltage across capacitor increases. When the current becomes zero the voltage across the capacitor is positive. At the point when current becomes zero TH1 is

load

batte

T1

T2C

L

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turned off. Now TH1 and TH2 are off. The voltage across capacitor is maintained at V+ Ec.. The across L is zero the time interval ab must be more than the Toff of the Thyristor. This is necessary to ensure that the stored charges in TH1 are reduced to zero so that at point 6 TH1 is in completely off state. At point 6 when the anode of TH2 is positive due to charge on the capacitor TH2 starts conducting. The capacitor discharges and the current I flows in the opposite direction as of the previous current. He current reaches its negative peak value and the decreases to zero at point C when TH2 is turned off. The above sequence is repeated in the next cycle when TH1 is turned on. The frequency of output voltage is

f = [1/2*(T/2 + Toff)]

T is the time period of oscillations. Also

T = 2*TT/(1/LC – R2/4C2)

When TH1 is turned on the equations are

V + Ec = Ri + Ldi/dt + i/c * INTEGRATION(I dt)

Taking Laplace:

V/s + Ec/s = I(s)[R + Ls + 1/Cs]

Since the circuit is under damped the solution is

I = (V + Ec)/(w+L)*exp(-Rt/2L)sin(wrt)

3.5 Procedure: To begin with switch on the power supply to the firing circuit. Check the trigger pulses by varying the frequency. Make the connections as shown in the diagram. Now cancel trigger o/p s from the firing circuit to gate and cathode of SCR. Observe the waveform across the load. Vary the frequency and again observe the waveform. Repeat the same for different values of L and C.

3.6 Discussion: Waveform of output voltage obtained from CRO are shown below in fig 3.2. Wave is not pure sine wave, but have discontinuity in between. These discontinuities can be reduced by selecting appropriate values of L and C. The wave shape obtained are same as obtained from psim simulation of series inverter in exp1. Hence, the results are verified.

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Fig 3.2 Output voltage wave shape.

3.7 Results: Single phase series inverter have been studied and the results are verified from exp1, as wave shapes in both cases are same.

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E xperiment 4

4.1 Aim: “To study single phase parallel bridge inverter with feedback diode.”

4.2 Apparatus:

1. Bridge in inverter firing circuit unit (Gate drive current 200mA) 2. Parallel bridge inverter power circuit (30V,2A)3. Power supply 230V AC4. CRO5. Load

4.3 Circuit Diagram:

Fig 4.1 single phase parallel bridge inverter with feedback diode

4.4 T heory : The bridge current with feedback diode and load is shown in figure. Initially SCR T1 and T1’ conducts, T2 and T2’ are fixed. T1 and T1’ are turned off due to reverse biasing by the capacitor voltage. The capacitor is now charged up by the sum of the current in commutating inductance and load current. The commutation process starts with damped oscillations and the capacitor voltage tends to exceed Vdc at this instance t2, when D2 and D3 becomes forward bias and begin to conduct. The voltage across them drops almost to a

battery

D1 L/2

L/2

D3

D4D2

T2

T1

T2’

T1’

C(C1/C

load

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zero as a result the voltage across the commutating reactors is also zero, and the capacitor voltage and capacitor current cannot change. Therefore, the reactor current is divided into Z load. Current Vdc/RC flows through SCR2 and SCR3 and the diode current flowing through D2 and D3.

Power Circuit:

a) Vdc(in) – Terminal to connect i/p dc supply.b) ON – Switch to ON the dc supply to tha power circuit.c) Fuse – 2A glass fuse for protection.d) T1,T1’T2,T2’ – SCR’s TYN 6/6.e) D1,D1’,D2,D2’ – free wheeling diodesf) L/2 – Commutation inductance g) C1 – Commutation capacitanceh) C2 – Commutation capacitance

4.5 P rocedure : Make all the connection of power circuit is built 1 phase bridge inverter. Connect the firing pulses from the bridge inverter firing circuit. Connect the loads at load points. Initially connect i/p DC supply to the power circuit. Apply firing pulse from firing circuit by off/on switch. Observe output waveforms across the load whose voltage can be varied by varying dc input to power circuit. If the commutation fails there is no voltage at the output. This can be visible by sudden voltage drop and current rise in regulated power supply. To switch off the inverter, switch off the DC supply first and then trigger output. don’t switch OFF the trigger output without switching OFF the DC supply. It will lead to commutation failure.

