Re: [CAD_CAM_EDM_DRO] Re: Confused over My own Calculations!

mayfieldtm wrote:

> Thanks Dave:
>
> I understand what your saying, but it just does not make sense that a
> the name plate rating would be specifying any particular mode such as
> bipolar or series or whatever unless it specifically says so.
>
> I take the amp rating to be the maximum amperage on any one winding.
> All the other specs such as ohms refer to only one winding.
>
> If a motor has 8 leads or 4 windings, then each winding can handle up
> to 3.5 amps, and if you tie two of the winding together in parallel,
> then that pair can handle 7 amps. Then you tie the other 2 together,
> then that winding pair can handle 7 ampers also. So the total motor
> current is 14 amps.
> Take this same motor and connect in series, and each winding pair can
> handle 3.5 Amps for a total motor current of 7 amps.
>
> Is what your saying is that the stated maximum current is the TOTAL
> maximum motor current in any configuration? Now that just might
> penetrate my thick skull. But, in that case, you would think a better
> spec would be maximum wattage.

No. Wattage is a better way to calculate it for a chopper-type drive.
Each winding is, say, 2 V at 4 A, so that is 8 W. So, the real power
delivered to each winding at idle is 8 W, or 16 for both. Now, you have
to add some factor for losses in the motor and the driver. This might
be 5 W or so at idle. So, 21 W as a rough guess. With a 65 V power
supply, the draw should be I=P/V = 21/65 = .32 Amps.

Now, this is ONLY at idle, and will vary due to the design of the motors
and drivers. When the motor is producing substantial torque at some
speed, the power drawn will increase dramatically, and therefore the
current mist also rise. But, the peak power may occur only during acceleration,
and therefore be transient. So, you should not need a power supply
rated for continuous output of 8 A per motor in the above example.
Probably 4 A/motor would work in most cases, except where you know
all motors would be accelerating simultaneously.

Jon