UNIT II - PHASE-CONTROLLED CONVERTERSPART A

1. What is Dual converter? Mention its functional mode of operation?

(M/J14)A dual converter can be of a single phase or a three phase. A dual

converter consists of two bridges consisting of thyristors in which one for rectifyingpurpose where alternating current is converted to direct current which can begiven to load. Other bridge of thyristors is used for converting D.C to A.C.2. Explain the inversion mode of fully controlled bridge rectifier?

(N/D12) /(M/J12) When firing angle α is greater than 90˚, then the voltage is maximum in

negative is called inversion mode.3. Mention the effect of source inductance in converters. (M/J15)(M/J16)

The source inductance causes the outgoing and incoming SCRs to conducttogether .When the source inductor increases, the overlap angle increases and asa consequence the average output voltage decreases.4. What is overlap angle?

(N/D15) When both incoming and outgoing thyristors conductsimultaneously causing short circuit of dc load.The period is referred as overlapand angle corresponding to it is known as overlap angle.5. Compare half controlled and fully controlled rectifier.

(M/J14)

S.No

Half controlled converter Fully controlled converter

1. It uses one thyristor. It uses four thyristor

2. One quadrant converter Two quadrant converter

3.At time one thyristor isconducting

At a time two thyristor areconducting

6. Define - THD (M/J12)

The total harmonic distortion, or THD, of a signal is a measurement ofthe harmonic distortion present and is defined as the ratio of the sum of thepowers of all harmonic components to the power of the fundamental frequency

7. Why is power factor of semi converter better than full converter? (N/D12)(N/D14)(N/D15)

Semi converter has only 2 thyristors where else full converter has 4thyristors. Therefore semi converter uses thyristors and diodes for rectificationwheras full converter has no diodes. Henceforth it uses 4 thyristors for rectification.

8. What is displacement factor for two pulse converter? (M/J13)The phase angle between sinusoidal supply voltage vs and fundamental

component Is1 of supply current Is is Φ , this angle is called as displacement anglethe cosine of displacement angle is called as displacement factor.

9. What is circuit turn-off time for single phase full converter? (M/J13)After the current in a thyristor has extinguished, a finite time delay must

elapse before the anode can again be positively biased and retain the thyristor inthe off-state. This minimum delay is called the circuit commutated turn off time ().

10. Mention the disadvantages of dual converter with circulating currentmode of operation.

(N/D13)The disadvantage of the circulating current mode of operation is that a

current flows continuously in the dual converter circuit even at times when the loadcurrent is zero. Hence we should connect current limiting inductors (reactors) inorder to limit the peak circulating current within specified value. The circulatingcurrent flowing through the series inductors gives rise to increased power losses,due to dc voltage drop across the series inductors which decreases the efficiency.Also the power factor of operation is low. The current limiting series inductors areheavier and bulkier which increases the cost and weight of the dual convertersystem

11. A single phase full converter feeds power to RLE load with R=6ῼ and E =60 v. the load inductance value is very large so as to maintain the loadcurrent continuous and ripple free. The ac source voltage is 230 V and 50 Hz.

Find the average value of the output voltage for firing angle delay of 50⁰.(N/D 13)

Vo=(2*Vm/ ᴨ) cosἀVo=(2*(230/sqrt(2))/ᴨ) cos50 = 208.90V

12.What is meant by phase control? (N/ D 14)(M/J 15)

Phase control is the most common form of Thyristor power control. TheThyristor is held in the off condition -- that is, all current flow in the circuit isblocked by the Thyristor except a minute leakage current. Then the Thyristor istriggered into an “on” condition by the control circuitry.13. What is the function of freewheeling diode and state its advantages.

(N/D16) (M/J16)i) Increase DC voltage for a given firing angle due to the elimination of negativeportions of the instantaneous dc waveform in a SCR phase controlled converter,ii) Will reduce the generated ripple voltage on the DC side of a SCR phasecontrolled converter due to same reason as in above, reducing the filteringrequirements.iii) Will improve the input PF in an SCR phase controlled converter due to endingthe input current waveform earlier by permitting internal free-wheeling.

