Re: Mariss' power supply circuit

John,

Thank you for posting your thoughtful questions. I will address the
circuit one later, this will be about DC brush motors.

This is a situation where both sides are right; the seeming
difference has to do with the size of the motor and the application
it is optimized for.

Motors can be optimized either for efficiency or for power. An
electric powered model airplane would be an extreme example of a
motor optimized for power. Here the power to weight ratio of the
motor is of paramount importance, motor life and efficiency is
secondary. A 100HP industrial motor would be the other extreme. Here
conversion efficiency and motor life is paramount, size and weight is
secondary.

A DC brush motor delivers peak power at 1/2 no-load speed
(coincidentally 1/2 stall torque), and its theoretical efficiency
cannot exceed 50% at that operating point. This is where you would
operate the model airplane motor.

The same motor develops peak efficiency at an infinitesimal load,
where its efficiency can approach 100%. Because of economic
realities, the actual load is 2 to 10% of stall torque, where
theoretical efficiency will be 98 to 90% respectively.

Let's say the motor develops 100W mechanical in the first instance.
It will also dissipate 100W as heat, being 50% efficient. In the
second instance, the same motor will deliver 4W mechanical and 0.08W
of heat, being 98% efficient. It all depends on the designed
operating point.

Motors used in the context of this list fall between those two
extremes. They generally are below 1kW in power output and are NEMA
42 frame size or smaller. In this class size motor, stall current is
rated and is equal to rated voltage divided by terminal resistance.
Rated voltage can be applied abruptly without demagnetization and the
motor can be shorted while turning at its rated no-load speed without
harm.

Generally their continuous operating torque is between 10% to 25% of
stall torque and is limited primarily by thermal considerations. They
deliver between 85% to 70% efficiency respectively. It is permissible
to operate these motors at or beyond the peak power point briefly.

If anyone is interested, I can write up a short "primer" that
explains why the numbers mentioned are so.

Mariss

--- In CAD_CAM_EDM_DRO@y..., "jmkasunich" <jmkasunich@y...> wrote:
> --- In CAD_CAM_EDM_DRO@y..., bjammin@i... wrote:
> > At 03:14 AM 6/20/02 +0000, you wrote:
> > >> Shorting the armature of a motor while it is spinning at high
> > >> speed will also cause 50-100 times rated current to flow.
> >
> > I'd like to see the math on that.
> >
>
> OK...
>
> 25HP industrial DC motor with 500V armature, rated current
> is 25HP * 746 watts/HP / 500Volts = 37.3amps
> The nameplate values will be a little more, maybe 38-39A
>
> The motor is 96% efficient, and about half the losses are
> iron and field winding losses. That means armature copper
> loss is about 2%, so the IR drop at rated amps is 2% of 500V.
> So the armature resistance is 2% * 500V / 38A = 0.26 ohm.
>
> Motor is spinning at rated speed (1750 RPM), so the armature
> counter-EMF is 500V. Short the armature, and the current is
> limited only by armature resistance.
> Current = 500V / 0.26 ohm = 1923A
>
> 1900 amps through a 38 amp motor is gonna bust something!
>
> > >> with a short, there is no current limit.
> >
> > Sure there is; the windings themselves have resistance.
>
Snip
>
> John Kasunich