Re: [CAD_CAM_EDM_DRO] Re: It's more VOLTAGE needed to overcome back emf in the coil

turbulatordude wrote:

>--- In CAD_CAM_EDM_DRO@yahoogroups.com, "Steve Stallings"
><stevesng@n...> wrote:
>
>
>>As I stated in my original post, it is the VOLTAGE to the motor coils
>>that will increase as the motor spins faster and develops more back
>>EMF. The chopper circuit will continue to control the maximum
>>current, but the net result is that more power (Voltage times
>>Current) will be delivered to the motor as it spins faster under load.
>>
>>The 1.8 times factor relates to the heat buildup in the transformer
>>when connected to a full wave bridge rectifier and capacitor filter.
>>Most transformer vendors will recommend that the RMS rating of the
>>transformer be 1.8 times the CONTINIOUS current that you intend to
>>draw from the DC output of the supply. CNC motor drivers applications
>>seldom draw maximum current, so it is possible to design for a lower
>>current and base your ratings on the expected duty cycle of the load.
>>
>>
>
>I have seen the 1.8, but it was associated with capacitors. when
>using a transformer for a DC supply and using capacitors, derate the
>trasnformer by using the 1.8. but, if using inductors, do not derate.
> That was one of those simple charts on a transformer page. I'll read
>the link and get a better idea of what is going on.
>
>I had thought the 1.8 was to select the current from the transformer
>so the rectified current was avaiable. a 10 amp AC will drop by 1.414
>to 7.07 amps DC. 10A times 1.8 or 18 amps AC will drop to 12.7 amps
>when reftified and filtered with a cap......
>
>Dave
>
>

It looks like you're going in the wrong direction with that 1.8 factor.

My understanding is that you take the VA rating of the transformer, and
divide it by 1.8 to get the appropriate DC wattage rating.

Once you have the wattage rating, you can divide that by the peak
voltage (which will be 1.414*the AC voltage) to get DC amps.

So - for a 1KVa transformer with 24VAC output:
Divide 1KVa by 1.8 = 555 Watts
Multiply 24VAC by 1.414 = 33.9 V DC output
Divide 555W by 34V = 16.37 A DC

Note that the voltage drop in the diode bridge doesn't affect output
current, only final output voltage.

- Steve