4.6 O bservation:

Voltage = 12.2V

Current = 0.04A

(T) Time period = 5ms

(N) No. of div. = 4

Frequency = 1/(N*T) = 1/(.005*4) = 50 Hz

4.7 Discussion: Waveform of output voltage obtained from CRO are shown below in fig 4.2. Wave is not sine wave, is a square wave. Different wave shapes are obtained by selecting appropriate values of L and C. The wave shape obtained are same as obtained from psim simulation of series inverter in exp2. Hence, the results are verified.

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Fig 4.2. Waveform of output voltage obtained from CRO

4.8 P recautions: If commutation fails, switch OFF the input DC supply and try again checking all the connections, devices, and firing pulse.

4.9 R esult: The output wave obtained is a square wave which has been traced from CRO using a tracing paper and has been compared with the waveform Fig 2.2 and has been found to be approximate. Hence results are verified.

Experiment 5

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5.1 A im : “To study speed control of DC motor using single phase half and fully controlled bridge converter”

5.2 Apparatus:

Firing circuit: DC Chopper firing circuit.

Power circuit: DC Chopper firing circuit.

Source : 220V De power supply.

Load : 1HP/220V DC shunt motor with wading and speed measuring arrangement

Using optical culoder pickup.

Testing Equipment: Power Scope/CRO with 10:1 probe.

5.3 Circuit Diagram:

Fig 5.1 D.C. Jones Chopper

battery

C

T1

T2

D1

Dm

L L

load

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5.4 Theory: Jones Chopper is an eg. Of class D commutation in which a changed capacitor is switched by an auxillary SCR to commutate the main SCR. In this circuit , SCR is the main switch and SCR2 is the auxillary switch which is of lower capacity than SCR , and is used to commutate SCR1 by the reverse voltage development across the capacitor C The special feature of the circuit is tapped autotransformer through a portion of which load current flows. Since 4 and 0.5 are closely coupled , the capacitor always get sufficient energy to turn SCR1 off. When SCR1 is turned on , the capacitor C discharges resonantly through SCR1 ,4 and D1. This discharge current does not flow through L2 and back to the battery because of the transformer action of T. The load current is picked up by SCR1 and the flywheel diode D1 is reverse biased and its current reduced to zero. As the capacitor voltage swings negative, the reverse bias on diode D2 decreases. This continues to time (L1C)^1/m. The capacitor voltage assumes a polarity as shown in fig. When SCR2 is turned on, the negative voltage on capacitor C is applied current which is nominally constant starts to flow in SCR2 and capacitor C. The di/dt of SCR2 is limited by the circuit stray inductance. The capacitor C is charge positivelyt at first up to a voltage equal to the supply voltage Vdc. The flywhell diode becomes forward biased and begins to pick up load current and capacitor current starts to reduce. After this, the enrgy ½ LI2 in inductance L2 is forced into capacitor C changing it +ve (Y2CV2). The capacitor current continues to decrease increasing the current os SCR2 till the capacitor current reduces below the holding current of SCR2 when it is turned off. The cycle repeats when SCR1 is again turned on.

ADVANTAGE: It allows the use of higher voltage and lower mf commutating capacitors. This is because the trapped energy of the inductor L2 can be forced into the commutating C range simply changing C by supply voltage. Voltage rating of SCSs should also be high. There is no starting problem , any SCR can be turned on initially. Since both the on and off time can be varied individually , it is flexible.

The Vpeak across SCRs and capacitor = supply voltage.

Vpeak = Vde +I1 √4/C

The commutation capacitor however doer not retain all its energy and is –vely charged at the instant of commutation to n Vpeak where n lies betn 0.5-0.9. The commutating capacitor C must divert the load current long enough so that before C is changed to ov , the SCR is off.

