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Electric Drives1 ELECTRIC DRIVES Ion Boldea S.A.Nasar 1998.

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Presentation on theme: "Electric Drives1 ELECTRIC DRIVES Ion Boldea S.A.Nasar 1998."— Presentation transcript:

1 Electric Drives1 ELECTRIC DRIVES Ion Boldea S.A.Nasar 1998

2 Electric Drives2 6. CHOPPER-CONTROLLED D.C. BRUSH MOTOR DRIVES Table 6.1. Single phase chopper configurations for d.c. brush motors

3 Electric Drives3

4 4 Figure 6.1. First quadrant chopper operation a. ) continuous modeb.) discontinuous mode The voltage equation for constant speed is: (6.2) (6.3)

5 Electric Drives5 The average output voltage for the discontiuous mode may be determined noting that the motor voltage is zero: (6.5) The output current expressions are obtained from (6.2) - (6.3): (6.6) (6.7) The continuity condition is i a (t on ) = i a ’(t on )(6.8)

6 Electric Drives6 For the second quadrant chopper (table 6.1.b) the d.c. motor e.m.f. e g with S 2 on produces a current rise in the inductance L a : (6.10) When S 2 is turned off the energy stored into the inductor is sent back to the source as long as V 0 > V a : (6.11) with the solution: (6.12) (6.13) The boundary conditions are: i a (t on ) = i a ’(t on ), i a (0) = i a0 and i a ’(T) = i a0 (6.14)

7 Electric Drives7 It is thus possible with e g

8 Electric Drives8 Figure 6.3. Source current waveforms a.) first quadrant operationb.) second quadrant operation

9 Electric Drives9 An LC input filter (figure 6.4) will provide a path for the ripple current such that only (approximately) the average current is drawn from the supply. The n th harmonic current i n in the supply (figure 6.4.b) is: (6.15) Figure 6.4. First quadrant chopper with LC input filter a.) basic circuitb.) equivalent circuit for n th harmonic

10 Electric Drives THE FIRST QUADRANT (STEP - DOWN) CHOPPER A d.c. brush motor with permanent magnet excitation with the data R a = 1 , K e p = 0.055Wb/rpm, is fed through a first quadrant chopper (table 6.1.a) from a 120Vd.c. supply at a constant (ideal) armature current of 10A. Determine: a. the range of duty cycle  from zero to maximum speed; b. the range of speed Solution: a. The average output voltage V a is: (6.16) At standstill n = 0 and thus: (6.17) For maximum speed the voltage is maximum: 120V (  = 1). Consequently  varies from 1/12 to 1.

11 Electric Drives11 b. The voltage equation for maximum speed is: (6.18) (6.19) The speed range is thus from zero to 2000 rpm. c. For the d.c. brush motor and the chopper as above and  = 0.3 calculate the actual armature current waveform, its average value, and voltage average value at n = 1600 rpm, for the chopping frequency f ch = 50Hz. Determine the chopping frequency for which the limit between discontinuous and continuous current is reached at the same t on as above. Solution: We now apply the armature current expressions (6.6) - (6.7) first for f ch = 50Hz. The turn - on time interval (6.20) with

12 Electric Drives12 (6.21) Assuming discontinuous current mode: (6.22) Also: (6.23) The current i a ’ becomes zero at t = t 1 : (6.24) Thus indeed the current is discontinuous. Figure 6.5. Discontinuous current

13 Electric Drives13 The average current i av is: (6.25) The chopping frequency for which the limit between the discontinuous and continuous current is obtained for: (6.26) In this case  becomes (6.27) As the current is discontinuous the average voltage is from (6.5): (6.28)

14 Electric Drives THE SECOND QUADRANT (STEP - UP) CHOPPER FOR GENERATOR BRAKING A d.c. brush motor with PM excitation is fed through a second quadrant chopper for regenerative braking (table 6.1.b). The motor data are: R a = 1  ; L a = 20mH, e g = 80V (given speed). The supply voltage V 0 is 120Vd.c., and t on = s. Determine: a. The waveform of motor current for zero initial current b. The waveform of source current c. The maximum average power generated Solution: The current waveforms are as shown on figure with i a0 = 0. (6.29) b. (6.30)

15 Electric Drives15 Figure 6.6. The step second quadrant chopper The boundary conditions are: (6.31) The unknowns are B, B’ and T. Consequently: (6.32)

16 Electric Drives16 (6.33) (6.34) Note that the source current occurs during the S 1 turn - off and is negative, proving the regenerative operation. The average source current i av is: (6.35) The average power regenerated P av is: (6.36)

17 Electric Drives THE TWO QUADRANT CHOPPER Consider a two quadrant chopper (figure 6.7) supplying a d.c. brush motor with separate excitation. The load current varies between I max >0 and I min < 0. Figure 6.7. Two quadrant chopper supplying a separately excited d.c. brush motor Determine: a. The voltage and current waveforms for the load current varying between an I max >0 and I min |I min |. b. Derive the expression of the conducting times t d2 and t d1 of the diodes D 1 and D 2 for case a.)

