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.)