im trying to understand led's and i need to know how i can find out what the ones i have lying around's voltage or necessary information to get them to work with out burning them up....ive got 12v power that i want to run and ive got a couple of power switches that run things in my car i.e. my antena and i want the lights to light up when the power is on but dont know where to begin to test them. thanks
You can usually guess, as long as you know the colour. If you have a multimeter you can find the forward coltage using the diode test if you want to be mroe scientific, but you'll still have to guess the current.
so with a mulitmeter in hand and one of the big red leds out of a nintendo what is necessary? to figure out what it is
If the multimeter has a "Diode Test" position just switch to it and put the probes to the led wires. One way the led should glow and you get a reading, other way nothing. One problem, the meter tests at a very low current, so you get a very dim light and a low forward voltage reading. But it's OK for testing if the led is OK and which lead is which. Safe figures for 99% of unknown leds: Red, yellow, amber, orange, most green : 2.1V, 15-20mA Blue, white, true green: 3.6V, 10-15mA.
thank you that helps tremendously....no say i have 12volts i want to use as my power what do i need to do to drop it down to say 2.15v ? thanks
You don't look at it that way - you start with 12V, the led takes 2.15V leaving 9.85, and you set the led current with a series resistor so Resistor = Volts/Current (Ohms Law). For Current=20mA (0.02A), thats 9.85/0.02 = 492, a 470R or 510R resistor is near enough. You don't need to know the exact led voltage to pick a suitable resistor, 2V would give 500R, 2.3V would give 485R needed, so near enough is good enough.
Ohms Law I=Current (Amps) R=Resistance (Ohms) E=Voltage(Volts) P=Power (Watts) E=IR R=E/I I=E/R Find out the LED's current rating usually 20ma Milliamps for most standard led's Find out the Bias voltage for the LED (Usually depends on color) Subtract the bias voltage from the source voltage, and divide that by the current rating for the LED. This will give you the value of the resistance that you will have to put in series with your LED in order to control current flow. EXP. 12V Supply minus 3.4V to bias the LED = 8.6V to drop across the resistor. LED has a current rating of .02Amps (20ma). These are your known Variables. plug them into the formula R=E/I to find the Resistance value and you get 430 Ohms rating for your resistor (always go next Highest standard value of resistor. It's always better to allow less current to flow instead of more than the LED is rated for, because more means less as far as your LED's lifespan is concerned..... Sometimes much less To calculate the Wattage that your Current Limiting Resistor should be just take the voltage drop across the resistor exp.(8.6) and multiply it by your Current (.02Amps), which gives you .17 Watts of power being dissipated by the resistor. A 1/4 watt should work in most cases, unless you are hooking several LED's in parallel, or using a Luxeon emitter or somthin, also if you are using an extreamly high supply voltage. Note. Current is a constant in a series circuit. Voltage is Constant in a parallel circuit. Sorry if this confusing, just use BitTech's calculator. http://www.bit-tech.net/article/68/
cool ok so basically from what im understanding is that its the resistor that basically "filters" out the voltage for the led....so if i tie it into a 12v powersupply using the proper resistor it will only allow that 2-3v through.?
NO, Think of voltage as pressure Think of Current as rate of flow think of resistance as a kink in the hose. By kinking the hose you only allow a certain ammount of current to flow through the circuit wich also drops the pressure. The LED wich is essentially a diode... A one way shutter that needs a small amount of pressure to open once the shutter is open current wil flow through at the rate you limmited it to by the kink in the hose. Once the shutter is open the diode is essentially a wire allowing current to flow. If you increase resistance (kink the hose harder) pressure and current supplied to the diode (shutter) will decrease. I could go on. but I think I'll just confuse you.
Without the resistor, there would be a 12v potential between the cathode and anode. That's too much difference for the LED, so it burns out. What the resistor does is removes some of that potential. Even though there is still 12v across the two supply rails, the resistor will take 8.6v of it. If you wire V+>LED>Resistor>ground and use your multimeter to measure the potential between the two leads of the LED, it would only be 3.4v across it, which won't burn it out. If you measure between the two leads of the resistor, you would find that there is 8.6v potential across it. The resistor just takes some of the potential away, so that the LED gets the 3.4v it needs to light and not burn out.
But, voltage does not go through the hose at all. It's more like you have two tanks of water, one at 100psi and one at 3psi, with a hose inbetween. The tanks are voltage. But the voltage does not go through the hose.
so yah then thanks for your guyses help im just getting into modding and stuff....i know electronics and wiring just still learning the functions of transistors/resistors etc.
Go to your library and get a book about beginning in electronics. You can learn a lot from them, not only the uses of semis and whatnot, but also the theoretical stuff, like why we're saying that the resistor isn't filtering anything.
