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I have a 16ma 9volt current source which I connected 3 LEDs to in series. I measured the current flowing through the positive terminal and it was 12ma. Next I connected 3 LEDs to separate 16ma 9volt current source. I measured the total current and found it to be 48ma. All the LED brightness were very similar.

I am assuming that the brightness is a function of power (watts) and that as the power increases, either the voltage or the current must go up. If the voltage stays the same (9 volts) and the current increases by a factor of 3, then the power (total LED brightness) must increase by a factor of 3. However, this is not the case. Maybe there is a heat factor that I am not considering.

How can 15ma illuminate 3 LEDs in series as brightly as 36ma can illuminate 3 LEDs in parallel?

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  • \$\begingroup\$ Because LED brightness is loosely proportional to current. \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 9 '16 at 0:01
  • \$\begingroup\$ LED's have a 'clamp' voltage at which they begin to draw current. Increasing the current makes them brighter, yet the clamp voltage rises very little. Sometimes they are used as a 'poor mans' voltage regulator. Clamp voltage is typically 2.9vdc to 3.5vdc. For red LED's the clamp voltage is about 1.8vdc. For Infra-red LED's the clamp voltage is about 1.05vdc. \$\endgroup\$ – Sparky256 Jul 9 '16 at 0:25
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    \$\begingroup\$ Not much of a current source if it's putting out 3 times rated current (75% might be running into voltage compliance issues, particularly with white or blue LEDs, but there's no excuse for 300% if it's a "current source.") \$\endgroup\$ – Ecnerwal Jul 9 '16 at 1:00
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If you use a multimeter, you will find that the voltage across the current source with 3 LEDs in series will be around 5V, but if you measure the voltage across 3 LEDs in parallel, the voltage will be around 1.5V.

The answer to your question is that if you push the same amount of current through each LED (15mA in both cases!) the voltage will settle so that the same amount of power is used.

schematic

simulate this circuit – Schematic created using CircuitLab

This property where current splits up in parallel devices and is equal in series devices is related to Kirchoff's Current Law, which you can read about. It is something like the law of "conservation of current"... current is never created or destroyed, it just loops around and comes back to where it started.

However this,

Next I connected 3 LEDs to separate 16ma 9volt current source. I measured the total current and found it to be 48ma.

can't happen. A 16mA current source will put out 16mA. A 9V voltage source will put out 9V.

A current source will adjust its output voltage* to ensure the current stays where it's set. A voltage source will adjust its current to make sure the voltage stays where it's set.

*Unless it's a really crappy current source, like a battery and a resistor!

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  • \$\begingroup\$ sounds like I need to go back and test with a volt meter... There is a lot of unexpected voltage changes. \$\endgroup\$ – Hoytman Jul 9 '16 at 1:12
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The light output of an LED in the range you are looking at is proportional to current. I would not say loosely proportional, I would say that it is a good linear relationship. So you are comparing two cases: case 1: 3 LED's with current = 12 mA case 2: 3 LED's with current = 16 mA

If you used an instrument to measure the light output of the two cases, you would probably find that the case 2 LED's were putting out more light (33% more). But that is not a huge difference when comparing them by eye, so they look about the same to you.

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