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?

 

[BigBlue1]

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 ATV components from this current. It's the only thing that makes sense, because there doesn't seem to be any other path. This is surprising.

The other 0.9V went into dropping across the resistance of everything else in the circuit between the two voltmeter probes. From the negative battery terminal to the battery plates to the positive terminal to the positive cable to everything else in the closed circuit to the vehicle's ground/frame to the negative cable to the negative cable clamp. All those inches of wire (small resistance) and all the minor resistance at each connection drops the voltage some.

For a simplistic example that demonstrates, go get a 100' extension cable, fashion connection with one of the leads to your (+) battery post and then measure between the (-) battery post and the other end of the extension cord. It won't be the same as the voltage between (+) and (-) posts. Now plug in a couple smaller ext cords and watch voltage drop further, from additional length and each imperfect connection. Same concept applies to your vehicle test.

Rob

 

[MossRoad]

I measured the voltage between the battery neg (-) post and the neg (-) cable, with the cable disconnected.

You still have the resistance of every piece of wire, connector, electrical components, etc... in the circuit from the positive post, down the wire, into the starter, into any other components, through the frame, and back up the negative wire.

All of those components have resistance and drop the voltage.

 

[TCJatko]

Have you checked the connection to the frame or other ground point on the battery negative cable? Sounds like you have a poor connection there.

 

[CobyRupert]

You still have the resistance of every piece of wire, connector, electrical components, etc... in the circuit from the positive post, down the wire, into the starter, into any other components, through the frame, and back up the negative wire.

All of those components have resistance and drop the voltage.

Yes, but remember that you should only get that voltage drop when there is current flowing. (Remember, the neg (-) cable is disconnected from the battery post, so there's no path for current to flow)

I would of never guessed that the mere act of measuring voltage with a voltmeter across this open connection (between battery and cable) would allow enough current to flow through the meter that I get such a significant voltage drop through the paths on the ATV. (i.e. I would of never believed that two (2) different Voltmeter's internal resistances are low enough to allow so much current through)

 

[Industrial Toys]

The resistance of a meter, particularly a digital one is VERY high. Like millions of ohms. If not, you would put a load or even a short on the circuit being measured. I seem to recall that old cheaper analog meters might have had 10 or 100, 000 ohms resistance. The more expensive meters had higher reistance and the really expensive one, higher still with transistor amplifiers.

 

[MossRoad]

OK I got ya. You're thinking your meter is letting current through?

 

[WranglerX]

The voltage drop is not important, what is important is the drain....

If you are seeing a drain with everything shut off the means you still have a loop circuit, which you should not (except if you have ECM) if you don't find device that is causing current loop you are not going to solve the running down of battery...

Keep disconnecting things till what is causing the drain is found....Fix that item...

Easiest fix may just put a battery disconnect in either cable and at end of work session just "disconnect"....

Battery Switches

Probably find one down the street at FLAPS...

Dale

 

[CobyRupert]

The resistance of a meter, particularly a digital one is VERY high. Like millions of ohms. If not, you would put a load or even a short on the circuit being measured. I seem to recall that old cheaper analog meters might have had 10 or 100, 000 ohms resistance. The more expensive meters had higher reistance and the really expensive one, higher still with transistor amplifiers.

Exactly! This is why it doesn't make sense.

OK I got ya. You're thinking your meter is letting current through?

...or maybe I've stumbled upon a tear in the space-time continuum where Kirchoff's Voltage Laws doesn't apply and I can somehow capture this unexplained/mystery 0.9 Volts to power a perpetual motion machine.

Yes, I've put a disconnect switch on the cables, but when it's open: I still get 12.6 volts across battery terminals, and 11.7 volts across the open switch!
Kirchoff, where are you!!

 

[RandyT]

ECU as well as a voltage regulator if battery voltage is not disconnected by the key in the off position all the time. This is normal and anything under 100 milliamp is considered normal when measuring between the negative post of the battery and the disconnected negative battery cable with an amp meter.

 

[Ford850]

Exactly! This is why it doesn't make sense.



...or maybe I've stumbled upon a tear in the space-time continuum where Kirchoff's Voltage Laws doesn't apply and I can somehow capture this unexplained/mystery 0.9 Volts to power a perpetual motion machine.

Yes, I've put a disconnect switch on the cables, but when it's open: I still get 12.6 volts across battery terminals, and 11.7 volts across the open switch!
Kirchoff, where are you!!

But you are bridging that open circuit with the meter. Similar to you bridging the open circuit between the posts with the meter.