5.5 Procedure :

1)TESTINGOD DC – Chopper firing circuit. Connect 3 phase mains cable to unit switch ON the main supply to the unit. Keep the INT-EXT switch in INT position. Observe the test point signals. Observe the trigger o/p pulses. Main pulse and auxillary pulse. The duty cycle of mani pulse should vary from 10-90% approx. Make sure firing circuit are proper before connecting to power circuit.

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5.6 O bservation table :

S .no Input voltage (Vin)(V)

Output voltage (Vo)(V)

Duty Cycle

(%)

Speed

(rpm)

1. 210 124 10 1192

2. 210 151 20 1469

3. 210 168 30 1623

4. 210 184 50 1787

5. 210 189 70 1837

6. 210 194.6 90 1893

Table 5.1 Observation table for D.C. Jones Chopper

5.7 PRECAUTIONS:

(1) Make sure that the firing pulse is proper before connecting to power circuit.

(2) Check the continuity of pulses in back panel.

(3) A small value of resistance in series with the armature should be used for commutation process.

(4) Interconnection should be tight.

5.8 Discussion: From observation table, we can observe that output voltage and speed are directly varying with the duty cycle or firing angle. Graphs have been plotted, depicting the variation of o/p voltage and speed with duty cycle.

5.9 Result: Speed control of DC motor using single phase half controlled bridge converter have been studied successfully.

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Experiment 6

6.1 Aim: “To study three phase inverter on psim.”

6.2 Apparatus: Psim and Pc.

6.3 Circuit diagram:

Fig.6.1 Three phase inverter

6.4 Theory: Three phase inverters are used for providing adjustable frequency power to industrial applications. A basic 3 phase inverter is a six step bridge inverter, using

Vin

T

T

T3

T

T5

T6

Vao

Va

Vbo

Vc

22

minimum of six switches. As shown in fig 1, 3 phase inverter consists of 3 half bridge inverters arranged side by side. Three phase load is assumed to be star connected. The topology of circuit is shown. There are two possible patterns of gating the thyristors. In pattern, each thyristor conducts for 180 degrees and in other for 120 degrees, discussed below:

Mode 180°

As shown in fig 6.2, each thyristor conducts for 180° of a cycle. Thy pair in each arm, i.e. T1,T4; T3,T6; T2,T5 are turned on with a time interval of 180°. Thy in upper group, i.e. T1,T3,T5 conduct at an interval of 120°. Same is true for lower group also. Diodes fed back current to dc source in case of RL load, there is no use of diodes in R load.

T1 T4

T6 T3 T6 T5 T2 T5 I 5,6,1

II 6,1,2

III 1,2,3

IV 2,3,4

V 3,4,5

VI 4,5,6

Fig 6.2 Switching sequence for 180 mode

For above shown firing strategy, the output waveforms of line to phase and line to line voltages are shown in fig6.3 for R load respectively.

0° 60° 120° 180° 240° 300° 360°

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Fig.6.3 180° mode R load

Mode 120°As shown in fig 6.6, each thy conducts for 120° of a cycle. Thy pair in each arm, i.e. T1,T4; T3,T6; T2,T5 are turned on with a time interval of 180°. Thy in upper group, i.e. T1,T3,T5 conduct at an interval of 120°. Same is true for lower group also. T1 conducts for 120° and for next 60° neither T1 nor T4 conducts. Firing strategy is shown in fig 6.5 0° 60° 120° 180° 240° 300° 360°

T1 T4

T6 T3 T6 T2 T5 I 6,1

II 1,2

III 2,3

IV 3,4

V 4,5

VI 5,6

Fig 6.4 Switching sequence for 120 mode

For above shown firing strategy, the output waveforms of line to phase and line to line voltages are shown in Fig 6.5 for R load

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9 2 . 9 5 7 7

V i n

0- 2 0

2 0

V a o

0- 2 0

2 0

V b o

0- 2 0

2 0

V c o

0- 4 0

4 0

V a c

0 0 . 0 0 2 0 . 0 0 4 0 . 0 0 6 0 . 0 0 8 0 . 0 1T i m e ( s )

0

I a

Fig 6.5 120° mode R load

6.5 Discusion: 1.On comparing the waveforms for 180° and 120° modes, we can clearly see that harmonic component in phase voltage for 180° mode is more than that of 120° mode. Whereas for line to line voltage, 180° mode has less harmonics. 2. Current waveform is better for RL load.3. While using 180° mode, there is a chance of direct short circuit across supply in case T1 and T4 turn on simultaneously, due to absence of commutation interval between turning off of T1 and turning on of T4.In 120° mode, there is an interval of 60°, hence no chance of short circuit.