14. Classify the different types of controlled rectifier. (N/D16)

Classification of controlled rectifier are .

i)Semi controlled or half controlled rectifier. the semi converter further classified into symmetrical and asymmetricalconfiguration.ii)Fully controlled rectifier

PART-B1.Explain the operation of 3 phase fully controlled bridge rectifier with relevantwaveforms. (M/J12)CIRCUIT OPERATION

The operation of a 3-phase fully-controlled bridge rectifier circuit is described inthis page. A three-phase fully-controlled bridge rectifier can be constructed usingsix SCRs as shown below.

The three-phase bridge rectifier circuit has three-legs, each phase connected toone of the three phase voltages. Alternatively, it can be seen that the bridge circuithas two halves, the positive half consisting of the SCRs S1, S3 and S5 and thenegative half consisting of the SCRs S2, S4 and S6. At any time, one SCR fromeach half conducts when there is current flow. If the phase sequence of the sourcebe RYB, the SCRs are triggered in the sequence S1, S2, S3, S4, S5, S6 and S1

and so on.The operation of the circuit is first explained with the assumption that diodes areused in place of the SCRs.

Let the three-phase voltages be defined as shown below.

It can be seen that the R-phase voltage is the highest of the three-phasevoltages when is in the range from 30o to 150o. It can also be seen that Y-phasevoltage is the highest of the three-phase voltages when is in the range from 150 o

to 270o and that B-phase voltage is the highest of the three-phase voltages when is in the range from 270o to 390o or30o in the next cycle. We also find that R-phase voltage is the lowest of the three-phase voltages when is in the rangefrom 210o to 330o. It can also be seen that Y-phase voltage is the lowest of thethree-phase voltages when is in the range from 330 o to 450o or90o in the nextcycle, and that B-phase voltage is the lowest when is in the range from 90 o to210o. If diodes are used, diode D1 in place of S1 would conduct from 30o to 150o,diode D3would conduct from 150o to 270o and diode D5 from 270o to 390o or30o inthe next cycle. In the same way, diode D4would conduct from 210o to 330o,diode D6

from 330o to 450o or90o in the next cycle, and diode D2would conduct from90o to 210o. The positive rail of output voltage of the bridge is connected to thetopmost segments of the envelope of three-phase voltages and the negative rail ofthe output voltage to the lowest segments of the envelope.

At any instant barring the change-over periods when current flow gets transferredfrom diode to another, only one of the following pairs conducts at any time. Period, range of Diode Pair in conduction

30o to 90o D1 and D6

90o to 150o D1 and D2

150o to 210o D2 and D3

210o to 270o D3 and D4

270o to 330o D4 and D5

330o to 360o and 0o to 30o D5 and D6

If SCRs are used, their conduction can be delayed by choosing the desired firingangle. When the SCRs are fired at 0o firing angle, the output of the bridgerectifier would be the same as that of the circuit with diodes. For instance, it isseen that D1 starts conducting only after α = 30o. In fact, it can startconducting only after α = 30o, since it is reverse-biased before α = 30o. The biasacross D1 becomes zero when = 30 o and diode D1 starts getting forward-biasedonly after α =30o. When vR(α) = E*Sin (α), diode D1 is reverse-biased before α =30o and it is forward-biased when 30o. When firing angle to SCRs is zerodegree, S1 is triggered when α = 30o. This means that if a synchronizing signal isneeded for triggering S1, that signal voltage would lag vR(α) by 30o and if thefiring angle is α, SCR S1 is triggered when

α = α + 30o. Given that the conduction is continuous, the following table presentsthe SCR pair in conduction at any instant.

