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The LED I am using requires a higher voltage to light than I have supplied, and as a result, it doesn't light at all.

I would expect at least a dim light, but light is not generated.

Why this behavior of "if there isn't the required voltage level there's no light"? What is going on inside the LED?

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    \$\begingroup\$ I would expect at least a dim light Why would you expect that? The process that generates the photons only starts above a certain voltage/current. So how can photons be generated when the required conditions are not met? Do you understand how an LED works? If not: research it. It will then become obvious what is needed for the LED to produce light. \$\endgroup\$ – Bimpelrekkie Nov 18 '17 at 12:22
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    \$\begingroup\$ @Bimpelrekkie It's reasonable to assume an LED works like an incandescent light bulb if you've never encountered anything that would tell you otherwise. \$\endgroup\$ – Todd Wilcox Nov 18 '17 at 19:52
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    \$\begingroup\$ @ToddWilcox: The next step after reaching that point is to open a book and read about what LEDs are and how they work, surely? That's "my assumption is apparently wrong" management 101. \$\endgroup\$ – Lightness Races with Monica Nov 19 '17 at 3:06
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    \$\begingroup\$ @LightnessRacesinOrbit That would be a reasonable approach for someone that's not a complete newbie in EE. Since on this site we allow questions from newbies and enthusiasts, sometimes you come across questions like this one, which are clearly posted by people that don't even know how/where to begin with EE. After all, there are even some manufacturers that market their LED as a "SOLID STATE LAMP" (as they were some kind of light bulb of the past) in their datasheet. Confusion in a newbie is understandable in this case. \$\endgroup\$ – Lorenzo Donati supports Monica Nov 19 '17 at 7:28
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    \$\begingroup\$ @LightnessRacesinOrbit "The next step after reaching that point is to open a book and read about what LEDs are and how they work, surely?" I think the 2017 version of that process is to get on electronics.stackexchange.com and ask how LEDs are different from incandescent bulbs. I've actually put almost all of my books in storage, since I literally never use them any more. The Internet is one of the new "basics of learning". This is absolutely the right place for such a question. \$\endgroup\$ – Todd Wilcox Nov 20 '17 at 22:04
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LEDs don't work like ordinary (incandescence) light bulbs.

Main differences (a bit simplified for very beginners):

  • They have a polarity, hence they must be powered using DC respecting that polarity. Reverse the polarity and they won't work. You may also damage them if you apply more than ~4V-5V in the reverse direction (these are safe values; the exact maximum tolerable value depends on the specific device).

  • Light emission begins only if a certain voltage is reached (threshold voltage), under that voltage the emission is negligible. Therefore, if you have a battery whose voltage is below the threshold voltage of the LED, you are out of luck, unless you use a more complicated circuit (e.g. a Joule thief or a boost DC-DC converter) to power the LED.

  • After the threshold voltage is reached, any very slight increase in voltage makes the LED conduct heavily, i.e. absorb a huge current. Hence you need a resistor in series to limit that current to a safe limit. There are other questions/answers on this site explaining how to calculate the value of the limiting resistor.

  • Once conducting, the light intensity emitted is roughly proportional to the current (not voltage) that flows in the diode (so you get a brighter LED if you decrease the value of the limiting resistor). This up to the maximum current limit of the LED. After that limit has been reached, the device goes POOF!

You also ask why all this happens, but the answer is fairly complex, since it depends on the physical structure of the semiconductor crystal inside the diode. The physical explanation lies in quantum mechanics and solid-state physics, really tough subjects.

The Wikipedia article on LEDs only scrapes the surface of the internal workings of the LEDs and still is fairly complex.

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    \$\begingroup\$ Yes. The interesting thing is that until the threshold voltage is reached, an LED is effectively an open circuit - no current. Once it is passed, it's effectively a short circuit - meaning infinite current (assuming ideal wires, of course). Which is why circuitry using LEDs will always have a resistor in series with the LED. The resistor determines the current (voltage/resistance) and the current determines the brightness of the LED. In general always use a resistor with an LED and use the largest one that will produce the brightness you require, to minimize power consumption. \$\endgroup\$ – David C. Nov 20 '17 at 22:23
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    \$\begingroup\$ I wouldn't call it "absorb a huge current", absorb sounds like the current vanishes in the LED - which it does not, it passes through it. \$\endgroup\$ – Arsenal Nov 21 '17 at 13:27
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    \$\begingroup\$ @Arsenal please, reread my post, (1) I explicitly marked my points as "simplified for very beginners"; (2) current absorption is a perfectly valid engineering term, although maybe not so common. See the abstract of this IEEE article. \$\endgroup\$ – Lorenzo Donati supports Monica Nov 21 '17 at 14:43
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    \$\begingroup\$ @Arsenal Of course also "the device goes POOF" and "emitting magic smoke" is standard jargon! :-) \$\endgroup\$ – Lorenzo Donati supports Monica Nov 21 '17 at 14:51
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    \$\begingroup\$ @LorenzoDonati seems like we have different ideas about educating people. I like to introduce them to the words we actually use to make researching and expressing oneself easier in the future, but I can also see your point. Current and voltage is just a hard thing to explain. \$\endgroup\$ – Arsenal Nov 21 '17 at 16:19
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I see Lorenzo has already answered your question directly (+1). Here is what you can do to light your LED and see what you've got.

