DC Battery Fault

What happens when a 24V filament lamp is connected, rather than a high-impedance DVM?

What voltage were you expecting?

Iwas expecting 0 volts as cells were not connected to any external metal parts battery bank output voltage between positive and negative was24V

What happens when a 24V filament lamp is connected between battery and earth, rather than a high-impedance DVM?

What is the ohms reading between the -ve terminal and earth?

I'm sorry.......did you say something? I hate it when they won't listen.

Due to leakage from battery back to earth.

Check the cell corresponding to 17V w.r.t. -ve terminal of bank.

It may be giving trouble.

I also faced similar problem of leakage voltage which varies with change in humidity. The leakage tend to increase with high level of humidity in atmosphere.

If the above-mentioned 24V filament lamp between the battery and earth won't light, then the problem is insignificant, and the OP has nothing to worry about.

Going back to the OP: not connected to earth, perhaps, but what is the battery sitting on, and where are you connecting your voltmeter to earth?

Re: "the problem is insignificant"

While the problem is probably insignificant, it does represent a leakage path. When I used to maintain lead acid batteries, a leakage path like that usually indicated a need to clean the batteries, specifically around the terminals.

If that doesn't stop the leakage, there may be an actual acid leak from a cell.

If the goddam' bulb won't light, then the OP doesn't have an electrical problem to solve.

<sigh>

How about using 10 gauge copper wire instead of the bulb? If it does not get red hot there is no problem, right?

My point being: that the bulb will not light if there is insufficient current available, but that the very small current available can nevertheless be enough to cause problems such as corrosion, battery drain, etc., even though it cannot cause a bulb to light.

This is probably nit picking, but I would rather advise that the voltage be measured across the bulb, rather than using bulb illumination alone as an indicator.

Clearly, if the bulb lights, there's a problem. If it does not light you cannot rule out a smaller problem, such as the battery cases being wet on the outside and sitting on wet ground.

Quite.

Davy, we assume you're using a DVM. You say you have a reading of 17V from –ve to earth, what voltage do you get from +ve to earth?

One problem with the DVM it imposes vertualy no load during a test, spurious voltages can show. Try using an analogue meter.

from positive to earth there was 9v Iwas using a fluke DVM

Does the goddam' filament bulb light between -ve and earth?

You can lead a horse to water, but...

Well, there's enough here to drown in!

If you have 17V from one terminal to earth, 9V from the other to earth and 24V between terminals - something doesn't add up, right?

So the 17V and 9V readings are very little, if anything, to do with leakage and a lot to do with pickup from the local radio station etc., right?

Oh, and by the way, does a 24V filament lamp light when you connect it between -ve and earth?

You are wrong in thinking that you have anything here on Earth that is not connected to Earth. Everything around your battery provides a current path to Earth. The air between your battery terminals to Earth is the simplest path. A more complicated path is through the shell of your battery, to the paint of your stand, to the metal of the stand to the tile floor to the concrete of your building to the building ground. Now the air and the shell of your battery are likely a very high impedance path but they are still a path. Now the instrument that measured 17V from battery to Earth also has an input impedance. If the instrument's impedance and the battery voltage were known then one could make a calculation of what the minimum isolation resistance path that you have.

Compounding your problem of working with an isolated circuit is all of the interference coupling signals from AC power lines and radio transmitters that are all very tiny currents. But because these tiny currents are traveling through a high impedance path the voltage is measurable.

OK. Consider this hypothetical scenario:

• A brand new 240VAC - 24VAC transformer is energised for the first time, and even if the 24VAC terminals are not connected to anything, a Fluke DVM says 17V from one terminal to earth. On de-energising and switching to ohms, the thing goes >20MΩ. Is there a problem?
• Take the transformer away and substitute the OP's battery. Same thing happens on both volts and ohms. Is there a problem?

<sigh>

To reiterate,

"One problem with the DVM it imposes virtually no load during a test, spurious voltages can show. Try using an analogue meter."
The analogue meter will impose a slight loading and stop any spurious voltages. If needs be connect a lamp between the leads of your DVM to impose a load. This follows what PWSlack has been saying you need a load to remove the spurious voltages.

Hey!!! What IF... and bear with me here... we connect a filament bulb between the negative and ground and see if it lights up?!!

That a good idea.

Hey, what if we connect a 24V lamp across the positive and negative terminals to see if it lights up.

If the bulb lights up maybe we can now see if any other debris is connecting one of the two terminals to ground. I wonder what the voltages are across the battery with the light bulb ON and from each terminal to ground?

Maybe we should just glue a few cockroaches to this battery, put it in a missile just off the California coast, launch this to Mars and then measure the voltage to ground.

Geese... what a great idea. I am surprised no one else suggested that course of action.

LOL...

That's not Stinky Pete is it?

