# Ohm's Law & LED circuit basics



## T-Man (May 16, 2008)

You be da judge.

First I am using some different color 1.8 tower LEDs.

Here I used yellow with 620 ohms good for 12 volts DC then I used a 9 volt battery that put out around 7 volt as measured with a meter.










If I took another resistor and LED and tied it across the battery like the first one, so both work, that is a parrallel circuit.



The I used 5 LEDs set up in seires pos to neg, etc, the long lead aimed at the Pos battery terminal.

So 7 volts did this with no resistors added. Yellow red blue and white. The intensities are different because each has it's own specifications. Slightly different.










Now I used 5 yellows and got a dim light The intesity looks good but the camera distorts it. 









Four worked better.


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## Massey (Apr 16, 2011)

There is also a voltage drop across the LED. that is why 4 are brighter than 5.

Massey


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## gunrunnerjohn (Nov 10, 2010)

I would recommend never connecting an LED to a power supply with no current limiting. They don't inherently limit the current, and a change in the supply voltage of half a volt can go from normal operation to meltdown!


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## T-Man (May 16, 2008)

That makes sense, until now I have always added up the number of LEDs to below the voltage supply and then calculated the resistance.

I also don't have the specs for the 1.8 mm.

Guess I got carried away.


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## gunrunnerjohn (Nov 10, 2010)

Truthfully, I normally just wing it as far as resistance is concerned. I rarely run LED's at the maximum rated current. For my marker lights on TMCC stuff (12VDC feed, I typically use a 2.2k resistor, which gives me around 5ma. I find at 20ma, they're way too bright.


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## T-Man (May 16, 2008)

The LED just doesn't have a photogenic profile. Showing results are deceiving, I hear you on the resistance. I used them to get the different colors to balance out in intensity. The bread board is handy to fine tune a use for the LED. I has only recently come to my attention that using over 1k resistance should be considered .


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## gunrunnerjohn (Nov 10, 2010)

I have some running with 4.7k being fed with 16 volts, it all depends on the effect I'm trying to achieve. The light output isn't linear with current, you don't get four times the light going from 5ma to 20ma for instance. I attribute that to the logarithmic response of the eye to light, though I may be all wet.


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## waltr (Aug 15, 2011)

LEDs are current devices with a fairly fixed voltage drop. The voltage drop is dependent of the chemistry used to make to LED. Reds is typically 1.4V, Yellows can be higher and White & Blue are around 4V. You should get the voltage drop from the LED's data sheet or you can measure the voltage across the LED with a largish resistor is series.
Then you can calculate the LED's current by subtracting the LED's voltage drop from the supply voltage and divide by the resistor value (Vsouce - Vdrop) /R or calculate the resistor values from the desired LED current (Vsource - Vdrop) / Iled. 

You can run LEDs in series. Just add the voltage drop on each LED before subtracting from the Voltage source for the calculation of the resistor value. This works best if all the LEDs are exactly the same since different LEDs will be brighter or dimmer for the same current. This is good to do if the supply voltage is high (for example 12V). The reason is that LEDs require current with a fixed voltage. 
If you run 3 LEDs with each having their own current limiting resistor (say 10mA) they each need to connect to the 12V supply (parallel connection) for 30mA total current draw from the power supply. Then each draws 10mA and each limiting resistor dissipates the excess power as heat ( Vacross resistor * Ithrough resistor = Power loss). 
Now if you need three LEDs that are all the same wire them in series with a smaller resistor value (since the resistor doesn't need to drop as much voltage) and you now only draw 10mA from the power supply and the resistor dissipates less power.

Examples:
Red LEDs, Vdrop = 1.4V, Max current = 20mA, desired current = 10mA
Power supply = 12V

One LED and resistor:
12V - 1.4V = 10.6V
10.6V / 0.01A = 1060 Ohm
Power in resistor = 10.6V * 0.01A = 0.106Watts 
which means the resistor should be rated at 1/4Watt
Three in parallel draws 30mA from the supply and the total power loss is 318mW.

Now three LEDs in series and one resistor:
12V - 3*1.4V = 7.8V
7.8V / 0.01A = 780 Ohm
Power in resistor = 7.8V * 0.01A = 0.078Watt

When wiring LEDs in series always be sure that the total voltage drop across the LEDs is less than the power supply voltage by a few voltages so that the resistor value be large enough to limit current.

Hope this helps.

===

Mod edit, for future Search hits (great post!) ...

LED voltage drop
LED circuits
LED info
LED basics
LED wiring


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## T-Man (May 16, 2008)

I think that sums it up. 

I have noticed different intensities because of color and recently have taken different resistance to balance it out. Buying a lot at Eadio Shack leaves me a a disadvantge because of the diferent colors. So I don't have the precise specs, Even ebay I bought a lot but did't get all the good specs becuse five colors were offered.

Thanks for the good write up.


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## gunrunnerjohn (Nov 10, 2010)

I've bought some LED's in bulk and have pretty good luck with intensity matching. My only really bad experience was the recent purchase of the red/white bi-color ones, the white side died like flies, even when I didn't abuse them. I don't know if it was just a defective batch, or a bad design, but I was bummed, since it made me start over on my lighting I was using them for!


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