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I have some super-bright LED components that I'm using in a breadboard project. To my surprise, I found that if I ground one leg of the LED and touch the other (unconnected) leg with my finger, it lights up very dimly. Polarity doesn't matter. Why does that happen? There's no source of electrical power that I know of in this "circuit".

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  • \$\begingroup\$ I modified my answer in a pretty substantial way. Let me know if you need any clarification. \$\endgroup\$ – Kit Scuzz Nov 12 '13 at 10:06
  • \$\begingroup\$ Ground one leg? To an actual Earth connection? Your "ground" might only be a power-supply common terminal, and floating at many tens of volts AC wrt actual earth.Test it with a DVM set to AC volts, measuring between an Earth connection and your LED's "ground leg." (True earth or "green ground" appears at metal screws of AC power outlets, or metal plumbing, etc.) Typical DC supplies will have roughly 60VAC on either terminal wrt earth, caused by capacitance between the power-transformer's pri. vs. sec. It's a 1:2 capacitive voltage divider connected across AC line & neutral, with few-uA output \$\endgroup\$ – wbeaty Sep 26 '18 at 2:33
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EDIT:

It's been pointed out in the comments that I've phrased this rather misleadingly, let me take another stab at this with the following caveat: while I'm confident I have the following set of interactions correct, but I'm uncertain of what exactly the terminology should be.

As a person, you are largely a radio wave sink. In addition to the myriad of electron sources I've pointed out below, you're also continually rectifying radio waves into extremely small AC currents/eddies. In effect: you are a big, really ineffective antenna.

So after the electrons on the surface of your skin which have built up drain off as DC, the difference between yourself and ground will fluctuate in time with the radio waves you rectify. In the US, this is dominated largely by 60HZ, the frequency of the AC current running to all your sockets.


Original answer:

"Super bright" LEDs or other high-efficiency LEDs are designed to turn a very small amount of electricity into light. Since the human body acts like a big capacitor, and you exist in an environment awash in (harmless) radio waves and electrons making a mad leap through the air, or being pulled from your clothes or fuzzy carpets: you tend to build up a bit of charge!

In most environments this bleeds off through your contact with less charged substances, and sometimes simply through the less-charged molecules in the air (especially moisture in the form of humidity).

If you give that charge a different route to earth ground, such as through and LED leg, you'll bleed off that charge a bit more quickly and put a small amount of electricity through the leg. This is enough to get that extremely high-efficiency element to glow very dimly!

Now as for why it work on either leg: there's a threshold of voltage you have to put across a diode to get any electricity through it (silicon is generally about 0.7V at room temperature, gallium arsenide is around 0.3V, etc) and the electricity building up on your skin is AC. What end up happening is that the LED is getting a tiny AC current around this threshold, and the diode part of the Light Emitting Diode means that the current only flows the "right" way. Thus, light!

Hope that helps.

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  • \$\begingroup\$ "the electricity building up on your skin s AC" - this is wrong (or at least very confusing) you seem to be describing a build up of static charge - that is not AC. \$\endgroup\$ – RedGrittyBrick Nov 2 '13 at 11:18
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    \$\begingroup\$ This answer doesn't make sense. AC cannot 'build up on the skin' and is totally different to a static electrical charge which, by its nature is DC. High static voltage is most likely to just punch a hole through a sensitive PN junction (hence anti static precautions using electronic devices). The most probable cause (IMHO) is simple pickup from the mains power field (aka hum) which is being rectified by the PN junction of the LED. BTW the material used in electronic devices is silicon not 'silicone' which is an elastic silicon based polymer used to seal gaps in bathrooms etc. \$\endgroup\$ – JIm Dearden Nov 2 '13 at 19:29
  • \$\begingroup\$ @JImDearden I added a clarifying note at the top. You're right that I conflated two ideas which are not terribly similar, and I did it in a way which is pretty misleading. Give me a heads up if you still think this is wrong. \$\endgroup\$ – Kit Scuzz Nov 12 '13 at 10:02
  • \$\begingroup\$ @RedGrittyBrick Same comment as I aimed it at JimDearden, let me know if you still disagree \$\endgroup\$ – Kit Scuzz Nov 12 '13 at 10:05
  • \$\begingroup\$ @Kit: I do appreciate you making the effort to revise your answer, but I still find it confusing. I have distilled your points and those made by Jim D into a "Community Wiki" answer (CW because I don't want to steal rep points, just to arrive at an answer that seems clearer to me at least). \$\endgroup\$ – RedGrittyBrick Nov 12 '13 at 12:06
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There's no source of electrical power that I know of in this "circuit".