Period, range of SCR Pair in conduction

α + 30o to α + 90o S1 and S6

α+ 90o to α + 150o S1 and S2

α + 150o to α + 210o S2 and S3

α + 210o to α + 270o S3 and S4

α + 270o to α + 330o S4 and S5

α + 330o to α + 360o and α + 0o to α + 30o S5 and S6

The operation of the bridge-rectifier is illustrated with the help of an appletthat follows this paragraph. You can set the firing angle in the range 0o< firingangle < 180o and you can set the instantaneous angle also. The applet displaysthe SCR pair in conduction at the chosen instant. The current flow path is shownin red colour in the circuit diagram. The instantaneous angle can be either set inits text-field or varied by dragging the scroll-bar button. The rotating phasordiagram is quite useful to illustrate how the circuit operates. Once the firing angleis set, the phasor position for firing angle is fixed. Then as the instantaneousangle changes, the pair that conducts is connected to the thick orange arcs. Oneway to visualize is to imagine two brushes which are 120o wide and the device inthe phase connected to the brush conducts. The brush that has "Firing angle "written beside it acts as the brush connected to the positive rail and the other actsas if it is connected to the negative rail. This diagram illustrates how the rectifiercircuit acts as a commentator and converts ac to dc. The output voltage isspecified with the amplitude of phase voltage being assigned unity value.

SYNCHRONIZING SIGNALS

To vary the output voltage, it is necessary to vary the firing angle. In order to varythe firing angle, one commonly used technique is to establish a synchronizingsignal for each SCR. It has been seen that zero degree firing angle occurs 30o

degrees after the zero-crossing of the respective phase voltage. If thesynchronizing signal is to be a sinusoidal signal, it should lag the respective phaseby 30o and then the circuitry needed to generate a firing signal can be similar tothat described for single-phase. Instead of a single such circuit for a single phaserectifier, we would need three such circuits.

When the 3-phase source supply connected to the rectifier is star-connected, theline voltages and the phase voltages have a 30o phase angle difference betweenthem, as shown below.

The line voltage can also be obtained as:

This line voltage lags the R-phase voltage by30o and has an amplitude which is1.732 times the amplitude of the phase voltage. The synchronizing signal forSCR S1 can be obtained based on vRB line voltage. The synchronizing signals forthe other SCRs can be obtained in a similar manner.

To get the synchronizing signals, three control transformers can be used, with theprimaries connected in delta and the secondaries in star, as shown below.

For S1, voltage vS1 is used as the synchronizing signal. Voltage vS2 is used as thesynchronizing signal for SCR S2 and so on. The waveforms presented by thesynchronizing signals are as shown below. The waveforms do not show the effectof turns ratio, since any instantaneous value has been normalized with respect toits peak value. For example, let the primary phase voltage be 240 V and then itspeak value is 339.4 V. The primary voltage is normalized with respect to 339. V. If the peak voltage of each half of secondary is 10 V, the secondary voltage arenormalized with respect to 10 V.

2. Explain the operation of 3 phase half controlled bridge rectifier or semi converter(N/D12)

3. Explain the functional modes of dual converter with necessary diagrams(M/J13))(N/D 13)(N/D14)(M/J15) )(N/D16)

A separately-excited dc shunt motor can be operated in either direction ineither of the two modes, the two modes being the motoring mode and theregenerating mode. It can be seen that the motor can operate in any of thefour quadrants and the armature of the dc motor in a fast four-quadrant driveis usually supplied power through a dual converter. The dual converter canbe operated with either circulating current or without circulating current. Ifboth the converters conduct at the same time, there would be circulatingcurrent and the level of circulating current is restricted by an inductor. It ispossible to operate only one converter at any instant, but switching from oneconverter to the other would be carried out after a small delay. This pagedescribes the operation of a dual converter operating without circulatingcurrent. As shown in Fig. 1, the motor is operated such that it can delivermaximum torque below its base speed and maximum power above its basespeed. To control the speed below its base speed, the voltage applied to thearmature of motor is varied with the field voltage held at its nominal value. Tocontrol the speed above its base speed, the armature is supplied with itsrated voltage and the field is weakened. It means that an additional single-phase controlled rectifier circuit is needed for field control. Closed-loopcontrol in the field-weakening mode tends to be difficult because of therelatively large time constant of the field.

