OPERATION of a 3 Phase Rectifier

7/27/2019 OPERATION of a 3 Phase Rectifier

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OPERATION OF A 3-PHASE FULLY-CONTROLLED RECTIFIER

CIRCUIT OPERATION

SYNCHRONIZING SIGNALS

MATHEMATICAL ANALYSIS

SIMULATION MATHCAD SIMULATION

CIRCUIT OPERATION

The operation of a 3-phase fully-controlled bridge rectifier circuit is described in this page. Athree-phase fully-controlled bridge rectifier can be constructed using six SCRs as shown below.

The three-phase bridge rectifier circuit has three-legs, each phase connected to one of the three

phase voltages. Alternatively, it can be seen that the bridge circuit has two halves, the positivehalf consisting of the SCRs S 1, S3 and S 5 and the negative half consisting of the SCRs S 2,S4 and S 6. At any time, one SCR from each half conducts when there is current flow. If the

phase sequence of the source be RYB, the SCRs are triggered in the sequence S 1, S2 , S3 , S 4,S5 , S 6 and S 1 and so on.

The operation of the circuit is first explained with the assumption that diodes are used in placeof the SCRs. The three-phase voltages vary as shown below.

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

phase voltages when is in the range from 150 o to 270 o and that B-phase voltage is the highestof the three-phase voltages when is in the range from 270 o to 390 o or 30o in the next cycle.We also find that R-phase voltage is the lowest of the three-phase voltages when is in the

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range from 210 o to 330 o. It can also be seen that Y-phase voltage is the lowest of the three- phase voltages when is in the range from 330 o to 450 o or 90o in the next cycle, and that B- phase voltage is the lowest when is in the range from 90 o to 210 o. If diodes are used, diodeD1 in place of S 1 would conduct from 30 o to 150 o, diode D 3

would conduct from 150 o to 270 o and diode D 5 from 270 o to 390 o or 30o in the next cycle. In the same way, diode D 4

wouldconduct from 210 o to 330 o, diode D 6

from 330 o to 450 o or 90o in the next cycle, and diodeD2

would conduct from 90 o to 210 o. The positive rail of output voltage of the bridge isconnected to the topmost segments of the envelope of three-phase voltages and the negative railof the output voltage to the lowest segments of the envelope.

At any instant barring the change-over periods when current flow gets transferred from diode toanother, only one of the following pairs conducts at any time.

Period, range of Diode Pair in conduction

30o to 90 o D1 and D 6

90o to 150 o D1 and D 2

150o

to 210o D2 and D 3

210 o to 270 o D3 and D 4

270 o to 330 o D4 and D 5

330 o to 360 o and 0 o to 30 o D5 and D 6

If SCRs are used, their conduction can be delayed by choosing the desired firing angle. Whenthe SCRs are fired at 0 o firing angle, the output of the bridge rectifier would be the same as thatof the circuit with diodes. For instance, it is seen that D 1 starts conducting only after = 30 o.

In fact, it can start conducting only after = 30o

, since it is reverse-biased before = 30o

. The bias across D 1 becomes zero when = 30 o and diode D 1 starts getting forward-biased onlyafter =30 o. When v R ( ) = E*Sin ( ), diode D 1 is reverse-biased before = 30 o and it isforward-biased when 30o. When firing angle to SCRs is zero degree, S 1 is triggeredwhen = 30 o. This means that if a synchronizing signal is needed for triggering S 1, that signalvoltage would lag v R ( ) by 30 o and if the firing angle is , SCR S 1 is triggered when = +30o. Given that the conduction is continuous, the following table presents the SCR pair inconduction at any instant.

Period, range of SCR Pair in conduction

+ 30 o to + 90 o S1 and S 6

+ 90 o to + 150 o S1 and S 2

+ 150 o to + 210 o S2 and S 3

+ 210 o to + 270 o S3 and S 4

+ 270 o to + 330 o S4 and S 5

+ 330 o to + 360 o and + 0 o to + 30 o S5 and S 6



The operation of the bridge-rectifier is illustrated with the help of an applet that follows this

paragraph. You can set the firing angle in the range 0 o < firing angle < 180 o and you can setthe instantaneous angle also. The applet displays the SCR pair in conduction at the choseninstant. The current flow path is shown in red colour in the circuit diagram. Theinstantaneous angle can be either set in its text-field or varied by dragging the scroll-bar

button. The rotating phasor diagram is quite useful to illustrate how the circuit operates. Once

the firing angle is 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. One way tovisualize is to imagine two brushes which are 120 o wide and the device in the phase connectedto the brush conducts. The brush that has "Firing angle " written beside it acts as the brushconnected to the positive rail and the other acts as if it is connected to the negative rail. Thisdiagram illustrates how the rectifier circuit acts as a commutator and converts ac to dc. Theoutput voltage is specified 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 vary the firingangle, one commonly used technique is to establish a synchronizing signal for each SCR. Ithas been seen that zero degree firing angle occurs 30 o degrees after the zero-crossing of therespective phase voltage. If the synchronizing signal is to be a sinusoidal signal, it should lagthe respective phase by 30 o and then the circuitry needed to generate a firing signal can besimilar to that 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, the line voltagesand the phase voltages have a 30 o phase angle difference between them, as shown below.

The line voltage can also be obtained as:

This line voltage lags the R-phase voltage by30 o and has an amplitude which is 1.732 times theamplitude of the phase voltage. The synchronizing signal for SCR S 1 can be obtained based



on v RB line voltage. The synchronizing signals for the other SCRs can be obtained in a similar manner.

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

For S 1, voltage v S1 is used as the synchronizing signal. Voltage v S2 is used as thesynchronizing signal for SCR S 2 and so on. The waveforms presented by the synchronizingsignals are as shown below. The waveforms do not show the effect of turns ratio, since anyinstantaneous value has been normalized with respect to its peak value. For example, let the

primary phase voltage be 240 V and then its peak value is 339.4 V. The primary voltage isnormalized with respect to 339. V. If the peak voltage of each half of secondary is 10 V, thesecondary voltage are normalized with respect to 10 V.


Analysis of this three-phase controlled rectifier is in many ways similar to the analysis of single-phase bridge rectifier circuit. We are interested in output voltage and the sourcecurrent. The average output voltage, the rms output voltage, the ripple content in outputvoltage, the total rms line current, the fundamental rms current, THD in line current, thedisplacement power factor and the apparent power factor are to be determined. In this

section, the analysis is carried out assuming that the load current is a steady dc value.AVERAGE OUTPUT VOLTAGE

Before getting an expression for the output voltage, it is preferable to find out how the outputvoltage waveform varies as the firing angle is varied. In one cycle of source voltage, six pairsconduct, each pair for 60 o. This means that the period for output waveform is one-sixth of the

period of line voltage. The output waveform repeats itself six times in one cycle of inputvoltage. The waveform of output voltage can be determined by considering one pair. It is



seen that when v R ( ) = E* Sin ( ), SCR S1 and S6 conduct when varies from 30 o+ to90o + , where is the firing angle. Then

The waveform of output can be plotted for different firing angles. The applet below takes inthe firing angle as an input and plots the output. The peak line-to-line voltage is marked as 'U'and the applet starts with the instant an SCR is fired and displays the output waveform for oneinput cycle period.

The average output voltage of the bridge circuit is calculated as follows, with a change in

variable, where = + 60 o.

In the expression above, U is the peak line-to-line voltage, whereas E is the amplitude of phasevoltage of 3-phase supply.

