How do I hook up LED's to a battery?
First of all, I don't have much electronics experience apart from doing a semester of electronics in yr 12 (and they didn't really teach us how any of it works, just said 'put this resistor here and that there' etc).
I pulled 3 LED's out of a pc which my neighbour was going to throw out, so I don't really know what the voltage range is.. (If that makes it impossible to use them, please tell me which ones to buy because I still want to do the following)
Now, I want to use them in a craft project and I don't have a clue how to hook them up properly. It doesn't really matter to me whether I hook them up to a battery or a power point, but I'm on a low budget so I figure battery would be cheaper, right?
Now, how on earth do I go about making the circuit board for them? Do I need to make a proper circuit board at all (I don't have that liquid for getting the copper off etc)? All I want is for those three LED's to be lit up solid.
Yellow, red, and green LED's have forward voltages of somewhere around 1.5 volts. What this means is, when the proper amount of current is flowing through them (usually about 20 milliamps) then the anode of the LED will be aboout 1.5 volts higher than the cathode.
The cathode is usually marked by a small flat spot on the rim of the LED.
You don't need a circuit board, all you need to do is solder the three LED's together end to end, so that the cathode of LED #1 goes to the anode of LED #2, and the cathode of LED #2 goes to the anode of LED #3.
Next you need to solder a resistor to the cathode of LED #3. I'll tell you how to calculate an exact value in a moment, but you can use 240 ohms to start and you will be very close.
Lastly, solder a 9-volt battery snap so that the positive lead (red) goes to the anode of LED #1, and the negative lead (black) goes to the resistor lead which is NOT soldered to LED #3.
Now, you can get a nice little experimenter's PCB from Radio Shack for about $2 that you can sodler these LED's and resistor to if you really want a circuit board.
FInally, about that resistor value: if you have a multimeter with a diode checker function, use it to find the forward voltage of the three LED's. Hooking it up to the LED one way will give a reading somewhere around 1.5 (this is the LED's forward voltage), and the LED will light up a little. Hooking it up the other way around will produce an "open" or "infinite" or "error" reading of some kind, and the LED will not light up. Anyway, write down the three forward voltages you measure for the three LED's, and then add them all together. Subtract that total from 9, and divide the result by 0.02 (you can multiply the result by 50 if you don't have a calculator, that will be easier than doing a long division by 0.02).
What you are doing is this: You are makking a series circuit with a total of 9 volts (the battery) across it. Some of the voltage is going to be across the three LED's (this is the total of the three forward voltages you measure) and the rest will be across the resistor. So by subtracting the total of the LED forward voltages from 9, you get the number of volts that will be across the resistor. Next you are dividing that voltage by the current you want, which is 0.02 Amperes (20 milliAmperes). Because Ohm's Law tells us that Voltage = Current times Resistance (E = I*R) and we can re-arrange that using algebra to show that R = E / I. Since we just calculated what the voltage across the resistor will be (E), and we know what current we want (I, = 20 milliAmperes or 0.02 Amperes), we can calculate howm many Ohms resistsnace (R) is needed to make it work that way.
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