Electro Mechanical System



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High Speed Operation

  • Each coil of the Stepper motor is switched through a transistor based circuit
    • L and R represent the inductance and resistance of the coil
  • Current pulse in winding is assumed to be ideal
  • Practically it is not ideal due to the time constant To = L/R of the coil

High Speed Operation

  • The performance of the stepper motor is effected
  • Due to the time constant initial current i1 rises slowly
  • Initial torque developed is low
  • Rotor moves slowly
  • When transistor is switched off
  • Current i2 continues to circulate through diode

High Speed Operation

  • Effective duration of pulse is increased to Tp+3T0
  • Switching time of coil is prolonged
  • Shortest possible pulse can be achieved if the current is switched on and immediately switched off when it reaches its maximum value
  • Duration of shortest pulse current I is 6To sec
  • Time constant of stepper motor coils is in between 1 ms (min) to 8 ms (max)

High Speed Operation

  • Therefore duration of one step can not be less than 6 ms (6To = 6 x 1 ms = 6 ms)
    • Assuming the lowest time constant of 1 ms
  • Max. stepping rate is 1000/6 or 166 sps
  • 166 sps is considered to be slow
  • To achieve higher stepping rate the coil time constant has to be reduced
  • The stepper motor coil itself can not be changed
  • So how can the Time constant be decreased?

Modifying Time Constant

  • Time constant can be decreased by adding external resistance and raising supply voltage so same rated current I flows
  • 1000 sps become feasible
  • What is the disadvantage of raising the supply voltage?
  • The power supply becomes more expensive
  • Considerable power is wasted in external resistor

Bi-level Drives

  • Bi-level drives allow fast switching times without using external resistors
  • The switching circuit is modified
  • Instead of a single transistor switch Q1
  • Two transistor based switches Q1 and Q2 are used
  • Switch Q1 is initially closed
  • Q2 is closed to initiate pulse
  • Time constant is L/R = 8 ms

Bi-level Drives

  • When Q2 is closed the switching circuit becomes a simpler circuit as shown
  • Current reaches 200 A in 8 ms
  • Current rises @ 200 A/8 ms = 25000 A/s
  • Rated current of 10 A is reached in 0.4 ms
  • Switch Q1 opens at 0.4 ms

Bi-level Drives

  • The circuit after opening of switch Q1 becomes as shown
  • Current of 10 A is maintained
  • The current remains at 10 A until the step pulse is to be terminated
  • To terminate pulse after 5 ms Q2 is opened

Bi-level Drives

  • The circuit after opening of switch Q2 becomes as shown
  • 57 V source drives current against i
  • Time constant remains 8 ms
  • Current decreases @ 57/60 x 25000 = 23750 A/s
  • Current reaches 0 A at 10/23750 = 0.42 ms
  • Q1 is closed when current reaches 0 A

Learning Outcome: Bi-level Drive operation


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