Explain battery voltage?

[CobyRupert]

So I put a new battery in my ATV. I'm also installing a battery disconnect switch on the negative (-) post of the new battery because I have a 30mA (.03 amps) drain on the battery.
The drain is either through the voltage regulator, or possibly the starter "solenoid" (which might be a semi-conductor "switch"? Not sure). I've unplugged just about everything else (winch relay, wires to key switch, etc..) Doesn't matter, the drain HAS TO have a path from the positive post to the negative post of the battery. Otherwise it's an open circuit and no current can flow.

But here's my mystery:

With only the positive battery post connected to the ATV wiring, I measure 12.6 Volts across the battery. (Ok, check.)
But from the unconnected negative cable (which is connected to the frame, etc..) I measure around 11.7 volts from the cable to negative battery post.

So why don't I measure 12.6 volts from the (ground) cable instead of 11.7V to the negative post? There can't be any current flowing (leaking) from the positive battery post to the negative post because the cable isn't connected.
Nothing is connected to the negative battery post.
Where is the 1 Volt going (i.e. how can there be a 1V drop) if no current is flowing from battery pos (+) to neg (-)?

 

[BigBlue1]

Completing the circuit with the voltmeter allows the current to flow. Whatever is connected in the ATV's circuitry has resistance and is the reason for the voltage drop you see.

Rob

 

[AWSubie]

Ur battery could have an Internal short between cells

 

[WranglerX]

First there should be no drain on battery with all devices off ...UNLESS you have a computer (ECM) in vehicle, most vehicles with computers will have a drain to keep it alive, you can not get away for it... ..... If no computer you have a defective device some place, you need to locate it to stop drain, ...

About out your voltage from cable to battery post you You are measuring through device in system (your 30 ma. drain) so voltage reading will be skewed some.... Turn ignition on and what is your reading from post to cable?

Want to learn more then you ever wanted to know about batteries.....

Car and Deep Cycle Battery Frequently Asked Questions (FAQ) 219

Dale

 

[CobyRupert]

Completing the circuit with the voltmeter allows the current to flow. Whatever is connected in the ATV's circuitry has resistance and is the reason for the voltage drop you see.
Rob

I thought of that, but didn't think a voltage meters allowed enough current to flow.

Doing the math (see below) that's saying that both of meters I used (that gave me 11.9 Volts from neg (-) post to neg (-) cables only have a resistance of around 5460 Ohms? Isn't that really low?

Math:
If I have a 30mA leak, whatever is leaking has a resistance of (R=V/I, R=12.6V/30mA) R= 420 Ohms.

I measure 12.6V across battery and 11.7V from neg(-) post to neg(-) cables. A 0.9 voltage drop.
So if I have a 0.9 Vdc voltage drop across that 420 Ohm "leaking device" when measuring with the meter (and saying meter allows current to flow), this means:
2.14mA current is flowing through the voltmeter and the "leaking device" when the voltmeter completes the circuit loop. (I=V/R, I= .9Vdc/420 Ohms = 2.14mA)

This means the resistance of the "leaking device" AND the voltmeter combined would be (R=V/I. R=12.6V/2.14mA) R= 5888 Ohms.

This means the voltmeter only has a resistance of: (5888-420=) 5460 ohms.
Is it that really low? I thought it would be in mega-ohms?


I checked that my open circuit voltage (from neg (-) cable to neg (-) post) of 11.7 volts with 2 meters.

 

[CobyRupert]

First there should be no drain on battery with all devices off ...UNLESS you have a computer (ECM) in vehicle, most vehicles with computers will have a drain to keep it alive, you can not get away for it... ..... If no computer you have a defective device some place, you need to locate it to stop drain, ...

Yes, but my inquiry is regarding: There should be no drain if the negative post of the battery isn't even connected to the ATV. How is there any current loop? BigBlue suggested that the meter I measure with across the unconnected cable to battery post creates a significant current loop?

About out your voltage from cable to battery post you You are measuring through device in system (your 30 ma. drain) so voltage reading will be skewed some.... Turn ignition on and what is your reading from post to cable? ...but there is no drain when the cable is disconnected. You can't turn ignition on with cable disconnected, or just relying on the meter providing "connection"

Want to learn more then you ever wanted to know about batteries.....

Car and Deep Cycle Battery Frequently Asked Questions (FAQ) 219

Dale

Thanks for your reply.

 

[Industrial Toys]

Big Blue is exactly right.

Either a computer or I am told, leaking diodes in the alternator will discharge your battery.

