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

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CDRIVE

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

Here's a picture of a swan instead.

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.
After viewing your schematics it confirms my suspicions. You're reply though leaves me speechless.

Chris
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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That's because the question has already been answered so there's no point any more.

But to be fair, my guess was a lucky one because if I had done the calculations before I made it, I may not have risked the embarrassment :)

I think this has been an important lesson in how even a small resistance can have a dramatic and largely unexpected impact on the current through a LED, and also yet another pitfall of placing LEDs in parallel. For me it is a lesson in expecting the unexpected in burden voltage.
 

hevans1944

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A perfect example of the GIGO principle: Garbage In, Garbage Out.

Failure to consider ALL the parameters affecting experimental results has led more than one person down the primrose path. But to seriously question Kirchoff's Laws requires either some real chutzpah or extreme naivety about how the real world works.

Hows about getting three "identical" ammeters: one in series with the supply, second one in series with one LED, and the third one in series with the second LED. Add the currents up on ammeters two and three and compare to the reading on ammeter one. Sum of ammeters #2 and #3 not exactly equal to reading of #1? Welcome to the real world of uncertainty in measurements and calibration inaccuracy. Or is it close enough for government work? Either way, WAFWOFT IMHO. Pretty pictures add nothing.
 

BobK

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You schematics show a very different story than you told in the original post. In the OP, you mentioned putting in parallel LEDs (number not stated) that would individually draw 60mA and getting 22mA, or 36% of the expected current. Your later posting shows two in parallel that draw 16mA individually and 26mA when paralleled, thus expected 32mA actual 26mA or 81% of the expected current. Note that this is at a higher current than the original post, so it should have shown a larger decline if it was due to the drop in the ammeter.

I stated that I did not believe the first set of numbers, and still don't. The actual measurements posted are a different story, showing much less effect when parallelling the LEDs. One that is believable.

So, I conclude that your original measurements were just plain wrong, and could not be repeated.

Bob
 

CDRIVE

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A warning for anyone that stumbles upon this topic in the future!

Don't expect any of these results if simulated in Spice. For one thing (unlike the real world) identical components in Spice are Absolutely IDENTICAL unless you edit their properties.

Secondly Spice Voltmeters and Ammeters are perfect instruments too! Neither of them represent the internal resistance of a real world meter unless you also edit their properties. IE, for a voltmeter representing a modern DVM it should be shunted with a 10MΩ resistor. For a VTVM it should be 11MΩ; unless you own an old HP410B. For that the shunt would = 100MΩ. I still love mine!

For an Ammeter series resistance must be inserted. The value of which should mirror the spec sheet of your meter for the current range selected.

Ohms Per Volt Meters

For youngsters that inherited Grandpa's old Simpson 260, simulating its loading effect is more complicated. These meters didn't present a constant input resistance like a DVM or VTVM. It changes on each voltage range. The Ohms Per Volt rating is printed on the meter face. To calculate the meter's loading (burden) resistance you must multiply the voltage scale your using by the Ohms Per Volt spec. For example: If the meter is a 20,000 Ohms Per Volt then on the 1.5V scale the meter will present 30KΩ of loading resistance. On the 150V scale it will load the circuit under test with 3MΩ.

Chris
 

seanspotatobusiness

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You schematics show a very different story than you told in the original post. In the OP, you mentioned putting in parallel LEDs (number not stated) that would individually draw 60mA and getting 22mA, or 36% of the expected current. Your later posting shows two in parallel that draw 16mA individually and 26mA when paralleled, thus expected 32mA actual 26mA or 81% of the expected current. Note that this is at a higher current than the original post, so it should have shown a larger decline if it was due to the drop in the ammeter.

I stated that I did not believe the first set of numbers, and still don't. The actual measurements posted are a different story, showing much less effect when parallelling the LEDs. One that is believable.

So, I conclude that your original measurements were just plain wrong, and could not be repeated.

Bob

The measurements in the original post were made using different LEDs and a different power supply and a different meter and I don't have access to that equipment because it's not mine (it belongs to Edinburgh Hacklab and I can only visit on Tuesday evenings without being a paying member). I also don't remember which colour LEDs I was using. In any case, the numbers are what I measured with their multimeter. Maybe mine has lower resistance than theirs (they do have several).
 

BobK

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You also did not tell us you used different color LEDs! This is a total no-no. Different color LEDs operate at different voltages. If you ever were to parallel LEDs, which you should not, they should be as similar as possible.

Bob
 

seanspotatobusiness

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You also did not tell us you used different color LEDs! This is a total no-no. Different color LEDs operate at different voltages. If you ever were to parallel LEDs, which you should not, they should be as similar as possible.

Bob

They were the same colour; I just don't know which colour.
 

hevans1944

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If you ever were to parallel LEDs
NEVER parallel LEDs, matched or not. It just isn't worth wasting time trying to solve the problems that can occur. Connect two or more LEDs in series with appropriate current-limiting resistor if powering from a constant-voltage supply or power them from a constant-current supply.
 
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BobK

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Agree, but try telling that to the makers of cheap Chinese LED flashlights.

Bob
 

hevans1944

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Agree, but try telling that to the makers of cheap Chinese LED flashlights.

Bob
Like they really care if the LED brightness matches. And isn't this just a scheme to sell more cheap batteries?
 

davenn

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OK answers have been given and before this thread really goes off track, I will close it


Dave
 
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