RMS OUTPUT VOLTAGE

The rms output voltage is calculated as follows:

The ripple factor of the output voltage is then:

The applet below displays the average output voltage, the rms output voltage and the ripplefactor for the case of continuous conduction through the load.

It is seen that the average output voltage is negative when firing angle exceeds 90 o. It meansthat power flow is from the dc side to the ac source. When the firing angle is kept in the region



0o < < 90 o, this circuit is said to be operating in the rectifi er region . When the firing angle iskept in the region 90 o < < 180 o, this circuit is said to be operating in the in ver ter r egion .When the circuit operates in the rectifier region, the net power flow is from the ac source to thedc link. In the inverter region, the net power flow is in the reverse direction. To operate in theinverter region, it is necessary to have a dc source present in the dc link which can provide the

power that is fed back to the ac source.

RMS LINE CURRENT

The rms value of line current is relatively easy to find out if dc link current is ripple-free andsteady. The load current is ripple-free if inductance in the dc link is relatively large. Tomaintain load current at any firing angle, it is assumed that the dc link contains a voltagesource. Given that the resistance of the load circuit is zero, the voltage source should equal theaverage output voltage of the bridge circuit, if dc link current remains steady at some value.The waveforms shown below are based on the assumption that these conditions are met. It has

been shown that if v R ( ) = E*Sin ( ), SCR S 1 conducts when varies from + 30 o to +90o and that SCR S 4 conducts when varies from + 210 o to + 270 o. If the amplitude of dc load current is assigned to be unity, the line current waveform is then a rectangular pulse,remaining at + 1 from + 30 o to + 150 o, at - 1 from + 210 o to + 330 o, and zeroelsewhere. The amplitude of fundamental in line current is then 3.464/ which evaluates tonearly 0.78) and the amplitude of other odd harmonics is 3.464/n , where n is the oddharmonic number. When dc load current is steady and has a magnitude of unity, the rms linecurrent is obtained as shown in equation (5). The rms value of the fundamental is obtained asshown in equation (6). Equation (6) is based on how trigonometric Fourier coefficients aredefined for waveforms with quarter-wave symmetry. When the line current is a rectangular andsymmetric, the phase current is the same as the line current and the fundamental component of

phase current lags the phase voltage by an angle equal to the firing angle. Hence thedisplacement power factor is expressed as shown by equation (7). Since the line current is notsinusoidal, the apparent power factor, usually referred to just as the power factor in most of thetexts, is less than DPF and is represented by equation (8). Since the line current is notsinusoidal, the distortion component in the line current has to be computed. This component,called the THD( Total Harmonic Distortion ), is calculated as shown in equation (9).



SIMULATION

The applet displayed below simulates the circuit in the form of animation. The only parameter to be set is the firing angle and the program can be run for a single-cycle or stepped through.You can click on Pause button to pause and view the display.



OPERATION WITH A RESISTIVE LOAD

CIRCUIT OPERATION


SIMULATION

PSPICE SIMULATION SUMMARY

CIRCUIT OPERATION

The three-phase fully-controlled bridge rectifier with a purely resistive load is shown above.With any single-controlled rectifier circuit, the load current is unidirectional. Hence when theoutput voltage of the bridge tends to become negative, the load current is zero and conductionis discontinuous, with the output voltage clamped to zero volts. The operation of this circuithas been described in detail in the previous page.


Since the load current tends to be discontinuous, two expressions for the average output voltagecan be derived, one for the continuous mode and the other for the discontinuous mode. In thecontinuous mode, the average output voltage is as shown in equation (1). The conduction

becomes discontinuous when the firing angle exceeds 60 o and remains less than 120 o. Then theaverage voltage is obtained as shown in equation (2).







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In equations (1) and (2), the amplitude of phase voltage is designated as E and the amplitude of line voltage is designaed as U.

The rms voltage is computed as follows. When < 60 o, the conduction is continuous and theexpression for the rms voltage is presented in equation (3), whereas equation (4) expresses therms voltage obtained when firing angle > 60 o.

The ripple factor can be found out as defined in the previous page.

The line/phase current can be defined as follows. Let R-phase voltage be defined to be

vR ( ) = E*Sin ( ).

Then the R-phase current i R ( ) is defined as follows, when the conduction is continuous.

When the conduction is discontinuous,



From this definition of phase current, the rms line current, the rms of the fundamental in linecurrent and its THD can be found out.

The applet below displays the average output voltage, the rms output voltage, its ripple factor,the rms line current, the fundamental rms content in line current and its THD as a function of firing angle. The peak average output voltage, (3U/ ), is taken to be unity and (3U/ R) is setequal to unity, where R is the load resistor.

SIMULATION

The applet shown below displays the source voltage, the output voltage, the current in R-phaseand the voltage across SCR S 1. In addition, the relevant statistical details are also displayed.To run the applet, key-in the firing angle and then click on Start Button.

PSPICE SIMULATION

The Pspice program for simulation of this three-phase rectifier circuit is presented below. Themodel used for the SCRs is the same as defined for the single-phase fully-controlled bridgerectifier. The three-phase brdige rectifier contains six SCRs and it is necessary to define six

pulse sources, one for each SCR. The pulse sources have been defined for a firing angle of 30o and the frequency of the three-phase source is 50 Hz. At any time two SCRs need toconduct, one from the top half and another bottom half and hence only if one SCR is triggeredat a time, conduction may never get established. To overcome this problem, two SCRs are

triggered at the same time. For example, when SCR S 2 is to be triggered, SCR S 1is alsotriggered. In the same way, when SCR S 3 is to be triggered, SCR S 2 is also triggered and soon. In order to effect this in program, one voltage-controlled voltage source is defined for eachSCR. The dependent source defined for SCR S 1 is dependent on two sources, the pulse sourcethat defines when SCR S 1 is to be triggered and the pulse source that defines when SCR S 2 is to

be triggered.

* Three-phase Full-wave Fully-Controlled Bridge RectifierVA 1 0 SIN(0 340V 50Hz)



VB 2 0 SIN(0 340V 50Hz 0 0 -120)VC 3 0 SIN(0 340V 50Hz 0 0 -240)XT1 1 4 11 4 SCRXT3 2 4 13 4 SCRXT5 3 4 15 4 SCRXT4 5 1 14 1 SCRXT6 5 2 16 2 SCR

XT2 5 3 12 3 SCRRP 4 0 100KRN 5 0 100KR1 4 5 10

VP1 21 0 PULSE(0 10 3333.3U 1N 1N 100U 20M)VP2 22 0 PULSE(0 10 6666.7U 1N 1N 100U 20M)VP3 23 0 PULSE(0 10 10M 1N 1N 100U 20M)VP4 24 0 PULSE(0 10 13333.3U 1N 1N 100U 20M)VP5 25 0 PULSE(0 10 16666.7U 1N 1N 100U 20M)

VP6 26 0 PULSE(0 10 0M 1N 1N 100U 20M)RP1 21 0 100KRP2 22 0 100KRP3 23 0 100KRP4 24 0 100KRP5 25 0 100KRP6 26 0 100KEP1 11 4 poly(2) (21,0) (22,0) 0 1 1EP2 12 3 poly(2) (22,0) (23,0) 0 1 1EP3 13 4 poly(2) (23,0) (24,0) 0 1 1

EP4 14 1 poly(2) (24,0) (25,0) 0 1 1EP5 15 4 poly(2) (25,0) (26,0) 0 1 1EP6 16 2 poly(2) (26,0) (21,0) 0 1 1

* Subcircuit for SCR.SUBCKT SCR 101 102 103 102S1 101 105 106 102 SMODRG 103 104 50VX 104 102 DC 0

VY 105 107 DC 0DT 107 102 DMODRT 106 102 1CT 106 102 10UF1 102 106 POLY(2) VX VY 0 50 11.MODEL SMOD VSWITCH(RON=0.0105 ROFF=10E+5 VON=0.5 VOFF=0).MODEL DMOD D((IS=2.2E-15 BV=1200 TT=0 CJO=0).ENDS SCR



.TRAN 10US 60.0MS 0.0MS 10US.PROBE.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV).END

The responses obtained for a load resistance of 10 are presented below.

