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Electric Drives1 9.16. FEEDBACK LINEARIZED CONTROL Vector control was invented to produce separate flux and torque control as it is implicitely possible.

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Presentation on theme: "Electric Drives1 9.16. FEEDBACK LINEARIZED CONTROL Vector control was invented to produce separate flux and torque control as it is implicitely possible."— Presentation transcript:

1 Electric Drives1 9.16. FEEDBACK LINEARIZED CONTROL Vector control was invented to produce separate flux and torque control as it is implicitely possible with d.c. brush motors. It is also known that, with constant flux, the torque is proportional to torque current and linear torque - speed characteristics may be obtained. Such an artificial decoupling and linearization of induction motor equations may, in principle, be done with some other nonlinear transformations. Feedback linearization control [5 - 7] is such a method. Still equations ((9.116) - (9.118)) - in stator coordinates - with stator current and rotor flux dq components, together with the equation of motion, are required. The new output variables are the rotor flux squared  1 (x) = r 2 and the rotor speed  2 (x) =  r.

2 Electric Drives2 Figure 9.52. Feedback linearization control It should be noted that the computation effort is greater than for vector control or DTFC and a thorough knowledge of motor parameters is necessary. Dynamic performance quite similar to both advanced vector control or DTFC has been obtained [32]. It seems too early to forecast the industrial prospects of feedback linearization control in competition with advanced vector control or DTFC.

3 Electric Drives3 9.17. SCALAR (V 1 /f 1 ) CONTROL For pump and ventilator like applications the speed control range is only from 3 to 1 up to 10 to 1. Motion (speed) sensors are avoided in such drives. Traditionally scalar, V 1 / f 1, open loop control has been used for such applications. In essence the voltage amplitude V 1 and its frequency f 1 are related by: (9.120) V 0 * is called voltage boost and is required to run the motor properly at low speeds. The primary frequency is ramped as desired and, based on (9.120), an open loop PWM procedure is used to control the PWM inverter (figure 9.53).

4 Electric Drives4 Figure 9.53. V 1 / f 1 open loop (scalar) control

5 Electric Drives5 Figure 9.54. Instability zones for V 1 / f 1 open loop scalar control a.) i mr /  1 planeV  / f 1 plane In general [33]  1min is: (9.121) where  r is the rotor time constant.

6 Electric Drives6 Also it has been shown [33] that instabilities occur if the figure of merit f m is: (9.122) where  m is the mechanical time constant. The slope of two asymptotes in figure 9.54.a are directly related to  r and  m. The values of f m for a few 2, 4, 6 pole motors [33] are shown in figure 9.55. Figure 9.55. The figure of merit fm (9.122)

7 Electric Drives7 Compensating the slip frequency is attempted for steady state when, in fact, the rotor flux is constant in time and thus the torque T e (9.15) is: (9.124) with (9.125) The motion equation is: (9.126) Assuming that the mechanical transients are slow the rotor flux may be considered constant. The solution of  r for such slow transients is: (9.127) (9.128)

8 Electric Drives8 The principle of slip frequency compensation method [34] consists of increasing the reference frequency  r * by the estimated slip frequency S  1 (figure 9.56) to move, for given torque, from B to C and thus make  r independent of load. Figure 9.56. Principle of slip frequency compensation

9 Electric Drives9 The signal flow diagram (figure 9.57) of such a scheme illustrates the estimation of the rotor flux from the voltage model. Figure 9.57. V 1 / f 1 scalar control with feedforward slip frequency compensation

10 Electric Drives10 The voltage model is adequate only above 2Hz but this is acceptable in V 1 / f 1 drives. Equations (9.124), (9.126) and figure 9.57 suggest the signal flow diagram of figure 9.58. Figure 9.58. Signal flow diagram for slip frequency compensation

11 Electric Drives11 The  r /  r * transfer function G 0 is simply (from figure 9.58): (9.129) G 0 offers a stable response only if: (9.130) On the other hand the transfer function G 1 between  r and T load is: (9.131) Equation (9.131) suggests that for steady state (s = 0 in (9.131))  r for unit step load torque is: (9.132) So:(9.133)

12 Electric Drives12 For full compensation and thus (point C in figure 9.56). In this case (9.130) is automatically met. In special cases, at low speeds, ( ) in order to provide a stable operation in case of random friction torque perturbations. Good performance has been demonstrated down to 100rpm [34].

