Anyone Seen This? Fully Charged, But Lights Don't Come On.
- Thread starter Orion
- Start date
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If it's a 'protected' 18650, could be the protection circuit tripped for whatever reason.
In some cases, that condition will be 'reset' when first installed in some chargers. I've experienced this, and my particular charger reset them.
Others have had success by momentarily connecting a working cell in parallel (+to+ and -to-) to the 'zero volt' one, and that's been known to reset them. I don't personally know if that's completely 'Kosher' or not.
Is it a protected cell, and do you by chance have a voltmeter?
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I had one that was completely discharged that wouldn't start the charger. I did the + to + with a charged battery for IDK fifteen seconds? and then it would take a charge. Most of my cells are protected, so I imagine that it was a protected cell.
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It's showing 4 volts, so does it power anything at all? Or did you mean the cell depletes very quickly?
Dave_H
Flashlight Enthusiast
Sounds like something wrong with the cell, I would dispose of it.
Dave
Dave
Just a FYI -
I just discovered that the SkyRC MC3000 doesn't do this when a 0V cell is loaded. It has 0V across the output terminals when idle. The XTAR VP2s I use will do this, but they have ~4.75 VDC across their output terminals when idle.
I learn something new every day as long as I'm still breathing.
This!!!
Absolutely, get rid of it! Anytime one of these types of cells starts acting weird, you need to immediately and properly dispose of it. You KNOW something is wrong. No need to play Sherlock Holmes and investigate deeply into it. Not worth it!
Dave_H
Flashlight Enthusiast
Just put it in a small sealed bag e.g. zip sandwich baggie to keep from shorting, or if something should leak. Some retail stores here take batteries, not all accept Lithium-ion. Home Depot might, they take regular alkalines etc. but
may vary by country/location.
Dave
JimIslander
Enlightened
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- 683
+1 dispose of it.
No need for protected cells. Most lights have protection built in, and simple management will protect your cells for you in unprotected lights.
Unprotected cells are all less than $10, with some exceptionally high quality OEM cells less than $5.
No need for protected cells. Most lights have protection built in, and simple management will protect your cells for you in unprotected lights.
Unprotected cells are all less than $10, with some exceptionally high quality OEM cells less than $5.
kerneldrop
Flashlight Enthusiast
It's odd that your multi-meter reads 4 volts, yet has no output.
Protected batteries with a tripped circuit will show 0 volts.
Interesting.
I'm assuming you don't have an LED with two leads that you can touch the + and - ends of the battery to see if it lights up.
Just for curiosity.
Protected batteries with a tripped circuit will show 0 volts.
Interesting.
I'm assuming you don't have an LED with two leads that you can touch the + and - ends of the battery to see if it lights up.
Just for curiosity.
That's what confuses me. That it is showing 4 volts,...yet not producing any output.
Is it possible that the issue is with your multi-meter?
Nope. It works well.
I completely understand your confusion, @Orion. Here's how the story goes:
Modern voltmeters have very high input impedance (in this case, the DC resistance component of that is what's relevant), and present almost no load to the circuit under test (which is ideal). The cell in question has a very high internal resistance (due to unknown failure / degradation). Your voltmeter, as it draws practically no current, will see the difference of potential (voltage) across the output terminals of the cell, so it will read the voltage of the cell despite the excessive internal resistance (effectively open-circuit / no-load conditions).
The sum of the voltage drop(s) across all resistive elements in a circuit will always equal the source voltage. When the circuit is completed, as when a load is placed across it (the light is switched on), all / most of the voltage provided by the cell will be dropped across that (abnormal) internal resistance of the cell. That internal resistance is essentially acting as a large 'current limiting' resistor, so the cell can provide little or no actual current, and is not sufficient for the load (flashlight) to operate on, so it will be appear 'dead'.
When the circuit is completed (a load is placed across the cell), if you were to connect your voltmeter across the cell (or load connection) in parallel, you would measure zero (or very little) voltage, because that voltage is being mostly dropped across the internal resistance. The load can therefore draw little / no current from the cell. The potential difference (voltage) of the cell can be measured when there is no load on it (open circuit), therefore you will measure that across the cell. As soon as the circuit is completed and a load is placed across the cell, most / all of the voltage you measured 'open circuit' is then being dropped across that internal resistance, and thus your voltmeter would measure little / no voltage, the load will not see it, and cannot operate on it. If you could practically connect your meter in this way, that's what you would see. If you instead connected a current meter in series with the circuit, you would likewise measure little / no current.
If the light you're using is 'known-good', and works OK with a known-good cell, the cell you're having problems using is defective and should be disposed of. Do not attempt to 'post mortem' the cell (to try to determine 'root cause'), as that could easily end in tears. Just dispose of the cell safely without further thought on the matter, as previously recommended. Now that you know what's 'wrong with the cell', I would suggest that you not worry further about what caused it internally. Cells can / will fail, and in this case, it 'failed safe', which is how we hope thay will fail (vs some 'china syndrome' type event
Modern voltmeters have very high input impedance (in this case, the DC resistance component of that is what's relevant), and present almost no load to the circuit under test (which is ideal). The cell in question has a very high internal resistance (due to unknown failure / degradation). Your voltmeter, as it draws practically no current, will see the difference of potential (voltage) across the output terminals of the cell, so it will read the voltage of the cell despite the excessive internal resistance (effectively open-circuit / no-load conditions).
The sum of the voltage drop(s) across all resistive elements in a circuit will always equal the source voltage. When the circuit is completed, as when a load is placed across it (the light is switched on), all / most of the voltage provided by the cell will be dropped across that (abnormal) internal resistance of the cell. That internal resistance is essentially acting as a large 'current limiting' resistor, so the cell can provide little or no actual current, and is not sufficient for the load (flashlight) to operate on, so it will be appear 'dead'.
When the circuit is completed (a load is placed across the cell), if you were to connect your voltmeter across the cell (or load connection) in parallel, you would measure zero (or very little) voltage, because that voltage is being mostly dropped across the internal resistance. The load can therefore draw little / no current from the cell. The potential difference (voltage) of the cell can be measured when there is no load on it (open circuit), therefore you will measure that across the cell. As soon as the circuit is completed and a load is placed across the cell, most / all of the voltage you measured 'open circuit' is then being dropped across that internal resistance, and thus your voltmeter would measure little / no voltage, the load will not see it, and cannot operate on it. If you could practically connect your meter in this way, that's what you would see. If you instead connected a current meter in series with the circuit, you would likewise measure little / no current.
If the light you're using is 'known-good', and works OK with a known-good cell, the cell you're having problems using is defective and should be disposed of. Do not attempt to 'post mortem' the cell (to try to determine 'root cause'), as that could easily end in tears. Just dispose of the cell safely without further thought on the matter, as previously recommended. Now that you know what's 'wrong with the cell', I would suggest that you not worry further about what caused it internally. Cells can / will fail, and in this case, it 'failed safe', which is how we hope thay will fail (vs some 'china syndrome' type event
PhotonWrangler
Flashaholic
This is exactly correct. Even the old-school passive VOMs might have an input impedance in the 20,000 ohms per volt range, which is still not low enough to produce a valid test for a batter with high internal resistance.
