I'm confused...(musings on LEDs and capacitors) - Page 1
Forum section: electronics / beginners

author: GadgetBoy
author_rank: Regular Contributor
timestamp: December 23, 2018, 04:24:16 am
content: 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

author: rrinker
author_rank: Super Contributor
timestamp: December 23, 2018, 05:23:56 am
content: 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).

author: spec
author_rank: Frequent Contributor
timestamp: December 23, 2018, 05:47:27 am
content: Hi GadgetBoyQuote from: GadgetBoy on December 23, 2018, 04:24:16 amJust to confirm that your conclusion is absolutely correct.

author: GadgetBoy
author_rank: Regular Contributor
timestamp: December 23, 2018, 05:50:22 am
content: Quote from: rrinker on December 23, 2018, 05:23:56 amThat 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

author: james_s
author_rank: Super Contributor
timestamp: December 23, 2018, 06:08:45 am
content: 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.

author: IanB
author_rank: Super Contributor
timestamp: December 23, 2018, 06:10:05 am
content: Quote from: GadgetBoy on December 23, 2018, 04:24:16 amYou're on the right track--more or less.One mistake is that an LED does not have "a" forward voltage. It has a V-I curve, where the current through the LED is a function of the voltage, much like a resistor. But unlike a resistor, where the V-I curve is linear, the V-I curve of an LED is exponential. For every small increase in voltage you get a big increase in current. This means the current is very sensitive to the applied voltage and therefore it is difficult to control the brightness of an LED by voltage regulation.In your experiment the parallel capacitor charged up until the voltage reached the balance point on the V-I curve--at which point the LED had a steady brightness and the capacitor had no current through it. When you disconnected the power the LED moved down the V-I curve as the capacitor discharged, gradually getting dimmer as the voltage decreased.

author: GadgetBoy
author_rank: Regular Contributor
timestamp: December 23, 2018, 06:34:02 am
content: I remember trying to get finer control the brightness of an LED using an ALOG potentiometer. It worked better than a linear one, but still not as well as I'd hoped.

author: not1xor1
author_rank: Frequent Contributor
timestamp: December 23, 2018, 07:56:54 am
content: Yeas. You are really confused.Didn't anybody tell you yet that a thread subject should provide a hint to the content?Do change the subject to something more meaningful!!!BTW a quick search on the net should suggest you that energy stored in capacitors is:1/2 * C * V2and in leds datasheet you should find what a led is supposed to tolerate.Probably more than I do tolerate meaningless subjects.

author: GadgetBoy
author_rank: Regular Contributor
timestamp: December 23, 2018, 08:42:23 am
content: Quote from: not1xor1 on December 23, 2018, 07:56:54 amFixed. Happy?Sent from my ONEPLUS A3000 using Tapatalk

author: GadgetBoy
author_rank: Regular Contributor
timestamp: December 23, 2018, 08:44:17 am
content: 24 hours later on the 6F bank. I'm intrigued to see how long this thing is going to glow.Sent from my ONEPLUS A3000 using Tapatalk

author: rrinker
author_rank: Super Contributor
timestamp: December 24, 2018, 12:51:52 am
content: With 6F, that LED is going to glow for a long time.One thing the smaller physical size supercapacitors has enabled is a "keep alive" circuit for model trains. I only have one locomotive with a keep alive in it, but with (I believe, I haven't taken it apart to see what value the caps are) a total of .1F available, this loco will keep rolling for nearly a minute if I take it off the rails and set it on my bench - like a windup toy. Motor turning, driving the thing, and the LED headlight is on.

author: RJSV
author_rank: Super Contributor
timestamp: January 07, 2019, 12:07:16 am
content: Wow, what a great thread, and all responses are helpful, there. I have always called the VI curve for an LED type as 'spongy', which is exactly wrong. How about this explanation: You can press on a glass window all you want, but its the movement that indicates breakage is emminant.Uh, maybe lame, but seriously many of (us) have a difficult time on the LED drive theory.One way to view it is only see the voltage as a 'reaction', and calculate using a 'nominal' voltage (crudely viewed as not changing).That means calculate the current using your applied voltage, minus the 'characteristic voltage'. You can measure that, likely it will be 1.8 volts, for a red LED. So use 9 volts (measure the battery) minus the 1.8 volts, thats what is driving your series connected resistor and LED.Adding a second LED just right directly in parallel with the first LED will cause the 'Characteristic voltage to only look a bit lower, like 1.7 volts (keeping in mind the LED pair should be very similiar or from the same manufacture batch, etc.).Now, I would argue: "You said no voltage, and now you are saying apply 9 volts, etc.". Well remember, placing that series resister creates amore linear look, from the source capacitor viewpoint, the LED is buried somewhat, or isolated from a pure applied 'voltage'.Now I'm confused! (kidding) RJ

author: GadgetBoy
author_rank: Regular Contributor
timestamp: January 07, 2019, 12:09:42 am
content: If anyone is curious, the 6F cap lasted a week and a half.Sent from my ONEPLUS A3000 using Tapatalk