How big a stepper motor and driver do you need?
"Pick the weight of the heaviest item you are pushing around. If it weighs 40lbs, use 40lbs. multiply it by the IPM [Inches Per Minute] you want. Say that's 1,000 IPM. Divide the result by the magic number "531". The answer is 75.3 Watts".
Watts = IPM * Lbs / 531
Watts = torque (in in-oz) * RPM / 1351
This is generally going to me a more accurate estimate of the motor required for a mill or other system where a lead screw is pulling a load because the additional friction and mass of the system that moves the load is also included when you measure the torque required to turn the screw.
Estimating Drive Amperage
Watts = amps * volts so you need a current and voltage rating that, when multiplied, reach more than your minimal required watts. In the example above, with a 75.3 watt power output requirement, you need a 4 amp motor driven at 24 volts or a 3 amp motor driven at 30 volt or a 2 amp motor at 36 volts...
Estimating Drive Voltage
Given a target wattage (which is the power you need from the motor), the voltage needed to drive the motor is that wattage, divided by the rated amperage of the motor.
DriveVolts = Watts / Amps
The voltage on the motor label, is the steady state voltage, or the voltage which the motor can support when it is sitting at rest, which will be much less than the voltage used when the motor is turning. The maximum voltage that the motor can take when turning, the maximum safe drive voltage for the motor, can be estimated from the motors Inductance rating (if known) by finding it's square root and then multiplying that by 32.
MaxVolts ~= SQRT( Inductance in mH ) * 32
In the example above, if the 4 amp motor had an inductance rating of 0.75mH or more, then we could feed its driver 24 volts and expect to get about 75 watts power out and drive our 40 lb load at 1000 IPM. That assumes a good lead screw and bearings, and no binding or other loads, of course. It also assumes 100% efficiency, which just doesn't happen, so build in an extra 20% or so as a "reality" factor.
In general, the motor needs more voltage to run at higher speeds, because the voltage pushes the current to the necessary level faster. Think of it like this: If you have a /really/ long garden hose, and you want a trickle of water out the far end, you can supply a trickle at the valve and just wait... eventually, the water will trickle out the other end. But if you want that trickle NOW, you turn the water value on full, then wait for the water to start coming out the end before you turn it down. Of course, if you fail to turn it down, you are going to swamp the motor at the other end of the hose! That is why the motor driver must regulate the current by turning down the drive voltage once the coil reaches it's maximum current flow.
Higher supply voltage allows faster changes to the current and field strength in the motor coils and so faster rotation. But there is also a balance; at higher PSU voltages the motor will get hotter (because of increased wattage) and it will jerk faster from step to step, so you get more resonance and more problems with stalling. A higher voltage can actually stop the motor getting through resonance bands.
Roman Black says: "I hate seeing the "bigger is better" mentality, good design is about balancing all the factors for the exact performance needed."
Also: Java simulation of a stepper motor drive circuit demonstrating the advantage of higher supply voltage combined with current regulation.
This little calculator can help find the maximum speed that a motor can rotate
given it's Inductance and steps/rev and the drivers amps and voltage. It's
based on the fact that the time it takes to build up a field in a coil is
T= (I * L) / V
To make one step, the field must build up, and collapse. This is for unipolar and bipolar parallel
T = 2 * ( L * I ) / V
For a motor connected bipolar serial, although you get more torque, the field must build from the maximum in one direction, through zero, up to the maximum in the other direction so:
T = 2 * ( L * I * 2 ) / V
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