The power circuit of the dual-converter dc drive is shown in Fig. 2.

Each converter has six SCRs. The converter that conducts for forwardmotoring is called the positive converter and the other converter is called thenegative converter. Instead of naming the converters as positive converterand negative converter, the names could have been forward and reverseconverters. The field is also connected to a controlled-bridge in order to bringabout field weakening. The circuit shown above can be re-drawn as shown in Fig. 3. Usually aninductor is inserted in each line as shown in Fig. 3 and this inductor reducesthe impact of notches on line voltages that occur during commutation overlap.

CIRCUIT OPERATIONThe operation of the circuit in the circulating-current free mode is not verymuch different from that described in the previous pages. In order to drive themotor in the forward direction, the positive converter is controlled. To controlthe motor in the reverse direction, the negative converter is controlled. Whenthe speed of motor is to be changed fast from a high value to a low value inthe forward direction, the conduction has to switch from the positive converterto the negative converter. Then the direction of current flow changes in themotor and it regenerates, feeding power back to the source. When the speedis to be reduced in the reverse direction, the conduction has to switch fromthe negative converter to the positive converter. It is seen that conduction hasto switch from one converter to the other when the direction of motor rotationis to change, so that regeneration can occur. During regeneration, thedirection torque developed by motor is opposite to that of the motoringtorque. Thus the regenerating torque acts as the breaking torque and themotor decelerates fast. At the instant when the switch from one converter to the other is to occur, itwould be preferable to ensure that the average output voltage of eitherconverter is the same. Let the firing angle of the positive converter be aP, andthe firing angle of the negative converter beaN . If the peak line voltage be U,then equation (1) should apply. Equation (1) leads to equation (2). Then thesum of firing angles of the two converters isp, as shown in equation (3).

In a dual-converter, the firing angles for the converter are changed accordingto equation (3). But it needs to be emphasized that only one converteroperates at any instant.

When the speed of the motor is to be increased above its base speed, thevoltage applied to the armature is kept at its nominal value and the phase-angle of the single phase bridge is varied such that the field current is set to avalue below its nominal value. If the nominal speed of the motor is 1500 rpm,then the maximum speed at which it can run cannot exceed a certain value,say 2000 rpm. Above this speed, the rotational stresses can affect thecommutator and the motor can get damaged. Next it is shown how the operation of motor can be represented by means ofa block diagram. This approach can be helpful in designing the closed-loopsystem.

4.(a)Explain the operation of single phase half bridge rectifier with RL load andfreewheeling diode.(M/J13)

Freewheeling diodes are added to improve the load current waveform

Thyristor T is turned ON at angle ωt = α

At ωt = π, Vs is 0

At ωt > π, Vo is negative causing D to turn ON. Thyristor T turns off

Load current is transferred to diode

Load current goes down from ωt = π to ωt = 2π + α

Thyristor T conducts from (ωt = α) to (ωt = π), (ωt = 2 π +α) to (ωt = 3π)and so on..

During these periods Diode D is reverse biased

5. Explain the operation of single phase fully controlled rectifier with R load with neatwaveforms in both rectification and inversion mode (N/D 15)(M/J 15) ) (M/J16)

