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I am new to this and have what is probably a newbie question. I am using SMD 5630 LED's from a strip that I am cutting off and wiring each LED individually. They will be wired in parallel and I am trying to understand how many I can power with a 12V 1AMP power supply. (Edit: The power supply converts to DC current.)

From the spec's I found each LED requires 3.6V at 50 MA and using an Ohm's Calculator I found I need 22 ohm resistors. I just don't understand how to figure out the maximum number that I can support. Could someone explain, speaking slowly and using small words :)

Thanks!

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  • \$\begingroup\$ Hi, I am not using the strip, I am taking off the individual LED's and wiring them for a project. amazon.com/gp/product/… \$\endgroup\$ – Scott Dec 21 '15 at 3:17
  • \$\begingroup\$ how are you planning to power them? 12V to 1 led to 1 resistor? Regulator? \$\endgroup\$ – Passerby Dec 21 '15 at 3:38
  • \$\begingroup\$ Trying this again.... Yes 1 LED to the powers supply in parallel with a resistor for each LED. \$\endgroup\$ – Scott Dec 21 '15 at 4:04
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    \$\begingroup\$ that setup with 12V will mean the resistor needs to drop 9V * 0.05A = 0.45 Watts of power. You will need to use a 1/2 Watt resistor. If you don't mind the inefficiency of it that's okay. \$\endgroup\$ – Passerby Dec 21 '15 at 4:48
  • \$\begingroup\$ If you run the LEDs in paralell your driver circuit efficiency would be 3.6/12 or 30% .This would put you in the halogen lamp ballpark .Even if you kept it simple and placed only 2 leds in series your expected efficiency would be 60% , \$\endgroup\$ – Autistic Dec 21 '15 at 10:33
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Adding to Passerby's answer: Replicate what's on the strip. It wastes the least current. The strip is made up of repeating circuits as shown below.

schematic

simulate this circuit – Schematic created using CircuitLab

You should do the same. Polarity of LEDs matters. The resistors can go anywhere in the series circuit - they're probably evenly spaced on the strip to spread the heat.

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IF the picture on the Amazon page is accurate to what you have, each segment has 2 27Ω resistors in series with 3 leds.

enter image description here

Using Ohm's law I = V/R to find out the current, Assuming 3V to 3.6V Forward Voltage drop on each led, that means the current at 12V is between

(12V Source Voltage - (3 Leds * 3.0V)) / 54 Ohms =  0.055 Amps (55 mA)

and

(12V Source Voltage - (3 Leds * 3.6V)) / 54 Ohms =  0.022 Amps (22 mA)

Since current is shared in series, this means each chip on that strip is only 22 to 55 mA.

So they are no where near your believed 150mA or 1/2 Watt.

To get the correct current, as far as the leds you have go, cut a single segment off, and using a ammeter or multimeter in current mode, measure the amperage of that single segment. That will tell you what it is using on the strip. Then I would adjust at least 10% down for a longer life. If you can't measure current, then measure the voltage across one of the resistors. Using Ohm's Law as above, take that voltage and divide by that resistance for the current flowing through it. That's what the LED should be using.

Update: It was suggested that the resistors are in parallel, and that the normal 5630 leds take 150 mA @ 3.4V. That may be true.

(12V Source Voltage - (3 Leds * 3.4V)) / 13.5 Ohms =  0.150 Amps (150 mA)

Only way to be sure would be to trace the circuit and measure the current or voltages with a multimeter.

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  • \$\begingroup\$ I am still confused though. From my vague recollections of college in the 1990's, amps are what drive the LED's? So if the LED uses .5MA, I should be able to support 20 per 1amp power supply? Sorry I know this is basic stuff, but I am a total newbie at this... \$\endgroup\$ – Scott Dec 21 '15 at 3:44
  • \$\begingroup\$ @Scott if each uses 0.05 Amps (50 mA) then yes. \$\endgroup\$ – Passerby Dec 21 '15 at 4:42
  • \$\begingroup\$ @Scott of course, you could run the leds at a lower current, say 30 mA, and get 33 leds instead. Or you could do as the led strip does, and wire 3 in series and get more bang for the buck. \$\endgroup\$ – Passerby Dec 21 '15 at 4:49

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