Re: Mariss' power supply circuit

Hi:

This is my first post to the group, but I've been reading it
on-and-off since I talked to Bill W and Jon Elson at NAMES.

A quick introduction: I have a Shoptask machine and one of
these days I'm gonna add CNC to it. I do power electronics
for a living - I design VFDs, and used to work on industrial
DC drives as well. Unfortunately I mostly work on drives
that are 200HP and up, way too big for home shop use. When
looking at CNC stuff, I have to keep reminding myself that the
rules are different here. Stuff that is absolutly neccessary
at 200HP is massive overkill and too expensive at 1/4 HP.

Now to the subject of the post:

--- In CAD_CAM_EDM_DRO@y..., "mariss92705" <mariss92705@y...> wrote:
> Hi,
>
> My assumption has been this: A motor has a stall (locked
> armature) current and torque rating. The current is equal
> to the motor's rated voltage divided by its armature
> resisistance; the torque is equal to the motor's torque
> constant multiplied by that current.

I have to contradict your assumption about current ratings and
armature resistance. I will qualify my statements by saying
that I know they apply to industrial sized DC motors (1HP and
up - my area of expertise). I believe that they also apply
to smaller motors, with the possible exception of very small,
inefficient motors as seen in toys, etc. Here goes:

The armature resistance is not equal to the motor's rated
voltage divided by it's rated current (or even it's peak current).
It is usually much smaller than that. If the armature resistance
were equal to rated volts over rated amps, then at rated amps,
all of the voltage applied to the motor would be dropped across
the resistance, and all of the energy supplied would be turned
into heat. Instead, the resistance is kept as small as possible,
so that the vast majority of the voltage across the motor is the
counter-EMF, and as little as possible is IR drop. In the ideal
case, R is zero, IR drop is zero, and armature voltage is directly
proportional to speed.

> The motor will momentarily draw that rated stall current
> as it accelerates from rest when the rated voltage is
> abruptly applied.
> That is exactly identical to shorting a motor running at
> its rated no-load speed. Both currents will be identical
> and within ratings.
>
> Mariss
>

In large motors, the IR drop at rated speed may be only 1-2%
of the total armature voltage. If you abruptly apply rated
voltage to a stopped motor, it will draw 50-100 times the rated
current, unless the power source limits the current. (Just
about all drives, both industrial and servo, do limit current.)
Typically, the drive limits current to 150-200%, and the motor
accelerates in current limit until the counter-EMF gets close
to the requested voltage. Then the drive comes out of current
limit and current drops to whatever level is needed to deliver
the steady-state torque reqirement of the load.

Shorting the armature of a motor while it is spinning at high
speed will also cause 50-100 times rated current to flow. And
with a short, there is no current limit. I am personally aware
of one case where the semiconductors in a DC drive failed,
shorting the armature of a 25 HP motor while it was spinning
a roller on a paper machine (high inertia load). The fault
current generated enough braking torque to snap the shaft of
the motor (around 2" in diameter)! That was a shunt wound motor,
not a permanent magnet one. I expect that the resistance of the
small PM motors used for servos is higher, but I still wouldn't
short the armature for braking. DB resistors used on industrial
motors are often sized to draw 150-200% of the continuous rated
current at the rated voltage. Servos have higher peak ratings,
maybe you might want to size it for 300-400%. But don't just
short it.

John Kasunich