It shows the circuit diagram of a single phase fully controlled half wave rectifiersupplying apurely resistive load. At ωt = 0 when the input supply voltage becomes positive thethyristor Tbecomes forward biased. However, unlike a diode, it does not turn ON till a gatepulse isapplied at ωt = α. During the period 0 < ωt ≤ α, the thyristor blocks the supply voltageand theload voltage remains zero as shown in fig 10.1(b). Consequently, no load currentflows duringthis interval. As soon as a gate pulse is applied to the thyristor at ωt = α it turns ON.Thevoltage across the thyristor collapses to almost zero and the full supply voltageappears acrossthe load. From this point onwards the load voltage follows the supply voltage. Theload beingpurely resistive the load current io is proportional to the load voltage. At ωt = π as thesupplyvoltage passes through the negative going zero crossing the load voltage and hencethe loadcurrent becomes zero and tries to reverse direction. In the process the thyristorundergoesreverse recovery and starts blocking the negative supply voltage. Therefore, the loadvoltageand the load current remains clamped at zero till the thyristor is fired again at ωt = 2π+ α. Thesame process repeats thereafter.

6. A single phase bridge converter is utilized to produce regulated dc output voltage. The input voltage is 230v and the load current is 8A for a firing and of 30 degree.[1] Calculate the dc output voltage [2] Calculate the dc output voltage and current if a freewheeling diode is used at the output for the same firing angle(June 2013)

7. Discuss the effect of series inductance on the performance of single phase full converter indicating clearly the conduction of various thyristors during one cycle. (M/J13)(M/J14)

Fig shows a single phase fully controlled converter with source inductance. For

simplicity it has been assumed that the converter operates in the continuous conduction mode. Further, it has been assumed that the load current ripple is negligible and the load can be replaced by a dc current source the magnitude of which equals the average load current. Fig. 15.1(b) shows the corresponding waveforms. It is assumed that the thyristors T3 and T4 were conducting at t = 0. T1 and T2 are fired at ωt = α. If there were no source inductance T3 and T4 would have commutated as soon as T1 and T2 are turned ON. The input current polarity would have changed instantaneously. However, if a source inductance is present the commutation and change of input current polarity can not be instantaneous. Therefore, when T1 and T2 are turned ON T3 T4 does not commutate immediately. Instead, for some interval all four thyristors continue to conduct as shown in Fig. 15.1(b). This interval is called “overlap” interval.

During this period the load current freewheels through the thyristors and the output voltage is clamped to zero. On the other hand, the input current starts changing polarity as the current through T1 and T2 increases and T3 T4 current decreases. At the end of the overlap interval the current through T3 and T4 becomes zero and they commutate, T1 and T2 starts conducting the full load current. The same process repeats during commutation from T1 T2 to T3T4 at ωt = π + α.

From Fig. 15.1(b) it is clear that, commutation overlap not only reduces average output dc voltage but also reduces the extinction angle γ which may cause commutation failure in the

inverting mode of operation if α is very close to 180º. In the following analysis an expression of the overlap angle “μ” will be determined.

8. Describe the working of a single phase full converter in the rectifier mode with RL Load. Discuss how one pair of SCRs is commutated by an incoming pair of SCRs. Illustrate your answer with the waveforms of source voltage, load voltage and source current. Assume continuous conduction. Also derive the expressions for average and rms output voltage. (M/J 13) )(N/D 16)

