Resistored setup battery "overpowering" problem

[Vazan Maceu]

Hi people

I´m working on a couple of hilts with resistored setups. They´re powered by 4 AA battery packs made from 1.2v NiMH 2500Mah cells that should be outputting 4.8v... And that´s the problem: They´re giving out over 6 volts! This makes the resistors (that were planned to work with an input of 4.8v, and are of course the proper ones) to get extremely hot to the touch, in fact, burning fingertips and all... The question is, could my battery charger be the cause of this "overcharge"? Tried two different sets of batteries with same specs with same results, volcanic resistors

Any hint/clue on this? Thanks in advance

[xwingband]

Your charger is making them charge way too much. 5V maybe 5.2V is the max they should be at.

[Vazan Maceu]

Your charger is making them charge way too much. 5V maybe 5.2V is the max they should be at.

Yeah, that´s probably the cause, then... It seems that it´s time to search for a new battery charger, thanks xwing

[erv]

nope, your charger isn't necessary overcharging. Standard nimh cells charge at 1.3 to 1.4 v, sometimes more.
Before thinking that it's overcharging : what kind of charger is it ? smart charger ? or simple charger (wall mount) that charges individual cell in 8 hours or more ?
If the charger is a delta peak (intelligent charger), it might detect the peak at the wrong moment OR it can be due to the pack itself.
So, to answer properly :
- link to your battery charger, model, kind etc
- battery you are using : brand, place where you bought them, and age

it depend also on the current used to charge the battery / pack : it should not exceed the capacity of the battery. But for a smart charge, it should not be too low neither, or the delta peak won't be detected at the right moment, and it can lead to overcharging.

Erv'

[Vazan Maceu]

So, to answer properly :
- link to your battery charger, model, kind etc
- battery you are using : brand, place where you bought them, and age

it depend also on the current used to charge the battery / pack : it should not exceed the capacity of the battery. But for a smart charge, it should not be too low neither, or the delta peak won't be detected at the right moment, and it can lead to overcharging.

Thanks for helping , here´s the info on the battery charger in English and French. It´s from a brand called Energy Sistem, and this is an image of the back label on the charger:


There are two battery sets that I´m using: The most recent one was bought from Lidl (VERY cheap ), brand is Tronic (yikes!), and they´re Ni-Mh - 1.2v - 2500 mAh, size AA, HR6 / HRM 15/51 (not sure if this is useful, anyway). They´re no older than six months or so. The other set came with the charger brand is Energy Sistem, , and is around three years old, being also Ni-Mh - 1.2v - 2000 mAh, size AA R6.

Thanks in advance for your help and interest

[Dark Helmet]

draganfly makes a 5.1 volt regulator that keeps a constant 5.1 v output with up to a 32 v input, it's WAY more expensive than a few resistors (about $30), but it has built in charge port leads prewired and is compatable with all rechargeable battery types, even li-poly. i'll be using this setup with a 7.4 v li-ion pack. the big drawback is spending over 100 bucks for the charger/battery pack/regulator,and not knowing how reliable this is going to be. GOOD LUCK!

[Vazan Maceu]

Thanks for the info Dark Helmet ... The thing is that I´m trying to keep this setup on a budget. First I was planning to get a couple MR Boards to do the trick, but no luck finding them. And I´m using resistors because the next best thing (the pucks) are slightly beyond my reach

Anyway, thanks for your advice

[Eandori]

Erv' was touching on this, an unconnected battery can have a much higher voltage then it's rating. A 1.2v "rated" cell can output up to 1.5v when measured with nothing connected. When you start pulling current out of the battery, that's called "loading" the battery. The more you load the battery, the lower it's measured voltage is.

So...

4AA's might read as 6v when disconnected, but I seriously doubt they still will when you are yanking 1 to 3 amps out of them. Most likely, when your battery is connected your cells are operating CLOSER to the 1.2v rating per cell, for a total of 4.8v.

No.... my thoughts to your heating resistor problem is most likely the resistor you chose, and the LED it's driving.

Power = voltage x current. Most small resistors are only rated for 1/4 watt. If your LED in your saber is using 1.2 amps of current, and has a forward voltage of 2.9v, then your LED in your saber is using 3.48 watts of power. (1.2 x 2.9 = 3.4

Now.. if something like that is the case... and your cells are putting out 4.8v with a load... and your LED is dropping 2.9v that leaves (4.8 - 2.9 = 1.9v) 1.9volts across the resistor!!! With a current of 1.2 amps that would be 2.28 watts dissipated across that resistor.

Now... to come full circle....

a 1/4 watt resistor, with 2+ watts dissipated across it is GOING TO GET HOT. Even a resistor rated at 10 watts would heat up some. So.. for you, step #1 is to measure the voltage drop across your resistor WHEN THE SABER IS ON. Then use Ohm's law to figure out your current.

Resistor = R, Current = I, Voltage = V, V = I*R or I = V/R

So, measure your voltage across that resistor, divide it by it's resistance, that's your current. Now multiply the voltage and current... that is your power dissipation across that resistor.

Now.. compare that MEASURED power dissipation to the RATED power dissipation for that resistor. Is that the issue?

To lower the power dissipation across your resistor, you must pick an input battery voltage VERY CLOSE to the forward voltage of your LED. Or... use a Buckpuck from TCSS.

[Vazan Maceu]

Mmm, I see now, Eandori, great explanation, now it´s crystal clear. Thanks a lot