Operating Voltage

[jhwatts]

What is operating voltage range for the 3 axis 200 oz-in package? The HobbyCNC website says 24v 10A. In another post I see 28v 14A. If I have a range I will not what to look for in a transformer. I don’t want to over drive my equipment and I want it to operate consistently but also meet the necessary power requirements.

[ger21]

Originally Posted by jhwatts What is operating voltage range for the 3 axis 200 oz-in package? The HobbyCNC website says 24v 10A.

Quoted from the website:

42VDC maximum input voltage, 12VDC minimum input voltage. 24VDC Minimum Recommended Voltage.

If you go to their "kits" page, it says 35V recommended.

[spalm]

Gerry is right. Just remember if you are picking your own transformer that they are in AC volts. You will have to add a diode bridge and capacitor to get DC. AC volts times 1.414 = DC volts. This is why HobbyCNC picked a 24 volt transformer (34 volts DC, actually you will get a little more than that as most transformers are under-rated, or are rated at full load). Do not go much over 40 volts DC or you will fry the driver chips.

Again transformers are usually under-rated so a 10 amp transformer will be fine.

Don’t forget to add a heatsink to the driver chips. The tabs are all at ground potential so one bar of aluminum bolted to all of them will work fine. Also add a fan.

Steve

[jhwatts]

I have modified an old ATX computer power supply produce 24 volts at a little over 8 amps. I will be taking the positive 12 volt tap at 8 amps and the negative 12 volt tap at .5 amps to produce 24 volts at 8.5 amps. This seems all good in theory but the power supply is rated at 110 watts and ohms law tells me that (W=I*V) 24*8.5=204 watts. I’m thinking that this probably is not going to work. Any thoughts on this?

[Al_The_Man]

Current is not additive, for that setup the operating current would be the lowest only, which is .5amps.
Also Computer supplies typically have a overated wattage rating.
Al.

[jhwatts]

Is the current determined from adding the reciprocal of the currents and then taking its reciprocal. (1/(1/8+1/.5)=1/(.125+2)=.470 A) In other words scrap that idea.

[Al_The_Man]

Originally Posted by jhwatts In other words scrap that idea.

That part is right.
The analogy is Voltage is pressure and current is flow, so the flow will be restricted to the to the minimum rate. So if you have a current capacity of 8 amps in series with a capacity of .5 amps, then the maximum will be restricted to the lowest rating, being .5a.
Al.