Why am I melting battery holders?

I made up a little homebrew bike light using canibalized parts from a TaskForce 3W 2C flashlight.

Here is what I did:
http://forums.mtbr.com/showthread.php?t=251089&highlight=proof+of+concept

While I think that having it run on two - 2C batteries hooked in parrallel will be just fine, I thought I would just try to run a 4AA holder connected to the light. So basically the equivalent of 4 in series. The battery holder started to get very warm at the negative terminal of one of the batteries; not the negative terminal of the 9V connector mind you, but on the butt end of the battery.

I know this unit has a DC step up and is designed to run on 3V, but why would the actual batteries get so hot as to melt the plastic battery holder.

Sorry, if this is a stupid questions, but I just don't know why.

Thanks.

Usually melting occurs in two conditions...
1) shorting
2) high current drain + high resistance in your battery path

Give you an example, I used to run Mag85 with some 3AA-to-1D holders. When I did that, it took about one minute to melt all three holders in my Mag because the resistance in those holders were high and the drain was high.

ernest

Heat = Resistance

Clean up the contact areas

Also I would'nt think it was a good idea to be running 4 cells when the original electronics package was designed for 2.

Mac

Agreed that running 4 cells in a unit designed for 2 is not the best long term. I just wanted to see if a higher input voltage to the light engine would produce a brighter light, even if it were at a cost of shortened LED life.

I got the resistance = heat thing, but I can't figure why there is resistance at the butt end (negative) of one battery in a 4AA holder. In my very basic understanding of all this, I was thinking that the DC-DC step up was taking the 3V (or less as batteries go down) and increases to something greater then 3V or it may just be a 3V maintenance over time. So, given that, if I fed it 4.8V with Li rechargables, the increased input voltage would negate the need for step up, or just overload it (I don't know).

All contact point are clean and voltage out is measured at 4.8V for fully charged batteries. If it is not a resistance between point of contact, where is it.

If as the above example of the 4A D-cell converter, it is a drain issue, why? Again, I must be an idiot because I can't think of why feeding 4.8V into a circuit designed for 3V would cause increased drain. I would have expected less drain since the step up would not have to do anything.

Or.....

Is this the gatekeeper? the step up circuit prevents higher voltages and is the point of resistance, but the weak point is the battery that is pushing curent (voltage) but the circuit is pushing back (resistance).

Does any of the above make any sense or should I just crawl back into my cube?

ZipLED said:

I know this unit has a DC step up and is designed to run on 3V, but why would the actual batteries get so hot as to melt the plastic battery holder.

Click to expand...

Many (all?) boost regulators need to have input voltage below some threshold voltage in order for the output current to be regulated. If you exceed the
voltage it may not be in regulation anymore (may go into a direct-drive mode). That may be the problem here -- you may not be in current regulation anymore and much more current is flowing than you'd like. (The only thing that may be saving you is that there is some resistence to limit the current?)

ZipLED said:

All contact point are clean and voltage out is measured at 4.8V for fully charged batteries. If it is not a resistance between point of contact, where is it.

Click to expand...

I assume you are running AA NiMh, voltage for 4 fully charged batteries should

be around 5.6 - 5.8vdc, 4.8vdc would be the nominal rating.....

If the battery holder is melting its probably POS, and needs to be replaced the way to go is a solered battery back. But you cant charge the cells in their normal cradle charger you need a RC type charger.

The power output is a reult of the voltage times current. The current is determined by the voltage and the resistance.

If you double the voltage while the resistance (the converter) stays the same the power goes up 4 times.

Batteries and battery holders have maximum ratings. You've exceeded that rating.

Some converters short out when too much voltage is applied. They pass the voltage straight to the LED, which can cause a LOT of current to flow.

Daniel

Thanks for the explanations.

The battery holder is just a simple Radio Shack 4AA with a 9V terminal connector.

Makes perfect sense about the boost becoming a floodgate and not a regulator. This would allow the LED to be overdriven and the batteries just pumping out as much as they can. Hence the heat.

I rechecked the batteries and yes, about 5.6V is what I am getting.

Thanks again everyone.

ZipLED said:

Thanks for the explanations.

The battery holder is just a simple Radio Shack 4AA with a 9V terminal connector.

Makes perfect sense about the boost becoming a floodgate and not a regulator. This would allow the LED to be overdriven and the batteries just pumping out as much as they can. Hence the heat.

I rechecked the batteries and yes, about 5.6V is what I am getting.

Thanks again everyone.

Click to expand...

Don't know for sure that you have a boost reg. in those lights, but here is more info on boost regulators:
http://en.wikipedia.org/wiki/Boost_converter
and the technical description of the MaxFlex boost converter has a blurb on unregulated output if Vin > Vout, here:
http://www.taskled.com/techmaxflex.html