What exactly happens when LEDs are placed in parallel? Does one just swamp the other out current wise? I know what happens visually, they either have similar drops and have roughly the same brightness, or one works while the other doesn't.

 

One takes more current than the other, and that one is likely to fail.

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If you put a red LED in parallel with a blue LED the red LED with glow but not the blue.

If you put two red LED in parallel then one still might have a slightly higher voltage drop than the the other so the one with the lowest voltage drop will be the brightest.

In short don't do it, always put LEDs in series unless you use separate resistors.

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Even if you do use separate resistors for parallel LEDs, yoru efficiency winds up in the toilet, since the currents through the LEDs are additive. If you wire them in series, it's much more efficient.

 

What exactly happens though, since diodes try to have a specific voltage drop, yet each one is fighting for that specific voltage drop.

 

It's not that complecated, the one with the lowest voltage drop just takes the most current.

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Assuming a 2Volt LED with 10mA current consumption, if you have 10 in parallel that's 2V x 0.1A = 0.2W.

If you have 10 in series it's 20V x 0.01A = 0.2W

Power is exactly the same.

Brian

 

Look at it in terms of ohm's law. If we were to assume a constant current source of 10mA for a minute, and two LEDs one of which has a forward voltage drop of 2V and another which has a forward voltage drop of 2.2V (both of which have a max current of 10mA).

Doing the math, the resistance for both these components is;

2V LED: = 2V / 0.01 = 200Ohms.
2.2V LED: = 2.2 / 0.01 = 220Ohms.

If you were to stick those in parallel with a current limiting resistor in series to clamp the current at 20mA (10mA for each LED) then what happens in practice is a bigger proportion of the current will flow through the smaller resistance. Therefore the LED with the lower forward voltage drop takes a bigger proportion of the available current and the LED with the higher voltage drop takes a smaller proportion of the available current. This effectively means that the 2V LED will be passing more current than it is designed to handle. In time it is possible that it will fail, and when that happens you've got 20mA being passed solely to the 2.2V LED (because the 2V LED has gone o/c and no longer takes its split of the current). Then that one fail quite soon afterwards!

Brian

 

When applying Ohms Law to LEDs in parallel, you need to include the internal resistance of the LEDs. Flashlights (torches) put LEDs in parallel; although their reliability is not optimum, it seems to be acceptable for the price paid.

The internal resistance is NOT V/I since diodes aren't linear. You can approximate it with dV/dI in a small range around the operating point. The remaining drop can be approximated as a voltage source.

 

The above information isn't completely correct because in parallel both LEDs will have the same potential difference across them. But because one has a lower forward voltage drop it means it has a lower resistance than the higher voltage drop LED and therefore it will pass more current.

Brian

 

mneary;

When I said the above information I actually meant my own post but you beat me to it with one of your own.

Brian

 

I said efficiency. If you have a 10-volt source, you can put five 2-volt LEDs in series with no limiting resistor. All your power goes into lighting the LEDs. If you have the same 10-volt source and you use 5 LEDs in parallel, each with its own limiting resistor, where does most of your power go?

 

Ah ok, yes in that case you would indeed consume more power as 8 volts will be dropped across each current limiting resistor and so when a current flows, power will be dissipated in the resistor (via heat).

If you connected the LEDs as you suggested, what happens if the 5 LEDs you connect have a combined forward voltage drop of 9.7 volts? Or what happens if your voltage source outputs 10.2V instead of exactly 10.0V?

Brian