# LED Circuit Design?

I'm a Mechanical working in the LED field. I'm trying to design a LED circuit with the little knowledge I have.

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. Does the resistor go at the end of the series circuit? If this is the case then I understand how variance in the circuit would be able to use available voltage - and then the resistor is at the end to disipate what is left over. I've heard concerns that I'm not allowing for more head room #i.e. doubling the resistor value and wattage requirements). If I double the voltage dropped don't I take this away from the LED circuit possibly starving it? I have always assumed the resistor goes at the beginning of the circuit - regulating the current/voltage prior to it entering the LED series circuit.?

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. Does the resistor go at the end of the series circuit? If this is the case then I understand how variance in the circuit would be able to use available voltage - and then the resistor is at the end to disipate what is left over. I've heard concerns that I'm not allowing for more head room #i.e. doubling the resistor value and wattage requirements). If I double the voltage dropped don't I take this away from the LED circuit possibly starving it? I have always assumed the resistor goes at the beginning of the circuit - regulating the current/voltage prior to it entering the LED series circuit.?

Steve