Electronics > Beginners
I'm confused...(musings on LEDs and capacitors)
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GadgetBoy:
So, I was preparing to do a video demonstrating the importance of putting a current limiting resistor in series with an LED.

So, I set up a simple project, switch, 22k resistor, then a 220 uF capacitor in parallel with an LED (with a 22k resistor in series with the led). Push the switch, LED fades on. Release switch, LED fades out. This part of the demonstration worked great, and let me demonstrate how capacitors start as a short circuit, then increase in resistance until they appear as an open circuit. That went great.

Then I wanted to dump the capacitor into the led without current regulation, so I pull the resistor that's in series with the led, so it's just the inrush resistor, then the led in parallel with the capacitor.

My theory was that as long as you hold the switch, the capacitor is fine (because of the inrush resistor), but as soon as you interrupt the power, the capacitor dumps an unregulated pulse of current into the LED, and kills it.

So, I push my switch, wait for the voltage to peak, then release the switch. The LED fades out, and nothing interesting happens. So I figure the 220uf cap is too small and can't dump enough current, so I upgrade to a 470uF. Same result. Next I try I 6F super cap. Still nothing interesting (side note, it's almost 24 hours later and the led is still lit).

So, based on these observations, I'm drawing the conclusion that you can't get runaway current on an LED unless the source voltage is higher than the forward voltage of the LED, and this circuit rather precisely charges the capacitor to the exact forward voltage of the led.

Am I on the right track here?

(I'm currently on my couch buried in cats, but I'll upload diagrams a little later if you guys want)

Sent from my ONEPLUS A3000 using Tapatalk

rrinker:
 I'd say you are on the right track here.

 I've seen enough purpose-designed circuits that work like this, by never exceeding the LED Vf, and then use no current limiting. I'm not a fan of doing this on purpose though, I just think of all the things that could go wrong, especially if you run right to a specific LED's Vf - next time you build the same circuit maybe the LED is from a different batch and a slightly different Vf - to the low side. Poof.

 Telling - never saw a commercial circuit designed this way. Common voltages of 3.3 and 5 volts are above most any LED's Vf rating, so some form of current limiting is always included. Or the LED is driven by a constant current source so the voltage doesn't matter (up to the limits of the current source component).

spec:
Hi GadgetBoy


--- Quote from: GadgetBoy on December 23, 2018, 04:24:16 am ---
I'm drawing the conclusion that you can't get runaway current on an LED unless the source voltage is higher than the forward voltage of the LED, and this circuit rather precisely charges the capacitor to the exact forward voltage of the led.

Am I on the right track here?

(... I'll upload diagrams a little later if you guys want)

--- End quote ---
Just to confirm that your conclusion is absolutely correct.
GadgetBoy:

--- Quote from: rrinker on December 23, 2018, 05:23:56 am --- I'd say you are on the right track here.

 I've seen enough purpose-designed circuits that work like this, by never exceeding the LED Vf, and then use no current limiting. I'm not a fan of doing this on purpose though, I just think of all the things that could go wrong, especially if you run right to a specific LED's Vf - next time you build the same circuit maybe the LED is from a different batch and a slightly different Vf - to the low side. Poof.

 Telling - never saw a commercial circuit designed this way. Common voltages of 3.3 and 5 volts are above most any LED's Vf rating, so some form of current limiting is always included. Or the LED is driven by a constant current source so the voltage doesn't matter (up to the limits of the current source component).

--- End quote ---
That was the interesting thing, and I learned something else as well (although in retrospect, it made sense). I was putting 9V into the circuit, but the capacitor never charged above 2.879v (I was able to get some extremely accurate Vf readings with this project), which makes sense, as it wouldn't see any higher voltage than what was being dropped across the led. If I put a current limiting resistor in series with the led, the cap would charge to the combined voltage drop across the led and resistor. It was all quite educational.

Sent from my ONEPLUS A3000 using Tapatalk

james_s:
The LED acts as a shunt regulator and limits the voltage the capacitor can charge to. The capacitor can't "dump" it's charge, you need a load that can draw the charge out of it according to Ohms law. Where it might be a little confusing is that an LED does not behave like a resistive load, it has a relatively high resistance until the voltage exceeds a certain threshold at which point the it behaves as a much lower resistance.

Now if you charge the capacitor up on its own to a sufficiently high voltage and then connect it across the LED, it is going to discharge very quickly and a large enough capacitor will damage the LED.
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