The Output Voltage Waveform

The Line Current(phase A) Waveform



The Voltage(SCR S 1) Waveform

SUMMARY

This page has described the operation of the three-phase fully-controlled bridge rectifier with aresistive load. The next page describes how this rectifier functions when the load contains aninductor also.



OPERATION WITH RL LOAD

CIRCUIT OPERATION


SIMULATION

PSPICE SIMULATION

MATHCAD SIMULATION

SUMMARY

CIRCUIT OPERATION

The circuit of a three-phase fully-controlled bridge rectifier circuit with an RL load and avoltage source has been shown above. The purpose of providing a dc source in the dc link is to

illustrate how two-quadrant operation can take place. When the firing angle is held between0o and 90 o, the circuit operates in the rectifier region. When the firing angle is held between90o and 180 o, the circuit can operate in the inverter region if the source E is sufficientlynegative. In the inverter region, power flow is from the dc link to the ac 3-phase source. Theinductor reduces ripple in dc link current.


When current flow in the dc link is continuous, the average output voltage can be calculated asoutlined in the previous page. It is difficult to estimate what the average output would be if there is a source present in the dc link and the conduction is discontinuous. If there be nosource in the dc link, the average output voltage for discontinuous conduction is expressed byequation (1). The analysis of the circuit is along the lines described for the single-phasecontrolled rectifier circuit. In order to get an expression for the line/phase current, it isnecessary to get an expression for the load current. The expression for load current is obtainedfrom the expression for output voltage. The output voltage is described by equation (2).

The differential equation that describes the load current is expressed by equation (3). Thesolution is of the form expressed by equation (4). The impedance of load is Z and the loadangle is . They are defined as shown in equation (5).





















In the equation above, w is the angular frequency in radians/second corresponding to thesource frequency. When the load current is continuous, equation (6) is valid. Using equation(6), equation (7) is obtained. Solving for A, we get equation (8). When the conduction isdiscontinuous, i L(0) = 0 and then A is evaluated as shown in equation (9).



Once the value of A is known, the load current can be found out. From the load current, anexpression for the line current can be obtained. When SCR S 1 is ON, the line current equalsthe load current. The line current is the negative of load current when SCR S 4 is ON, and it iszero when neither S 1 nor S 4 is ON. From the expression of the load current, the rms value of line current, the rms value of the fundamental component of line current, the THD in linecurrent, the harmonic spectrum of line current, the DPF and the apparent power factor can bedetermined as outlined in the earlier pages.

SIMULATION

The parameters to be keyed in are the firing angle, the value of source in the dc link (between1.0 and -1.0 preferably) and the ratio of load reactance to load resistance. The load reactanceis wL, where w is the angular frequency in rad/s corresponding to the ac source frequency.Click on Start Button for the program to respond.

PSPICE SIMULATION

* Three-phase Full-wave Fully-Controlled Bridge RectifierVA 1 0 SIN(0 340V 50Hz)VB 2 0 SIN(0 340V 50Hz 0 0 -120)VC 3 0 SIN(0 340V 50Hz 0 0 -240)XT1 1 4 11 4 SCRXT3 2 4 13 4 SCRXT5 3 4 15 4 SCRXT4 5 1 14 1 SCR



XT6 5 2 16 2 SCRXT2 5 3 12 3 SCRRP 4 0 100KRN 5 0 100KR1 4 6 10L1 6 5 31.8M

VP1 21 0 PULSE(0 10 3333.3U 1N 1N 100U 20M)VP2 22 0 PULSE(0 10 6666.7U 1N 1N 100U 20M)VP3 23 0 PULSE(0 10 10M 1N 1N 100U 20M)VP4 24 0 PULSE(0 10 13333.3U 1N 1N 100U 20M)VP5 25 0 PULSE(0 10 16666.7U 1N 1N 100U 20M)VP6 26 0 PULSE(0 10 0M 1N 1N 100U 20M)RP1 21 0 100KRP2 22 0 100KRP3 23 0 100KRP4 24 0 100K

RP5 25 0 100KRP6 26 0 100KEP1 11 4 poly(2) (21,0) (22,0) 0 1 1EP2 12 3 poly(2) (22,0) (23,0) 0 1 1EP3 13 4 poly(2) (23,0) (24,0) 0 1 1EP4 14 1 poly(2) (24,0) (25,0) 0 1 1EP5 15 4 poly(2) (25,0) (26,0) 0 1 1EP6 16 2 poly(2) (26,0) (21,0) 0 1 1

* Subcircuit for SCR.SUBCKT SCR 101 102 103 102S1 101 105 106 102 SMODRG 103 104 50VX 104 102 DC 0VY 105 107 DC 0DT 107 102 DMODRT 106 102 1CT 106 102 10UF1 102 106 POLY(2) VX VY 0 50 11

.MODEL SMOD VSWITCH(RON=0.0105 ROFF=10E+5 VON=0.5 VOFF=0)

.MODEL DMOD D((IS=2.2E-15 BV=1200 TT=0 CJO=0)

.ENDS SCR

.TRAN 10US 60.0MS 0.0MS 10US

.PROBE

.FOUR 50 V(4,5) I(VA) I(L1)

.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV)



.END

The results of the Fourier series analysis are presented below.

The Fourier Series Spectrum of the output voltage

The Fourier Series Spectrum of the Line Current (Phase A)



The Fourier Series Spectrum of the Load Current

It can be seen that the lowest harmonic frequency in the output voltage is at 300 Hz, the sixthharmonic, whereas there is hardly any harmonic present in the load current because of therelatively large inductance in the load circuit. On the other hand, the 5-th and the 7-thharmonics are visibly high in the line current.



CIRCUIT OPERATION


SIMULATION

MATHCAD SIMULATION

SIMULATION USING C

SUMMARY

CIRCUIT OPERATION

The circuit of a three-phase fully-controlled bridge rectifier with source inductance is presentedabove. The presence of source inductance introduces an additional mode of operation when thefiring angle is less than a certain value. Let us assume that SCRS S 1 and S 2 are in conductionwhen SCR S 3 is triggered. Then current from the 3-phase supply does not transfer from S 1 toS3 instantaneously, and the transfer of current, called commutation, takes a while. During thiscommutation overlap, both S 1 and S 3 conduct in addition to S 2. SCR S 1 continues to conduct tillthe current through S 3 rises to equal the dc link current.The effects of commutation overlap are:









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i. A slight reduction in output voltage,ii. A notch in the supply voltage to the circuit during commutation overlap.