Consider a crystal radio receiver, there is, similarly, no obvious source of power - no batteries, no mains-connected AC power supply unit. Yet it generates sound waves through an earpiece. In this case, there is energy used to power the circuit and to create pressure waves in air. That energy all comes from the aerial.

It actually doesn't take much energy to illuminate a LED very dimly. I used to keep a 2 AA aluminium LED torch in my car. It doesn't get much use. One day I noticed it wasn't working very well. Then I noticed it was glowing dimly when it was switched off. I opened the torch to find that the AA batteries had leaked and the leaked electrolyte had bridged the switch in the endcap. Obviously the torch was ruined and I suspect the batteries were not putting out much in the way of voltage. To me this demonstrated that a nominal 2 v LED can actually be very dimly illuminated with a very small amount of power. I now wish I'd made measurements.

if I ground one leg of the LED and touch the other (unconnected) leg with my finger, it lights up very dimly

Most likely, your body is acting as an aerial and is picking up a strong local source of electromagnetic radiation, most likely "mains hum" from your nearby electrical appliances and cables.

Mains hum is a major nuisance for developers of audio applications and many people will have experienced the injection of mains hum into a circuit when parts of the device are touched.

Polarity doesn't matter.

This tells us that the signal being injected is essentially AC not DC.

Mains hum is AC

Any kind of static charge is DC - it is either a build-up of surface electrons or of positive charge carriers - it doesn't alternate (it wouldn't be described as static if that were so). These surface carriers are typically produced by a mechanical process involving friction. Generally there isn't a continuous supply of charge if you have stopped rubbing the source of static (cat, carpet, etc).

As Jim noted, an LED is a diode and a diode will half-wave rectify any AC signal. It will produce light during half of the AC cycle. It doesn't matter which way round it is. Its orientation only affects whether it produced light during the negative part of the cycle or during the positive part of the cycle.

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WARNING: possible electrocution hazard! Old question, but that LED-lighting effect is suspicious. Whenever you see this effect, perhaps check for danger.

If a DC power supply is seriously faulty, then a dangerous AC voltage may exist on one or both DC output terminals. This easily will light an LED connected between the power supply and your body (where your body provides a couple hundred picofarads capacitance to nearby earthed metal conduit, metal in floors, etc.)

To test for a hazard, use your DVM set to measure AC line voltage. Very cautiously connect it between your supply's "ground" connection and a true "earth" or "green ground" connection. A true connection to earth is available at the metal screw on any AC power outlet, also on exposed metal conduits in garage or basement.

If you find 120VAC (or 220VAC UK) between earth and your "supply ground," the DC supply is probably internally connected to full AC line voltage. Your "ground" is actually live, and presents an electrocution hazard.

On the other hand, if you find a much lower voltage than 120VAC, then probably it's not hazardous.

Many DC power supplies have floating output terminals where AC is concerned. But they aren't truly floating, because their transformer's low-volt secondary usually has some capacitive coupling with the 120VAC primary winding. A typical "floating" voltage on the transformer's secondary will be around 40-70VAC, where the transformer is acting as a capacitive voltage-divider, with some picofarads of capacitance "connecting" the entire secondary coil between AC-neutral and AC-line. With such small capacitance in series with the ~60VAC signal, at most it can only provide some microamperes, with respect to an earth connection. (This is more than enough to drive a DVM voltmeter input to detect the large voltage. But far too low to be hazardous, or even detectable by human touch. No shocks.)

Also: if the capacitance between the primary and secondary of a DC supply transformer is fairly large (many hundreds of pF,) and then you connect an LED between a green-ground and one of your DC supply terminals, the LED may light up dimly. This happens because the LED flashes on the forward cycle, then experiences breakdown on the reverse cycle, so can conduct some AC via that between-windings unwanted capacitor. (An ideal LED would just flash once when first connected, as it charged up the series capacitance, which would then block any further DC signal. Real LEDs will break down if exposed to tens of volts in the reverse direction.)

Other stuff: measure your DC and AC body volts wrt earth. In pre-1940s houses without metal conduit or even ground wires in the walls, the 60Hz appearing on humans can be many volts. Also, I lived a mile from an AM radio tower, and it caused all sorts of stuff like this. Cell towers might do the same (check your body RF voltage with a scope?) Now if you only see pure DC on your body, and the LED won't light if hooked reverse, that's impossible! Your body must be pumping out ions into the air, hundreds of microamps worth.

