Current limiting on homemade driver

[MrBean]

I'm building a stepper driver using 5804B IC's which are rated at 1.25A. The steppers I have are rated 5.1 Volts, 1.3 Amps. Which I think means they would have a resistance of 3.923 Ohms (I'll call this 4 Ohms).

I have run one of the steppers from the +5V of a computer PSU and it worked ok, but really needs to use a higher voltage.

If I use a 12V supply the motor would draw 3 Amps blowing itself and the 5804 up, so I need a current limiting resistor. I can use a 6.8 Ohm resistor in series with the motor for a total of 10.8 Ohms. 12/10.8=1.11Amps (nice and safe), but how do I calculate the power rating of the required resistor. Do I use the total resistance of 10.8 Ohms.

e.g. 12V/10.8 Ohms=1.11A. > 12V*1.11A=13.32 WATTS

Or just the resistance of the resistor.

e.g. 12V/6.8 Ohms=1.765A. > 12V*1.765A=21.18 WATTS

If I go higher with the voltage the difference in the calculations gets much larger, along with the cost of the resistors.

Maybe I have done this totally wrong, but I'd really appriciate any help on the subject.

I plan on eventually building a PWM circuit for current limiting, but that's a way off yet (if ever).

[abasir]

If I recall correctly, P=V*I=I*I*R

Thus for your case, taking the latter;
P = 1.11 * 1.11 * 6.8 = 8.37W (minimum)

regards,

[balsaman]

it's the voltage that is dropping across the resistor times the current.

So, its 12 volts minus motor volts times current

Basically 7 volts times 1.1 or 7.7 watts.

You can download Steppercalc at www.e-zflight.com/files/stepcalc102.zip

This figures it all out for you.

Eric

[MrBean]

balsaman,

This does not seem right. Both the motor and resistor will see 12Volts. The resistor dumps current not voltage.

My guess would be to view the motor and resistor as one resistor as they are wired in series. Thus

12v/10.8 Ohm= 1.11A = 12v*1.11A= 13.32WATTS

But as I'm unsure, I've asked others for help.

3 posts, 3 different figures. Anyone care to garantee that theirs are correct?


Oooouuuhhh. What if you then divide 13.32W by 10.8 Ohms giving 1.23 ohms per watt. (opw ???). multiply 1.23 opw by 4 ohms (motor), giving 4.93 WATTS consumed by the motor, leaving 8.39 Watts of the total 13.32 Watts, to be consumed by the resistor.

That's only 0.02 Watts away from abasir's calculation.

Anyway, thanks for the comments. Perhaps I'll never know the real answer and use over rated resistors, like 25Watts just to be safe.

Thanks again. Regards...

[abasir]

Mr Bean,

The current in the circuit is 12V/10.8ohm=1.11A.

The voltage drop across the resistor is V=IR=1.11A*6.8ohm=7.48V
Using P=VI=7.48V*1.11A=8.3W

and voltage drop across motor is V=IR=1.11A*4.0ohm=4.44V
Using P=VI=4.44V*1.11A=4.9W

The motor & resistor see the same current but the voltage drop is different.

Make sense?

[MrBean]

abasir,

Thanks, I understand the math now. One thing I am still unsure on though.

I read somewhere that steppers perform better at a higher voltage than that which they are rated at, and that they can be run up to around 10* their voltage rating. (10* may be incorrect, I can't remember the exact number).

However, the current must be limited to prevent damage to the motor.

In your last post you state:

"and voltage drop across motor is V=IR=1.11A*4.0ohm=4.44V"

does this mean that the motor is still running from 4.44V and not 12V. If so, I may as well just run them at the rated 5V and not worry about current limiting.

Maybe I have missunderstood the article I read a while back.

Thanks for the help. I know more today than I did yesterday, so at least I have made some progress.

[abasir]

The article you read is correct. I'm running my motor at 12x rated voltage now.

The calculations shown is the 'steady-state' current, i.e., the coil has been switched on for some time. The motor coil (inductor) will resist the _initial flow_ of current when switched on, thus the higher voltage will speed-up the process. Subsequent to this, the inductance effect reduces because current already flowing; this is when the power resistor come into action to limit the current (above calculations).

I'm no expert in the subject so other Gurus here may explain to you better. You may want to read the white paper at www.geckodrive.com for some additional info.

Another informative site (Warning: Nice charts but lots of maths )
http://www.cs.uiowa.edu/~jones/step/current.html Check out Figure 4.2; specific to your question above.

Regards,

[balsaman]

My calculation was different because I was "rounding" a little. Our posts were being written at the same time...

Eric

[MrBean]

Thanks guys. I've been surfing and reading information like a madman and the stuff that made no sense has become understanable. A lot of it still looks like rocket science, but I feel I understand enough to continue building my driver board.

Many thanks...