When the source has inductance, other loads connected to this source along with the controlledrectifier are supplied voltages with notches in them and some of these loads can be sensitive tothese notches and they may operate improperly. Hence in order to reduce the magnitude of notches, it is mandatory in some countries for the rectifier to be provided with an inductance in

series with each of its three-phase input lines. If these inductors are much larger than the sourceinductance, the notch voltages are absorbed by these inductances and the other loads connectedto the same 3-phase source are not supplied with distorted voltages. The internal inductancesconnected in series with the source are sometimes referred to as 4% inductances. If the inductor is such that the voltage drop across it is about 4% of the phase voltage at rated current, it isnormally sufficient to reduce the notches at the source terminals to an acceptable level.


When there is an inductor in series with each input line, it is necessary to find out its effect. Weneed to find out:

a. The reduction in output voltage. b. The duration of commutation overlap.c. The relationship between the firing angle and the commutation overlap.

REDUCTION IN OUTPUT VOLTAGE

Calculations by hand are carried out assuming that the dc link current remains steady withoutany ripple. The source voltages at its terminals and the output voltage appear as shown below,

assuming that the inductances belong to the source.

It is seen that there are six notches in one input cycle. The reduction in average output voltagecan be found out as follows. Let SCR S 1 be in conduction and let S 3 be triggered. Let currentthrough the dc link be I DC . Then current through Y-phase has to rise from zero to I DC , whereascurrent through R-phase has to fall from I DC to zero. On the other hand, loop currentiLOOP marked in the sketch below has to rise from zero to I DC. This means that duringcommutation current through Y-phase would rise from zero to I DC and the volt-second area theoutput misses out is L 2IDC , that absorbed by the inductor in the Y-phase.



From the volt-seconds lost per commutation, we can find the total volt-seconds lost in oneinput cycle. Since there are six commutations per cycle, the total volt-seconds lost per cycle isexpressed as shown by equation (1). Dividing this area by the time corresponding to one cycle,we get the average voltage reduction in output. The time corresponding to one input cycle is1/f, where f is the line frequency. Then the average reduction in output voltage is obtained asshown in equation (2). In equation (2), we make use of the relation that the angular frequency,w = 2 f. It is to be noted that commutation overlap occurs only when there is continuousconduction through the load and the average output voltage is expressed by equation (3). Inequation (3), U is the peak line-to-line voltage and is the firing angle.

COMMUTATION OVERLAP ANGLE

The commutation of commutation overlap depends on:



a. the firing angle, b. the dc link current andc. the source inductance or the inductance in series with each phase.

To find out the commutation overlap, it is sufficient to analyse one commutation. Let SCR S5 be in conduction and let SCR S 1 be triggered at a firing angle of . As seen earlier, the loopcurrent i LOOP builds up from zero to I DC . This means that the current in the inductance in R-

phase builds up to I DC , whereas it decreases to zero in the inductance in the B-phase. Let thecommutation last for an angle . Then during commutation, the voltage across the sourceinductance is expressed as shown in equation (4). It is assumed that the three phase voltages aredefined as follows:

vR = E. Sin ( ), v Y = E. Sin ( ), and v B = E. Sin ( ).

The amplitude of phase voltage is denoted by 'E' and the amplitude of line voltage is denoted by 'U'. In equation (4), the loop current is denoted as ‘i’. Since = wt, the equation for commutation overlap can be represented as shown in equation (5). During this interval, theloop current changes from zero to I DC . Hence equation (6) defines how current in R phasechanges.

The solution of equation (6) is presented in equation (7). Equation (7) can be re-arranged and presented as shown in equation (8). Solving for , we obtain equation (9).



It is seen that overlap angle increases

a. as the firing angle moves closer to either 0 o or 180 o , b. as the dc link current becomes larger, andc. as the source inductance gets larger.

The above equation has been obtained based on the assumption that the dc link current remainssteady, which happens only when the dc link inductance is relatively large. In practice, it is nottrue and hence the above equation yields only an approximate result.

The rest of mathematical analysis follows the familiar route. The items of interest are:

a. RMS output voltage of bridge circuit, b. Average output voltage of bridge circuit,c. Ripple Factor of output voltage of the bridge,d. RMS output voltage/voltage across load resistor,e. Average output voltage (across load resistor),f. Ripple factor of output voltage (across load resistor),g. RMS line current,h. RMS value of fundamental component in line current,i. THD in line current,

j. Displacement power factor,k. Apparent power factor andl. Harmonic analysis.

In order to simulate it is necessary to have an expression for line current and load current. Letus consider one output cycle, starting from the instant SCR S 1 is triggered till SCR S 2 istriggered. During this period, the R-phase voltage is defined as shown in equation (10). WhenSCR S 1 is triggered, we have commutation overlap till the load current is transferred from SCR S5 to SCR S 1 and let the commutation overlap angle be . During commutation overlap, thecurrent through SCR S 1 rises from zero to load current. At the end of commutation overlap, linecurrent is obtained from the expression in equation (11). During the period of commuationoverlap, the voltage source that is seen by the dc link circuit is described as shown in equation



(12), where v OL represents the source during overlap period.

On simplifying equation (12), we get equation (13). During this period, the source appears tohave a source inductance equalling 1.5L 2 , as viewed from the load. We get this value because

the path through SCR S 6 contains L 2 whereas the path through both S 1 and S 5 appears to havean equivalent inductance of value equal to 0.5L 2. The current through the load at the instantwhen SCR S 1 is triggered can be expressed as shown in equation (14).. The values of Zand in equation (14) are expressed in equation (15).

The constant A in equation (14) is to be evaluated and2

= tan ( ). At the end of commutation,this current would be equal to the line current. Hence we obtain equation (16). The terms usedin equation (17) are defined in equation (18).



Equation (16) is the first of the three equations we need in order to obtain a solution and thisequation equates the line current to the load current at the end of commutation. Equation (17)expresses the load current at the end of commutation, using another constant B and B is also to

be evaluated. Another equation can be formed as shown in equation (19), which equates theline current at the end of commuation to the load current. Unlike equation (16), equation (19)makes use of constant B. Equations (16) and (19) are two of the three equations required tosolve for A, B and . The third equation is obtained as follows. The load current at the instantwhen = /3 can be computed in two ways, one from equation (16) and the other fromequation (19) and these two expressions can be equated to yield equation (20).

From these three equations, the three unknowns, A, B and , can be obtained. These equationshave been used in the program written for simulation.

SIMULATION

Three applets are presented in this section. The first applet animates the circuit. The only purpose is to illustrate the sequence of operation of this circuit. It does not take in any parameter. To run it, click on the RUN button. The SI NGLE STEP button allows the user tostep through, the PAUSE button allows the user to stop the program and the RESET buttonallows the user to view the simulation once more. To view the simulation once more, click on



the RESET button and then click on the RUN button. During the simulation, the user can ask the program to pause for a while, then step through for a while and allow it to run through tothe end of its cycle.

The second applet takes in these parameters:

a. the firing angle in degrees,

b. the ratio of load reactance to load resistance (load reactance evaluated at line frequency),

c. the ratio of source reactance to load resistance(preferably below 0.1 p.u), and

d. the value of dc link source.

The program allows you to view either the waveforms or the statistics. The values of load andline reactance are to be entered in per unit. For example, if the line reactance is entered as 0.05

p.u., rated load current would cause a drop of 4% of phase voltage across the line reactance.

An example is presented now to explain how the per unit values can be set. Let a 3 phase, 415V, 50 Hz source supply power to the converter. Then the maximum average voltage that can beobtained is obtained as shown below. Let the nominal rated dc link current be 100 A. Then thenominal load resistance or the base impedance for the system is computed as shown below. Itis also shown how the line reactance can be obtained, if its value in p.u. is known. Given thatthe current through the load is free of ripple, the rated RMS line current is obtained asillustrated below. From the total rms line current, inclusive of both the fundamental componentand the harmonic components, the fundamental rms component is obtained as illustrated

below.



