Need Help!- Question about 'fast decay' mode while designing bipolar stepper drivers

[lxwservo]

Hi,everybody.

I want to thank MechanoMan,Mariss, Kreutz, Dandumit,etc firstly, I learn much from the topic you have discussed.

Now, I designing an two phases hybird stepper driver with DSP2802, IR2104 and H bridge. I sample the current in during the time of slow decay(shorting the motor winding through keeping low Hbridge ON ), and reverse voltage fast decay is adopted as fast decay mode.

If the current is 1.5A and the next microstep commanded requires the current to be 1.25A, through calculation, it need nearly 26us reverse voltage(Assumes the inductance of motor is 0.0085H, resistance is 3.3Ω, stepangle is 1.8° and bus voltage is 90V) . And it need nearly 138us to drop current from 1.5A to 0A.

138us is too large for microstep control while speed large than 3000rpm.

Questions:
1. Did I have a misunderstanding about fast decay mode?
2. The value of L/R is too large for high speed application?


Regards

[lxwservo]

I suppose I get the answer, because 1.5A is the rated current, in 3000rpm, the current maybe 0.5A or even more smaller.

[H500]

At high speeds, you will not be able to achieve the lower current in the required time even with fast decay. That's an unavoidable limitation of the system.