So basically what you're saying if I understand you is that drivers have certain amount of on and off time within the pulse time sent to the from the micro controller. Also that you can predict the current rise by the 1-(1/e^(t/tao)) equation but the decay is something I would have to look into on a driver spec sheet. Essentially if I can determine the on time, the off time and the decay equation(s) then I can create an equation to model the current in the motor through both the on and off times and consequently the torque. I'm assuming this makes more of a difference when the on time is used to push the current as high as can be had based on the frequency but doesn't reach the current limit. When the on time is long enough to reach the current limit then the voltage just kicks off and back on very quickly to try and produce an average current equal to the current limit or trip current as you put it. Correct me if I misunderstood or made any wrong assumptions.
One thing I would like to know is lets say I have a motor that produces 'X' holding torque at 1 amp, if the motor current rises to .5 amps can I expect X/2 torque?
OMC-online sells stepper motors and they specify a holding torque rating and a 'rated torque' rating which is 20% less than the holding torque which is why I asked if the holding torque value should be de-rated. In other words is the holding torque or 'rated torque' rating the appropriate one to use in calculating torque based on current in the motor?
To give an example to run the numbers here is a motor 4.2kg-cm Nema 17 circuitspecialists.com
Step angle : 1.8°
Current : 0.5 A
Resistance : 19 ohms
Inductance : 32 mh
Holding Torque : 4.2 kg-cm
Supply Voltage : 24 v
Pulley radius : 5 mm
Pulses per sec. : 20,000
Calculated
Pulse time : 0.00005 seconds or 0.05 milliseconds or 50us
Current based on supply voltage : 24/19 = 1.263 A
Current % based on pulse time : 1-(1/e^(.05/(32/19))) = 1-(1/e^(.05/1.684))) = 1-(1/1.03) = .02925 or 2.925%
Current based on % : .02925 * 1.263 = .036949 A
% of rated current : .036949/.5 = .073898 or 7.3898%
Torque produced at motor : .073898 * 4.2 = .31037 kg-cm (best case scenario where there is no off time)
Torque produced at pulley : .31037 kg(force) * 1 cm = x force * .5 cm = .62074 kg(force) *.5 cm = .62074 kg force * 9.80665 (kg m/s2)/kg force * .5 cm = 6.08738 kg * (m/s2) * .5 cm
Distance traveled per step : (pi * 2 * .5 cm)/200 = .015708 cm or 0.15708 mm
Acceleration : Torque at pulley = Torque from load (mass * acceleration * distance traveled), 6.08738 kg * (m/s2) * .5 cm = 1 kg * acceleration * .015708 cm, divide out the masses and distances = 193.767 m/s2
Top Speed : Distance per step * pulses per second (assuming full step) = .15708mm * 20,000 steps/second = 3141.59 mm/second
Clearly the acceleration and top speed are way beyond what the machine could probably handle let alone the extruder being able to keep up. Even if you cut that in half assuming the on time is half of the pulse time and the decay mode is as fast as possible, those numbers are still high.
One thing I would like to know is lets say I have a motor that produces 'X' holding torque at 1 amp, if the motor current rises to .5 amps can I expect X/2 torque?
OMC-online sells stepper motors and they specify a holding torque rating and a 'rated torque' rating which is 20% less than the holding torque which is why I asked if the holding torque value should be de-rated. In other words is the holding torque or 'rated torque' rating the appropriate one to use in calculating torque based on current in the motor?
To give an example to run the numbers here is a motor 4.2kg-cm Nema 17 circuitspecialists.com
Step angle : 1.8°
Current : 0.5 A
Resistance : 19 ohms
Inductance : 32 mh
Holding Torque : 4.2 kg-cm
Supply Voltage : 24 v
Pulley radius : 5 mm
Pulses per sec. : 20,000
Calculated
Pulse time : 0.00005 seconds or 0.05 milliseconds or 50us
Current based on supply voltage : 24/19 = 1.263 A
Current % based on pulse time : 1-(1/e^(.05/(32/19))) = 1-(1/e^(.05/1.684))) = 1-(1/1.03) = .02925 or 2.925%
Current based on % : .02925 * 1.263 = .036949 A
% of rated current : .036949/.5 = .073898 or 7.3898%
Torque produced at motor : .073898 * 4.2 = .31037 kg-cm (best case scenario where there is no off time)
Torque produced at pulley : .31037 kg(force) * 1 cm = x force * .5 cm = .62074 kg(force) *.5 cm = .62074 kg force * 9.80665 (kg m/s2)/kg force * .5 cm = 6.08738 kg * (m/s2) * .5 cm
Distance traveled per step : (pi * 2 * .5 cm)/200 = .015708 cm or 0.15708 mm
Acceleration : Torque at pulley = Torque from load (mass * acceleration * distance traveled), 6.08738 kg * (m/s2) * .5 cm = 1 kg * acceleration * .015708 cm, divide out the masses and distances = 193.767 m/s2
Top Speed : Distance per step * pulses per second (assuming full step) = .15708mm * 20,000 steps/second = 3141.59 mm/second
Clearly the acceleration and top speed are way beyond what the machine could probably handle let alone the extruder being able to keep up. Even if you cut that in half assuming the on time is half of the pulse time and the decay mode is as fast as possible, those numbers are still high.