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