4 x 2V = 8V. From a 9V battery, I should get 1V/0.03A = 33Ω resistor for the circuit. Which I dont have, so I left in a 220Ω resistor which I already wired into it. If anything it should make the LEDs light up even less.
Anyway, Ive temporarily left them on (5s tops) and they light up pretty well. How do I calculate to make sure they wont burn up?
30mA is the absolute max rating, and you should not get near that value if you want the LEDs to last.
IMHO 20mA is already a lot for a 1206 smd LED.
1-10mA should be enough to see (look at) signal LEDs without blinding you.
If you want "light", then use e.g. 1watt star base LEDs.
Vf @ 20mA could be 2-2.4volt, meaning you should only use three in a string on a 9volt supply.
And calculate current for a Vf of 2.2volt.
Leo..
If you look at the schematic of the LED module, you will see that it already has a 100Ω resistor in series with the LED. (This can be clearly seen on the photos on the page you linked to).
So if you have 4 of them in series, and a 220Ω as well, then you have 620Ω limiting the current.
The voltage across each LED will be less than the typical 2V shown in the datasheet,as that is at a current of 20mA.
Taking that into account, My guess is that the current will be around 5mA. Well within the LEDs limits.
Have you thought about what happens as the battery ages... Goes down to 8.5V.. 8V... 7.5V...?
In general, it's a good idea to keep the voltage across the resistor greater-than or equal-to the voltage across the LED(s). That makes the brightness a bit less dependent on supply voltage.
@JohnLincoln
Good catch. Missed that buildin 100ohm resistor.
One of those LEDs can be connected directly to a 3.3volt (~13mA) or even 5volt source (~30mA)
Three LEDs in series can connect directly to a 9volt battery.
Leo..
Marciokoko:
Ok So 1 led:
3.3v = i * 100 => 0.033 amps
No, the LED itself drops ~2volt and the resistor drops the remaining (3.3-2) = 1.3volt.
1.3volt across the resistor is 1.3volt / 100ohm = 0.013Amp = 13mA
Same for three LEDs on 9volt.
The three LEDs drop 3*2 = 6volt.
9volt - 6volt LED = 3volt across the the three resistors in series.
LED current = 3 / 300ohm = 10mA.
Leo..