# LEDs in parallel don't consume sum of individual currents??

Discussion in 'LEDs and Optoelectronics' started by seanspotatobusiness, Sep 20, 2016.

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1. ### seanspotatobusiness

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Sep 11, 2012
I checked the currents that passed through several LEDs at a particular voltage. They varied from 12.0 to 15.1 mA at 3.0 V. I then connected them all in parallel and the current running through them was about 22 mA. How can this be? I thought it would be the sum of the currents that passed through the LEDs connected individually. All the LEDs should have the same voltage across them so they should all be passing the same current they passed when tested individually??

2. ### davennModerator

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Sep 5, 2009
Yes because there is a voltage drop across each LED, none of them individually are getting 3V unless you are feeding them with a total voltage that equals
the added up drops across each LED

when in parallel each LED is directly across the voltage supply, each LED will draw that approx. 22mA when there is an appropriate series resistor
and IF you put the ammeter in series with the supply before or after all the LED's you should see ~ 60 - 70 mA drawn

3. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
The alternative explanation is that each led has a slightly different forward voltage for a given current. This can also be expressed as them having significant differences in forward current at the same forward voltage.

The upshot of this is that placing several LEDs in parallel will cause an unequal current to flow through each of them.

Combine this with a less than stiff voltage source and you will get the behavior you noted.

Our LED resource goes into this in some detail.

The practical solution is twofold

1. Don't operate LEDs from voltage sources (without a resistor)

2. Don't place LEDs in parallel without individual resistors (exceptions can be made for matched LEDs in thermal contact).

Last edited: Sep 21, 2016
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4. ### seanspotatobusiness

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Sep 11, 2012
I think there is a misunderstanding. The ammeter was in series with the LEDs which were in parallel to each other. I expected to see about 50 mA drawn but only saw 22 mA.

5. ### seanspotatobusiness

193
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Sep 11, 2012
I understand that the currents running through different LEDs in parallel will be different for each LED due to manufacturing variations. What I don't understand is why the current isn't the same for each LED in parallel as it is when tested on its own. I was using a bench power supply and confirmed that the voltage was the same in both tests using a separate multimeter.

6. ### BobK

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Jan 5, 2010
If your voltage source was perfect, and you measured the current through each LED at the same voltage, then you should get the sum of the currents when you parallel them at that voltage. So you are (theoretically) correct that the result is unexpected.

What is your voltage source?

I suspect that the voltage drops when loaded and that is why you are not seeing the currents add. Measure the actual voltage across the LEDs when they are paralleled and indvidually.

Bob

7. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
Schematics usually clear up textual confusion.

On another note many nubes don't factor in insertion loss (internal resistance) of an Ammeter and what effect it will have on the current being measured or voltage drops too. The more sensitive the meter range the more effect it has.

Chris

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8. ### davennModerator

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Sep 5, 2009
yeah there is
you are still not stating exactly where the meter was positioned

9. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Aka "burden voltage".

If you have a bench supply with separate sense inputs then you can regulate the voltage at a point other than the output terminals allowing you to cancel out issues caused by the burden voltage, lead resistance, etc.

10. ### BobK

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Jan 5, 2010
No. I don't believe any of it. With a bench power supply and a circuit requiring only 60mA the voltage is not going to drop enough to lower the current to 22mA. We need a schematic, showing how everything is connected, including the meters. And you need to redo the measurement using two multimeters, one for the voltage across the LED(s) and one for the current between the supply and and LED(s) measured simultaneously.

Bob

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11. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
I missed this quote. This discrepancy is far beyond burden voltage issues. Only a schematic is going to clear this up. I hope that when seanspotatobusiness says parallel LEDs he doesn't mean Anode to Anode, Cathode to Cathode!!!

Chris

12. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Ok, I'm going to attempt a thought experiment involving public math and an undetermined outcome. Lets see how well I foul it up. (and I'm doing it on a phone, so beware autocratic ions)

Lets assume three identical LEDs with a constant dynamic resistance of 0.1V/5mA (20 ohms) for currents between 5 and 20mA. Other than the LEDs being identical, I think this is a reasonable model of reality.

If one LED with an unknown resistance in series with it draws 15mA, what does that value of resistance need to be to cause three in parallel to draw 21mA?

So, we see a drop in current through the LEDs of 8mA, suggesting a drop in voltage of 0.16V.

So what value resistor will drop an additional 0.16V when the current increases by 7mA?

That would be about 23 ohms.

Assuming the meter is on a 100mA range, this would imply a burden voltage of around 2.3V, which is atypical for this sort of range.

Interestingly, it also implies (I think) that in this case the series resistance is very close to the dynamic resistance of the LEDs.

If you have two multimeters and can measure the voltage across the ammeter while measuring the current through the LED(s), or measure the voltage across the LED(s) while measuring the current through them we can get more information on both the burden voltage and the dynamic resistance (you will need 2 measurements at different current for the latter)

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13. ### BobK

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Jan 5, 2010
Wow, my gut agrees with Steve's brain!

Bob

14. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
That was impressive! I hope Steve wasn't driving while computing and texting that!

Chris

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15. ### seanspotatobusiness

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Sep 11, 2012
I don't have a second multimeter at this time but in any case I'm confident that the source of the problem is indeed the multimeter. The LEDs get noticeably brighter and draw more current (measured by bench power supply) when I short the leads of the multimeter.

16. ### AnalogKid

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Jun 10, 2015
If you don't have individual current limiting resistors for each LED, your results are exactly correct.

ak

17. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
You have 77 posts at the time of this posting. Not all of them are in this topic but after that many posts we shouldn't have to get on our knees an beg for a damn schematic!

Sorry for the rant but you've been asked repeatedly and as far as I can see you've totally ignored our pleas.

Chris

18. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
And while you're answering Chris' request for a schematic and a photo (which I'm sure the rest of us would like for confirmation), can you also tell us the make /model of multimeter and tell us what range it was switched to when making the measurements. A photo of it would be good too.

19. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
Reading back on my post I think I elevated the request to a demand.

Chris

20. ### seanspotatobusiness

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Sep 11, 2012
That's because the question has already been answered so there's no point any more.

The multimeter is a Precision Gold Academy PG10B and it was set to 200 mA. It will soon be replaced with a Uni-T because it's no longer accurate and some functions cause the display to switch off.

[mod edited to show schematic]

Last edited by a moderator: Sep 24, 2016
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