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I am trying to send power from a system to a load that is almost 200 meters aways. Now, the supply from the battery is approx 70V or so. At the load end the voltage received is 55V. Now, I understand that the power cable will have its own losses due to which there will be a drop.

But I want to understand the mathematics of this drop and the equations involved in it. Can anyone direct me to an app note for this please ?

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    \$\begingroup\$ This is simple Ohm's law. Find out the resistance/meter and multiply resistance * length * amperage-draw * 2 (for the two conductors) to get the voltage drop. Conversely, you can do voltage-drop / length / amperage / 2 to get resistance/meter. \$\endgroup\$ – DoxyLover Jul 23 '16 at 5:52
  • \$\begingroup\$ Is there a reason you can't use a remote sense here to compensate out the drop? \$\endgroup\$ – ThreePhaseEel Jul 23 '16 at 6:02
  • \$\begingroup\$ What do you mean remote sense ? \$\endgroup\$ – Board-Man Jul 23 '16 at 6:11
  • \$\begingroup\$ @ThreePhaseEel. Imagine, my input to the dc dc is 30V. The output is 24V. At the load end it is 18V. Now, the 18V being detected by the sense pins will make my dc-dc conv to pump up the Vout from 24 to say 28V. This, is inside the 30V. But, suppose, the sense detects it at say 12V, due to a long cable. The module will require more than 30V Vin to compensate for this drop,right ? How is this taken care for ? \$\endgroup\$ – Board-Man Jul 23 '16 at 6:34
  • \$\begingroup\$ 1. I'm confused. If your battery voltage is 70 volts or so and the voltage at the other end of the cable is 55 volts, isn't that due to cable losses? 2. Please edit your question with a schematic or block diagram showing the entire system, and with text describing the maximum current through the cable and its resistance per unit length or conductor diameter. \$\endgroup\$ – EM Fields Jul 23 '16 at 6:58
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As others have explained, it is a simple calculation using Ohm's Law. Two methods of transmitting power over a long distance include...

  1. Remote Sense. Where the power supply uses an additional wire to sense the voltage at the far end (where the load is). Then the power supply adjusts its output to compensate for the losses through the long distance.

  2. Remote voltage regulator (or converter). Where a higher voltage is sent through the line and a regulator is added at the distant end where the load is. This has the advantage over remote sensing because it doesn't require another wire way out to the end.

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  • \$\begingroup\$ Imagine, my input to the dc dc is 30V. The output is 24V. At the load end it is 18V. Now, the 18V being detected by the sense pins will make my dc-dc conv to pump up the Vout from 24 to say 28V. This, is inside the 30V. But, suppose, the sense detects it at say 12V, due to a long cable. The module will require more than 30V Vin to compensate for this drop,right ? How is this taken care for ? \$\endgroup\$ – Board-Man Jul 23 '16 at 6:35
  • \$\begingroup\$ If you use the remote-sense method, you must plan that your source has the range to accomadate the worst-case voltage drop. Modern digital switching power chips make the remote voltage regulator method much more popular than the remote-sense method. Because copper costs much more than silicon these days. \$\endgroup\$ – Richard Crowley Jul 23 '16 at 6:40
  • \$\begingroup\$ Could you kindly give me an example of the remote voltage regulator method please ? My chip, is the remote sense method \$\endgroup\$ – Board-Man Jul 23 '16 at 6:42
  • \$\begingroup\$ There are dozens and scores of examples on Ebay. Search for: voltage regulator board \$\endgroup\$ – Richard Crowley Jul 23 '16 at 6:46
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The mathematics is Ohm's Law, V=IR.

It's exactly proportional, voltage drop proportional to current, and to resistivity, which is inversely proportional to copper cross section.

I use a rule of thumb for the resitivity of copper at room temperature, which is 17mohm per meter of 1mm2 cable. Copper increases resistance quite markedly as it warms, +10% for a 25C temperature rise, so that 17mohm is given to all the accuracy needed.

Let's say you were transporting 1 amp to your load through 200m of 1mm2 cable, and back of course.

400m of 1mm2 cable, by my rule of thumb, will have a resistance of 400*17m = 6.8 ohms. It will have a drop of 6.8 volts at one amp, 13.6 at two amps, over 20v at 3 amps.

Ordinary house wiring 1mm2 cable is actually rated for 10A. However, that's a heating spec, not a voltage drop spec. It's expected that runs will be short, and at 240v, a bit of voltage drop is small compared to the supply voltage.

If you use (say) 4mm2 wire instead, your resistance and voltage drop will be one quarter of previously, but your cable cost will probably be not quite 4 times.

The easy solutions for low drop are therefore ...

a) to use less current (which may not be possible)

b) to use more copper cross section (which will cost money)

The more complicated solutions for low drop are ...

c) step the voltage up before you send it, and step it down at the far end (the economics of this get better when you have a very long distance, just ask the power companies, when the cost of the conversion (money and electrical efficiency) is less than the cost of a thicker cable)

d) step the voltage up at the receiving end so it's right

e) step the voltage up at the sending end, so it's right at the far end (slightly more efficient than (d) due to the lower current, but you'll need remote sensing)

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