🚀 go-pugleaf

I think so. The high voltage drop across R is more accuracy for
white/blue LED.
But it will lost so many power. 

^_^



John Woodgate wrote:
> 
> >Yes, I mean in series.
> >But two strings can not use the same R values.
> >So sad it very hards to calculate the current and R.
> >the forward voltage drop does not identical in each LED.
> >and it changes when current change too.
> >
> >ex:
> >
> >strings 1 :   ( 12 volt - ( 3.453 + 3.6324 + 3.833 ) ) / 40 ohm = 27 mA
> >
> >Now I want 20 mA ,  How to calculate the resistor ?
> >Since forward voltage change when the current change too.
> 
> A very slight change in temperature would change your 4-figure voltages
> significantly. You are looking for a degree of precision which is
> meaningless. Consider the voltages to be 3.45, 3.63 and 3.83 V. Between
> 27 and 20 mA, they will hardly change. So your resistor has to drop 12 -
> 10.91 V at 20 mA, and is thus 54.5 ohms. A standard 56 ohm would be OK.
> While the current won't be *exactly* 20 mA, it will be close enough not
> to make a significant difference to the light output.
> 
> However, this resistor value is much too low to keep the current
> substantially constant if the (12 V - just 2-figure accuracy) supply
> voltage varies. Furthermore, 1.09 V is not really enough voltage to
> operate a simple electronic constant-current circuit. With just two LEDs
> in series, the simple series-resistor technique is much more
> satisfactory. 12 - (3.63 + 3.83) = 4.54 V. 4.54/20 mA = 227 ohms - use
> 220 ohms, giving 20.6 mA. Then at, say, 12.2 V supply, we get 4.74 V
> across 220 ohms, which means that the current is 21.5 mA. If that is not
> 'constant' enough, 4.54 V is quite enough to operate a simple electronic
> constant-current circuit. Just a single transistor, two resistors and
> two diodes (or a red LED) would do.