Joe Perez

iTrader: (8)

A list member recently PM'ed me with some electrical questions, and I thought the message might make good archive material. We all know that bigger is better when it comes to wiring, but perhaps not the reasons why.

Thank you. My ego is plentiful and boundless.

Well now, those are some intelligent, lucid, well thought out questions. They're also ones that force me to refer to some of the old reference texts.

But you're absolutely correct.

The resistance of a piece of wire varies with temperature and material, but those factors being equal, the resistance increases as the size decreases. The reason you couldn't read it on your multimeter is because we are dealing with very small values- on the order of milliohms (thousandths of an ohm). Meters that can measure accurately at that range are pricey, to say the least.

And you're also on the money that the practical implications of this are directly proportional to the current being carried across the conductor. Specifically, we are interested in voltage drop- the amount of voltage lost as the current flows through the conductor.

In simple terms, Ohm's Law states that E = I x R, where E is voltage, I is current, and R is resistance. If two of those values are known, we can solve for the third.

Here are the per-foot resistance values for stranded copper wire at 20°C in some common automotive sizes:

AWG ohm / ft.
24 - 0.0257
22 - 0.0161
20 - 0.0103
18 - 0.0065
16 - 0.0041
14 - 0.0026
12 - 0.0016
10 - 0.0010

So, let's take a ten foot length of 24ga wire and run 200ma through it. I(.2) x R(.257) = 0.051, or just over 50 millivolts of drop. So if your working voltage going in is 14 volts, you'd get 13.95 volts out. Not bad.

Now, let's say that we're trying to pass 20 amps through that same piece of wire. 20 x .257 = 5.14 volts. Eek! We've lost over a third of the voltage. Not only that, but all the electricity that we "lost" is actually being dissipated as heat in the wire. Specifically, P = I x E, so solving for P we get 103 watts of heat. Basically, the insulation will be melting off the wire, probably in flames.

Now, let's say we increase the wire to 12 gauge. This reduces R to .016 ohms for a ten-foot run. So, 20 x .016 = .32 volts. Acceptable.

Note that the amount of voltage drop is dependent only on the amount of current flowing- it does not care what the voltage on the line is. 100ma at 12 volts and 100ma at 10,000 volts will produce the same voltage drop. This is why power companies use extremely high voltages (tens of thousands of volts) for their transmission lines. By using step-up transformers to raise the voltage, they decrease the current flowing through the wire for a given amount of power. Remember P = I x E? If I is 1000 and E is 120, then P (power) is 120,000 watts. If they increase E to 1,200 then I can drop to 100 for the same P. The latter will result in 1/10th the amount of loss over a given line, for the same load. Then at the step-down transformer, whatever voltage drop occurred gets divided by the windings ratio. So if a distribution line operates at 12,000 volts and has 25 volts of drop by the end, the transformer that converts that down to 240 volts will also reduce the 50 volt loss to a mere 0.5 volt.

Magic.

Not really. There is some resistance any time you have a junction, but of all the methods available to us for joining two pieces of wire together, soldering is second only to fusion-welding for low resistance. And at the power levels you are likely to encounter on a car, the resistance of a good solder joint is so small as to be essentially meaningless.

I never put much research into them, but I'd assume that, at a minimum, they're using separate ground wires for:
1- Injector drive
2- Sensors
3- General ECU circuitry

For the reasons described above, there's always going to be some voltage potential across a ground wire. If you have four injectors constantly pulsing on and off, at the same time, accounting for several amps total, then there's probably going to be a couple hundred millivolts of noise on that particular ground. If something else (like an A/D converter) were also trying to ground through that line, the noise would appear to be a pulsating offset in the signal. By giving the injectors their own ground, this noise is prevented from interfering with the more sensitive circuits in the system.

Of course, the MS makes no such provision- they use a single ground plane for everything. And frankly, this seems to work OK. It would be feasible to lift the legs of the FETs that drive the injectors and ground them separately, but most people find it sufficient to just do a good, thorough job of using what's there. Personally, I connected four ground wires. Two of them connect to the factory ground returns, and two of them run directly to the head.

That's another important point; when the car is running, the battery isn't supplying the power- the alternator is. Thus, the return path doesn't care about the long run to/from the battery. This is why the factory ECU grounds run to the cylinder head- they're back near the fuel pressure regulator. That's where I ran my additional grounds, using 16ga wire.