Re: [CAD_CAM_EDM_DRO] stepper amperage question

Hi Kevin,

All stepper motors want a limited amount of current. 3A in your case.
Too much, and you fry the stepper (yes, doubling the voltage usually
doubles the amperage)!

The faster you get the motor current to 3A, the faster the motor will be
able to accelerate (L/R time constant). SO, we can use MORE Voltage to
get us to 3A faster. As long as we don't exceed 3A. The simplest way
is to limit the current, with a BIG ol' resistor. This is often done in
inexpensive unipolar drivers. If we want to use 12V to "get us there"
faster, then we need to drop 7V across the resistor (12-5=7). E=I*R, so
2.3 ohms ought to do it. This is called an L/R driver. 21 watts will
need to be dissipated in the resistor (3*7), the motor gets 15W (notice
a problem here?). A better way is to "chop" the current, such that it
never exceeds 3A. Most all bipolar and some unipolar drivers do this.
No big resistors! The current is rapidly turned on and off, and only
allowed to reach 3A.

So, if if you have an L/R drive (formula above), either add, or increase
the resistance AND wattage of the L/R resistors (the big HOT ones). A
chopper driver would probably already be running at or near it's rated
voltage; which is typically limited by the ratings power devices. An
L/R driver would also be limited by ratings.

HTH

Alan KM6VV


kevinagilent wrote:

>
> I would like to know if someone out there can shed some light
> when you have a unipolar stepper motor and a unipolar stepper
> controller the motor is rated at say 5v. and 3. amps if you want to
> increase the motor voltage to say 12v. does the amperage also double?
> lets say i want to increase the voltage to 24v. what would the
> amperage need to be ? is the a formula that someone could tell me?
> thanks for the help
> kevin
>