LEDs

Oh yeah. I would really like to know how to calculate voltage drop, it would help w/ other projects also :lol:

zachtheterrible said:

Oh yeah. I would really like to know how to calculate voltage drop, it would help w/ other projects also :lol:

Click to expand...

It's just simple ohms law - V=IxR or rearranged R=V/I

You need to know the voltage drop off the LED, say 2V (for easy figuring), you've got a 9V battery - so the resistor has to drop 7V. You now need to decide what current you want, say 10mA, so all you have to do is divide 7V by 10mA, which gives 700 ohms.

In all these sorts of calculations you always need to know two of the variables, you can then find the third. Often, as in this case, one is a known constant, and the second is a decision you have to make, based on what you want to do.

Alright, i think i got it down, thanx :lol:

According to your diagram, and my basic knowledge on electronics,
I think you should considere that connecting directly the diode to you voltaage source will damage you DIODE, so you need to limit the current
A resistor will do that (calculate and connect it in series with the led)
for example

your diode can work at 3.5 volts @ 15mA then
voltage source 9V

(9-3.5)/0.015 = R = 366 ohms, but thi value it just in calculation, the nearest value is 330 or 680 ohm's

so you will need to take into account spending more current with 330 or make an array of resistors.

Thanx sardineta :lol:

sardineta said:

(9-3.5)/0.015 = R = 366 ohms, but thi value it just in calculation, the nearest value is 330 or 680 ohm

Click to expand...

330 or 680?
There are some more resistor values closer to 366 then 680 in the E12 series.
390 or 470 ohm