Hey PWS, barbiturates usually work for me

The fault always goes away, until tomorrow

Let's look at this one step at a time, starting with the obvious:

1) Is your fluke set to monitoring DC voltage? If it is set to AC it can give you bogus readings.

2) You say you get -17V at one terminal and +9 at the other. This is a difference of 26V, which would be reasonable for 24V battery being float charged.

3) With all due respect to the previous poster, in a practical sense, there is such a thing as not being grounded to earth. He is right only in the theoretical sense. What I suspect you have is a floating ground. The voltage difference in the battery only makes sense referenced between the terminals. If the negative terminal is not connected to ground, then earth ground and battery ground can and will have a random voltage difference.

4) When I hear "battery bank" I immediately suspect it is connected to something, such as your house wiring. Now you can get into all sorts of problems such as nicks in the wires or ground faults. One of the most common is an improperly installed neutral wire. Stick your meter (check both AC and DC) into the ground and neutral connections. If you get a voltage, you probably have a problem with your neutral or with your ground.

The problem with trying to diagnose these things remotely is that we can't see exactly what you are talking about, and the info you provided was sketchy.

31 comments and nobody has seen the light. There is no battery fault. If the battery is really disconnected from earth, the DVM/earth test forms the only link, all be it very high impedance. The negative terminal would float up to earth potential and 0v would be recorded. Equally,when testing the positive terminal, it would float down to earth potential and 0v would again be registered.

So why are you getting "consistent" (How many times did you do the test?) readings of 17v and 9v. This suggests that there is a different resistance between the +v/earth test and the -v/earth test. The DVM is common so that can be eliminated. The only things left are the terminals (are they clean?, have you tried scraping off any oxide layer?) and the earth(are you testing to the same earth point?)

One last question. Why is the battery voltage floating? Floating voltages are notoriously difficult to fault find and there is rarely any advantage to using them. All you need is 24V between the two terminals. Deliberately tie one terminal to ground and the "problem" goes away.

<...31 comments and nobody has seen the light....>

Would that light be coming from a 24V filament lamp?

I cannot explain this!

There are so many unknown and you have received a lot of replies indicating procedures to follow and return with some answers.... You did not reply to any of the suggested tests!!

One More: Disconnect the bank terminals from any connected charger or load, just the battey by itself, and then do the voltage test with the DC range on your DVM, also the lamp test checking the current drain even if the lamp does not light up. --> you should get no readings except some very negligeable (~1V dc??) which will not show any current flow (microamps??). The negligeable will be due to stray pickups that your DVM will get when the leads will act as antenae...

With the above test, if you pick up any substantial current (any where near 10mA and above), then you might be just pulling our legs..

A battery is a pretty good capacitor - try thinking in those terms

Yes in terms of holding some charge. But it works differently since the 2 poles are immersed in an electrolite that is conductive. Yes we can sometimes look at it as a big capacitor in terms of charges.

In this case, the only capacitor will be between the internal part with the conductive electrolite, and the outside of the casing if we stic some conducting film etc.

But the charging will be very small due to the thickness of the dielectric (Case plastic) and the 24V potental. Even this voltage is only between the extreme poles as mentioned by the OP.

I do not think that there will be enough charge to give a voltage of 17 DC between the Negative and an isolated earth / ground point.(In fact between any of the 2 poles for this matter... think of it).

In any case, if it is a capacitive charge, it will discharge quickly and zero if any pole is shorted (via a lamp foe ex.) to the ground... The OP is not giving a clear complete picture of the circuit he is looking at to obtain such voltages as mentioned!

Most battery banks are wood pallets or tables because batteries discharge quite quickly sitting on concrete. That sort of capacitance and a bit of acid would explain this - I feel.

The only battery system I've ever maintained all resided in my automobile. But it makes no sense to me that concrete would make any difference to the charge of a battery. Now I can see that the low cost and low reactivity to acids could make wood a more suitable support material than concrete or metal. I've recently seen a large capacitor bank that all of the batteries sat in long Nalgene® tubs as a secondary containment vessel.

A similar effect you may be more familiar with, is a sheathed element (as in stove, or kettle, or water heater)

If you don't ground the sheath, it will float at half the supply potential (with enough current available to wreck sensitive instruments)

Obviously the dielectric strength of the filler is adequate to resist (several times) the supply potential, so it's not a 'leakage' problem. And it's 'one shorted turn' magnetically.

Yes, but the battery is DC and does not induce any current in any metal nearby.

The Capacitance will have to be: The internal Plates as one pole (Positive and negative), thwe electrolite as Dielectric (-?-) and the plastic container as a further dielectric, and then any nearby conductor as the external Capacitor pole. I don't think this is going to give any similar result as by the OP.

Most probably, he has some devices connected to the battery (charger, others as loads...) and the leak to ground is though them.