Given that the line inductance is 1 mH, its p.u. value is obtained as shown below.Given that the dc link inductance is 10 mH, its p.u. value is computed as illustrated below.Usually the line inductance is called as the 4% reactor, implying that when the line current is atits rated value, the rms value of the fundamental component of voltage across the line reactor is4% of the phase voltage. For example, if the rms phase voltage is 240 V, the drop across 4%reactor at rated current would be 9.6 V. When the line voltage is 415 V, the phase voltage iscalculated as shown below.The drop across the line inductor can now be stated as a fraction of

the phase voltage as given below.

This means that if the drop across the line inductor is to be 4% of phase voltage, the inductanceshould be 0.4 mH and not 1 mH.

The third applet takes in these parameters:

a. the ratio of load reactance to load resistance (load reactance evaluated at line frequency), b. the ratio of source reactance to load resistance(preferably below 0.1 p.u), andc. the value of dc link source.

Within the program the firing angle is varied and the plots of rms bridge output voltage, theaverage bridge output voltage, its RF, the rms line current, the fundamental rms component of line current, the THD in line current, the average output current, its RF and the overlap angle asa function of firing angle are displayed. The value of dc source in the dc link is assigned to bezero in this applet.

MATHCAD SIMULATION

The MathCad file can be downloaded by clicking on the image below.



SIMULATION USING C

A C-program called, ss_resp.cpp, can be downloaded by clicking on the image below. It can

be compiled as a C program and executed.

The program was run with the following values:Time constant of the Link inductor in radians: 1.00Time constant of the Source Inductance in radians : 0.03Firing Angle = 30 o.

The results obtained are presented below.

http://services.eng.uts.edu.au/~venkat/pe_html/ch05s4/ss_resp.cpp

http://services.eng.uts.edu.au/~venkat/pe_html/ch05s4/mathsim/fw3phbr3.mcd











Another C-program called, harm3ph.cpp, can be downloaded by clicking on the image below.It can be compiled as a C program and executed.

The results obtained for the same parameters are presented below.

harmNo LoadCur OutVolt harmNo LineCur

http://services.eng.uts.edu.au/~venkat/pe_html/ch05s4/harm3ph.cpp





0 1.6874 1.6872 1 0.93105

2 2.69E-08 1.16E-08 3 2.31E-08

4 2.75E-08 2.94E-08 5 0.19975

6 0.029517 0.1743 7 0.1142

8 2.85E-08 2.82E-08 9 1.85E-08

10 2.85E-08 3.00E-08 11 0.080921

12 0.00639 0.074642 13 0.061509

14 2.87E-08 2.90E-08 15 1.82E-08

16 2.86E-08 2.97E-08 17 0.048484

18 0.0023099 0.040382 19 0.040233

20 2.87E-08 2.91E-08 21 1.84E-08

22 2.87E-08 2.94E-08 23 0.032741

24 0.0009234 0.021504 25 0.02832

LoadIAvg 0.84371 LoadIRMS 0.84398

RFCur 0.021433

VoAvg 0.84362

VoRMS 0.85546

RFVolt 0.14184

ILineAvg 0.56248

ILineRMS 0.68481 THD 0.684811

OverLapAngle 0.085085

SUMMARY

This page has explained how source inductance leads to commutation overlap and how itaffects the output voltage. Next page how a DC power supply can be built using the 3-phasefully-controlled bridge rectifier.



AN APPLICATION: A DC POWER SUPPLY

CIRCUIT OPERATION


SIMULATION

CIRCUIT OPERATION A three-phase fully-controlled bridge circuit is a much more suitable circuit to be used for generating a variable dc output voltage than the single-phase fully-controlled bridge circuit, onaccount of two reasons, which are:a. reduced ripple content in its output and

b. much higher ripple frequency.Both these factors lead to an LC filter which is relatively small and economical. This pagedescribes how such a power supply can be built and controlled.

An inductor in the dc link reduces ripple in the output current of the bridge circuit, whereas thecapacitor absorbs the ripple in output voltage. The inductor has to be designed such that it doesnot saturate even when it carries the maximum current. This means that it should have anairgap in the path of flux. The ripple current through the capacitor can also be significant.Hence it needs to be checked from the datasheet that the capacitor chosen has the requiredripple current rating. For such an application, an electrolytic capacitor is normally chosen andits voltage rating should also be adequate.












We can have a block diagram to describe the operation of this dc power supply obtained usinga three-phase fully-controlled bridge rectifier. The output voltage V o is varied by varying thefiring angle . The firing angle in turn is controlled by voltage V C, which is the output of a PIcontroller. The inputs to the PI controller are a voltage named V ref representing the desiredoutput voltage and the output voltage V o of the bridge circuit. If V o is less than the desiredoutput voltage, the resultant error causes the output, V C, to increase, which in turns shouldadvance firing angle. As the firing angle is advanced, the output voltage of the bridge circuitincreases. The next section describes how the block diagram can be analysed, leading tosimulation of the system.


The simulation program is based on the pseudo-code displayed below.

Start block: Set the values of load reactance, line reactance, capacitor, load fraction. Set the desired output voltage, PI controller parameters Set the current firing angle to be 120 o. Set base reference angle to 60 o. Set Commute = 0. ( Indicates no commutation overlap exists at start. ) Go to Loop Routine. Set theta to zero.

L oop Routine Call Compute routine.

Increment theta. If {(theta + base reference angle) = (next firing angle) }

[ current firing angle = next firing angle. base reference angle = next firing angle - 60 o.



Set Commute = 1 . ( Indicates next SCR is triggered) ]

Execute Loop Routine

Compute Routine: If (Commute == 0)

[

compute next value of link inductor current. ]

else [ compute next value of triggered SCR current. compute next value of link inductor current. if (SCR current link inductor current) Reset Commute to 0.

] Compute next value of capacitor voltage. Compute next value of PI controller's output.

Compute next firing angle.

The equations used in the Compute Routine are obtained as follows.

When there is commutation overlap, the output voltage behind the source inductance isexpressed by equation (1). In equation (1), is the overlap angle and U is the amplitude of line voltage. If the output voltage be v o( ) during this period, the differential equation for thecurrent through the dc link inductance is presented as equation (2). During commutationoverlap, the current in the SCR just triggered on is described by equation (3). Thecommutation overlap ends when the current through this SCR equals the dc link inductor current. When there is no overlap, the bridge output voltage is described by equation (4).



The differential equation that describes the dc link inductor current is then described byequation (5). The differential equation for the capacitor voltage is easily obtained and it isexpressed as equation (6). Next the equations relating to closed-loop control are described. Letthe output of PI controller be v C( ) and it is expressed by equation (7). In equation (7), A is aconstant to be evaluated, K is the proportional gain of the controller and T is its time constant.The above equation is represented as equation (8) , which is more convenient for use in

simulation. In equation (8), V ref is the desired output voltage. The output of the controller isnormally checked to ensure that it is within the set limits. From the output of the controller,the firing angle, can be obtained. The maximum output voltage of the controller shouldcorrespond to zero degree firing angle and the minimum to 120 o firing angle. Hence we get thefollowing equation for firing angle. This means that the range for v C( ) is from 0 to V Cmax .



