i am modifying my 18v rechargable torch by doing away with the hot 18v bulb and replacing it with a (12x) LED light that im using from a 4.5v headlight torch (the donor torch). so i need to cut the voltage from 18v down to 4.5v.... can i do this with aresistor, and if so, what value should that resistor be? thanks kind people!
2007-12-10 12:31:30 · 5 answers · asked by philo 2 in Science & Mathematics ➔ Engineering
If you don't use a solid state voltage regulator, you will waste 75% of the power in the battery in the resistors.
You can get a simple TO-3 adjustable regulator that will handle the amps although you may have to search a bit to get one that will be happy cutting down from 18 volts because that is a bit high. The whole design would be neater if you could change the wiring of the head so that it was divided into 4 sections of 3 LED's each (presumably still 4.5 volts if all 12 are in parallel) and wire those in series to get 18 volts. Of course, the head may not be built that way and if you are asking here in this way, you don't know enough to do it.
2007-12-10 13:13:51 · answer #1 · answered by Mike1942f 7 · 2⤊ 0⤋
One way to do it:
First, measure the current used in the 4.5V torch. Then select a resistor where R = 13.5 / i where i is the current you measured. For example, if you measure 100mA, R would be 13.5/0.1 or 135 ohms. The number 13.5 is used to find a resistor that will drop the voltage from 18 down to 4.5 under the expected current condition.
If you are using a wimpy resistor that gets too hot, you cat put more resistors in parallel. If you use 2 in parallel, use twice the R value. If you use 3, use three times, and so on. Good luck.
2007-12-10 12:47:06 · answer #2 · answered by Anonymous · 0⤊ 0⤋
I'm assuming that a "torch" is a flashlight. I'm not sure how much space you have for the electronics but your best bet might be to get an inexpensive 5 volt regulator chip. You may still need a current limiting resistor if the LEDs are not self limiting. If you want to use just resistors, then it could be done with two. One would go inseries with the LEDs and one in parallel with them to ground. The ratio of the resistors would be 13.5 for the series to 4.5 for the parallel. You may need to experiment with values to find the right current flow but I would start with something like 1.2k and 470 ohm.
2007-12-10 12:46:56 · answer #3 · answered by Anonymous · 0⤊ 1⤋
Go to your local electronics repair shop (Everywhere is full of them !!)
Ask the engineer to make you a LM317 Circuit, its more stable than a resistor and runs cooler - a resistor will get so hot it will probably melt the case !!. Only 3 components are needed.
I used this to charge a 6 volt battery from a car lighter socket.
You can adjust the voltage output too.
Have Fun !! Jim Maxwell
2007-12-11 16:24:09 · answer #4 · answered by Jim Maxwell 3 · 0⤊ 0⤋
#1 Powerful Tactical Flashlight : http://FlashLight.uzaev.com/?DfQd
2016-07-11 03:51:32 · answer #5 · answered by Major 3 · 0⤊ 0⤋