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Voltage Dropping a Circuit
Posted to Technical Theory Forum on 1/13/2017 66 Replies

A voltage drop occurs when current flows through a UNINTENDED resistance.

It might be a bad ground, a wire with only 1-2 strands that are unbroken, burned relay or switch contacts, a corroded connector, etc.

Say you've got a car with dim headlights. You backprobe those headlight connectors with the engine running and the lights on, and you only measure 11.8 volts.

Put those same DMM leads across the battery posts and you get 13.8 volts.

You sir, have a 2.0 V Voltage Drop.

What happens if you unplug those headlights and stick your DMM leads in the socket?

You'll measure 13.8 Volts. Huh? What? Why?

Because no current is flowing! Your meter doesn't consume enough amperage to shock a flea. A circuit must be LOADED to measure a voltage drop.

Remember V=IR? If current (I) is darned near zero, so will the VOLTAGE DROP be.

So back to our headlights.

Let's say these are 60 watt headlights. To make the numbers easy let's say the battery is 12V instead of 12.6 (engine off) or 13.8 (engine running)

60 watts = 12V times 5 Amps.

So we've got (approximately) 5 amps flowing through that headlamp.

If we've got a 2.0 volt VOLTAGE DROP, that means we have:

V= IR 2.0V = 5.0A times .4 ohms

SOMEWHERE imbetween the battery posts and that headlamp there is .4 ohms of UNINTENDED, UNEXPECTED, UNWANTED resistance that shouldn't be there.

Oh, in HEALTHY wiring, switches, relays, connectors you might have a total of .1 ohms, but this is an unhealthy amount of resistance.

To make the headlights bright, you need to locate and eliminate that .3 - .4 ohms of unwanted resistance.

You can NOT locate it by using your ohmmeter. The resistance is too low, your metere isn't accurate down there. You do it by measuring the voltage drop at each node along the way from the battery to the bulb.

That includes the fuse, the headlight switch, the dimmer switch, the connector, everything on the plus side.

And everything battery, cable, chassis ground, headlight ground on the negative side.

You put one lead of your meter on the battery post itself, and then "step" along the circuit until you see WHERE you are losing xs voltage.

Once again, the headlight MUST BE ON! There must be current flowing in the circuit or that undesired resistance will have no, zero effect.

A whole lot of times you'll find a ring terminal, providing a ground for the headlights, screwed to the body, either corroded, on top of fresh paint from the body shop, whatever.

It might have .1 or .2 ohms of resistance, IF you could actually measure it accurately. You can't.

If you TRY, how hard you press your test leads against the terminal will change the reading...

But you CAN measure the fact that on one side of it you have 0.00 volts, and on the other, 0.5 volts or more.

The question you SHOULD be asking is "How much voltage drop is acceptable?"

In a headlight circuit, .2-.3 volts isn't going to make much difference.

In an ECM, you want .02V or less. Otherwise all your sensors are going to read wrong

So let's voltage drop that ECM!

First of all, START THE CAR! Unless the ECM is awake, powered up, and using current, you can't measure a voltage drop!

DMM across the battery terminals reads 13.8V

DMM across the B+ and GND leads on the ECM reads 13.785 volts. Good enough! (voltage drop of .015 V)

DMM across the IGN and GND leads reads 13.5 volts.....NOPE!

You've got a voltage drop in your IGNItion switch circuit you need to hunt down and kill

It's obviously NOT a bad ground because when you checked GND versus B+ all was good. It's on the positive leg.

Yeah, okay, you're either getting it, already got it, or someone else is going to have to 'splain it to you in some other fashion.

If you still don't get it, PAY SOMEONE to teach you. Voltage dropping is the single most important skill you can have in this industry short of "righty-tighty, lefty-loosey!"

Wade from Colorado

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