![]() ![]() It may be convenient to start with a commercially available lamp driven by 12 Vdc, which could be supplied by about 8-10 alkaline cells. By adjusting the number of LED's in series and parallel and the number of batteries in series, you should be able to get an efficient combination. An alkaline cell's voltage depends on how much current it's supplying, so the voltage can be adjusted a bit depending on how many parallel LEDs are being driven. It's important to match the power supply voltage to the optimum operating voltage of the LEDs. This is safe, because there are no high voltages at any point in the circuit.Īccording to a Wikipedia article, new white-light LEDs are about as efficient as fluorescent bulbs. You can directly get plain LEDs and power them off the batteries. Perhaps you'd do best using some LED lights. A dc-ac converter (inverter) would add substantially to the weight. Those compact fluorescent lights are designed to be used with ac power, not the dc power supplied by batteries. ![]() Ordinary alkaline cells may not supply 75W for very long at all - a good car battery will last much longer. You can easily find 40W halogen bulbs at most hardware stores that run on 12 volts of electricity (AC or DC). The headlights in a car have exactly the parameters you desire - about 50 watts per, and there are two of them, driven in parallel on a 12V car battery. I would instead find a bulb that works at lower voltage- say 15V or 25 V. Stringing together 120V of batteries (say 9 or 10 car batteries in series) leaves a very dangerous voltage, easily enough to kill someone, without the relative safety provided by standard wiring outlets and plugs. The batteries will provide dc voltage, not the ac provided by wall outlets, but the effect in heating the bulb filament is the same. Sure, any arrangement of batteries that provides the right voltage (about 120 V, if it’s a standard household bulb) will do. The light you do give off is infrared, which can be detected but not directly by our eyes. Thats because your temperature is too low to give off visible light. You yourself generate about 60 W of heating power, the same as a 60 W bulb, but Ill bet you dont visibly glow very much. If you use a bit too low a voltage, the bulb will glow orangeish, because it can still put out some colors of light but not the blue part of the spectrum. Thus using one fourth of the power will give much less than one fourth of the light output. Second, the amount of visible light produced in the bulb is virtually zero until the filament temperature gets close to the standard operating temperature. First, the heating power in the bulb goes as the square of the voltage, at least until the voltage gets big enough for the bulb to heat up and increase its resistance. You might think that using a lower voltage would only slightly dim the light, but actually the effect is much more severe. Bulbs from cars are usually designed to work with about 12V, the output of a car battery or of eight standard battery cells in series. Ordinary flashlight bulbs are designed to work with about 3V, easy to obtain with two batteries in series. Standard bulbs are designed to work with a voltage of around 120 V, which is an unusual range for batteries. If the battery has too high a voltage, so much current will flow that the filament will get too hot and vaporize. ![]() If the battery has too low a voltage, the current flowing through the bulb will be small and the bulbs filament wont get hot enough to visibly glow. Its important to select a bulb which matches what your battery can put out. The spring contacts in flashlights work much better (but even they are troublesome from time to time). Its notoriously hard to get good electrical contact on batteries and bulbs by soldering wires on. Other bulbs will have metal prongs sticking out. ![]() Many bulbs have one electrical contact with screw threads on it, with the other contact as a round dot on the end of the base. Just connect the positive terminal of the battery with one electrical contact of your light bulb and the negative terminal with the other electrical contact of the bulb. ![]()
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