14
\$\begingroup\$

So I understand that LED's have a maximum current (like 20mA for instance), but scientifically why is this?

Using the water analogy it seems like a high voltage would be the thing that would mess up something (I like to think of it like a huge amount of "pressure" blowing out a pipe or something). Why would a rate of electron flow damage something?

\$\endgroup\$
  • 3
    \$\begingroup\$ @IgnacioVazquez-Abrams What? Voltage is every bit as physical as pressure. It is in fact a very good analogy because they are each physical potentials (energy per unit stuff): pressure could be measured in joules per cubic meter, and voltage in joules per coulomb. \$\endgroup\$ – Phil Frost Jan 4 '15 at 13:39
  • 9
    \$\begingroup\$ That's the minimum current at which the magic smoke has enough energy to escape. \$\endgroup\$ – A E Jan 4 '15 at 19:37
  • \$\begingroup\$ Sticking with pipe analogy - how about the rate of waterflow causes attrition on the pipe, wearing the sides down and eventually causing a break - like an electrical fuse? \$\endgroup\$ – OJFord Jan 4 '15 at 23:13
  • \$\begingroup\$ " Why do LEDs have a maximum current?" makes them sound exceptional. Pretty much everything has a maximum current. (although isolators are more easily described as having a maximum voltage) \$\endgroup\$ – MSalters Jan 5 '15 at 12:25
  • \$\begingroup\$ Everything has a maximum current, voltage (...and power, dV/dt, etc.), but some devices (or elements thereof) are "voltage devices" (e.g. microcontrollers, capacitors, MOSFET gates) and others "current devices" (e.g. diodes, LEDs, BJTs, MOSFET drain-source). It's far more informative to couch maximums in the device-relevant terms rather than "convert" them into something else (e.g. Igs(max)? :P). \$\endgroup\$ – Nick T Jan 5 '15 at 16:54
32
\$\begingroup\$

It's difficult to come up with an analogy because the usual analogies for electrical systems are fluid systems. A great thing about fluid systems is that the working fluid is also good at cooling things, and most people's practical experience with fluid systems involves rates of flow where heating is not very significant.

So let's try a different analogy: a string being pulled through the resistance of your fingers. Your fingers are the LED, and the voltage drop of the LED is analogous to the difference in tension of the string on either side of your fingers. Current is analogous to the rate at which the string is being pulled.

Will your fingers be damaged if the string is pulled too fast? Yes: we call it "rope burn". This will happen even if you adjust the resistance of your fingers to maintain a constant difference in tension on the rope regardless of its speed (analogous to the approximately constant voltage drop of the LED).

The reason is that the rate of work done, and thus, the heat generated, is the product of the force your fingers apply to the rope and the rate at which the rope is moving through your fingers. You can get a rope burn by squeezing too hard, or moving the string too fast.

"Rate of work" or "rate of energy" is called power. One way to define it, for mechanical systems, is the product of force (\$F\$) and velocity (\$v\$):

$$ P = Fv $$

Since power is a rate of energy it should be in units of energy per time. In SI units, thats joules per second, also known as the watt. So, however fast the rope is moving, and however much force your fingers are applying to it, you are doing work at the rate of some number of joules per second. This energy can't vanish: it becomes heat in the rope and your fingers. Once you exceed your body's ability to transfer heat away from your fingertips your skin gets too hot and you are burned.

The analogy for electrical systems is that power is the product of voltage and current:

$$ P = VI $$

\$V\$ is approximately constant for an LED, but if you increase \$I\$ enough, you generate heat faster than it can radiate to the ambient environment. The LED gets too hot and is damaged.

