Tesco 3w c cell (again)

[fire2727]

Picked up one of these http://direct.tesco.com/q/R.203-8191.aspx Now that ive got my first light up and running http://www.candlepowerforums.com/vb/showthread.php?t=219020 im looking to start a led mod. I just want to take out the battert tube so that it can be fitted onto the bars of my mountain bike and fit a battery pack.

The mod to the battery tube should be easy enough, just need a point in the right direction for the battery pack would like it to run for a couple of hours.

A pack of AA,s a couple of C cells or D cells which is best? did see these http://cgi.ebay.co.uk/ws/eBayISAPI....ksid=p3907.m29&_trkparms=algo=LVI&its=I&otn=1

But not got a clue thought I'd check before I bought them.

 

[fire2727]

Do i work out the Ah for a 3w cree in the same way i'd work out the amh for my 12v 50w mr16 bulbs ?

 

[Ogg Vorbis]

Some simple formulas might be of help

Power (W) = Voltage (V) * Current (A)

So for the 3W Tesco torch, we know that the input should be 3V, or two alkaline @ 1.5V ea so current should be 1 Amp

3V * 1A = 3W

I just measured the current draw at the battery end on mine and got 0.85A, but my batteries aren't that new etc

then Ah is just a measure of capacity, or how long it can supply a current for...

So a 5Ah battery like the ones you linked (1000mA = 1 A) would provide

Run Time = Capacity / Current draw
= 5000mAh / 1000mA
= 5 hours +

But maybe some AA's would be better as they won't weigh as much? I'm not sure what you plan to do with the tesco torch so it's hard to say what would be best.

Hope this is helpful? and good luck

Dan

 

[fire2727]

Thanks ogg, so I can run two of these batteries on this torch and get 5 hours runtime wooo hooo:rock: the plan is to use it as a bike light, im going for a second one to fit along side it.


Again thanks for your help

 

[fire2727]

Right up and running now very happy with the lights lovecpf
Now I need to know how I can charge these http://cgi.ebay.co.uk/ws/eBayISAPI....ksid=p3907.m29&_trkparms=algo=LVI&its=I&otn=1 batteries, they are now in a sealed pack with a jack plug connection, any help appreciated. A link to a charger would be great thanks again all

 

[VidPro]

assuming you have these 2 cells put in series.

one easy cheap way would be with a 3v DC wall wart with a max of about 400ma output. that way you make up a slow dedicated overnight charger that would rebalance them.
you would have to check things a lot, so the wal wart didnt run to hard for itself (burn itself up) or the batteries, mabey toss a diode or a bit of resistance in, after checking with ammeter and voltmeter. if you have a charge jack on it already (from what you said) it would be cheap easy and convienient. and getting a 3v 1AMP, as the starter then having it run at half that, would keep it alive for a long time, so when i say 400ma, i mean in the final result.

a transformer or car cord with 5V output, and use an adjustable voltage regulator and adjust it down to about 2.9-3v which would charge as fast as it could at first, then as the voltages reached similarity would reduce the charge and finish by topping off, once the cells are mostly charged, you would set the max voltage , to control the max current at the end. this would charge a bit faster, still be cheap, and would slow down at the end when set right.
same thing reconfigure adjustable regulator to control current, and lock it out so max current was within spec for overcharge of the cells, but then you dont get the faster initial charge , which is very usefull even on a 8+hr charger.

any of the RC ni-mhy chargers , which would be capable of very fast charging, and would top off and balance , and do everything perfectally, and much faster, costs range from $79 and up.

with these notes: if the actual capacity of the cell is lower the ammounts would be lowered for the slow charge, these are the claimed specs "Standard charger Current(mA):500" the numbers i used were just safely under them.
and to use the first 2 ideas, you gotta check and test and be sure that at the end of the charge when the batteries are (combined) at about 2.8v that the current is below about 400ma.

use a holder where you can remove the cells, and charge them on a single channel charger that handles C-D cells, like a Maha or something, this give you the advantage of the best way to charge it, by doing so without being in SERIES, and also the advantage of replacing , and having spares for longer runtimes.

 

[fire2727]

Thanks for the advice VidPro, I've got an old phone charger, output is 3.7v / 355ma, could I use that or will it need a resistor to drop it down to exactly 3v? What should i check for as im charging.

Charging time is not a problem, I have no need to charge them up quickly.

I relies these are probably basic questions so thanks for your time.

 

[VidPro]

you could use it if you follow the specs, so it will need to be DC first , that is obvious, and even though the output says its max,
it might be able to run for 2 times that , depending on the build of the walwart, and that would burn out the wal wart, and be over spec.

lots of the wal warts will fail over time if driven to hard, that is a main factor, even if you get the connection to stay within spec, the wall warts life itself.

so check the ACTUAL amps, put some FAT diodes inline, or a FAT resister, no use having the "resistance" item break over time due to the heat either.
and remember it's the Amps or ma when the battery is at a Charged state, that are most important, to keep below specs.
the total amps out of the wall wart are important to the wal warts life itself.

your better off with a walwart or transformer that is CAPABLE of more amps, then driving it much slower, and your better off somehow keeping the voltage lower, so the power tapers down as the batteries get higher in voltage. faster charge at first good, slower charge as it is finished charging good.

it should be noted that with a 3.7v charger resistance will not change the VOLTAGE itself, only the amperage that goes in at various battery voltages, which would be sufficent as long as it is within spec. but a Fat 3 amp cheap "barrel" diode (for example) with .7v drop will reduce the voltage AT certian currents, and provide a better taper at the end of charge.
Then that gets notes too the diodes voltage drop will vary depending on the current, but it would work out good in reality, lowing the max voltage at the higher currents and getting a nice taper at the end of charge.

shoot, wouldnt it have been easier to say toss in a big fat .6-.7v dropping diode, and test it once, and it should be good to go

 

[fire2727]

Thanks again VidPro im trying to digest your much needed information. Might be a bit too much to take in.

 

[fire2727]

Is there a off the shelf way of charging my 2 x c cell 5000 mah and 3 x aaa 1600 mah batteries ( in separate lights of cause ) while they are in series or do I have to remove them from my lights? I would like to plug into a jack plug fitted to the lights.

Ive charged a 6v sealed lead acid light with just a car battery charger ( set to 6v ) and jack plugs, it's the lower voltages i cant get my head around.

The above posts are much appreciated but are way over my head to work with :thinking:

Thanks again for the help so far .........