13 Electric Drives13 9.18. SELF - COMMISSIONING By self - commissioning we mean here the parameter estimation and controller closed loops calibration by the drive itself, when working on initialization mode, on the site of application, before actual operation starts.

14 Electric Drives14 Figure 9.59. a.) Direct vector current controlb.) current model rotor flux estimator

15 Electric Drives15 However in order to calibrate the current, flux and speed controllers:  first, the stator resistance r s and the transient time constant  ’ = L sc / (r s +r r ) have to be estimated and, based on these values, the current controllers are to be calibrated;  second, with the current controllers on, the rotor time constant  r is estimated and used in the rotor flux estimator (figure 9.59.b). Further, the flux controller is calibrated.  third, the reference rotor flux level and the corresponding magnetizing current are calculated;  fourth, through a no load acceleration test, the mechanical time constant  m (9.128) is calculated. Various methods to estimate the induction motor parameters r s, L sc, L m,  r, J may be classified into:  step voltage response tests at standstill;  frequency response tests at standstill;  dynamic tests (non zero speed).

16 Electric Drives16 To account for the inverter dead time effect, measurements are taken for two different modulation indexes m d1 and m d2 : (9.134) Up to this point the current controllers are only grossly calibrated. The current levels are 50% and 100% of rated current and only a voltage vector is applied (V 1 for example). To estimate the motor transient time constant  ’, the inverter is controlled through a binary port (with the modulator inhibited) for a few microseconds noting the time t peak and the peak current reached, i peak (figure 9.60). Due to the short time interval the main flux does not occur and thus the motor equation is: (9.135) (9.136)

17 Electric Drives17 The average vaue of  ’ from a few tests is computed. Figure 9.60. Step voltage response at standstill Based on the above results, the current controllers (figure 9.61) may be callibrated. T c is the sampling time. Figure 9.61. D.c. (synchronous) current controller

18 Electric Drives18 Based on optimal gain method [36] the PI controller constants are: (9.137) Further, we have to estimate the rotor time constant. We feed the inverter at standstill with d.c. current (applying V 1 with PWM) and then turn it off and record the stator voltage: (9.138) (9.139) Approximately the rotor time constant  r is: (9.140)

19 Electric Drives19 To do so we use the drive in the V 1 / f 1 open loop mode at 10% of rated frequency and the motor no load current is transformed into synchronous coordinates. The rotor time constant  r is changed until the i q current in synchronous coordinates is zero. The d axis current will be the rated magnetizing current i mo * (provided V 1 / f 1 ratio is the rated one), [36]. The same test is performed above rated frequency for rated voltage to get the value of i mr (  r ). Also Lm = r / i mr has to be found in the process. The flux controller (figure 9.59) may be calibrated (figure 9.62). Figure 9.62. Rotor flux loop controller

20 Electric Drives20 The mechanical time constant  m may be calculated from constant torque (i mr, i q constant) rotor field orientation acceleration tests. Noting that the speed reached  a after the time t a, the mechanical time constant is obtained from the motion equation: (9.141) (9.142) With  m from (9.125) and J from (9.142) we get: (9.143)

21 Electric Drives21 The timing of self - commissioning operations may be summarized as in figure 9.63. [36] Figure 9.63. Self - commissioning timing The whole process lasts about 60 seconds. After that the drive is ready for full performance operation. For sensorless operation slightly different tehniques are required for self - commissioning. In high performance drives the motor parameters are tuned through MRAC, on - line adaptation or through on line estimators. These aspects are beyond our scope here. [37, 38]


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