18 Electric Drives18 Solution: a. Let us first draw the load current which varies from a positive maximum to a negative minimum (figure 6.8). The conduction interval for each of the four switches T 1, D 2, D 1, T 2 are: T 1 ; for t d2 < t < t c ; T 2 ; for t d1 < t < T;(6.37) D 1 ; for t c < t < t d1 ; D 2 ; for 0 < t < t d2 ; b. The equations for the current are: (6.38) (6.39) with the solutions: (6.40)

19 Electric Drives19 (6.41) with the boundary conditions i a (0) = i min, i a ’(T) = i min and i a (t c ) = i a ’(t c ). Figure 6.8. Voltage and current waveforms of two quadrant chopper fed - d.c. motor

20 Electric Drives20 The unknowns A, A’ and t c are obtained from: (6.42) (6.43) (6.44) Consider a d.c. brush motor whose data are: e 0 = 120V, R a = 1 , L a = 5mH, e g = 80V, I max = 5A, I min = -2A and the chopping frequency f ch = 0.5kHz. For the two quadrant chopper as above calculate: a. The t c / T =  on ratio b. The conducting intervals of the 4 switches

21 Electric Drives21 Solution: The constants A, A’, t c expressions developed above ((6.42) - (6.44)) yield: (6.45) (6.46) (6.47) (6.48)

22 Electric Drives22 The conducting interval of D 2, t d2, corresponds to i a = 0: (6.49) Thus the main switch T 1 is conducting for a time interval: (6.50) To calculate the conducting time of the diode D 1 we apply the condition i a ’(t d1 ) = 0: (6.51) (6.52) Consequently the diode D 1 conducts for ms. Finally the static switch T 2 conducts for the time interval (6.53)

23 Electric Drives THE FOUR QUADRANT CHOPPER Figure 6.9. D.c. brush motor fed through a four - quadrant chopper

24 Electric Drives24 Figure Four quadrant chopper supplying a d.c. brush motor a.) Third quadrant: i av 0, V av <0;

25 Electric Drives THE INPUT FILTER A first quadrant chopper with an L - C input filter supplies a d.c. brush motor with PM excitation under constant current start - up. a. Demonstrate that maximum rms ripple current in the chopper current i ch occurs at a duty cycle  = 0.5. b. For f ch = 400Hz, I a = 100A, rms fundamental (a.c.) current allowed in the supply is 10% of d.c. supply current. Capacitors of 1mF which can take 5A rms ripple current are available. Determine L f and C f of the filter for f ch >2f r. c. For case b.) calculate the d.c., first and third harmonics of the supply current. Solution: Thus the d.c., I sdc rms I srms and ripple I sripple components of chopper currents are: (6.72) (6.73)

26 Electric Drives26 The maximum ripple current is obtained for which leads to  = 1 / 2. The filter design will be performed for this worst case. (6.74) Figure First quadrant chopper with L f C f filter a.) basic circuitb.) chopper input current The chopper configuration (figure 6.11) provides square current pulses for i s of width .

27 Electric Drives27 To design the filter we need first the chopper current harmonics content which, for  = 0.5, is: (6.75) with(6.76) (6.77) (6.78) (6.79) The input (source) a.c. current is not to surpass 10% of d.c. input current, that is.

28 Electric Drives28 According to equation (6.15) for n = 1 we obtain: (6.80) Also the fundamental capacitor current I C1 is: (6.81) As each 1mF capacitor can take 5A, 10 such capacitors in parallel are needed and thus C f = 10mF. On the other hand the reactance x L is:  (6.82) (6.83)

29 Electric Drives29 The filter resonance frequency f r is: (6.84) The ratio between the chopper switching frequency f ch and the filter resonance frequency f r is: The same a.c. current components (6.15) (after filtering) are: (6.85) (6.86) (6.87) As noticed the L f C f filter produces a drastic reduction of source current harmonics.

30 Electric Drives DIGITAL SIMULATION THROUGH MATLAB-SIMULINK Figure D.C. brush motor drive - block diagram

31 Electric Drives31 The integration step (10  s) can be modified from the Simulink’s Simulation / Parameters. The chopper frequency is 20kHz and this block input is the t c / T e ratio. To find out the structure of each block presented above unmask it (Options/Unmask). Each masked block contains a short help describing that block (inputs / outputs / parameters). The block diagram of the electric drive system is presented in figure The motor used for this simulation has the following parameters: V dc = 120V, I n = 20A, n n = 3000rpm, R s = 0.5 , L a = H, J = 0.001kgm 2, K e p = 2.2Wb. The figures ( ) represent the speed (figure 6.13), torque (figure 6.14) responses and current (figure 6.15) and voltage (figure 6.16) waveforms, for the starting process and load torque (8Nm) applied at 0.2s, and reversal with no load at 0.5s and load torque applied at 0.6s. Fast response with rather low current ripple is obtained grace to the rather high (20kHz) switching frequency for a 5 ms electrical time constant d.c. brush PM motor four quadrant drive.

32 Electric Drives32 Figure Speed transient response Figure Torque response

33 Electric Drives33 Figure 6.15.Current waveform (i a ) Figure 6.16.Voltage waveform (V a )


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