LED light strip help

[mcgyverdc]

Building a 30 piece, 3mm led strip. Blue LED's, 3.2-3.5V forward voltage, 20mA DC forward current. I have a few different AC to DC wall convertors and understand the building in series or parallel and the current or voltage requirements. I'm proposing to do 10 leds in series and parallel 3 of these to make up the 30 leds. My convertor that i would like to use ishas a DC output of 12V @ 1.25A. Suggestions? resistor sizes? would you build it as i am 3 legs of 10 leds in parallel? thanks

 

[vne147]

Building a 30 piece, 3mm led strip. Blue LED's, 3.2-3.5V forward voltage, 20mA DC forward current. I have a few different AC to DC wall convertors and understand the building in series or parallel and the current or voltage requirements. I'm proposing to do 10 leds in series and parallel 3 of these to make up the 30 leds. My convertor that i would like to use ishas a DC output of 12V @ 1.25A. Suggestions? resistor sizes? would you build it as i am 3 legs of 10 leds in parallel? thanks

You cannot place 10 LEDs in series. If each LED had a forward voltage of 3.2- 3.5V then 10 LEDs in series would have a voltage drop of 32 - 35V. Your voltage supply is only 12V so that won't work. With a voltage supply of only 12V you will have to place 10 banks of 3 LEDs each in parallel.

Each bank should have it's own current limiting resistor of

12 - 3*(3.2) = 2.2V

R = 2.2/.02 = 110Ω

The 110Ω is a minimum, I would go with 120 or 130 to be safe. Also, if your AC adapter is unregulated, you might have to go even higher.

 

[audioguru]

Ten 3.5V LEDs in series need 35V plus about 3.5V for the current-limiting resistor.
Three 3.5V LEDs in series need 10.5V plus 1.5V for the current-limiting resistor.

Use a 12VDC/200mA or more supply with 3 LEDs in series and in series with a current-limiting resistor to make one string and make 10 strings.
Use Ohm's Law to calculate the resistor value.

 

[mcgyverdc]

Thanks guys. Ummm feel kinda dumb on the 10 vs 3 legs thing (should have been evident ;-))..... so 200mA would be a min. and I base the value of the resitor on whatever mA my power output supplies, right? Refresh my memory but does each leg (10) use the power source value of the current? or is it divided by 10 legs.....? I've got so many power supplies around and i want to wrap my head around this

 

[vne147]

Thanks guys. Ummm feel kinda dumb on the 10 vs 3 legs thing (should have been evident ;-))..... so 200mA would be a min. and I base the value of the resitor on whatever mA my power output supplies, right? Refresh my memory but does each leg (10) use the power source value of the current? or is it divided by 10 legs.....? I've got so many power supplies around and i want to wrap my head around this

No, you base the resistor on whatever current you want to flow through the LED and also the voltage drop across the resistor. The voltage drop across the resistor is whatever is left over from the supply voltage after you've added up all the forward voltages of the LEDs. The equation you should use to calculate the resistor value is:

V = IR

Where V in this case is the voltage drop acroos the resistor, I is the desired current through the LEDs, and R is obviously resistance. This relationship is called Ohm's law.

If the current requirement for each LED is 20 mA and you have 10 legs in parallel then he minimum supply current should be 200 mA.

 

[mcgyverdc]

Thanks VNE.... so theoretically, with the 10 legs and a supply of 200mA, there should be no "need" for a resistor on any of the legs...correct?

 

[vne147]

Thanks VNE.... so theoretically, with the 10 legs and a supply of 200mA, there should be no "need" for a resistor on any of the legs...correct?

In theory, if you were able to supply 200 mA and ONLY 200 mA that would be correct. However, this will not work in real life becasue each leg of LEDs will not have exactly the same voltage drop. What would end up happening is that the leg with the lowest voltage drop would end up taking much more current than the other legs and it would burn out. Once the first leg burned out now the 200 mA is divided by 9 not 10 and the remaining legs would take even more current. The effect would cascade until all your LEDs were burned out. You still need resistors to balance out the legs.

 

[mcgyverdc]

;-) thanks, got it!