Presentation on theme: "Electric Drives1 ELECTRIC DRIVES Ion Boldea S.A.Nasar 1998."— Presentation transcript:
Electric Drives1 ELECTRIC DRIVES Ion Boldea S.A.Nasar 1998
Electric Drives2 5. CONTROLLED RECTIFIER D.C. BRUSH MOTOR DRIVES Table 5.1. Phase controlled rectifier circuits
PERFORMANCE INDICES As the d.c. motor current, when fed from phase controlled a.c. - d.c. converters, is not constant and a.c. supply current is not sinusoidal, adequate performance indexes for the motor - converter combination should be defined. The main performance indexes related to the motor are: the torque - speed characteristic; nature of motor current - continuous or discontinuous; average motor current I a : If the supply voltage is a pure sinusoid only the fundamental of input current will produce the mean input power and thus: (5.4) where: V - rms supply phase voltage; I - rms supply phase current; I 1 - rms fundamental component of a.c. supply current; 1 - angle between supply voltage and current fundamentals.
Electric Drives6 input displacement factor DF or the fundamental power factor: (5.5) harmonic factor HF: (5.6) I h - rms value of the net harmonic current.
Electric Drives SINGLE P.E.S. CONTROLLED RECTIFIER A d.c. brush motor with separate excitation with the data: K e p = 2Wb, R a = 5 , L a = 0.1H is fed through a thyristor (figure 5.1) from a single phase a.c. source whose voltage is. The motor speed is considered constant at n = 750rpm. a. Calculate the motor current i a time variation for a delay angle = ; b. How the motor voltage V a varies in time? c. How i a varies in time in presence of a free wheeling diode in parallel with the motor armature (figure 5.1)? The thyristor and the diode are considered as ideal switches. Solution: a. For t> (figure 5.1) with discontinuous current: (5.8)
Electric Drives8 Figure 5.1. A d.c. brush motor supplied through a thyristor with the initial condition i a = 0 for t The steady state solution is: (5.9) and finally: (5.10)
Electric Drives9 The complete solution: (5.11) For t = t 1 = / 1 = = s i a (t) = 0 and thus: (5.12) and (5.13) (5.14)
Electric Drives10 Finally(5.15) Equation (5.15) is valid for t > until the current becomes zero soon after t as shown in figure 5.1. b. The motor voltage is equal to the source voltage as long as the thyristor is on and becomes equal to minus the e.m.f.:,when the motor current is zero (figure 5.1) c. In presence of the free wheeling diode D, the latter starts conducting when V(t) becomes negative ( t ). From now on the motor current i a ’ flows through the diode until it becomes zero: (5.16) with i a ’(0) = i a ( ) (5.17) (5.18)
Electric Drives11 (5.19) A’ = A. Thus (5.20) The current i a ’ = 0 for t’ The motor voltage V a becomes zero during the time interval when the free wheeling diode is conducting (180 0 to ).
Electric Drives THE SINGLE PHASE FULL CONVERTER A d.c. motor of 7kW and 1200rpm rating with separate excitation is supplied through a single phase full converter as shown in figure 5.3.a. The other data are R a = 0.2 , rated current I ar = 40A, K e p = 10Wb, the a.c. supply voltage (rms) is 260V. a. Draw the voltage and current waveforms for steady state and finite motor armature inductance and = 45 0 and = b. For a firing angle of = 30 0 and rated motor current (rectifier regime) calculate: motor speed, torque and supply power factor, neglecting the motor current ripple. c. By reversing the field current the motor back e.m.f. E g is reversed; for rated current calculate: converter firing angle ’ and the power fed back to the supply.
Electric Drives13 a.) Figure 5.3. The d.c. brush motor fed through a single phase full converter a.) the converterb.), c.), d.), e.) - voltage and current waveforms for motor action ( = 45 0 ) f.), g.), h.) - for generator action ( = )
Electric Drives14 Solution: a. The waveforms of currents and voltages are shown on figure 5.3. It should be noted that because the motor inductance is not very high the armature current, and speed pulsate with time. Motor action is obtained for = 45 0 when both the motor and converter average input powers are positive. For generator action ( = ) the motor current remains positive but the motor average voltage V a is negative. Thus both the motor and converter input average powers are negative for generating. b. As the motor current is considered ripple - free we use only the average values of voltage and current while the speed is also constant. The average motor voltage V av is: (5.31)
Electric Drives15 The motor torque T e is: (5.32) The motor speed n is: (5.33) As the primary current is now rectangular (40A) the rms current I 1 = 40A. The power from the supply, neglecting the losses,. Thus the supply power factor: (5.34) c. For generator action the polarity of induced voltage should be reversed: (5.35) Thus the motor voltage V a becomes: (5.36)
Electric Drives16 Finally:(5.37) The regenerated power: (5.38) As for this regime, either the flux p or the speed n changes sign the motor torque opposes the motion providing generator braking. In the process the motor speed and e.m.f. decrease and in order to keep the current constant the firing angle >90 0 in the converter should be modified accordingly. For the motor as above, having L a = 2mH, calculate the following: d. For = 60 0 and constant speed n = 1200 rpm draw the voltage and current waveforms knowing that the motor current is discontinuous. e. Calculate the motor current waveform for n = 600 rpm.
Electric Drives17 Solution: Figure 5.4. D.c brush motor supplied from a full converter - discontinuous current mode As the motor electrical time constant e = L a / R a = / 0.2 = 10ms is small, when the voltage becomes zero (for 1 t = ), the motor current decays quickly to zero along the angle (from to ) with and thus the current is discontinuous. As increases, for a given speed n (and e.m.f. e g ), the average voltage V av decreases with respect to the case of continuous current. The discontinuity of current contributes to a sluggish dynamic response as during zero current motor internal torque control is lost. The situation occurs especially at low torques (and currents). Special measures such as flux weakening (reducing p ) for low torques leads to reduced at high speeds avoiding the discontinuity in current.