The simulation program uses the above equations and displays the results in a graphicalformat.

SIMULATION

The applet below can be run with the default parameters. To set any parameter, click on thearrow pointing downwards beside the Peak Source Vol tage, a menu would appear. The defaultvalue of the parameter highlighted appears in the textfield for Set Value . To change the

parameter, select the parameter and then click within the editable textfield for Set Value . Inorder to change this parameter, you must click on Set Value button. You can set the desiredresponse to be one of three responses.

An example is presented now to explain how the per unit values can be set. Let a 3 phase, 415V, 50 Hz source supply power to the converter. Then the maximum average voltage that can beobtained is presented in equation (10). Let the nominal rated dc link current be 100 A. Thenthe nominal load resistance or the base impedance for the system is assigned as shown byequation (11). Given that the current through the load is free of ripple, the rms line current isobtained according to equation (12) and this value includes both the fundamental componentand the harmonic components. The fundamental rms component is obtained as illustrated byequation (13). Given that the dc link inductance is 10 mH, its p.u. value is obtained as shown

by equation (14).



Given that the line inductance is 1 mH, its p.u. value is obtained as shown by equation (16).Usually the line inductance is called as the 4% reactor, implying that when the line current is atits rated value, the rms value of the fundamental component of voltage across the line reactor is4% of the phase voltage. For example, if the rms phase voltage is 240 V, the drop across 4%reactor at rated current would be 9.6 V. When the line voltage is 415 V, the phase voltage isobtained as shown by equation (16). The drop across the line inductor can now be stated as a

fraction of the phase voltage as given by equation (17).This means that if the drop across theline inductor is to be 4% of phase voltage, the inductance should be 0.4 mH and not 1 mH.

It is possible to set the load resistance to a value other than its nominal value. The nominalvalue of load resistance is 5.6 . If the load resistance is to be 10 , then set L oad F raction asshown by equation (18).



Some explanation is offered here to indicate how the values for different parameters can be setin the applet shown below. The rated dc output voltage corresponds to 1 per unit. If the rateddc output voltage is 400 V, the peak voltage of 415 V, 3 phase, 50 Hz supply is 1.467 p.u.. Letthe nominal load current through the load resistor be 100 A. Then R nom is 4 . If the dc link inductance is 10 mH, its time constant in radians is (2 fLLink /R nom ) . Given that frequency is 50Hz, it works out to be 0.7854 radians or p.u., since it expresses the ratio of the reactance over the base impedance. The per cunit value to be set for the line reactance is also obtainedsimilarly. Given the value of the filter capacitance, its per unit value is obtained as(1/2 fCR nom). The applet allows the dc output voltage to be set to a value other than unityand the load fraction can also be varied.



INTRODUCTION

CIRCUIT OPERATION

CURRENT LOOP

THE PER UNIT NOTATION

SPEED LOOP

PARAMETERS IN PER UNIT NOTATION

CURRENT CONTROLLER DESIGN

SPEED CONTROLLER DESIGN FIELD CONTROLLER DESIGN

LOGIC FOR FOUR-QUADRANT OPERATION

SWITCH-OVER OF CONTROL FROM ARMATURE TOFIELD

SIMULATION OF THE FOUR-QUADRANT DRIVE

SUMMARY

INTRODUCTION

This page describes how a separately-excited dc shunt motor can be operated in either directionin either of the two modes, the two modes being the motoring mode and the regenerating mode.It can be seen that the motor can operate in any of the four quadrants and the armature of the dcmotor in a fast four-quadrant drive is usually supplied power through a dual converter. Thedual converter can be operated with either circulating current or without circulating current. If

both the converters conduct at the same time, there would be circulating current and the levelof circulating current is restricted by an inductor. It is possible to operate only one converter atany instant, but switching from one converter to the other would be carried out after a smalldelay. This page describes the operation of a dual converter operating without circulatingcurrent.

As shown in Fig. 1, the motor is operated such that it can deliver maximum torque below its base speed and maximum power above its base speed. To control the speed below its basespeed, the voltage applied to the armature of motor is varied with the field voltage held at itsnominal value. To control the speed above its base speed, the armature is supplied with its ratedvoltage and the field is weakened. It means that an additional single-phase controlled rectifier

http://services.eng.uts.edu.au/~venkat/pe_html/ch05s6/ch05s6p1.htm#intro




http://services.eng.uts.edu.au/~venkat/pe_html/ch05s6/ch05s6p1.htm#current_loop


http://services.eng.uts.edu.au/~venkat/pe_html/ch05s6/ch05s6p1.htm#pu_form


http://services.eng.uts.edu.au/~venkat/pe_html/ch05s6/ch05s6p1.htm#speed_loop


http://services.eng.uts.edu.au/~venkat/pe_html/ch05s6/ch05s6p1.htm#params_pu


http://services.eng.uts.edu.au/~venkat/pe_html/ch05s6/ch05s6p1.htm#current_cont


http://services.eng.uts.edu.au/~venkat/pe_html/ch05s6/ch05s6p1.htm#speed_cont


http://services.eng.uts.edu.au/~venkat/pe_html/ch05s6/ch05s6p1.htm#field_cont


http://services.eng.uts.edu.au/~venkat/pe_html/ch05s6/ch05s6p1.htm#logic_fourq


http://services.eng.uts.edu.au/~venkat/pe_html/ch05s6/ch05s6p1.htm#switch_over























circuit is needed for field control. Closed-loop control in the field-weakening mode tends to bedifficult because of the relatively 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 forward motoring is called the positive converter and the other converter is called the negative converter. Instead of namingthe converters as positive converter and negative converter, the names could have been forwardand reverse converters. 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 an inductor is inserted in

each line as shown in Fig. 3 and this inductor reduces the impact of notches on line voltagesthat occur during commutation overlap.

CIRCUIT OPERATION

The operation of the circuit in the circulating-current free mode is not very much different fromthat described in the previous pages. In order to drive the motor in the forward direction, the

positive converter is controlled. To control the motor in the reverse direction, the negativeconverter is controlled. When the speed of motor is to be changed fast from a high value to alow value in the forward direction, the conduction has to switch from the positive converter tothe negative converter. Then the direction of current flow changes in the motor and itregenerates, feeding power back to the source. When the speed is to be reduced in the reversedirection, the conduction has to switch from the negative converter to the positive converter. Itis seen that conduction has to switch from one converter to the other when the direction of motor rotation is to change, so that regeneration can occur. During regeneration, the direction



torque developed by motor is opposite to that of the motoring torque. Thus the regeneratingtorque acts as the breaking torque and the motor decelerates fast.

At the instant when the switch from one converter to the other is to occur, it would be preferable to ensure that the average output voltage of either converter is the same. Let thefiring angle of the positive converter be P, and the firing angle of the negative converter

be N . If the peak line voltage be U, then equation (1) should apply. Equation (1) leads to

equation (2). Then the sum of firing angles of the two converters is , as shown in equation (3).

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

When the speed of the motor is to be increased above its base speed, the voltage applied to thearmature is kept at its nominal value and the phase-angle of the single phase bridge is variedsuch that the field current is set to a value below its nominal value. If the nominal speed of themotor 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 the commutator and themotor can get damaged.

Next it is shown how the operation of motor can be represented by means of a block diagram.This approach can be helpful in designing the closed-loop system.

