I want to turn on a device away from the voltage source to 2 kilometers. My device operates at 24VDC or 48VDC / 35W.
I need to know how is it possible to amplify the voltage from 24Vdc to 240Vdc or higher. In this manner the current needed is lower and loss due to resistance too.
I don't know how to specify the characteristics of my current cable.
Is there something to help me to reach this goal?
You need a boost converter to lift your 24V DC to 240V DC. Because these are not 100% efficient the power in to supply 240V DC at 35 watts is likely to be about 40 watts.
Then you need to convert it back down to (say) 24 volts at the far end and this will probably mean the full input power will be about 45 watts.
With 40 watts at 240VDC "pumped" onto the cable by the booster the current will be 167 mA and now you have to decide on the wire gauge: -
If you choose AWG 24 - it has 84.2 ohms per 1000m. Double this for the return path and multiply by 2 for 2km length and you get a total resistance of 337 ohms - this needs to provide 167mA to the load and will accordingly lose voltage on the way.
Simple ohm's law tells you that the volts lost are 0.167 x 436 ohms = 56 volts. This would need to be added to the 240V DC you generate from the boost converter so immediately you are looking to generate more like 296 volts rather than 240 volts.
That's not a problem for the load (or the cable) but your input power has risen to about 50 watts.
AWG 16 has a 2km loop resistance of only 53 ohms so you may choose this instead. With 167 mA flowing the volt drop will be about 8.8 volts so maybe this is a better choice.
Hope this helps you get your head around the iterative process of choosing the right cable to suit your power budget.
35 watts at 48VDC is about 730 milliamperes into the load with 48 volts across it, so to specify the cable you need to know the tolerance allowable on the input voltage to the load.
Running it at a higher voltage will allow you to use smaller diameter cable, so just for the sake of this exercise let's say the spec on the 48 volts is +/- 10% and the device is linear, so with 43.2 volts across it it'll draw 657 milliamperes from the source.
With a 48 volt source servicing the load and only 43.8 volts getting to the load means that the cable between the source and the load must be dropping the 4.8 volt difference and, with 657 milliamperes running through it, represents a resistance of 7.3 ohms over the 4 kilometer (out and back) run.
That's about 1.83 ohms per kilometer so, perusing a copper wire table, we find that #7 AWG, at 1.634 ohms per kilometer will fill the bill, but the down side is that it'll be expensive to do it that way.
Another less expensive and, perhaps, more sensible way to do it - if AC mains aren't available at the remote location - would be to send mains power over much smaller diameter wire from the local to the remote location and convert it to 24 or 48 volts DC there; the caveat being, of course, that you're running a moderately high voltage over a fairly long route where lots of untoward things could happen.
Like, for example, a baby chewing through the cable.
