You should design for the worst case extremes of LED Vf and supply voltage.

And to expand Alec_t's warning, "worst case" is assuming the battery is at its highest voltage while under charge (e.g. 15V), and Vf is at the lower end of its quoted range.

Also, if I was working with a 20mA LED, I might shoot for 15mA. It'll still be plenty bright and will last longer than if you run it near the maximum. It'll also add a bit to the battery time.

 

Ho Alec, I assume then that I should use the max forward voltage, rather than typical? And use maximum battery voltage for source input voltage?

If so, as the LEDs are 4v max Vf, and the battery 12v, having my string of 3 LEDs gives 12v, so would I use a tiny resister? And if so, what ohm we talking? 1, 10 etc. As the math says 0 would be needed. Or would I need to lower the current to say 15ma rather than the 20?

 

If you want to avoid the LEDs passing excessive current you should assume they have the minimum Vf and the battery is at max voltage, as Wayneh stated in post #21.

 

I doubt you are going to get 4 volt LEDs. This is going to end up custom made. Wait until you get the LEDs and measure them. If they really do turn out to be 4 volts, you will only use 2 per string.

and while we're at it, there is no law that says you have to run the LEDs at their maximum survivable current. 70% to 80% will work just fine.

 

Hi all,

Wayne, sorry, didn't see your post before posting my reply yesterday, didn't realise we were on page 3.

#12, how would I measure them? The sheet I downloaded for them, free Cree, said max of 4v.

So, I'm designing for max battery and lowest forward voltage. Roger that and will redo math when home as away this weekend, hence late reply. I'll also use 15ma for current rather than the 20.

Thanks all,

Upex

 

#12, how would I measure them? The sheet I downloaded for them, free Cree, said max of 4v.

Create a circuit that will send about 15 milliamps through a 4 volt LED. Connect them so that the LED is "on" and measure the real voltage across the LED.