I did that because i was under the impression that you are using the "pulse frequency" (that is the step frequency?) as an input variable. I meant to explain the frequency at which the coils have on-time and off-time, its not related to the step line. You started with 50uS value, where does that come from.
The current rises like I(t)=V/R*(1-e^(-t/tau)) and from this, we figure out how much time it needs so that I(t) = I(peak) where the I(peak)=0.5A, is the setting made by the driver potentiometer. So this equation needs to be solved first, and solved for (t) which is the variable. Then we know how much time this coil needs from the point when the voltage was applied, until the current has reached the 0.5A mark. For the example given is about Ts=8*10^-4. And if we make 1/Ts, that gives the maximum frequency which would accommodate that large rise time, if the rises would occur just one after another, and that is a little over 1Khz and very far from 30Khz. This is basically why we do not use 19ohms coils with large inductance values. Basically wrong motor type, but you need to do that match comparatively with a motor like 3mH and 2ohms, and will see in the latter case it reaches 0.5A in just 6*10^-5, which is well closer to that starting value anyway (again considering that only rise time exists, without off-time).
The current rises like I(t)=V/R*(1-e^(-t/tau)) and from this, we figure out how much time it needs so that I(t) = I(peak) where the I(peak)=0.5A, is the setting made by the driver potentiometer. So this equation needs to be solved first, and solved for (t) which is the variable. Then we know how much time this coil needs from the point when the voltage was applied, until the current has reached the 0.5A mark. For the example given is about Ts=8*10^-4. And if we make 1/Ts, that gives the maximum frequency which would accommodate that large rise time, if the rises would occur just one after another, and that is a little over 1Khz and very far from 30Khz. This is basically why we do not use 19ohms coils with large inductance values. Basically wrong motor type, but you need to do that match comparatively with a motor like 3mH and 2ohms, and will see in the latter case it reaches 0.5A in just 6*10^-5, which is well closer to that starting value anyway (again considering that only rise time exists, without off-time).