17. More Power to You

 Tech-Talk

Part 16

Welcome to Part 16 of our series.

I should have started this series on DC Power by mentioning the three cardinal safety rules for any wiring project.

1.  Make sure you put a fuse as close as possible to the power source.

2.  Make sure you put a fuse as close as possible to the power source.

3.  Make sure you put a fuse as close as possible to the power source.


Remember that any wire between the power source and the fuse is unprotected.  If a short circuit develops, the power supply or battery will pump out all the current it can, as long as the circuit exists.  When the wire finally melts, the current will cease to flow -- but by then, chances are good that something will be on fire!

Something that a lot of folks don't realize is that the primary function of the fuse is to protect the wiring, not the equipment.  If your device malfunctions and starts to draw too much current, it takes a comparatively long time (possibly a second or two) for the fuse to blow.  An overheating electronic component can fry itself in just a few milliseconds.

Last time, we started looking at DC Power in the shack.  Let's work through a practical example to determine size wire is needed.
We'll start with a typical 100W HF transceiver that draws 20A at full power, assuming 13.8V.  We'll further assume, for simplicity's sake, that we have a 10 foot run from the power source to the radio.   If we consider the radio as a single "lumped" resistance, we essentially have a series circuit of three resistors -- the radio, the positive power line and the negative power line.  Basic theory tells us that all of the current flows through all three of the resistors, and that the voltage divides among them according to their resistance.  The voltage developed across each resistor is often called "voltage drop".  Thus the total voltage drop across the three resistors in our example will total 13.8V, by definition.  Mr. Ohm informs us that the voltage is equal to the current times the resistance -- E=IR.  We know that the current is 20A.

We'll narrow down our choices to #14, #12, and #10 wire.  A quick internet search yields the following resistances per 1,000' of stranded copper wire:

#14 -- 2.57 Ohms per 1,000', or .0257 Ohms for a 10' wire.
#12 -- 1.62 Ohms per 1,000', or .0162 Ohms for a 10' wire.
#10 -- 1.02 Ohms per 1,000', or .0102 Ohms for a 10' wire.

Remember that we need to double the voltage drop to account for both the positive and negative supply wires.  So using #14 wire, our calculation is:

2 x .0257(Ohms) x 20(Amps) or 1.028V.  Similarly, #12 yields a .648V drop, and .408V for #10.

Then 13.8V (our supply voltage) - 1.028V drop in the #14 wire leaves 12.772V to run the radio.  If the radio requires 13.8V +/- 15%, a typical spec, then it should work fine on 11.73V.  But keep in mind that this assumes no additional resistances (voltage losses) in any of the other parts of the circuit -- fuse holders, power connectors, and the like.  Those apparently insignificant losses can add up quickly!  Our measly 1/2 ohm resistance in 20 feet of wire has already cost us a full volt.  In the situation described, I'd certainly use #12 wire.  It's also important to note that resistance in wire is a function of length.  If your run is 20 feet instead of 10, you double the resistance in each leg, and therefore the voltage drop per leg.  We may also wish to power another device using the same cable, too -- which of course adds to the current draw and thus the voltage drop.  In this case, I'd almost certainly err on the side of caution and use #12 wire, or even #10, since the difference in cost, weight, and flexibility is rather small.

That's it for this month.  Next time, we'll continue looking at DC Power in the shack.

73 for now
John Bee, N1GNV
Quicksilver Radio Products

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