max power of stepper motor

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eem2am

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hello

do you know what is the maximum power consumption possible with this stepper motor..........the UBB1

**broken link removed**

We will be driving it with NJM2673 h-bridge driver.
 

As far I can see, there is'nt mention any electrical upper limit for voltage or current.

Exept from that (but should still being taken in account) the thermal resistance will be the limited factor.

If you limit voltage to the motor's voltage ratio, you already have the windings resistance. As I can see, this is a bipolar motor, wich means it have a center connection and most probably four winding connections.

If some driver circuit fails and rated voltage is applied to all four windings at same time, the effect would be:
P = U^2/R = 4 windings * 6V^2/18,5ohms = 7,8W

So 7,8W is worst case scenario, provided that no higher voltage than rated voltage is used.
 

OK thankyou, ..though there are only two windings in this motor....so i presume i must halve your kindly calculated figure?
 

eem2am said:
OK thankyou, ..though there are only two windings in this motor....so i presume i must halve your kindly calculated figure?
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Do you have the motor in hand, or do you plan to buy? I ask because from what I can see, it looks like you can choose between unipolar or bipolar, but otherwise same ratings.

If you've got a unipolar stepper (two windings) that means you should multiply with 2 instead og 4.
 

Hi,

i notice that page 15 of this

**broken link removed**

....says that the UBB1 stepper motor has a maximum power of 1.8W.



Why is the max power so low?

Could it be true that the motor has a peak power and a maximum power and that the peak power for this motor woul dbe much higher than 1.8W
 

Why is the max power so low?
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I don't see a much higher peak power in the datasheet (although some thermal overload can be possibly applied to the motor without damage).

Round about, the said maximum power consumption is achieved, when one winding is supplied with nominal voltage. This would be the case in (slow) full step operation of the motor. Obviously, it's not designed for a failing driver, overvoltage or other faults.

However, if you intend half-step, you have to reduce the current respectively voltage, at least when powering two windings to keep the specification.

If you've got a unipolar stepper (two windings) that means you should multiply with 2 instead og 4.
Click to expand...
Apart from the point, that the design doesn't consider driver failure, you apparently confused unipolar and bipolar operation. The latter has two windings.
 

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