Re: Stepper Motor Power Calculations

--- In CAD_CAM_EDM_DRO@yahoogroups.com, "Phil Mattison"
<mattison20@c...> wrote:

>
> Hi, I'm new to this board so I've been watching for a couple of

days, and it

> looks like the discussion gets pretty technical, so I'll run this

one up the

> flagpole:
>
> I recently built a stepper motor control box for PC parallel-port

stepper

> control using some motor controllers that incorporate PWM chopper-type
> current regulators for the motor phases. The regulators operate at a

range

> of voltages, so I selected a power transformer within the voltage

range of

> the regulator with a current rating equal to the sum of the current
> requirements of the motor coils. But I think I may have overkilled

it using

> this approach.
>
> Now that I think of it maybe I should have calculated the power

consumption

> (watts) of the motors at maximum current, and used that to determine the
> current rating of the transformer, since the motors run at a lower

voltage

> than the transformer supplies.
>
> For example, if the motor coils are rated at 3 amps max, and the

resistance

> is 2 ohms, the coil will drop 6 volts when running at max current, which
> means they are dissipating 18 watts. If I have four of them that's

less than

> 80 watts total worst case. A transformer with a 20-volt secondary would
> therefore need only supply 4 amps, not 12 as in the simple-minded

approach.

> Oh-well, live and learn.
>
> I guess I was thinking the chopper acts like a linear voltage

regulator, but

> on reflection I think it acts more like a transformer, and so the power
> calculations would apply. Has anyone seen a difinitive answer to this
> question? Thanks,
>
> --Phil M.
>

Power is measured in watts and has a voltage, current and time factor.

The average wattage over time is a function of the voltage times the
current times to period of the waveform. In a chopper drive the
voltage is applied to the coil and due to reluctance it takes a finite
time to move the charge into the coil. The higher the voltage the
faster it moves into the coil. The drive pumps voltage into the coil
until the current is reached then the pulse is cut off. The energy is
only being transferred during the time the pulse is on. SInce the
current will reach the cutoff point before a 100% waveform the energy
is whatever the percentage of on time is. The less load the motor has
the less time it takes to reach the cutoff point.

You are correct about the wattage. If the motor would dissipate 18W
at it's stated current and voltage it would still dissipate that
amount with a higher voltage but limited to a waveform that gives the
required current. What changes is the fact that you have to put more
energy into a system than you get out so if the motor is working it's
going to dissipate more energy and needs more watts. No motor is 100%
efficient so you have to build in enough reserve power in the
calculation to compensate for loses and loads.

The (very) conservative number is about 60% of the total current
ratings of the motors BUT in most circumstances that is an extremely
simplistic approach. As the supply voltage goes up the average
current is going to diminish since the period of the waveform will be
less. In a lot of applications not all three motors are running
heavily loaded at the same time and since we are dealing with power
it's over time. Linear power supply components have a lot of
"inertia" and can handle 50 to 100% overloads for a certain duty
cycle. I find that even 50% of the total motor ratings running at
least 10X the motor voltage rating is sufficient in all but the most
demanding aplications.

There is a general belief that if you need a 300W supply then a 500W
is better and a 1000W is even better than that (arh, arh). In reality
it's just wasted money and weight siting there doing nothing. Sort of
like having a 300 hp go cart (:-)

The best way to design a power supply is to actually measure the
current of the machine under the most heavily loaded condition and
then add a 20% fudge factor. Of course that assume you have some way
to power it to begin with with most HSCNC'rs don't so they just build
a monster power supply and go on. Given the nature of components a
power supply that is double the size you need won't cost double the
price. The expensive componets like the transformer and filter cap
will cost at least 50 to 70 % more and since to the home builder space
is plentiful and labor is free those are not factored into the actual
cost.