# Power dissipation

Discussion in 'General Electronics Chat' started by OneBigOgre, Nov 23, 2012.

1. ### OneBigOgre Thread Starter New Member

Nov 23, 2012
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Hiya, playing with W = I`R. Where I = 12.5A(max) and R = 3 ohms leaves me with 486.75 watts is this correct?
Background: using a L297 to drive a mosfet H bridge to run a 2 phase bipolar step motor. Yes, I want full step (.0005" resolution per step). I want to use the chopper circuit to the sens inputs on the L297. My power supplies output 38.4V, 12.5A dc unregulated. Using 1 power supply per motor. thx for your help

2. ### Audioguru New Member

Dec 20, 2007
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896
Did you measure the resistance of your motor?

The only time the resistance of a motor is used is when it is stalled or is starting because then it is just a piece of wire. When it is running then its resistance is much higher and its current is much lower.

3. ### OneBigOgre Thread Starter New Member

Nov 23, 2012
3
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Hiya, I have not connected a motor yet. The spec sheet for the motor reads resistance/phase Rm ohms .17 also, inductance/phase (Mh @ 1000hz) 2.11 thx for the response

4. ### Audioguru New Member

Dec 20, 2007
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If the motor's resistance is only 0.17 ohms then its stalled and starting current is 38V/0.17 ohm= 224A. Where will that high current come from??

5. ### OneBigOgre Thread Starter New Member

Nov 23, 2012
3
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Hiya, thx for the response. The motor is rated @ 1.7V, 10A. I don't know jack, I'm a Ag mechanic by trade. But trying to learn.

6. ### Audioguru New Member

Dec 20, 2007
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896
Maybe the motor draws only 10A when it is simply spinning and not working hard.