# How to read resistance of 5mm LED



## ImGeo (Jul 6, 2009)

*How do I** get the resistance of a 5mm LED that should be 89ohms, but is not reading at ll? 

*BELOW INFO IS INACCURATE (there was a closed loop connected to a 6v lead acid battery. w/ an open loop, I still cant read the resistance)

I just got a Cen Tech multimeter (Harbor Freight), and tried reading the resistance of a LED (the regular 5mm kind). I tried it at 200ohm, assuming it would be less, but got a void reading, so I increased it to 2000k ohm setting, and it got a reading of 1330 which means 1.33 million ohms... which is not possible, considering there is a 3.47 volt drop when running 39mA (R=V/I=3.47/.0389=89.2ohm for the LED).

This LED is part of a 35w halogen/5mm LED hand torch. The leads are soldered, so I did not unsolder anything to test to voltage/current/etc. I did check to make sure there was no closed circuit. Also, I tested it both ways (red or black lead connected to one end of LED).


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## Lyndon (Jul 6, 2009)

*Re: Trying to read resistance of LED, getting 1330K ohm*

Why do you think the resistance should be 89 ohms? An LED is an active device; "resistance" in the ohms law concept doesn't apply in the way you think it does.

What are you really trying to do? There's probably a better way to accomplish it.


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## ImGeo (Jul 6, 2009)

*Re: Trying to read resistance of LED, getting 1330K ohm*

It should be 89ohm based on the above info (there is a 3.47 volt drop across the LED when running 39mA) therefore R=V/I=3.47/.0389=89.2ohm for the LED

*Or is the current regulated somewhere, and it is not a simple circuit based on V=IR only?
And what creates the ~3.5v difference between the two ends of the 5mm LED?*


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## ImGeo (Jul 6, 2009)

*Re: Trying to read resistance of LED, getting 1330K ohm*

More info:
the 6.3v valve-regulated lead acid battery > switch > 70ohm resistor > 5mm LED > back to battery. I don't think there is anything in between to regulate the current. If working with V=IR, then the two resistors (the 75ohm and the 5mm LED) works to limit the current to 38.9 mA


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## zipplet (Jul 7, 2009)

*Re: Trying to read resistance of LED, getting 1330K ohm*

You can't measure the resistance of an LED. What appears to be the resistance is a combination of the forward voltage and current consumed (and turned into light/heat). This varies with input voltage and current.

DMMs measure resistance by placing a voltage across the load in series with another resister. You will see crazy readings with an LED as the voltage is probably low and with a miniscule current. Again, this is not something you should be doing 

What are you trying to do or is this purely an educational exercise?


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## jtr1962 (Jul 7, 2009)

If you want to model a white 5mm LED as a combination of circuit elements, it's best to think of it as a voltage source of about 2.8 volts (roughly the voltage at which it turns on) in series with a resistor of around 10 ohms. At very low currents (i.e. microamps) you'll see about 2.8 volts across the LED. At 100 mA you'll see about a volt more, or around 3.8 volts. But in any case, an LED will not behave as a simple resistor. You can't use a DMM to measure resistance. There are instruments which could measure dynamic resistance (i.e. the slope of the current versus voltage curve). For a 5mm LED this would be on the order of ten ohms, give or take.


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## Wiggle (Jul 7, 2009)

As mentioned, you can think of the LED having a "resistance" in a sense, but it is dynamic, depending on the surrounding circuit conditions. Every multimeter I know of requires the circuit to be unpowered to measure resistance so you will not be able to bias the diode into the test range you want and get a measurement at the same time.

If you like, measure the current through the diode and the voltage across it to determine an "equivalent" resistance at that operating point.


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## ImGeo (Jul 7, 2009)

In that case, how can I adjust the voltage to make it run on a LED? I have a 6v lead acid battery (from a halogen lamp), and want to be able to power a 3.7v CREE. How would I know what resistor to connect in series, or do I need a more complex regulator?


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## znomit (Jul 8, 2009)

ImGeo said:


> In that case, how can I adjust the voltage to make it run on a LED? I have a 6v lead acid battery (from a halogen lamp), and want to be able to power a 3.7v CREE. How would I know what resistor to connect in series, or do I need a more complex regulator?



The LED is 3.7V, leaving 2.3V for the resistor.

V=IR ... R=V/I

Say you want 500mA, ... 2.3V/0.5A = 4.6 ohms. 
Power is VI = 1W.

You need to check the calcs with the max and min battery voltages to see what range of currents you'll get, design around the max current and be a little conservative.


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## ImGeo (Jul 8, 2009)

Thanks Znomit! that really helped (and I found/google how to choose a resistor for LED's)

and another question so I can understand better.

If I have a 6v input voltage, want to power the LED at 350mA and 3.3V (based on CREE specs), I would ideally need a 7.7ohm resistor ( based on R = V/I = (6-3.3)/.350 = 2.7/.350 = 7.7 )

However, supposing a put in a 8 ohm resistor, what would happen to the voltage and current through the LED? Mathematically, R=V/I so the voltage and current could just change proportionally as long as their ratio remains at 8.


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## Neondiod (Jul 8, 2009)

The voltage across the led will remain almost the same, wich means 2,7V/8ohm = 0,3375A. Then say that the voltage across the led will sink to 3,29V (a guess, I haven't check the specs) when driven at 338mA , then it will be 6V - 3,29V = 2,71V/8ohm = 0,33875A. The first calculation was accurate enough.

If we for ex take a 100 ohm resistor there will be a bigger difference. First I use the same voltage drop 3,3V -> 2,7V/100ohm = 0,027A. Checkin the specs (wich I haven't) the voltage may be around 2,6V over the led driven at 27mA. 6V - 2.6V = 3,4V over the resistor. 3,4/100 = 0,034A. The difference is only 7mA calculating with the LED's voltage drop at 350mA versus the voltage drop at 27mA. You can now do the calculation with the LED's voltage drop at 34mA, say 2,605V -> 3,395 / 100 = 0,3395A -> Only 0,5mA differense from the second calculation. 

But the voltage drop across a LED will not only change with the current through it, but also with temperature, wich is importent to have in mind for powerleds.


BR / Andreas


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## Marduke (Jul 8, 2009)

Or if you wanted to run a Cree at 350mA off a 6V battery, you could simply use a buck circuit.


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