6.6 Result : Waveforms obtain for R load for 180° and 120° as shown in Fig 6.3, Fig 6.5.

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Experiment 7

7.1 Aim: “To stimulate a single phase step down cycloconverter on psim software.”

7.2 A pparatus : Psim and Pc.

7.3 C ir c uit diagram:

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Fig 7.1 Single phase cycloconverter

7.4 S pecifications : • GP1

Frequency :16.6667hzSwitching point:0 60 120 180

• GP2Frequency :16.6667hzSwitching point:60 120

• GN1Frequency :16.6667hzSwitching point:180 240 300 360

• GN2Frequency :16.6667hzSwitching point:240 300

• SIMULATING CONTROLTime step:0.0001sec.Total time:0.2sec

• VOLTAGE SOURCE: Peak value:325 volt Frequency:50hz

7.5 Waveforms obtained:

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Fig 7.2 Output waveform for step down cycloconverter.

7.7 Result & Discussion : Circuit of step down cycloconvertor has been designed on Psim and wave obtain is shown above in fig 7.2 and results obtained are same as theoretically predicted, for 2 cycles of input one cycle of output.

Experiment 8

8.1 Aim: To study single phase step up chopper.

8.2 Apparatus: 8.2.1Control circuit:

1. Frequency : Potentiometer to vary frequency

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2. Duty cycle: Potentiometer to vary the duty cycle3. Driver output ON/OFF: From 0-70% approximately, switch for driver output4. + or - : Driver output points to be connected to the base and emitter of the

transistor5. Mains : Power ON/OFF switch to the unit with built in indicator

8.2.2Power circuit:

1. VDC IN: Terminals to connect DC input from 30V/2A regulated power supply2. ON: Switch for DC input3. Fuse: 2A glass fuse4. +: +ve dc supply point after switch and fuse5. L: Inductor -40mH/2A6. Transistor: TIP 1227. D1 & D2: Diodes 10A/200V8. C: Capacitor for filtering -470mf/200V

8.3Circuit diagram:

Fig 8.1 step up chopper

8.4 Theory: In this chopper, a large inductor is in series with source voltage. When the switch is ON, a closed current path enclosing switch, input voltage and L. Here, L stores the energy in the ON period. When the switch is OFF ,the current through L cannot fall down suddenly , this current is forced through the diode and the load. As the current tends to decrease the polarity of the emf induced in L is reversed. As a result the voltage across the load exceeds Vi n. Hence this circuit acts as a step up chopper.

8.5 Observation: Output waveform shown in fig 8.2 are obtained for the above setup. Varying the duty cycle, the output voltage can be varied as shown in fig, for different values of duty cycle.

20V

RL load

L

T

Dm

29

Fig 8.2 Output waveform across L

8.6 Precautions:

1. Duty cycle should not exceed beyond 75%.2. The total current drawn from the voltage supply must be monitored and should be

within the current rating of the power supply.

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8.7 Result & Discussion : Single phase step up chopper has been studied and voltage waveform obtain on CRO approximately same as theoretical expected.

Experiment -9

9.1 Aim: To study the SCR based Cycloconverter functions and also its associated waveforms.

9.2 Apparatus:

1. 2 channel power oscilloscope2. PE trainer with INV/CON panel

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3. Lab multimeter battery supplied with PE trainer

9.3 Circuit diagram:

Fig9.1 SCR based cycloconverter

9.3 Theory: The cycloconverter configuration is shown in the figure and uses all four SCR’s . cycloconverters are used for frequency divisions .The circuit can produce load voltage at 25 hz or 12.5 hz. In simplest form frequency divisions in multiples of 2 is possible.

• The circuit functions ac centre tapped positive rectifier (using Th1 and Th2) for integral number of A.C cycles

• Further the circuit functions as centre tapped negative rectifier (Th3 and Th4) for same number of input A.C cycles

• Then this process repeats.

