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I am struggling with a calculation where I'm either missing something really simple or the maths is advanced beyond my current capability. I have tried rearranging wattage / ohms law calculations but to no avail. Can anyone please assist?

I have a system with a 1km long cable. At 20m intervals is a lamp driven by a 20W constant wattage driver (this is a parallel circuit). The supply voltage at the start of the cable is 55V. The lamp drivers will maintain 20W power from any supply voltage between 55V and 30V. As such I expect the current to increase for each lamp along the cable as the V drops due to the cable.

I need to size the cable correctly so that the last lamp gets at least 30V and also know the total current supply requirement at the start. The only over variable I would be able to supply is the resistance per km of various sizes of cable which I could enter into an expression one by one.

Any help would be most appreciated as I'm stumped by this! Many tahnks.

Christopher Ringwood
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    Watts are not linear in voltage or current alone, which makes it non-trivial. Several options (1) just put the whole lot into a spreadsheet, replicating formulae down is easy, or write a simple program in your favourite language, time to learn python if you don't have a favourite. (2) compute current for the last lamp, assume all others take that, it will give you a conservative worst case. – Neil_UK Nov 08 '21 at 11:41
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    Use a sim tool. – Andy aka Nov 08 '21 at 11:44
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    Dataaheet on constant wattage driver? – StainlessSteelRat Nov 08 '21 at 12:01
  • There are obvious bounds on the problem: If all of the loads get 55V, you'll need 18.2A, and if they all get 30V, you'll need 33A (I'm assuming 50 loads, not 51). Recognizing that most of the voltage drop occurs in the first few segments, I would be conservative and use wire rated for at least 30A. But to keep the end-to-end voltage drop to less that 25V, you're going to need wire that has a resistance on the order of 1 mohm/meter, which would be something like AWG5 or 17 mm^2. – Dave Tweed Nov 08 '21 at 12:41
  • Many thanks for the comments and assistance with this. I will try these methods which should lead me to a good enough specification for the cable. I did wonder if it was difficult or I was missing something obvious! – Christopher Ringwood Nov 08 '21 at 12:48
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    @DaveTweed You're right with the bounds. OP -- 1km of 5AWG copper wire is gonna be very very expensive, and so thick it will be hard to work with. Can you run multiple parallel cables? Or maybe use different gauges - 5AWG at the beginning where the current is high, taper down as you go. – Kyle B Nov 08 '21 at 14:30
  • That's precisely why low voltages are not used to transmit significant amounts of power over long distances. – Dave Tweed Nov 09 '21 at 04:43
  • Many thanks for the comments. The project has SELV specified instead of 230V mains. As always the client ignores physics and thinks the SELV DC solution should cost the same as the 230V mains solution and asks if there is anything that can be done to reduce cable sizes! In this project 10mm2 cable is normal and already expensive as FP600 fire rated! – Christopher Ringwood Nov 09 '21 at 12:36

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