Electronic – Given metal dimensions and material, is it possible to determine the maximum current that the structure can handle? [semiconductor processing]

Make sure all peripherals are off, the watchdog and brownout are off, and the processor is asleep. That should minimize current draw of the PIC itself. Make sure all digital input pins are driven solidly high or low, and make sure all output pins are set so that they don't cause current in the rest of the system.

The stack overflow/underflow reset has nothing to do with power drain. It's just part of the digital logic of the processor. Turning off the processor clock (putting the device to sleep) will turn that off along with the rest of the processor logic.

Some subsystems are independent of the processor clock and need to be turned off separately. Examples include the watchdog timer and the brownout reset detect. The A/D can also be on if running from its internal R-C clock. The datasheet shows the incremental current drawn for all these things. Go thru that list and make sure everything is off.

You say you can't meausure current of individual components, but of course you can. You could even remove the PIC from the board and wire all its outputs high or low, then see what the board takes. That completely removes the PIC from the equation. Or you could do the reverse. Put the PIC on a breadboard by itself running the same code that should shut it down, and measure its current then.

As a general rule of thumb, the current determines the thickness of the wire, and the voltage determines the thickness (and/or material) of the insulation.

The power grid does pretty much what you propose - use much higher voltages to reduce cable size. The reduced current also means reduced line losses, which is very important over long distances.

For instance, if you had 1kW of power you you wanted to move from A to B, you could use, say 100A at 10V, or maybe 10A at 100V.

For 100A power transmission you'd need 1 AWG wire. That has a diameter of 7.34822mm, and a resistance of 0.406392Ω per km. So over a 1km distance you'd lose 0.406392 * 100 = 40.64V. Ouch. That would just plain not work! So although the cable could physically cope with that current, over that distance you'd loose all your voltage. So that would be a no-go.

Try at 100V, 10A.

10A can go through 11AWG cable. That's 2.30378mm thick, and with a resistance of 4.1328Ω/km. Much higher resistance, but much lighter cable. How much voltage would we lose over 1km? 41.328V. Factor in the return path, so you double the distance, you end up losing 82.656V, leaving 17.344V left for the load. Getting there. Still not workable, but getting there. That equates to 173.44W.

How about if we pump it right up to 1000V, at just 1A? At 1A we can use 21AWG wire, at 0.7239mm thick. 41.984Ω/km, which would be 41.984V lost there, and 41.984V lost back. So 83.968V lost from your 1000, leaving 916.032V. That's 916.832W coming out.

Now supposing you wanted to transmit 100A over a 1km distance, and limit the voltage drop to say no more than 1V. What thickness of cable would you need for that? Well, for a 1V drop at 100A you would have a resistance of 1/100 = 0.01Ω. So your cable must have no more than 0.01Ω/km. The table I use doesn't go that low, so we'd need to do some calculations.

If we use copper wire, that has a resistivity (\$\rho\$) of \$1.68×10^{−8}\Omega/m\$ at 20°C. For resistivity we have the formula: $$ R=\frac{\rho L}{A} $$ where L is the length (1km), R is the resistance (0.01Ω) and A is the cross-sectional area.

So we can re-arrange that for A: $$ A=\frac{\rho L}{R} $$ and substitute our values: $$ A=\frac{1.68 \times 10^{−8} \times 1000}{0.01} $$ $$ A = 0.00168m^2 $$ And of course, that equates to a wire diameter of 4.6cm.

Is a 5cm thick cable practical for that? Not if you can increase the voltage to decrease the current, no.

So for power transmission it is possible to reduce the cable size over very long distances and reduce the losses over the line.

Over shorter distances the losses are considerably less, but can still be a problem at higher currents. But is it worth it, or better to just use fatter cable?

You also have to factor in:

• The efficiency of the power conversion - step up / step down.
• The cost of better insulation if you have very high voltages.
• The reduction in cost by using thinner cable.
• Safety issues - higher voltages are dangerous.

So is it a "free" ride? No. There will always be losses and caveats you have to look out for. It can, though, get around problems of longer distance transmission of power.