CURRENT LOOP

Let the field excitation be assumed to remain constant at its nominal level. Let the voltageapplied to armature be v a volts, the back e.m.f. e b volts and rotor speed w r rad/s. The back emf is expressed by equation (4), where K m is the coefficient relating speed of motor to its back

emf. If R a be the resistance of armature and L a its inductance, then the applied armature voltageequals the sum of the motor back e.m.f, the drop across its armature resistance and the dropacross the armature inductance, as shown in equation (5). In equation (5), v a is the voltageapplied to the armature and i a is the current though the armature. The above equation can berepresented in terms of Laplace transform, leading to equation (6).



The block diagram shown in Fig. 4 represents equation (6).

Given that the excitation of the motor is constant and that the effects of armature reaction arenegligible either due to interpoles or series compensation winding, the torque output M ecan beexpressed as shown in equation (7). If the load torque be M L N-m, the combined polar momentof inertia of motor and load be J kg.m 2 and its friction coefficient be B N-m/rad/sec, then thetorque output of motor equals the expression on the right-hand side of equation (8). Equation(8) can be represented in terms of Laplace transform, as shown in equation (9), where theLaplace transform of w, the motor speed, is assigned to be (s). A block diagram, as shown inFig. 5, can now be drawn based on equations (4), (6), (7) and (9). It can be seen that unit for K m is N-m/A.



With the load torque set to zero, a transfer function linking current I a(s) and the input voltage

Va(s) can be obtained. It is expressed in equation (10).

The reason for obtaining this transfer function is to facilitate the design of a controller for controlling the armature current. The two output parameters of interest are the torque and thespeed. The armature current is selected as one of the state-variables to be controlled in closed-

loop, since the torque output varies linearly with it. It is preferable that the variable to becontrolled by negative feedback is a variable that reflects some energy stored in a system. Herethe armature current reflects the energy stored in the inductance in the armature circuit. If themotor has a compensating winding and/or a compound winding, the inductance of this windingshould be added to L a. In some drives, an additional inductor is used in series with the armatureand this value should also be added to L a. Let G 1(s) reflect the transfer function in equation(10) and equation (11) shows the expression for G 1(s). The part of the closed-loop system thatis usually used for controlling the armature current is shown in Fig. 6.



The block diagram in Fig. 6 is now described. If the armature current is to be controlled inclosed-loop, it is necessary to have a current reference signal, marked as I R (s) in Fig. 6. This

signal is internally generated, most often as the output of the controller for speed and it isshown later how that is achieved.

It is possible to use a controller other than a proportional plus gain(PI) controller. A PDFcontroller(a pseudo-derivative controller) or a PID-controller can be used. But a PI controller isoften sufficient, since the integrating part of the PI controller leads to zero steady-state error for a step input and the proportional gain can be adjusted to yield fast response and stability. Theoutput of the current controller is often a voltage which sets the firing angle for the fully-controlled bridge circuit. Since the gain K B( ) is negative, the sign of both the proportionalgain K I and the integrating time-constant T I should be negative in order to keep the loop-gain

of the system represented by block diagram in Fig. 6 as negative.

A variation in the output of the current controller does not change the firing angleinstantaneously since the SCRs in the bridge are triggered in a sequence at an interval of 60 o onthe average and there is a delay before the change in the output of the current controller has aneffect on the firing angle. This delay can be classified as a transportation lag and it can beapproximated by a first-order transfer function, as shown in equation (12). In equation(12), y has been used in place of sT D.

For a system with 50 Hz input source, one-sixth of a cycle is about 3.3 ms and then the delay

TD can be set to be half of that value, that is 1.67 ms. What is carried out is an approximation tofacilitate the design of current controller.



As the firing angle increases, the instant of triggering of each SCR is delayed more and morefrom its reference point corresponding to 0 o firing angle. When the current flow in the dc link iscontinuous, the average output voltage of the bridge changes from its positive maximumaverage value to its negative maximum average value, as is allowed to very from 0 o to 180 o.In order to ensure that the loop gain is negative, it is necessary that the gain due to controlledrectifier circuit is inverted. It is explained later how it can be brought about for practicalrealization.

Another point to be noted is that the gain of the controlled rectifier is not constant and it varieswith firing angle. Let the maximum average output voltage be V om. Then equation (13) showshow the average output voltage at any firing angle, is obtained. The actual gain of thecontrolled rectifier is defined by equation (14). If the rated armature voltage, V RA , is assignedto be the base voltage, then the gain of the controlled rectifier in per unit notation can bedefined as in equation (15).

In equation (15), K A defines the ratio of maximum average output voltage of the bridge to therated armature voltage. It is seen that the gain varies and hence the controller has to bedesigned such that it operates properly over this range of variation.

THE PER UNIT NOTATION

It is better to design the current loop first before the outer loop design is attempted. But it isnecessary to describe the per unit notation that is adopted here. Let the rated armature voltageVRA be the base voltage. Then equation (16) is valid. As shown in equation (17), the ratedarmature current I RA is chosen to be the base current.

Per Unit Value of Rated Armature Voltage = 1 (16)



Per Unit Value of Rated Armature Current = 1 (17)

Then the base impedance for the armature circuit is obtained as shown in equations(18)whereas equation (19) shows how the per unit value of armature resistance can be obtained if itis r a . Given that the inductance present in the armature circuit is L a H, the voltage across it isobtained as shown in equation (20). Equation (21) is obtained by dividing both sides of equation (20) by V RA . Equation (21) uses symbol a, representing the time constant of the

armature circuit and it is defined by equation (22).

For a 3-phase controlled-bridge rectifier circuit, the maximum average output voltage that can be obtained at 0 o firing angle is shown in equation (23). Then the amplitude of line voltage of 3-ph supply is described by equation (24). The per unit value of the peak line voltage isobtained from equation (25).



We have seen so far how voltages, currents and impedances related to armature circuit can beexpressed in per unit values. Next, it is shown how the torque developed, moment of inertia Jand friction coefficient B can be expressed in per unit values. Let the torque developed bymotor be M e N-m. Then when the motor is operating with the nominal or the rated flux, thetorque developed by motor is defined by equation (26), where i a is the armature current. Also,let w r be the armature shaft speed in rad/s. Then the per unit value of the torque developed isexpressed as shown in equation (27), where I RA is the rated armature current.

The per unit value of moment of inertia is obtained as follows. Let R be the rated shaft speedin rad/s, and the moment of inertia of motor and the coupled load be J kg-m2. Let the torquerequired to accelerate this moment of inertia be MJ. Then equation (28) can be used to relate Jand MJ. Dividing both sides of equation (28) by the rated torque, we get equation (29). Fromequation (29), it is seen how the per unit value of the moment of inertia can be obtained.Similarly, we can get an expression for friction coefficient, as shown by equation (30).

It is necessary to state how the parameters for the current controller should be specified. Thegain, K I , is just a ratio whereas the time constant, T I , should be specified in seconds.



SPEED LOOP

Before the design of speed loop is to be attempted, the current loop should be approximated bya suitable transfer function. The block diagram in Fig. 6 can be expressed by a transfer function, say G 2(s). Then G 2(s) = I a(s)/I R (s). Using G 2(s), the speed loop can be represented asshown in Fig. 7. Again a PI controller is used and it may be necessary to provide a filter in the

path of speed feedback signal. The time constants are to be specified in seconds. In per unitnotation, the value of K M marked in Fig. 7 would be 1.

PARAMETERS IN PER UNIT NOTATION

At first, typical per unit values are obtained from the datasheet of a dc motor. Then it is shownhow the current loop can be designed. Next, the design is verified by simulation. Finally, thespeed loop design is illustrated.

