SpudFiles

I am almost finished with my ignition box for my combustion, but there is one part of the box I can't seem to figure out. It has a keyed saftey switch and a momentaary firing switch. What I want to do is install a LED to indicate that the saftey switch is on. The power supply is 2 rechargeable 9v batteries, each supplying 170mA. I just don't know how to lower the voltage enough to supply power to the LED. If I use a voltage reg, the lowest I can get down to is 5v. The LED I plan on using is rated for 1.7v at 20mA. Any help/suggestions will be greatly appreciated!

use a relay and a resistor. When the arm switch is flipped, it will close the relay (closing the circuit between the led and battery) but also complete the circuit for the momentary. there might be an easier way but I'm kinda blanking on the electronics thing at the moment.
I think for a nine volt battery a 400 ohm resostor works well, 320 ohm should work to.

Can't you get a LED at the right voltage?

you shouldn't need to lower the voltage for the led just lower the current, put a 1k resistor before the led and it should be fine (works out to be 18mA at 18 volts)

Almost all LEDs are ~1.4V devices. There are a few as low as 0.7V and some as high as several volts, but those are pretty rare.

Just wire a resistor and the LED in series after the key switch.

The value of the resistor is;
R = (power supply voltage - voltage drop of LED) / (current draw of LED)

"voltage drop of LED" is usuall 1.4V
"current draw of LED" is usually 20 to 30 milliamps

If your two 9V batteries are in series then;
R = (18V-1.4V)/0.025A = 664 Ohms

EDIT:
Ooops, left something out. LEDs are not usually run from athis high a voltage. So, you need to worry about the amount of power dissipated in the resistor and make sure the resistors power ratings isn't exceeded.

Power = (Voltage)(Current)

If the power supply is 18V (2x9V) and the current is 0.025A then the power dissipated in the resistor is (18V-1.4V)(0.025A) = 0.42 watts. The most common size resistors are only rated to 0.25 watts. At 0.42W a 0.25W resistor will get very hot. So, you should either use a 0.5 watt resistor, or you can use two 330 Ohm resistors in series, this give the need 660 Ohms, a voltage drop across each resistor of 8.3V and the power dissipated in each resistor is (8.3V)(0.025A) = 0.21 watts, well within the spec. for quarter watt resistors.

iPaintball wrote:I am almost finished with my ignition box for my combustion, but there is one part of the box I can't seem to figure out. It has a keyed saftey switch and a momentaary firing switch. What I want to do is install a LED to indicate that the saftey switch is on. The power supply is 2 rechargeable 9v batteries, each supplying 170mA. I just don't know how to lower the voltage enough to supply power to the LED. If I use a voltage reg, the lowest I can get down to is 5v. The LED I plan on using is rated for 1.7v at 20mA. Any help/suggestions will be greatly appreciated!

Actualy, LED's don't really care about voltage as long as it's in the right direction, its just current that's the problem. You shouldn't need a voltage reg, some people on the other forum I visit uses LEDs as a spark detector on a Tesla coil. find some way to lower the current instead of voltage.

It's here on the person's page: http://www.tesladownunder.com/HighVolta ... 0indicator

You have to scroll down a bit to find a pic w/ two LEDs soldiered w/ some resistors.

I used an LED+Resistor connected in parallel to a fan in my combustion to indicate that the fan was on. The supply was 9V.

Spedy wrote:Actualy, LED's don't really care about voltage as long as it's in the right direction, its just current that's the problem.

Almost but not quite.

LED's do care about voltage, in particular they need a minimum voltage in order to light. Most LEDs neead at least ~1.4V. If the voltage is much below 1.4V then it doesn't matter how much current you put into'm they won't light.

You are right that you must limit the current to the LED. Though with a 1.4 or 1.5V supply you can frequently get away with not limiting it.

If you string ~120 LEDs in series you can plug it directly into a 120V AC wall outlet without any current limiting, rectification or voltage regulation.