Stepper motor doesn't create enough power

Matthias09

Hey guys,

I'm using a SX28 to drive a stepper motor (see details below). I'm therefore using an external transistor chip (ULN2065B, Quad Darlington, one for every winding, one array can hold up to 500mA) and power the motor with 12V, 1.5A supply. I drive the stepper in half step mode. Everything works find, besides one thing: I've the feeling the stepper is much weaker than it can be, means I have the feeling it does not run under full power. I can too easily prevent the motor from turning just with my hand.

Now my question: What can I do to get more performance (and actually the performance it should be able to create) out of the motor?

Thank you very much! I appreciate your help!

Matthias



JAMECO VALUEPRO 155433, Stepper
UNIPOLAR, 4 windings
42BYG228-R
Voltage: 12V DC
Step Angle: 1.8 Deg
Phase Resis. 30 ohm
current: 400 mA
Phase Inductance: 23 mH
Detent Torque: 220 g cm

Post Edited (Matthias09) : 10/8/2009 10:20:17 PM GMT

Franklin

It looks like that stepper only has 3.8 inch oz of torque. Not much.

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- Stephen

Phil Pilgrim (PhiPi)

The holding torque (2000 gm-cm, or 27.7 oz-in) is the relevant spec here, not the detent torque, which is the torque required to rotate an unpowered motor.

You can improve running torque (but not holding torque) by doubling your supply voltage and connecting 30-ohm 5W resistors in series with the windings. They will get warm and waste power, but this will cut the inductive time constant for bringing the coils up to their rated current when they are switched. For maximum performance, though, a motor with a lower coil resistance and voltage spec is a better choice, since it can then be driven with a current-regulated chopper driver from a high-voltage supply. This provides the maximum efficiency and running torque.

-Phil

Matthias09

Phil,

the idea sounds pretty cool!

Three questions I have:
- how did you determine the holding torque?
- from this point on how did you find out what resistors I'd need (30ohm, 5W)?
- I need the resistors to cut down the current, but still supply the necessary voltage, right?

So you basically say I should change the 12V, 1.5A power supply to a 24V, 1.5A? I need the higher voltage to 'press' the current faster into the windings?
(actually now 5 questions, haha )

Thanks Phil. I do appreciate your comment!

Matthias

Phil Pilgrim (PhiPi)

- how did you determine the holding torque?
It's shown on the Jameco catalog page.

- from this point on how did you find out what resistors I'd need (30ohm, 5W)?
30 ohms is the phase resistance, so by doubling the voltage, you need to double the resistance of the load. Also, 12V² / 30 ohms = 4.8W.

- I need the resistors to cut down the current, but still supply the necessary voltage, right?
It works like this: When the coil is first energized, its resistance appears to be infinite, due to the inductance. Therefore there won't be a voltage drop across the series resistor, and the coil will see the full 24V. This, as you say, will "press" the current into the coil faster than if it saw only 12V. Once the coil begins to conduct and draw current, the series resistor will start dropping voltage, until the coil is saturated and drawing its fully-rated current from what has become, in essence, a 12V supply, due to the 12V drop across the series resistor.

-Phil