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Electro Mechanical System
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| Date | 16.12.2020 | | Size | 458.83 Kb. | | #54565 |
| Lecture 46 - 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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