And finally ...all this stuff is why we use expensive "electrostatically shielded" power transformers in instruments and audio equipment. These provide some internal metal foil between the windings, and the foil can be grounded (to actual earth.) We don't want 60Hz line signal being injected via the transformer secondary windings, even if it's only microamps coming through some picofarads of capacitance.

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  • \$\begingroup\$ That's a good point, thanks. I'm the original poster and my power supply measures only 0.8V AC between earth and supply ground, but I can definitely see how a bad power supply could produce the effect I was seeing. \$\endgroup\$ – Eric Lee Sep 30 '18 at 7:14
  • \$\begingroup\$ @EricLee Heh, then instead measure your body wrt earth. (Also check for RF with a scope. Is your location right near an AM tower, or cell antenna?) Even plain old 60Hz can be quite large on human bodies, in pre-1940s houses without any metal conduit or even ground-wires in the walls. Also, for maximum AC pickup, hook two LEDs in parallel pointing opposite. That way you can light LEDs even with a capacitor in series (it discharges the cap, not just charges it. They actually flash alternately.) \$\endgroup\$ – wbeaty Sep 30 '18 at 18:26
  • \$\begingroup\$ It is a common show for radio operators that work close to their high-power radio station antennas to show off by lifting a fluorescent tube in the air while holding one terminal in their hand. The tube would light up. It takes a lot more power to light up a fluorescent tube than an LED. \$\endgroup\$ – Edgar Brown Dec 13 '18 at 15:57
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Radiated E-fields and conducted line filter currents on power lines and SMPS have a way of radiating near field voltages. THese high impedance low-frequency waves are perfectly safe. For AM and FM radios, you can even get a better signal if the antenna is poor.

Power lines nearby are perfectly safe levels of low-frequency radiation and with a long antenna wire around the room and your body as a capacitor, at least 50Vpp can be measured on a 10M probe without touching the earth ground clip or case. That drops somewhat on a 1M probe indicating the high stray impedance coupling of air to your dielectric body current and can result in the x to xx microamp range. You might feel something at 100uA if on a sensitive part of your knee or wrist.

THat xx uA current is just enough to illuminate small LEDs with high efficacy and brightness.

Due to floating power supplies and long cables to the power plug, the coupling capacitance of these tiny leakage currents is increased. ( Safety tests allow 250uA) Earth grounding your 0Vdc suppresses this effect.

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The human body is a giant capacitor. This is how cap sense works, this is why you can build up a static charge and zap people. You are the source of the electrical power, because of the capacitance of your body, and the difference in voltage potentials between your body and the ground you used.

Same reason why two electrical circuits that want to talk to each other typically need to have their grounds connected, because otherwise there is a different voltage potential between them.

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    \$\begingroup\$ Could you please explain how a typical 3kV-30kV static charge noticeably illuminates a 2V LED when "polarity doesn't matter" \$\endgroup\$ – RedGrittyBrick Nov 2 '13 at 11:20
  • \$\begingroup\$ Its also a resistor \$\endgroup\$ – Voltage Spike Aug 8 '17 at 5:23
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Go to this link and you'll see why a crystal set can easily light up an LED under the right conditions:

http://theradioboard.com/rb/viewtopic.php?f=2&t=6464

The LED of choice for me was originally a red Led too, but they are hard to see during daylight. I then used 'water clear' blue led, which lights up very brightly based on some DC component of the rectified signal and some residual carrier wave as well. The load is capacitive, seeing that crystal earphones are essentially a pietzo element (2 plates separated by an insulator - the pietzo crystal material.

I'm personally not satisfied that 50 or 60 cycle hum will effect an an Led. My highest voltage gained just recently from the crystal set output was 3.2 volts DC. It can be done...

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Electrons are always flowing and dancing about whether there in a circuit or not what you are seeing here is them exciting the component inside the LED as they go to ground all you are is a ground it's that simple

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  • \$\begingroup\$ This is nonsense. \$\endgroup\$ – Transistor May 6 at 10:10
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Life is electrical by nature. Nerve impulses are electrical. When death occurs, the electrical activity ceases. Seems logical that a small amount of electricity is bleeding off into the circuit.

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    \$\begingroup\$ Read the other answers to understand why the LED lights. "... a small amount of electricity is bleeding off into the circuit" is not correct. \$\endgroup\$ – Transistor Aug 8 '17 at 6:23

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