LEDs are diodes, so conduct only in one direction. Unlike a ordinary light bulb, orientation matters. If the LED doesn't light one way, flip it around and try again.

To safely experiment with pretty much any LED, use a 5 V supply with at least 180 Ω in series. Using a higher resistance works, but will light the LED more dimly. Even with 1 kΩ in series, you will still be able to see any visible-light LED light up indoors.

The reason to use a 5 V supply is to limit the reverse voltage across the LED when connected backwards. Most LEDs can stand at least 5 V reverse across them.

A visible-light LED will drop a minimum of 1.8 V. That leaves (5 V)-(1.8 V) = 3.2 V across the resistor. Just about any LED can handle 20 mA forward current. By Ohm's law, (3.2 V)/(20 mA) = 160 Ω. I said 180 Ω minimum for a little margin and because that is a common value.

LED forward voltage is dependent on color. Common green LEDs drop about 2.1 V, for example. "White" LEDs are usually really UV LEDs with phosphors that re-emits in the visible spectrum. Those can drop around 3.5 V.

With a 200 Ω resistor and a 3.5 V LED, you get (1.5 V)/(200 Ω) = 7.5 mA. Such a LED will still light quite visibly with 7.5 mA thru it, even if it could have handled 20 mA or more.

Once you get your LED to light, you can measure its forward voltage, then adjust the resistor to allow the maximum current with that forward voltage. Assume the maximum is 20 mA unless you have a datasheet and it says otherwise.

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    \$\begingroup\$ I love this answer. Lorenzo answered like a scientist, but you nailed it like an engineer! \$\endgroup\$ – polfosol Nov 20 '17 at 7:52
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    \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. Or perhaps start a Meta discussion \$\endgroup\$ – clabacchio Nov 21 '17 at 10:54
  • \$\begingroup\$ Worth bearing in mind that phrases like "with at least 180 Ω in series" could be very confusing to newbies who might be unsure whether you mean something advanced that they don't about or whether you mean to literally just whack a 180ohm resistor into the circuit. \$\endgroup\$ – Clonkex Nov 22 '17 at 0:42
  • \$\begingroup\$ I would have them start with 330 Ohm. Why blast the LED with so much current? \$\endgroup\$ – Jeff Wahaus Dec 2 '18 at 12:03
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Physics explanation

Light bulbs

An incandenscent light isn't really a light source so much as a heating element. Any current through a wire heats it up a bit; once the wire is above room temperature it emits net energy via black-body radiation. The rate at which this energy is emitted depends on the fourth power of temperature, i.e. the higher the temperature the brighter. And the more current (or equivalently more voltage), the higher the temperature of the wire.

The fundamental physical process behind blackbody light emission is this: the atoms in a warm piece of material are shaken around by thermal motion. This motion is completely chaotic, so even if the average energy per atom is rather low, every once in a while an atom at the surface will get a push from multiple neighbours and thus gather enough energy that it can emit a visible photon (at least \$2.6\times10^{-19}\$ Joules). But far more often, it will only have enough enough energy to emit an invisible infrared photon.

LEDs

By contrast, LEDs pump the atoms directly to the energy that's required for emitting visible light. They do this by cleverly exploiting the band gap of a semiconductor. That's a quantum-mechanical feature of crystals like silicon, which “forbids” electrons from having energies in a certain range. You then take one piece of semiconductor which has been doped so the conductance electrons are all above the band gap, and one where they're all below the band gap. Then, when a current flows across the junction, each electron loses just the right amount of energy to excite an atom to produce a photon with the right energy to be visible – again, for red light this is about \$2.6\ldots3.2\times10^{-19}\$ Joules.

Only... why would the electrons continue to go over the junction? After an electron has crossed the junction, it won't be inclined to climb across the band gap again; that costs energy which the electron doesn't have. ...Unless you give it the energy from an external source: each volt that you apply to a circuit can supply an electron with an energy of \$1.6\times10^{-19}\:\mathrm{J}\$, a quantity that physicists call just an electron volt. So when you apply a voltage of \$U\$ to an LED whose band gap has an energy of \$U\times1\:\mathrm{eV}\$, you can keep up a current. This voltage isn't really dependent on how much current actually passes throught the LED, therefore the brightness can't effectively be regulated by tweaking the voltage – you need to regulate the current instead. And if the voltage falls below the band gap, the current will just cease entirely, because the conductance electrons just won't go into the n-doped domain at all anymore.


