Re: [CAD_CAM_EDM_DRO] Re: ...power...

mariss92705 wrote:

> NCS,
>
> I guess it's my fault I haven't made my point clearly. Let me try
> again, using the "there is no such thing as a free lunch" method.
>
> Motors generate mechanical power. Dyno test data I have for a 4A, 3V
> double stack 34 motor (M092-FD08) is 108 Watts mechanical at 60VDC,
> or a little over 1/7 HP. While it is generating that power, it
> consumes 159W of electrical power for an efficiency of 108/159 or 68%.
>
> Let's get back to the 159W. Where does it come from if not the power
> supply? If you have 3 motors, then you will need 477W total. At
> 60VDC, 7.95A is needed to deliver 477W.

The trick here is that unregulated power supplies can usually be overloaded
for short intervals without damage. Most, in fact, have surge ratings
which are often about twice the continuous rating.

Also, the condition in which all 3 axes are accelerating or delivering high
power simultaneously is relatively rare in most 3-axis cartesian machines.
(Quite different in hexapods.) So, one can generally live with a somewhat
smaller power supply than the worst-case draw indicates. I would guess
that a 300 W continuous-rated power supply would probably work fine on
a hobby-use 3-axis cartesian machine where the motors were in fact
adequate for the intended use. If the motors were to be used right at their

full capacity, or the machine was for serious industrial use, then the
worst-case load capacity might be necessary.

> You say 0.5A at 60VDC is fine. That is 30W. How do you get 108W motor
> shaft power while heating the motor with an additional 51W from a 30W
> supply? Remember, there is no such thing as a free lunch ;-/

That is clearly too low. This must have been computed from the idling
current
in ONE winding, but the motor has 2 windings. So, each motor needs 24 W
just to idle (in full step mode). So, 3 motors idling would draw 72 W plus
driver losses. But, this is a meaningless calculation for power supply
sizing.

Jon