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Can you please explain how to calculate thickness of a copper conductor for the next task:

I would like to use a DC power supply instead of some batteries for my sensors. So we have V volts and AH Amp*hour. A sensor working T Hours with the batteries. My sensors will be connected in series with different distances, I do not know them for now (I suppose we will need the worst case). There will be X sensors, and L meter for total length of the conductor. Enough voltage to operate for a sensor Vmin.

So what is the formula? Thanks a lot.

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A couple of things we'll need to know: how much current does your sensor need to operate (peak current draw, not the average over 2 years)? What is the minimum voltage the sensor can operate on? – The Photon Dec 5 '12 at 18:49
Unfortunately I do not know this. The sensors' manufacturer not providing this information. – BBK Dec 5 '12 at 19:02
What does Cu thickness AH rating and unknown sensors have to do when using a DC power supply? Pls disclose full details if you want help. Otherwise watch what you are smoking. I mean the question reads like smoke. – Tony Stewart Dec 6 '12 at 5:12
It's not clear, but it looks like the current will be tiny; almost any wire that's large enough to handle will do. – pjc50 Dec 6 '12 at 10:05
2 pjc50 - that is good news, but main thing is to be able to calculate it by myself. For example: what if the numbers will change dramatically? – BBK Dec 7 '12 at 19:08
up vote 0 down vote accepted

The thickness of your conductor needs to be chosen to maintain two things:

  1. When your sensor is drawing its maximum current, the voltage drop across the wire can't be so much that the sensor doesn't have enough voltage to operate. To calculate this, you need the peak current draw of the sensor. You know that the average current is about 60 uA, from your observed battery life. But this could be because the sensor is only turned on for 60 us every second, during which time it draws 1 A.

    You can calculate the resistance of your wire from its length, the cross-section area, and the resistivity of copper (reduced increased by about half from the ideal value for the effects of being "worked").

    From the wire resistance and the sensor current draw, you can calculate the voltage drop. If the drop is too much, you need a fatter wire.

  2. You need the wire not to heat up so much it melts its insulation or starts a fire. For steady currents, you're worried more about the fire hazard, and you can look up allowed values in standard ampacity tables.

    For peaky currents, its much more difficult, because you need to know a lot about how the wire is cooled (airflow, contact with other materials, etc), to know how much it will heat up in a short time. Looking up the peak current in the ampacity tables will give you a conservative estimate for the minimum wire diameter.

If you do have a 60 uA steady current and the length of the wire is less than, say, 100 m, then as pjc50 said in the comments you can probably get away with just about any wire, even down to AWG 30 or smaller.

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lets say minimal voltage Vmin=3v. I notice pjc50's comment - that is good news. But if the numbers are about to change dramatically? – BBK Dec 7 '12 at 19:01
Then you need to do the calculations I outlined. Your question asks for a method of solution, not a worked out solution for a specific case. Also, if you don't know the peak load current, you still don't have enough information to get a specific answer. – The Photon Dec 7 '12 at 19:03
Can you please help me with calculations and sources. Where can I find the tables? – BBK Dec 7 '12 at 19:05
Did you try Googling for "ampacity table"? What did you find? – The Photon Dec 7 '12 at 19:05
Here's one I found by googling "pcb ampacity table": armisteadtechnologies.com/trace.shtml – The Photon Dec 7 '12 at 19:09

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