Re: O'scoping stepper OV, RH6.1Linux hardware compatibility
Posted by
Jon Elson
on 2000-01-30 21:28:49 UTC
Multi-Volti Devices wrote:
If the freq was 2.3 x the equivalent TC, I suspect the motor would have little
or no torque, since the current in the coils would only be about 5% of the
rated value.
Oh, wait a minute - what is this L **R** time constant!?! It sounds like
it might apply to RL stepper drivers only, not switching-type drivers.
In a plain RL stepper driver, the R is chosen such that the applied DC voltage
divided by the sum of the external resistor and the winding resistance (usually
small) is equal to the rated current. The second kicker is that there is some
finite current decay time when a winding is switched off. Anyway, the
RL time constant is dominant in this circuit, and as you step faster, the
winding current never reaches the rated value.
In a switching driver, with only the winding resistance in the circuit, (as well
as the inductance, of course) and a higher DC voltage available, the winding
current rises much faster. If the switching circuit did not switch off the
transistors, the current would greatly exceed the ratings.
Becuse of this difference, comparing the RL time constants can't be done so
easily. For the first case (non-switching), it is the simple time constant,
T = L/R. This T (really a Greek Tau) is equal to the time it will take
the current to reach 63% of the rated current. (Note the odd quirk that
it gets faster for larger R, but that will require a higher DC supply voltage.)
The current will rise exponentially, as i/Imax = (1-exp(-t/T)), where i
is instantaneous current, Imax is the rated current (as in I = E/R),
t is time, and T is Tau=L/R.
For a switching driver, T = L/R still holds, but the time it takes to reach
63% of RATED current (not maximum) is much shorter, due to reduced
R and increased DC voltage. Under these conditions, it is pretty clear
why a switching stepper driver allows the motor to run faster.
(Yeah, now everybody knows I'm an electrical engineer.)
Jon
> Someone made reference to a Ted Lin article in Test & Measurement WorldUmm, could it be that the time between pulses is 2.3 x the LR time constant?
> about stepper quirks, and a salient point I got from it was optimum torque
> is obtained when the drive frequency is 2.3 times the motor L/R time
> constant. Of course the frequency probably varies with applications (like
> speed variation?) but it gives one a quantitative reference for maximum
> speed and torque.
If the freq was 2.3 x the equivalent TC, I suspect the motor would have little
or no torque, since the current in the coils would only be about 5% of the
rated value.
Oh, wait a minute - what is this L **R** time constant!?! It sounds like
it might apply to RL stepper drivers only, not switching-type drivers.
In a plain RL stepper driver, the R is chosen such that the applied DC voltage
divided by the sum of the external resistor and the winding resistance (usually
small) is equal to the rated current. The second kicker is that there is some
finite current decay time when a winding is switched off. Anyway, the
RL time constant is dominant in this circuit, and as you step faster, the
winding current never reaches the rated value.
In a switching driver, with only the winding resistance in the circuit, (as well
as the inductance, of course) and a higher DC voltage available, the winding
current rises much faster. If the switching circuit did not switch off the
transistors, the current would greatly exceed the ratings.
Becuse of this difference, comparing the RL time constants can't be done so
easily. For the first case (non-switching), it is the simple time constant,
T = L/R. This T (really a Greek Tau) is equal to the time it will take
the current to reach 63% of the rated current. (Note the odd quirk that
it gets faster for larger R, but that will require a higher DC supply voltage.)
The current will rise exponentially, as i/Imax = (1-exp(-t/T)), where i
is instantaneous current, Imax is the rated current (as in I = E/R),
t is time, and T is Tau=L/R.
For a switching driver, T = L/R still holds, but the time it takes to reach
63% of RATED current (not maximum) is much shorter, due to reduced
R and increased DC voltage. Under these conditions, it is pretty clear
why a switching stepper driver allows the motor to run faster.
(Yeah, now everybody knows I'm an electrical engineer.)
Jon