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

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

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

The output current expressions are obtained from (6.2) - (6.3): (6.6) 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 ia(ton) = ia’(ton) (6.8) Electric Drives

The boundary conditions are: For the second quadrant chopper (table 6.1.b) the d.c. motor e.m.f. eg with S2 on produces a current rise in the inductance La: (6.10) When S2 is turned off the energy stored into the inductor is sent back to the source as long as V0 > Va: (6.11) with the solution: (6.12) (6.13) The boundary conditions are: ia(ton) = ia’(ton), ia(0) = ia0 and ia’(T) = ia0 (6.14) Electric Drives

Figure 6.2. Second quadrant chopper operation It is thus possible with eg<V0 to retrieve the energy back from the d.c. brush motor by using the inductor La as an energy sink (figure 6.2). Figure 6.2. Second quadrant chopper operation Electric Drives

Figure 6.3. Source current waveforms a.) first quadrant operation b.) second quadrant operation Electric Drives

The nth harmonic current in in the supply (figure 6.4.b) is: (6.15) 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 nth harmonic current in in the supply (figure 6.4.b) is: (6.15) Figure 6.4. First quadrant chopper with LC input filter a.) basic circuit b.) equivalent circuit for nth harmonic Electric Drives

6.2. THE FIRST QUADRANT (STEP - DOWN) CHOPPER A d.c. brush motor with permanent magnet excitation with the data Ra = 1W, Kelp = 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 a from zero to maximum speed; b. the range of speed Solution: a. The average output voltage Va is: (6.16) At standstill n = 0 and thus: (6.17) For maximum speed the voltage is maximum: 120V (a = 1). Consequently a varies from 1/12 to 1. Electric Drives

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 a = 0.3 calculate the actual armature current waveform, its average value, and voltage average value at n = 1600 rpm, for the chopping frequency fch = 50Hz. Determine the chopping frequency for which the limit between discontinuous and continuous current is reached at the same ton as above. Solution: We now apply the armature current expressions (6.6) - (6.7) first for fch = 50Hz. The turn - on time interval (6.20) with Electric Drives

Assuming discontinuous current mode: (6.22) Also: (6.23) (6.21) Assuming discontinuous current mode: (6.22) Also: (6.23) The current ia’ becomes zero at t = t1: (6.24) Thus indeed the current is discontinuous. Figure 6.5. Discontinuous current Electric Drives

The average current iav is: (6.25) The chopping frequency for which the limit between the discontinuous and continuous current is obtained for: (6.26) In this case a becomes (6.27) As the current is discontinuous the average voltage is from (6.5): (6.28) Electric Drives

6.3. 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: Ra = 1W; La = 20mH, eg = 80V (given speed). The supply voltage V0 is 120Vd.c., and ton = 5.10-3 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 6.2. - 6.3. with ia0 = 0. (6.29) b. (6.30) Electric Drives

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

The average source current iav is: (6.33) (6.34) Note that the source current occurs during the S1 turn - off and is negative, proving the regenerative operation. The average source current iav is: (6.35) The average power regenerated Pav is: (6.36) Electric Drives

6.4. 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 Imax>0 and Imin < 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 Imax>0 and Imin<0 with Imax > |Imin|. b. Derive the expression of the conducting times td2 and td1 of the diodes D1 and D2 for case a.) Electric Drives

T2; for td1 < t < T; (6.37) D1; for tc < t < td1; 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 T1, D2, D1, T2 are: T1; for td2 < t < tc; T2; for td1 < t < T; (6.37) D1; for tc < t < td1; D2; for 0 < t < td2; b. The equations for the current are: (6.38) (6.39) with the solutions: (6.40) Electric Drives

(6.41) with the boundary conditions ia(0) = imin, ia’(T) = imin and ia(tc) = ia’(tc). Figure 6.8. Voltage and current waveforms of two quadrant chopper fed - d.c. motor Electric Drives

The unknowns A, A’ and tc are obtained from: (6.42) (6.43) (6.44) Consider a d.c. brush motor whose data are: e0 = 120V, Ra = 1W, La = 5mH, eg = 80V, Imax = 5A, Imin = -2A and the chopping frequency fch = 0.5kHz. For the two quadrant chopper as above calculate: a. The tc / T = aon ratio b. The conducting intervals of the 4 switches Electric Drives

Solution: The constants A, A’, tc expressions developed above ((6.42) - (6.44)) yield: (6.45) (6.46) (6.47) (6.48) Electric Drives

The conducting interval of D2, td2, corresponds to ia = 0: (6.49) Thus the main switch T1 is conducting for a time interval: (6.50) To calculate the conducting time of the diode D1 we apply the condition ia’(td1) = 0: (6.51) (6.52) Consequently the diode D1 conducts for 0.303 ms. Finally the static switch T2 conducts for the time interval (6.53) Electric Drives

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

Figure 6.10. Four quadrant chopper supplying a d.c. brush motor a.) Third quadrant: iav<0, Vav<0; b.) Forth quadrant: iav>0, Vav<0; Electric Drives

6.6. 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 ich occurs at a duty cycle a = 0.5. b. For fch = 400Hz, Ia = 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 Lf and Cf of the filter for fch>2fr. c. For case b.) calculate the d.c., first and third harmonics of the supply current. Solution: Thus the d.c., Isdc rms Isrms and ripple Isripple components of chopper currents are: (6.72) (6.73) Electric Drives

The maximum ripple current is obtained for which leads to a = 1 / 2. (6.74) The maximum ripple current is obtained for which leads to a = 1 / 2. The filter design will be performed for this worst case. The chopper configuration (figure 6.11) provides square current pulses for is of width a. Figure 6.11. First quadrant chopper with LfCf filter a.) basic circuit b.) chopper input current Electric Drives

To design the filter we need first the chopper current harmonics content which, for a = 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 . Electric Drives

According to equation (6.15) for n = 1 we obtain: (6.80) Also the fundamental capacitor current IC1 is: (6.81) As each 1mF capacitor can take 5A, 10 such capacitors in parallel are needed and thus Cf = 10mF. On the other hand the reactance xL is: (6.82) (6.83) Electric Drives

The filter resonance frequency fr is: (6.84) The ratio between the chopper switching frequency fch and the filter resonance frequency fr is: The same a.c. current components (6.15) (after filtering) are: (6.85) (6.86) (6.87) As noticed the LfCf filter produces a drastic reduction of source current harmonics. Electric Drives

6.7. DIGITAL SIMULATION THROUGH MATLAB-SIMULINK Figure 6.12. D.C. brush motor drive - block diagram Electric Drives

The integration step (10ms) can be modified from the Simulink’s Simulation / Parameters. The chopper frequency is 20kHz and this block input is the tc / Te 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 6.12. The motor used for this simulation has the following parameters: Vdc = 120V, In = 20A, nn = 3000rpm, Rs = 0.5W, La = 0.0025H, J = 0.001kgm2, Kelp = 2.2Wb. The figures (6.13 - 6.16) 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. Electric Drives

Figure 6.13. Speed transient response Figure 6.14. Torque response Electric Drives

Figure 6.15.Current waveform (ia) Figure 6.16.Voltage waveform (Va) Electric Drives