Download Control of stepping motors : a tutorial by Douglas W Jones; University of Iowa. Dept. of Computer PDF

By Douglas W Jones; University of Iowa. Dept. of Computer Science

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Typically, R1 will have as low a resistance as possible, in order to avoid the high cost of a power resistor. 0 volts, so if R1 is 2 ohms, this circuit will limit the current to 1 amp, and R1 must be able to handle 2 watts. 3 must be sized in terms of the current gain of T1 so that sufficient current flows through R1 and R3 to allow T1 to conduct the full rated motor current. 3 is run in linear mode, and therfore, it must dissapate quite a bit of power. For example, if the motor windings have a resistance of 5 ohms and a rated current of 1 amp, and a 25 volt power supply is used, T1 plus R1 will dissapate, between them, 20 watts!

If both are low, both pull-down transistors will be off. If both are high, both pull-up transistors will be off. As a result, this simple circuit puts the motor in dynamic braking mode in both the 11 and 00 states, and does not offer a coasting mode. 14 consists of two identical halves, each of which may be properly described as a push-pull driver. The term half H-bridge is sometimes applied to these circuits! It is also worth noting that a half H-bridge has a circuit quite similar to the output drive circuit used in TTL logic.

This is the result of the forward voltage drop in the diodes that are used to shunt the current when the switches turn off, plus the external resistance used to speed the decay of the current through the motor winding. For large values of Ton or Toff, the exponential nature of the rise and fall of the current through the motor winding is significant, but for sufficiently small values, we can approximate these as linear. Assuming that the chopper is working to maintain a current of I and that the amplitude is small, we will approximate the rates of rise and fall in the current in terms of the voltage across the motor winding when the switch is closed and when it is open: Von = Vsupply - I(Rwinding + Ron) Voff = Vdiode + I(Rwinding + Roff) Here, we lump together all resistances in series with the winding and power supply in the on state as Ron, and we lump together all resistances in the current recirculation path when the switch(es) are open as Roff.

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