In the per-unit calculations, a per unit value of 1 is assigned to the rated armature voltage, therated motor speed and the rated armature current. From the values of the rated armature voltageand the rated motor speed, obtain K m, the coefficient for the motor. Then other per unit valuescan be obtained as outlined earlier. The applet displayed below computes the per unit valuesgiven the actual values. The textfields contain default values and the applet computes anddisplays the per unit values when the Compute Button is clicked.

CURRENT CONTROLLER DESIGN



The transfer function G 1(s) expressed as equation (11) can expressed in terms of per unit valuesand then V a(s) and I a(s) marked in Fig. 6 would be values in per unit notation. Conversion of equation (11) such that it conforms to the per unit notation is explained below.

Both the numerator and the denominator of expression in equation (11) can be divided by BR a.The ratio of J/B can be represented as the mechanical time constant, m. The resultingexpression for G 1(s) presented as equation (31). Then equations (32) and (33) explain how

equation (31) can be converted such that it is in per unit notation.

The block diagram shown in Fig. 6 conforms to per unit notation. Here G 1(s) is expressed byequation (34) and V a(s) and I a(s) marked in Fig. 6 are values in per unit notation. Using (34),the transfer function, G 2(s) representing the block diagram in Fig. 6 can be represented asshown below. In this case, the controlled-rectifier is assigned to have its highest gain, K A. It islogical to do so, because the system designed for stability at gain K A would be stable at lower gains too.



The second applet in this page finds the location of the poles and zeros of the closed-loopsystem in Fig. 6, given the necessary data. It is seen for a wide range of controller parameters,the zeros are located such that they cancel almost two poles. The other two poles are locatedaway from the origin. It is seen that the design of current controller is fairly easy.

SPEED CONTROLLER DESIGN

It is necessary to design the speed controller next. To design the speed controller, it is

necessary to represent the transfer function G 2(s) suitably. It is found that for a wide range of values, two of the zeros of G 2(s) are located near two of the poles and the other two poles areaway from the origin. Hence while designing the speed loop, G 2(s) is set equal to unity. Thesimulation of the drive presented later would show whether this simplification is justified.

The design of the speed controller is carried out based on the assumption that the motor is onno load. A variable drive system tends to exhibit oscillatory behaviour under no load conditionsand hence the design based on no load condition is assumed to be justified. Here the output of the speed controller is not clamped, whereas there would be limits on the output of speedcontroller. The output of speed controller corresponds to armature current and it is necessary to



limit the peak value of speed controller in order to protect the SCRs used in the bridge. Theapplet presented later for simulating the drive has limits imposed on the output of the speedcontroller.

FIELD CONTROLLER DESIGN

The block diagram for closed-loop operation with the field controller in action turns out to besomewhat complex. The interaction that occurs within a separately-excited DC motor is first

presented in Fig. 8.

The block diagram in Fig.8 is now described. The field current, marked as I F, producesmagnetic flux in the motor and the back e.m.f of the motor is then proportional to the productof the field current and the speed of the motor. This statement is based on the assumption thatthe field flux in the motor is not saturated and that the field flux varies linearly with the field

current. If the field current is in per unit notation, where the field current corresponding to therated current equals unity, then the back emf can be shown to be equal to K m × i F × w R , where both K m and w R are also in per unit representing the motor coefficient and the speed of themotor. Once the back e.m.f and the applied voltage are known, the armature current can beobtained as shown in Fig. 8. From the values of armature current and field current, the torqueoutput of motor is obtained and the speed of the motor changes as shown.

For design of field controller, the block diagram in Fig. 8 is too complex. The design is carriedout using a simplified or a simplistic block diagram and the performance of the controller is



evaluated using the final simulation program, which uses a model that is reasonably close toreal system.

The design of field controller is based on the block diagram shown in Fig. 9.

It is easy to represent the block diagram in Fig. 9 in per unit notation. The gain of thecontrolled bridge for the field circuit is K F. Its value equals the ratio of the maximum rate of

bridge output voltage to the rated voltage of the field circuit and normally the value of K F islikely to be near 1.2. The delay due to firing circuit is again approximated by T D2, and it is setequal to (1/4f), where f is the frequency of the ac source. Then the field current is obtained in

per unit value and it can be made equal to the torque, assuming that the armature circuit hascomparatively a small time constant and that the armature current stays at the rated value. Thefriction coefficient, the mechanical time constant and the time constant of the filter in the speedfeedback signal are the same signals used for design of the speed controller. The applet belowcan be used to design the field controller. This applet runs somewhat slowly. The poles andzeros are calculated for the block diagram shown in Fig.9, whereas the step response isobtained using the block diagram in Fig. 10.





For four-quadrant operation, the scheme outlined in this page makes use of two converters,called the positive converter and the negative converter. It has also been shown that the sum of the firing angles of the two converters should be radians. Hence the synchronizing signals for the SCRs in both converters can be obtained as shown in Fig. 11.

It is also necessary to find out when the switch from the positive converter to the negative

converter or vice-versa can be made. One possible method is outlined in Fig. 12. Based on the polarity of current reference signal, a logic signal, called W can be developed. A comparator can be designed to yield an output of 1 (W =1 ) when the current reference signal is positiveand an output of 0 (W =0 ) when the current reference signal is negative. Along with W,another signal can be derived based on the armature current. A signal, called Enable, can be

produced such that Enable is 1 when the armature current is zero. When Enable is 1, the outputof a flip-flop can be set. Output Q takes on the polarity of W signal. When both W and Q are atlogic 1, the positive converter is allowed to be triggered. When both W and Q are at logic 0, thenegative converter is allowed to be triggered.



The rest of the control arrangement for generating the firing signals is shown in Fig. 13. The block diagram shows that there are active limits on the firing angle. The voltage output of the bridge can be sensed and when it is at about 1.05 times the rated armature voltage, the firingangle may not be allowed to become any smaller.



The variation of firing angle towards either 0 o or 180 o can be blocked, avoiding further rise inthe output voltage. It is better to have a provision that would allow for varying the limitingvalues to accommodate changes in source voltages. It would not be difficult to implement sucha scheme for a micro-controller based control system.

SWITCH-OVER OF CONTROL FROM ARMATURE TO FIELD

Field control is necessary above the base speed of the motor and the field current has to varyonly over a limited range, say from 0.6 pu to 1 pu. The signal that sets the active limits on thefiring angle for armature control can be used also for field control. The block diagram relatedto field control is shown in Fig. 14.

When the active limit is not set, the firing angle of the field controller is set such that the fieldcurrent remains at the nominal value. When the active limits are applied to the armature controlcircuit, the firing angle is varied such that the field current gets adjusted to the value requiredfor the speed reference set.

SIMULATION OF THE FOUR-QUADRANT DRIVE

Before selecting the type of response, set the value of the selected parameter. When you selecta parameter, the textfield shows the default value set inside the program. Change the parameter value if you want to and then you must click on the SET VAL UE button for the change to takeeffect. You can go from one type of response to another after the present calculations arecarried out. When you have selected a new type of response, you must click on Click to Start .If you click on Reset button, initializing routine is carried out and the motor speed is set to



zero, and the other values are also reset. It is possible to see the effect of step changes in speedor load or unbalance in source or unbalance in firing circuit.

SUMMARY

This page has illustrated how a four-quadrant DC drive operates. The next section shows howsemi-controlled bridge rectifiers operate.

Documents

OPERATION of a 3 Phase Rectifier