That's a bit too simplistic: Stefan-Boltzmann describes the intensity integrated over the entire electromagnetic spectrum. Only a narrow band of that is actually visible (that's the reason incandenscent light is so much less efficient than LEDs). Since the wavelength of peak intensity also depends on the temperature, brightness is in fact related not just to \$T^4\$ but to a more complicated relation, but still: higher temperatures always correspond to brighter light.

Similarly, Ohm's law isn't completely correct here because the resistivity depends on temperature. But the qualitative dependency higher voltage ⇒ higher electrical power still holds true.

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You just got an object lesson in how LEDs are non-linear.

Incandescent bulbs are linear once they light. Linear means it acts like a resistor: current draw is in proportion to voltage: half the voltage, half the current, 1/4 the power. An incandescent light would do what you expect.

LEDs have a very steep voltage-current curve: a small change in voltage results in a large change in current draw. You are off the bottom of that chart, hence no light.

The steep curve makes the LED very jumpy, small voltage changes result in big (and damaging) current changes. Worse, the curve changes based on temperature, binning and age. So LEDs are rated at a specific current rather than voltage. For indicators, you can limit current with resistors. For lighting, where you need peak performance, it's best to use an active driver circuit to regulate the current to spec.

Such circuits also lend themselves to boosting or bucking supply voltage to suit the LED. The Joule Thief is a simple circuit that solves the problem of driving a lighting LED with a single 1.5V battery.


For what it's worth, it's even worse with the third kind of light, arc-discharge lighting: fluorescent, neon, metal halide, mercury vapor, and high/low pressure sodium. They are insulators up until a certain voltage when the arc strikes... After which they are nearly a dead short. Current limiting is mandatory.

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    \$\begingroup\$ I was taught that a filament bulb made a reasonable constant-current load, the hotter it burns, the higher the resistance. Maybe this is what you mean by 'once they light'. \$\endgroup\$ – Sean Houlihane Nov 20 '17 at 0:56
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    \$\begingroup\$ @SeanHoulihane Yes, until a filament lights, it's near a dead short, and still has a considerable "inrush current" until it reaches full power. If you use an incandescent in a current-limited mode, you can do a lot of stuff like that. For chuckles, put an incandescent on a CC supply and watch it warm up slowly. \$\endgroup\$ – Harper - Reinstate Monica Nov 20 '17 at 1:38
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A little background on semiconductors...

Pure silicon(or germanium) is an insulator. Impurities are added to create "P" or "N" type material. When these are next to each other (in a P N diode or LED) the impurities effectively cancel each other out, leaving you with a small "pure" layer - that acts as an insulator.

If you hook up power the wrong way, the layer gets thicker and stronger - right up until you damage the device. (Varicaps use this principle to create a variable capacitor - thickness of insulation acts like distance between capacitor plates)

When you hook up power the right way, the layer gets thinner until the device eventually passes current. When that finally happens, your LED will start to glow.

One final note: It is possible to light an LED with just a 1.5V source: use a voltage "step up" circuit. The most common circuit is called a Joule Thief.

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Their voltage drop (~2 V) is higher than the supply voltage (1.5 V). If the supply voltage is lower than the voltage drop, any diode (including LED) doesn’t conduct at all.

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    \$\begingroup\$ This doesn't answer the question in a way which is useful. Why do you get no light whatsoever in this case? OP is expecting a scenario similar to an incandescent bulb, and this answer does nothing to explain why that expectation is wrong. \$\endgroup\$ – Phil Nov 19 '17 at 23:39
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In addition to the answers here, it's also worth pointing out that every LED is different (even by colour). They all have slightly different 'activation' voltages and breaking limits.

The correct way to be sure that you're not going to blow up your LED while also ensuring that you can expect light is to look at the data sheet for your LED.

Using the datasheet will give you essential experience in

  1. Locating said datasheets
  2. Reading & understanding them

To get you started, here's a random one for a white LED from the top of google; which is perhaps a bit more complex than most LED's because "white" isn't a colour in LED land.

There's an incredible amount of information in them, and if you don't understand them I'd suggest that you try, and then post a question again asking about a specific part you don't understand.

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  • \$\begingroup\$ Not only are different colored LEDs different (red are often rated around 2V, while white and blue LEDs regularly operate at 3+V) but even within a color, and sometimes even within a batch they may vary. That's why datasheets often show "Min[imum]", "Typ[ical]", and "Max[imum]" Vf ratings. \$\endgroup\$ – Doktor J Nov 22 '17 at 16:04

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