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  Reply # 2129271 19-Nov-2018 10:41
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hio77:
andrewNZ: What are you getting 24ac from?


240v to 24vac transformer.

 

Why not a 240v - 12v switchmode powersupply. Really common, dirt cheap, 5A is then easy .

 

I would have thought DC for less power loss long cable runs. Probably wont notice any diff , assuming suitable cable .




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  Reply # 2129305 19-Nov-2018 11:06
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1101:

 

hio77:
andrewNZ: What are you getting 24ac from?


240v to 24vac transformer.

 

Why not a 240v - 12v switchmode powersupply. Really common, dirt cheap, 5A is then easy .

 

I would have thought DC for less power loss long cable runs. Probably wont notice any diff , assuming suitable cable .

 

 

Running 240V would require certain specs to be met, spade failure concerns etc.

 

While the cable is protected and of reasonable sizing, It's simply not where i'd run 240v

 

 

 

 

 

Based on feedback here, looks like i'll be sticking with DC, which i'm happy with..





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  Reply # 2129325 19-Nov-2018 11:25
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It may be that you have a 9-36V (which seem to be pretty standard) to 5V DC-DC converter at the end of your wire, so you could possibly not do any step-down to 12V at all. Just put 36VDC across the link.

 

One benefit of going to a higher voltage over the link is that you can use lighter weight wires = cheaper.

 

 


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  Reply # 2129351 19-Nov-2018 11:49
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Did you say what you are powering up ?   If you are looking for 12V @ 2A  at the end of 200 metres of cabling.  Is the load continuous or If the load is consistent & not overvoltage critical why not use a variable voltage supply & adjust the voltage at the primary source to give 12V at the load end when device(s) operating ?  




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  Reply # 2129379 19-Nov-2018 12:18
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clive100:

 

Did you say what you are powering up ?   If you are looking for 12V @ 2A  at the end of 200 metres of cabling.  Is the load continuous or If the load is consistent & not overvoltage critical why not use a variable voltage supply & adjust the voltage at the primary source to give 12V at the load end when device(s) operating ?  

 

 

There are cells to take boot up load, as i already know this does spike high for like a second or two.

 

 

 

Total continuous load is about 1.2A last i measured, Although there has been a minor change of equipment so that figure may move by 0.4A max.

 

Supply voltage for DC would be going back to my 56V 6A switch-mode supply, Step down at the far end.

 

 

 

 

 

Devices are a few modems and a switch + AP.

 

Solar would be ideal but not really an option due to forestry unless in injected it halfway (and even then i'd have to well overbuild for overcast etc)





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  Reply # 2129386 19-Nov-2018 12:29
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tripper1000:

 

I don't think you are going to find the perfect prebuild solution. You're going to have to build some of it yourself.

 

I've done similar things and I have found AC to be better than DC. Whilst on paper AC and DC have the same losses, I believe most on-paper calculation fail to consider that 24v AC is actually 24v RMS which means the peak voltage is closer to 41 volts, which obviously suffers lower losses than true DC 24 volts. If you are pulling max current the losses would be similar, but you have stated you are over-spec-ing it, so you won't be pulling that peak voltage all the way down to 24v.

 

 

 

 

 

 

It doesn't. While the peak voltage is ~41V, the AC waveform also has large dead zones where very little power is transferred. We use RMS because it provides the same power as DC of the same voltage (in resistive circumstances).

 

 

 

AC actually has higher losses due to parasitic capacitance and reactance.

 

 

 

I'd probably second the advice to run 48VDC.


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  Reply # 2129394 19-Nov-2018 12:48
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And there would also be more losses with AC, as the load is not a resistive load. The rectifier only draws power near the peaks of the AC waveform. While an actual resistive load will draw power at all parts of the waveform except the zero crossing points.

This means that those current peaks have to be larger, to still get the same total amount of power transferred. And larger peak current means more resistive losses.





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  Reply # 2129408 19-Nov-2018 13:16
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You should be able to search for a rectifier on a site that sells electronic components, which meets the specs you require.


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  Reply # 2129716 19-Nov-2018 18:07
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tripper1000:

 

Tracer: Take a look at long-range point-point links in power systems, they're not AC. ;)

 

Ah yes, but why? Long distance is AC unless it has to go underground or undersea for a long stretch (eg Cook Strait) - they use DC for long under-ground/under-sea runs because it reduces capacitive/inductive losses into the soil/water. At mere 200 meters this isn't going to be a problem.

 

 

At large scale, in an AC grid, you don't run it elsewhere because the cost of equipment at either end outweighs the benefit of lower losses, you need a lot of reactive support at the rectifier/inverter stations, and anything other than a point-point link is not yet a solved problem. None of those are problems here, especially as the load is DC.

 

Side note: the vast majority of the "Cook Strait cable" is overhead.


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  Reply # 2129723 19-Nov-2018 18:12
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How about this:

https://www.dx.com/p/ly-kree-ac-12-24v-to-dc-5v-3a-15w-power-supply-converter-black-428476

24Vac in, 5Vdc out at up to 3A.

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