12v 9ah battery for hid lights

this might not be the right place for this question, but i figure i would ask. I am installing a H4 digital ballast HID light in my snowmobile. Just 1 light and ballast. Ballast says that input is 8-35v, the HID bulb is 35w. My sled does not have a battery. It runs AC voltage for the stock lights and gauges. I have all ready installed a Koso rectifier/regulator into my sled for a DC v temperature gauge. The converter specs are Input= 12-30V AC/DC, Output is 12-15V DC, Max load 6 Watt. I know this is not powerful enough to get the HID to strike the arc. So i am going to put a 12v 9 ah battery into my sled and run my temp gauge and HID light off it. my question is if i run the DC + wire from the converter to the battery, if it will work to recharge the battery as i ride. Or what i have to do to make it work. I would appreciate any help/ideas

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Jeff

The converter will not run the HID - it's only 6W max and the HID is 35W output (probably 40W or more input).


A battery will do it. Assuming the converter can supply up to 6W (and 14.0-14.4V), then you have a (~40-6=) 35W plus gauge shortfall.

At 12V, that's 3A, and a 7AH battery should last 2 hours and ~160 cycles (using Yuasa NP7-12 data and assuming 2hrs @ 3A in 100% discharged). (1200 cycles if 30% discharged.)

Thank you for your knowledge and fast reply. Unfortunately i do not know very much,'' (~40-6=) 35W plus gauge shortfall.'' I have no idea what that means. Could you put it in lamens terms? And would the Koso converter charge the 12v 9ah battery? or do i need to install diodes, resistors or anything so it will charge that battery?


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Jeff

No probs. Sorry for my shorthand.

For your first question, it matters not. You have AT LEAST 35W being consumed, yet a supply that can only provide 6W, hence the converter (alone) will not power the HID.


Hence add a battery as you suggested.
But what size? And, how long will it last (both wrt reserve time (how long till it reaches a certain discharge or remaining-capacity point) and if lifetime (200 cycles at 2 cycles per day means after 100 days you may be seeking a new battery; or 1200 cycles once per day = 3 years).

Hence my "extra" calcs to give an idea = ie, total drain is converter 42W input (plus gauge) less converter - ie, 42W-6W or '(~40-6) =35W" that I used earlier.
[ These are only ball-park estimates where convenience and conservatism apply - eg, I'm likely to round figures to convenient numbers like 35V becomes 36V for a clean 36V/12V = 4A current. 35/12 = 4.16A - not much different, and although we rounded "down" (4A not 4.14A) instead of the usual conservative (overstated) up, it tends to balance out, and its is a rough eztimate anyhow.


Your HIDs are 35W. That's usually the output power (bulbs) and the have inefficiency as well (they may be 80-90% efficient hence 3.5W - 7W lost with 35W output => 35+ 3.5-7W = 42 Watts input.
Plus add your temp gauge which is probably negligible (3W to 10w?)



To charge a battery you need:
- copious voltage
- copious current

Firstly the current. The charger must be able to replace the charge lost from the battery (ie, recharge) PLUS supply the load.
You converter is 6W so it won't do both.
But 6W can recharge the battery with no load. (A rough estimate - for every hour the load and converter is on, the battery loses ~36W-hr. To recharge at 6W means 6 hours, plus battery inefficiency mans 8-10 hours.)   

But the voltage has to suit.
Usually battery chargers are set to 13.8 to 14.4V. (14.4V is a "maximum" long-term voltage for most lead acid batteries. Vehicles were once set to 13.8V but that was increased to typically 14.2V - 14.4V at of the 1980s for longer battery life.)
So you need to set your converter to (say) 14.2V, preferably maybe 14.4V, though the battery may charge with anything above ~13.2V but much slower.


Not all converters tolerate a battery on the output - the battery might feed back into to converter.
In such cases a diode needs to be added. Since the converter can only supply 6W, a 1N00x didoes should do. (1A rating, IN4004 & 1N4007 are common, but any 1N4001 upwards will do.)

If it were a supply both battery and load, 45W/12V = 4A => a 5A diode PLUS the battery recharge current which could be many Amps (that gets tricky...).
So maybe a 10A diode (or 2 5A in parallel,though one will probably take more current than the other and, if one blows, they both blow).

And if a diode is used, the voltage has to be increased to compensate for the diode's ~0.6V drop. For an output of 14.4V, I'd set the converter to 14.9V, hence 14.9V minus diode's typical 0.6V = 14.3V for the battery and load (0.1V under our "maximum" 14.4V for a battery).


But adding diodes etc won't otherwise help your "undersizing".
You either need to switch off your HIDs (for probably 10 times longer than you had the on) else get and alternative power source.

Or use LEDs. (They will soon take over from HIDs.)

Thank you for taking the time to explain all that. So bottom line is the little converter won't recharge the battery fast enough to keep it fully charged. So i guess i am taking a trip to the sled junk yard and finding a voltage regulator off a sled with electric start and use that to charge my battery. If it charges a 12v battery in a sled with a normal battery, it should charge the 12v 9ah battery i put in. Hopefully it has the diodes and stuff built into it so it doesn't over charge it. Thank you for your help

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Jeff

A sled with electric start may have different hardware - ie, if it is charging, it must have some charging coil - (an alternator in cars) - it is unlikely that the "magneto" that supplies the spark and maybe a bit extra also does battery charging and vehicle electrics, though I could be wrong.   

Check is anything else is different engine/alternator wise with the el-start sleds.