That 50uS assuming 20?kHz which the smoothie can output is not related to the coils, it perhaps related to the speed at which motor moves, because each step means 1 move, so ofc more moves per sec is more speed. But stepper driver does not care for that frequency, each step is processes one after another, means 1 move and nothing beyond that, the step frequency cannot be an input variable. Basically first assumption is wrong, and the approach is made from wrong direction. This part between uC and driver input, the step line, is just chip to chip communication, you can take it as a protocol, like I2C or SPI, or something like that, it serves just to send commands, but it does not directly influence what happens to the motor coils.Quote
MeltManBob
I[...] the capabilities of the micro controller to send pulses [...]
What happens to the motor is the job of the stepper driver IC which is the "chopping boss" - the local chip in charge with an axe and cuts down current. You need to apply the formula and see it how the driver sees the motor coils. At t=0 driver puts the voltage on the coil, and the coil current needs some unknown time to reach the set point of 0.5A. Because inductors fight against current changes, that current rise takes time. From the formula the time is the variable, so you find out exactly how much time it takes for the current to go from 0 to 0.5A. This is the key number, has several implications. From that you can deduct what actually happens, for example if the current does get chopped or not. Making 1/that would be a very gross approximation of frequency because it would assume that all periods are only on-time and the current magically instantly disappears in between, but well that is to give an idea of the frequency. When that math is actually put in practice, those rise times need to space out to make room to the fall times, so the current can also "decay" each time after it rises, and that decay is at different time interval than the rise time which was previously calculated.