Ok now we are getting somewhere.
@VE1BLL (Military)
I now understand your concern. And I'm as concerned as well
Here is my circuit in a nutshell:
3.2 Vf typ (3.6 Vf MAX)
.150 mA
24 volt CV PS supplying 3.3 A
7 series in parallel with 22 chains
I calculated using typical Vf as the following:
24/3.2 = 7.5 devices so I rounded down to 7 devices in a series chain.
7*3.2 = 22.4 volts required
24 supplied - 22.4 volts = 1.6 volts to dissipate.
Resistance = 1.6/.150A = 10.67 ohms
Wattage across resistor = 1.6*.150 = .24 W
So I found a SM resitor 10.7 Ohms at .25 W.
I have a concern with Vf variance. Would this circuit not work if I get Vf's higher than 3.2 Vf? So if I change my circuit to 6 devices per series chain I need 19.2 volts and need to dissipate 4.8 volts. This just seems like a waste.
I still do not understand Vf variance. If I design for 6 and dissipate 4.8 volts with a resistor - won't the series chain be starved if Vf goes higher? So how do you accommodate this variance?
I just have worries about this design (regardless if I design for 6 or 7). I know some about LEDs but I do not know what happens when I get Vf variance.
@MacGyverS2000 (Electrical)
By doubling the resistor value don't I starve the series circuit? I.e. I dissipate needed voltage (or wattage) that is required by the LED series circuit in order to light. If I double to 3.2 volts to dissipate I'm stealing this from the series chain.
In my design I'm being held to optimizing the efficacy of the entire system -> total lumens/total wattage. So I'm trying to maximize total lumens and minimize total wattage.