Electronic – LEDs in parallel WITHOUT Resistor and Constant Current

ledparallel

Before I go on, I should stress that I have read many many other forums and scoured all stackexchange recommended 'questions that may already have your answer' things, and none of them answer my question.

Let's say I decide to put 3x 20mA 3v LEDs in parallel.

That equates to a power consumption of 60mA @ 3v = 180mW

But not all LEDs are created equal, and alas, one of the LEDs want 40mA. That should then increase the power consumption to 40+20+20=80mA @ 3v = 240mW and it burns itself out. Let's assume the other 2 LEDs are properly manufactured and draw 20mA

Shouldn't the other 2 LEDs still draw 40mA @ 3v?. It's not like I have a constant current device that forces 60mA.

Here is the scenario everybody talks about:

The power supply is 3v at 60mA (CC/CV)

  1. One of the LEDs is faulty and draws 40mA. The other 2 draw 10mA each (Since there is 20mA left over. Let's assume they split the current evenly)
  2. The faulty LED burns itself out. That leaves 60mA for 2 LEDs. So each of the proper LEDs use 30mA. But oh no!, one of them overheats and dies!
  3. The sole last LED is forced to use all 60mA of the current, and it soon overheats and dies.

My scenario:

3v power supply directly from a battery. There is no current limiting, current control, or anything at all. No resistor.

  1. Again, the faulty LED draws 40mA and dies.
  2. Shouldn't the other two LEDs still draw 20mA each? This means that overall total consumption of the circuit is only 40mA @ 3v.
  3. Boohoo, one LED breaks. That should be all. Why not?

My questions are:

Why would the two LEDs have to use 30mA if one dies? Why would the two LEDs draw the current meant for 3?

My 5 torches I have all have their LEDs arrayed in parallel. THey are high quality torches, most from Arlec. One of the torches have 16 LEDs in parallel. They do not have any resistor or anything. THe LEDs are literally all slapped onto a cheapo PCB, directly connected to the batteries. These torches all run fine, don't have varying levels of brightness, and haven't died on me in years. Why can't I do the same?

Best Answer

The power supply is 3v at 60mAh (CC/CV)

A power supply can't force both the current and voltage to specific values at the same time. It can either force a certain voltage and let the load determine the current, or force a certain current and let the load determine the voltage.

Remember, the load has its own I-V characteristic that it must obey. For example, a resistive load obeys Ohm's Law \$V=IR\$.

What a 3-V 60-mA CC/CV supply does is force 3 V, unless doing so would require more than 60 mA, in which case it just provides 60 mA at whatever voltage (lower than 3 V) it takes to make that happen.

Assuming your LEDs' forward voltage is less than 3 V, your supply will operate as a 60-mA current supply, and the scenario will play out much as you have described it.

My scenario: 3v power supply directly from a battery.

A battery is more like a constant-voltage supply (with a series resistance), so this will indeed act differently than the constant-current scenario above.

Shouldn't the other two LEDs still draw 20mAh each?

Yes, they will continue to draw whatever it is they draw at 3 V. If they're the same LEDs as in the constant-current supply scenario (where we hypothesized that the forward voltage is lower than 3 V), then they will draw much more than 20 mA because you're driving them well above their rated forward voltage.

Also, as they heat or cool, their I-V curves change. So one might be okay running on a 3 V battery when first connected, but then start drawing more current as it warms up and eventually (or very quickly) burn itself out.

Boohoo, one LED breaks. That should be all. Why not?

With a constant voltage supply, one LED failing wouldn't affect the others.

With a real battery, you have to remember the internal resistance of the battery. If one LED fails, then that will tend to increase the actual output voltage of the battery after accounting for internal resistance, and that could cause the other LEDs to fail.