Originally Posted by rowan (Post 12054132)

I understand what you're saying, I just don't understand how it works. With every fraction of a volt above the combined forward voltage of the LED string, surely the dissipation will increase remarkably?

I know that those keychain torch LEDs don't usually have a resistor, but that's because the button cell batteries cannot supply more current than the LEDs can tolerate. A short circuited 9V battery would probably be able to push out more than 30mA...

If the forward voltage of each LED was (say) 2.8V and the battery puts out exactly 9V then the difference is 0.6V, and to stay at 30mA you'd need 22 ohms of series resistance. Surely a battery does not have such a high internal resistance?

Originally Posted by Vick! (Post 12053492)

I understand what you mean. But mate, when we say 'right' voltage -- Its mean we are limiting the voltage to a certain level which do not harm the circuit (LED). As long as concern to limit the current, we do not limit. It get limited on its own depending on the internal resistance of the circuit (LED).


For example, attach two batteries of 1.5 volts each to make 3 volts, right? Then attach an LED to it, LED will be ON.

Did we limit the current? No, current is automatically remain within limit of approx 15-20mA due to the internal resistance.


Ohm's Law comes here, at constant resistance current and voltage are directly proportional. So, limiting voltage is actually limiting current as well.
You are also right in one way. But again, to limit the current we actually limit voltage, assuming resistance is constant.

Originally Posted by Basic_man (Post 12054708)

If you put the LED right on a power source, the LED will fry. Why? The internal LED resistance is LOWWWWWW, so the current will be BIGGGGGG.

So what you have to do, is put a resistor in serie to limit the currrent. :warning

Originally Posted by Vick! (Post 12053950)

Well, If we connect 3 LEDs of 3v each in series with a 9v battery. Do we need resistor?

Originally Posted by Vick! (Post 12054597)

your calculation is somewhat right, but 22 ohms of resistance is so small that it can be ignored and we assume LEDs are working on 3v each.

No doubt these values vary depending on the type of LED and strength of 9v battery. A fully charged 9v battery have potential difference of up to 9.4 volts whereas a less charged can have 8.4 to 8.8

And to answer how it works, in the chain of 3 LEDs .. other two acts as simple resistor to the third. So, actually each LED get ~3v potential difference across its terminals.

I don't know if I got what you were really asking .. may be lol ..


Hey, have a look at this .. it might help http://www.theledlight.com/ledcircuits.html

Originally Posted by dozey (Post 12056801)