Fig shows the circuit diagram of a single phase fully controlled bridge converter. It is one of the most popular converter circuits and is widely used in the speed control of separately excited dc machines. Indeed, the R–L–E load shown in this figure may represent the electrical equivalent circuit of a separately excited dc motor. The single phase fully controlled bridge converter is obtained by replacing all the diode of the corresponding uncontrolled converter by thyristors. Thyristors T1 and T2 are fired together while T3 and T4 are fired 180º after T1 and T2. From the circuit diagram of Fig 10.3(a) it is clear that for any load current to flow at least one thyristor from the top group (T1, T3) and one thyristor from the bottom group (T2, T4) must conduct. It can also be argued that neither T1T3 nor T2T4 can conduct simultaneously. For example whenever T3 and T4 are in the forward blocking state and a gate pulse is applied to them, they turn ON and at the same time a negative voltage is applied across T1 and T2 commutating them immediately. Similar argument holds for T1 and T2. For the same reason T1T4 or T2T3 can not conduct simultaneously. Therefore, the only possible conduction modes when the current i0 can flow are T1T2 and T3T4. Of coarse it is possible that at a given moment none of the thyristors conduct. This situation will typically occur whenthe load current becomes zero in between the firings of T1T2 and T3T4. Once the load currentbecomes zero all thyristors remain off. In this mode the load current remains zero. Consequently the converter is said to be operating in the discontinuous conduction mode. Fig 10.3(b) shows the voltage across different devices and the dc output voltage during each of these conduction modes. It is to be noted that whenever T1 and T2 conducts, the voltage across T3 and T4 becomes –vi. Therefore T3 and T4 can be fired only when vi is negative i.e, over the negative half cycle of the input supply voltage. Similarly T1 and T2 can be fired only over the positive half cycle of the input supply. The voltage across the devices when none of the thyristors conduct depends on the off state impedance of each device. The values listed in Fig 10.3 (b) assume identical devices. Under normal operating condition of the converter the load current may or may not remain zero over some interval of the input voltage cycle. If i0 is always greater than zero then the converter is said to be operating in the continuous conduction mode. In this mode of operation of the converter T1T2 and T3T4 conducts for alternate half cycle of the input supply.

However, in the discontinuous conduction mode none of the thyristors conduct over some portion of the input cycle. The load current remains zero during that period.

• In three phase dual converter, we make use of three phase rectifier which converts 3 phase A.C. supply to D.C. The rest of the process is same and same elements are used. The output of three phase rectifier is fed to filter and after filtering the pure D.C. is fed to load. At last the supply from load is given to last bridge that is inverter. It do the Invert process of rectifier and converts D.C. into 3 phase A.C. which appears at output.

Dual converter with circulating current• In this mode of operation, both the converters are turned on simultaneously, the firing

angles are adjusted by the control voltage Vc in such a way that (α1 + α2) = 180º• The average voltage produced by both the converters will be same, but their

instantaneous values will be different. The circulating current will flow due to this potential difference.

• The circulating current is limited by a circulating current reactor. Advantages

• Both the converters remain in conduction continually, irrespective of whether the load current is continuous or not..

• The dual converter responds fast if the converters are in continuous conduction.Disadvantage The disadvantage is that reactor has to be used. The reactors are costly and bulky at high power levels.

OPERATION

• The operation of the dual converter with circulating current is explained with the following assumptions : 1. The rector is lossless. 2. The firing angles of the two converters are controlled so that their sum is 180º. [ (α1+ α2)=180º].

• The waveforms for the circulating current mode are as shown in fig.(5) The firing angle for the converter I is α1=60º, therefore firing angle for the second converter α2=180-60=120º.

• Because of the circulating current, both the converters are kept in the on state, at the no load and with load conditions. Hence the voltage waveforms are well defined, as shown in fig.(5)

• The supply voltage and the converter output voltages are as shown in fig.(5) The instantaneous dc output voltage is the average of the instantaneous converter voltages.

• This output voltage has a shape different from that of the converter output voltages. But the average value of the output voltage is the same as the average value of individual converter voltage.

The voltage across the circulating current reactor, as shown in fig.(5) is the different between the instantaneous converter output voltages(v01-v02).Dual Converter without circulating current mode

• In a dual converter without circulating current operating mode, the flow of circulating current is completely inhibited through automatic control of the firing pulses, so that onlythat converter which carries the load current is in conduction and the other converter is temporarily blocked. Since only one converter operates at a time and the other is in blocking state, no reactor is required between the converters.

At a particular instant, suppose converter 1 is operating as a rectifier and is supplying the load current while pulses to second converter are blocked. For the inversion operation, converter 1 is first blocked by removing its firing pulses and load current is reduced to zero. Converter 2 is made to conduct by applying the firing pulses to it. The current in converter 2 would now build up through the load in the reverse direction. So long as converter 2 is in operation, converter 1 is in the blocking state since the firing pulses are withdrawn from it.