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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?

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  • \$\begingroup\$ basically you can if you know how and where to specify what LEDs you buy. I dont need separate resistors on any of my LEDS and within a single batch the Vf is usually within 50mV, but I have 40 years of electronics experiences and 10 years just on LEDs having shipped over 1 million custom 5mm white and other color LEDs \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 12 '17 at 5:21
  • \$\begingroup\$ The other sources will unanimously say "don't do that". Which is the practical answer to your question. \$\endgroup\$ – Brian Drummond Aug 12 '17 at 10:38
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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.

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  • \$\begingroup\$ I'm going to leave it for somebody else to answer how are cheap LED torches designed. I'd guess they use high-power LEDs that can take some abuse and count on the internal resistance of the battery to limit current enough, or use LEDs with internal resistors. But probably there's somebody reading who knows for sure how it's usually done. \$\endgroup\$ – The Photon Aug 12 '17 at 4:56
  • \$\begingroup\$ RIght, OK, so let's say I magically stabilize the voltage to 3v no matter the load. I shouldn't have any issues then, would I? I understand the LEDs will increase their current as the temperature increases, but as long as I have a way to dissipate the heat I shouldn't have any issues. \$\endgroup\$ – Che0063 Aug 12 '17 at 4:56
  • \$\begingroup\$ Yes, as long as they don't blow up with 3 V, they'll be okay with 3 V. (Realizing that for real LEDs the forward voltage changes with temperature) \$\endgroup\$ – The Photon Aug 12 '17 at 4:57
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When that third LED dies, the total current being drawn from the battery will slightly increase, as the parasitic internal resistance of the battery will drop less voltage (due to a lower current flowing through it). When that happens, the voltage across the battery, and incidentally, the LEDs, will increase accordingly. Depending how much the voltage increases, it could cause the LEDs to draw enough current to become troublesome. Since the two LEDs aren't created equal, it's possible one of them will pull slightly more current than the other.

That said, the internal resistance of your battery will also act as a current limiter. However, without individual current limiting through each LED, runaway conditions are possible, but possibly unlikely.

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Your circuit of three different loads (LEDs) in parallel does not follow the Ohm's law of linear networks because the LEDs are non-linear components, i.e. the R is not linear.

If you use different LEDs, the I-V diagram for these diodes are different and their forward bias region starts at different voltages - standard red LED at about 2 V, blue at about 3 V, aso. So, if you connect different LED components in parallel, the LED with the lowest forward bias region will draw the most current and with your 60 mA current limit (if it is a current regulation which I do not believe with simple plug-in power supplies) that LED will thermally destroyed... (most standard LEDS are specified at 20 mA). Apart from the overload of the one LED, the others with higher forward voltages will receive too less voltage to light up brightly or they remain dark at all.

So it is a really bad idea to connect LEDs (different type) in parallel. Only if you know the I-V diagram and the specification of your power supply, you may connect these diodes in parallel, you should use a resistor for each LED, and they will shine as desired.

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In your scenario, there are a few problems. One, you ignore that an LED draws x ma at y forward voltage. So if your supply is 3V, then it draws 40mA, and the other leds only draw 20mA, then even if the first led goes out, the others will not.

But your supply isn't perfect. It has an internal equivalent series resistance. As the current drawn goes up, the WAR causes a drop on the output voltage.

And parallel elements at the same node will have the same voltage across them. Two identical leds may have one led with 22mA at 3V but the other at 20mA at 3V. But Since the voltage changes based on the load, it's all interplay that makes it more complex than simple ohms law would suggest.

But back to your flashlights. One, they likely have a series limiting resistor you can't see. If my 99 cent 9 led flashlight has it, your flashlight likely does too. It also depends on the ESR of the batteries used. They ship with heavy duty batteries which have a higher ESR than alkaline batteries. And are considered disposable if the flashlight dies.

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