Electric Drives18 With the speed considered constant the voltage equation is: (5.39) with i a = 0 for 1 t 1 = , the solution of above equation is: (5.40) for and thus: (5.41) (5.42) The current becomes again zero for 1 t 2 = (5.43) The solution for is.
Electric Drives THE THREE PHASE FULL CONVERTER - MOTOR SIDE A three phase full converter (figure 5.6) supplying a d.c. motor has the data: line voltage V L = 220V (rms), K e p = 10Wb, R a = 0.2 and neglect the ripples in the motor current. a. For L s equal to zero calculate the output average voltage as a function of the delay angle and; for = 30 0 determine the motor speed for I a = 50A. b. Considering the a.c. source inductance L s 0 determine the expression of the output voltage as a function of and I a and calculate the motor speed for I a = 50A, = 30 0 and L s = 11mH. c. Calculate the overlapping angle at commutation for case b.). Solution: The principle of operation is quite similar to that of the previous case. However only a 60 0 conduction of each pair of thyristors occurs as shown in figure 5.6. In the absence of a.c. source inductances L s (L s = 0) the commutation is instantaneous. The average output voltage V av is: (5.48)
Electric Drives20 Figure 5.6. D.c. brush motor fed from a three phase full converter Consequently V av 0) depending on . For 90 V av <0 and thus the inverter mode is obtained. The motor current remains positive irrespective of . Two quadrant operation is obtained.
Electric Drives21 The motor voltage equation is: (5.49) (5.50) b. The effect of a.c. source inductances L s on current commutation is shown in figure 5.7 for the case of constant motor current (similar to the diode full converter (chapter 3)). The commutation between T 1 and T 5 is not anymore instantaneous; an overlapping angle u occurs. The effect of actual commutation is in fact a reduction of output voltage V d determined by the area A u when a kind of shortcircuit occurs between phases a and c. (5.51) This is so since the current in phase a is zero for 1 t = and equal to I d for 1 t = +u. The average voltage is reduced by 3 / A u.
Electric Drives22 Finally the average voltage is equal to that already found for L s = 0 minus 3 / A u : (5.52) Figure 5.7. The current commutation in presence of a.c. source side inductances L s
Electric Drives23 As we can see the angle u is not required to calculate the output voltage V d. However it is needed to ensure reliable operation in the inverter mode ( >90 0 ). The second Kirchoff law for phase a and c during commutation provides the equation: (5.53) with, that is (5.54) Thus: (5.55) with (5.56) (5.57) and finally (5.58)
Electric Drives24 Thus knowing and I d we may calculate the overlapping angle u. In the inverter mode ( >90 0 ) the commutation should be finished before + u = in order to allow the turn off time t off required for the recombination of charged particles in the thyristors ( ) for negative voltage along the turning off thyristors. The rectifier voltage V d for / 6 and I d = 50 is: (5.59) (5.60) The reduction of output voltage for the same due to commutation with 18V has contributed to a notable reduction in speed, for 50A, from rpm to rpm. c. To calculate the overlapping angle u we use the expression derived above: (5.61) A considerable value for u has been obtained.
Electric Drives THE THREE PHASE FULL CONVERTER - SOURCE SIDE ASPECTS For the three phase full converter and d.c. motor in the previous example with L s = 0 for = 0 0, = 45 0 with I d = 50A determine: a. The waveforms of a.c. source current and its harmonics factor. b. The rms of source current fundamental and of the total source current. c. The displacement power factor for L s = 0 and L s = 1mH for = d. Calculate the line voltage and voltage distorsion due to L s = 1mH and = Solution: a. As the motor armature current is considered constant the a.c. source current is rectangular as shown in figure 5.8.a and b. The presence of a.c. source inductances L s 0 leads to the overlapping angle u 0. b. The rms value of the current fundamental, I a1, is: (5.62)
Electric Drives26 Figure 5.8. A.c. source currents in a three phase full converter and constant motor current a.) for = 0, L s = 0;b.) for = 45 0, L s = 0;c.) for = 45 0, L s = 1mH
Electric Drives27 For the total a.c. source current which is made of rectangular wide “blocks”: (5.63) In the absence of L s the displacement power factor (DPF) angle is equal to and thus the DPF is: (5.64) In the presence of L s the displacement power factor is approximately: (5.65) To calculate u for = 45 0 we use equation (5.58): (5.66) Finally:
Electric Drives28 The current commutation in the presence of L s produces a further reduction of displacement power factor. In general the DPF decreases with increasing which constitutes a notable disadvantage of phase delay a.c. - d.c. converters. Special measures are required to improve DPF with increasing. d. The a.c. source current overlapping during commutation produces notches in the line voltage. From figure 5.9. we may obtain V ab = V an - V bn of the waveform shown in figure 5.9. Figure 5.9. A.c source line voltage notches due to Ls (commutation)
Electric Drives29 Considering u as small, the deep notch depth is considered equal to and thus the notch width u is approximately: (5.67) The depth of shallow notches is considered half of that of deep notches. IEEEE standard suggests the limitation of line notches to 250 s (5.4. electrical degrees) and the deep notch depth to 70% of rated peak line voltage in order to perform satisfactorily. Special filtering is required to cope with more recent standards. The voltage distorsion due to notches depends on the harmonics currents I and the a.c. source inductance L s : (5.68) with due to the almost rectangular form of a.c. source current.
Electric Drives THE DUAL CONVERTER - FOUR QUADRANT OPERATION Figure Dual converter with circulating current supplying a d.c. brush motor