9.4 Procedure:

The required SCR’s are Th1,2 and Th3 and Th4.The inverter transformer is used in step down mode as 230V/12-0-12 V. The load used is 10 ohm/40 W resistor from RHS panel from main unit.Make the connections as per schedule given above. Keep the firing mode selector switch on position 2. Keep D.C input switch ON AC input to PE trainer. Now observe the waveform.

100Ω, 40 W

Thy1

Thy 3

Thy 4

Thy230V

32

9.5 Observation: Following waveforms are obtained from C.R.O. for above circuit. Fig 9.2 shows o/p waveform for step down cycloconverter with o/p frequency (i) 25Hz and (ii) 12.5 Hz, respectively.

Fig 9.2 Output waveform of step down Cycloconverter

9.6 Discussion : The SCR based Cycloconverter offers input A.C frequency divisions in multiples of 2, 4. Variable voltage output can also be offered by controlling firing angle of these SCR’s in each cycle. These results are same as obtained in expt 7 through simulation, hence the results are verified.

9.7 Result: The output wave obtained is a lower frequency wave which has been traced from CRO using a tracing paper and has been compared with the waveform Fig 7.2 and has been found to be approximate. Hence results are verified.

Experiment -10

10.1 Aim: To study single phase AC & DC power control.

10.2 Apparatus:

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a)firing circuit

b)power circuit

10.3 Circuit diagram:

Fig 10.1 A.C. Power Control

Fig 10.2 D.C. Power Control 1 phase Half Wave Converter

Fig 10.3 D.C. Power Control 1 phase Full Wave Converter

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10.4 Theory:

a)Firing circuit:- This unit,generates line synchronized to pulse transformer isolated trigger pulses .These trigger pulses can be used to trigger :-

1)1 -Φ AC phase control using SCR’s(antiparallel SCr’s )

2)1 -Φ AC phase control using triac

3)1-Φ half wave rectifier (single scr)

4)1- Φ full wave rectifier (2 scr’s )

5)1-Φ half controlled bridge rectifier(2 scr’s and 2 diodes )

The firing circuit is based on zero crossing detector,ramp generator ,op-amp.comparator and amplifier /pulse transformr isolation method

Front panel details:-

1)power:- main switch for firing crcuit with built in indicator

2)firing:-potentiometero vary the α from 0 to 180•

3)SCR/TRIAC:selection switch for trigger output for scr /triac

4)OFF/ON:switch for trigger output with soft start feature

5)trigger output s:-T1/T2:-trigger output for SCR1 or TRIAC

T2:-trigger output for SCR2

Power circuit:-

The power circuit consists of 2 SCR’s ,3 diodes and a TRIAC .The power devices are mounted on suitable heat sink for dissipation .The snubber circuit is connected for dv/dt protection .a fuse is also provided in series wuth the devices for short circuit or over current protection .in the input side a MCB is provided to switch ON/OFF the supply to the power circuit .a voltmeter is provided and an ammeter is provided to measure the input /output voltage and current .

Front panel details :-

1)AC input:-terminals to connect AC input

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2)AC output :-AC supply terminals after the MCB to be connected to power circuit

3)MCB :-A 6A/2 pole MCB for ON/OFF the AC supply to the power circuit

4) T1 and T2:- SCR’s -16 A /600V

5)D3 and D4:-diodes -16A /600V

6)Dm:-free wheeling diode

7)Tr:-TRIAC -10A/600V

8)vometer :-0-300 V moving iron meter to measure input /output volatage

9)ammeter :0-5 A moving iron meter to measure current

10.5 Procedure :

Make the interconnections in the power circuit as given in the citrcuit diagram .switch ON the firing circuit .Observe the triggering outputs .Make sure that the firing pulses are proper before connecting to the power circuit than connect the trigger output from firing circuit to corresponding SCR’s /TRIAC .In the power circuit initially set the AC input to 30 V .switch ON the MCB and trigger output switch .select the SCR’s /TRIAC ,select switch and observe the output waveforms across ‘R’ load by varying the firing angle of potentiometer .If the output waveform is proper than you can increase the input voltage to rated value i.e 0-230 V gradually .vary the fit=ring angle and note down the output voltage.