If a series of LEDs I adapted to work off the 5VDC coming from a USB outlet start to lose intensity and a couple have died out, what has happened? Is the voltage too high for the series? Or is the current too high? And would putting in a resistance prevent the remaining LEDs to completely die out too? (I'm guessing that once LEDs start fading out, there's no way to turn back the damage already done.)

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    \$\begingroup\$ Both, very likely, one causing the other? \$\endgroup\$ – Marcus Müller Dec 1 '16 at 23:47
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    \$\begingroup\$ I'm curious to see what you mean by "...adapted to work off the 5VDC..." would you be willing to provide a schematic? \$\endgroup\$ – ambitiose_sed_ineptum Dec 1 '16 at 23:50
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    \$\begingroup\$ You can't have too much current without too much voltage, and vice versa. \$\endgroup\$ – The Photon Dec 1 '16 at 23:53
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    \$\begingroup\$ When you say "in a series" do you mean all the LEDs are wired in series, or do you just mean a group of LEDs? \$\endgroup\$ – user253751 Dec 2 '16 at 0:47
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    \$\begingroup\$ LEDs are like zeners with Zzt or ESR . If Vin >>Vf then use Ohms Law with a current limiting R, otherwise. pffft \$\endgroup\$ – Tony Stewart EE75 Dec 2 '16 at 1:07

So if I am understanding you correctly, this is a schematic of what you have:


simulate this circuit – Schematic created using CircuitLab

This means that Marcus Müller was correct, you are putting too much voltage across them, and therefore too much current is passing through them. LEDs have a reccommended current range, and you have discovered what happens when you exceed that range- your LEDs dim quickly and fail.

The best way to light LEDs is to use what is called a constant current driver. These are usually only used whith high power LEDs. Low powered LEDs, like the ones you are using, are just fine using only a resistor to limit the current. Here is a typical graph of the voltage across any diode and the current: Diode Voltage V Diode Current Source

For a resistor, this line would be a straight line, crossing the origin, where the slope is equal to the resistance. As you can see, after a diode (and by extension, LEDs) turns on (by that I mean passes the red dotted line), a small change in voltage leads to a much larger change in current. This is why it is usually easier to control the LEDs current than control the voltage across the LED. The function on the graph above is also dependent on the temperature of the diode (for more info, read about the Schockley Diode Equation). But for the low power LEDs you are using, we wont need to concern ourselves with temperature.

To get to the question directly:

Is the voltage too high for the series? Or is the current too high?

Both, as Marcus Müller pointed out: the over voltage caused over current.

And would putting in a resistance prevent the remaining LEDs to completely die out too?

It will help. To calculate the resistor value you need, here's what to do:

Measure the forward voltage of the LEDs, or find it in a datasheet. To measure it, use the diode function on a multimeter, and you must do this with a single (working) LED, disconnected from the rest of the LEDs. This will probably yield a result of about - 3 to 3.3V.

You will also need the forward current of the LEDs, this can be found in a datasheet, or can be measured when the LEDs have the proper forward voltage (for instance, in the original circuit you used the LEDs from).

The last peice of information you need is the voltage across the string of LEDs, in your case it is the 5V from the USB power.

For the purposes of calculation, I am going to use the values calculated by Niel_UK in your previous question, How to adapt a string of 10 LEDs powered by 3 AA batteries to work off a USB port? (credit to SamGibson for finding that for me)

We are going to say the forward voltage of the LEDs is 3V. We are going to call the forward current of the LEDs at 10mA each.

Here is the simplified schematic:


simulate this circuit

So, given the above assumptions, the value resistor you needed for all 10 LEDs was:

$$ V/I = R $$

$$ 5V/(10mA * 10 LEDs) = 20 \Omega $$

adjust the number of LEDs in the equation and you should e able to exend the life of your remaining LEDs a little longer!


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