Or, just a bridge of crap somewhere between close positive and negative polarities, like switch contacts. More likely to happen if the thing is within a hundred miles of salt.

For what it's worth, i have been involved with electronics a very long time. I have never in my whole life had to calculate the resistance of my meters or internal battery resistance. Just sayin. If I get into too much math, I am just as likely or more to make an error in that and be mislead.

 

[BigBlue1]

I thought of that, but didn't think a voltage meters allowed enough current to flow.

Doing the math (see below) that's saying that both of meters I used (that gave me 11.9 Volts from neg (-) post to neg (-) cables only have a resistance of around 5460 Ohms? Isn't that really low?

Math:
If I have a 30mA leak, whatever is leaking has a resistance of (R=V/I, R=12.6V/30mA) R= 420 Ohms.

I measure 12.6V across battery and 11.7V from neg(-) post to neg(-) cables. A 0.9 voltage drop.
So if I have a 0.9 Vdc voltage drop across that 420 Ohm "leaking device" when measuring with the meter (and saying meter allows current to flow), this means:
2.14mA current is flowing through the voltmeter and the "leaking device" when the voltmeter completes the circuit loop. (I=V/R, I= .9Vdc/420 Ohms = 2.14mA)

This means the resistance of the "leaking device" AND the voltmeter combined would be (R=V/I. R=12.6V/2.14mA) R= 5888 Ohms.

This means the voltmeter only has a resistance of: (5888-420=) 5460 ohms.
Is it that really low? I thought it would be in mega-ohms?


I checked that my open circuit voltage (from neg (-) cable to neg (-) post) of 11.7 volts with 2 meters.

I think you're running a few things together that aren't really related. The resistance of the voltmeter is not what is causing the voltage drop. The resistance of whatever components are on the ATV between the two points you tested are what is using power and dropping the voltage. Turn your meter to ohms and measure the resistance of the same path. You'll see there is some.

The fact that some circuit on your ATV is still drawing current when it is switched off is a function of the battery voltage and the resistance of whatever circuit is closed. I believe that's also what is providing the resistance you would measure above because it is the circuit that is closed. If no part of the circuit was closed then you wouldn't be able to measure any voltage and the resistance would be infinite.

All in all, your voltage measure observations are not helping determine your issue. What you need to look for is what component on the ATV is not fully off/open. That's where the power drain is coming from.

Rob

 

[CobyRupert]

I think you're running a few things together that aren't really related. The resistance of the voltmeter is not what is causing the voltage drop. The resistance of whatever components are on the ATV between the two points you tested are what is using power and dropping the voltage. Turn your meter to ohms and measure the resistance of the same path. You'll see there is some.

The fact that some circuit on your ATV is still drawing current when it is switched off is a function of the battery voltage and the resistance of whatever circuit is closed. I believe that's also what is providing the resistance you would measure above because it is the circuit that is closed. If no part of the circuit was closed then you wouldn't be able to measure any voltage and the resistance would be infinite.

All in all, your voltage measure observations are not helping determine your issue. What you need to look for is what component on the ATV is not fully off/open. That's where the power drain is coming from.

Rob

I'm not really looking for where the leakage is, my disconnect switch will fix that, and provide a key-less security feature. I'm just wondering why the voltage around "the loop" doesn't match (add up). (It HAS to. Kirchoff's Voltage Law)

I would expect that the resistance of the meter would be high enough so that no current flowed through the meter when I measured the voltage between the battery neg (-) post and the neg (-) cable, with the cable disconnected. I would expect this voltage to be 12.6V. Exactly the same as across the battery posts (12.6V).
But I only measure 11.7V.
So where did the other .9V go?

I wouldn't expect a Voltage Meter to pass that much current through itself. However, if the meter's resistance is low, enough current must flow through it when measuring voltage, that I get a voltage drop (0.9V) across the (leaking) ATV component from this current. It's the only thing that makes sense, because there doesn't seem to be any other path. It's surprising that a volt meter passes so much current.

 

[CobyRupert]

View attachment 600479

For example: On the above picture, let's say the V-source is my 12.6 volt battery, and the load (420 Ohms), allows 30mA to leak, just like my ATV does.
If one removes the "'b' circle" from the diagram (similar to disconnecting the neg (-) cable from the neg (-) battery terminal), there will be zero leakage current (i=0), as no current can flow around the loop.
One would expect that if I take a Voltage Meter and measure across where the 'b circle' was removed (from one break in the line across to the other), the meter should read 12.6V, exactly the same as V(source).
But my meter only reads 11.7V?