\$\endgroup\$
  • \$\begingroup\$ So a LED could be damaged if it had an extremely high voltage but low current then? \$\endgroup\$ – user3073 Jan 4 '15 at 19:56
  • 6
    \$\begingroup\$ The voltage of the LED is fixed to a couple of volts. If you increase the voltage beyond that, the current will increase to absorb all the power your power supply is capable of feeding into it - until it burns and becomes an insulator. \$\endgroup\$ – PkP Jan 4 '15 at 20:04
  • 1
    \$\begingroup\$ @Mercfh The function that relates current and voltage for resistors is Ohm's law, and there are several models for diodes, depending on how accurate you need to be. \$\endgroup\$ – Phil Frost Jan 4 '15 at 23:05
  • 7
    \$\begingroup\$ If you'd like to stick to the water analogy then a water wheel or turbine is a good model. Move the water too fast and you'll damage the bearings (they overheat) \$\endgroup\$ – slebetman Jan 5 '15 at 3:52
  • 3
    \$\begingroup\$ +1. I've never heard of the string/tension analogy before, and quite like it! \$\endgroup\$ – Shamtam Jan 5 '15 at 4:31
16
\$\begingroup\$

There's a forward voltage drop of a couple volts across the LED. This Voltage drop times the Current is the Power dissipated in the device. It creates light, but also heat. It's the heat that kills the LED.

\$\endgroup\$
  • \$\begingroup\$ If you drop a standard LED across a voltage source while holding it by the head, it sometimes generates enough heat before it dies to get noticeably warm. Don't do this with a high-power LED! \$\endgroup\$ – Warren Young Jan 4 '15 at 13:43
12
\$\begingroup\$

TL;DR: Flowing current creates heat, and for LEDs, heat kills the part.

Whenever electrons flow through a conductor, Joule heating happens. This is partly because of what heat really is, particles that makes up the object moving around, and having electrons pulled through it guarantees that some electrons will collide into something and have its energy transferred into that particle, heating it up.

When the LED is overloaded the excessive heating will cause the fragile bonding as well as the die itself to change. None of those change is constructive and eventually the heat destroys the part. For LEDs they burn out and maybe blow apart, for some other parts they can burst into flames.

\$\endgroup\$
2
\$\begingroup\$

Here's another way of looking at what some others have said:

The conversion of current to light is not 100% efficient, so the remainder of the energy not converted to light is heat.

Each electronic component has something called a "thermal resistance" measured in degrees Kelvin/Watt which says how easily the "waste energy" above gets out of the die to the PCB (typically the cathode for an LED) as heat. This is specified in the datasheet.

In addition, each electronic component has a maximum junction temperature, Tj at which it can operate according to the rest of the specified parameters in the datasheet.

With this information, given a constant thermal resistance, Rth, a fixed maximum power rating LED, Pdiss_max, and a constantly increasing power source driving the LED, what will happen is you will drive the junction temperature above its maximum rating and likely de-solder the wire bonds from inside the chip, rendering it inoperable.

Good question!

\$\endgroup\$
1
\$\begingroup\$

There's only a certain amount that that size of material can handle. Take a filament for example. Just the right size to glow but not burn up. It is being overwhelmed with electricity, there's only so much it can handle before it burns out. Same with LEDs. Depends upon the amount and type of material.

\$\endgroup\$
  • \$\begingroup\$ While this is true as far as it goes, I don't think it's helpful to think of it in terms of size. The leads on a standard 3mm LED probably have a fusing current of 20-30 amps. A high-powered LED isn't all that much bigger than a standard one; mainly, it has a better heat sink. Size is only loosely correlated with current-handling capacity. \$\endgroup\$ – Warren Young Jan 4 '15 at 13:48
  • \$\begingroup\$ Comparing an LED burning out to an incandescent filament implies they may share a common theory of operation, when in fact, they are about as opposite as is possible. At a fixed temperature, diodes are non-linear, while a filament is (ignoring any parasitic effects) purely linear. Furthermore, a filament has a negative thermal coefficient, and this negative feedback allows for stable operation over a range of voltages. In contrast, semiconductors generally have a positive thermal coefficient, and exhibit thermal runaway when provided with excess energy. \$\endgroup\$ – bcrist Jan 6 '15 at 7:49
1
\$\begingroup\$

All nice answers. I just wanted to add that if there was no non-radiative recombination in the LED's, then there would be much less heat and one could push more current through before it heats up... (Think newer high efficiency LED's)

\$\endgroup\$
0
\$\begingroup\$

Actually I found the water analogy pretty effective. A pipe will break apart If a massive quantity of water would transit into it. More specifically, it will melt down as a fluid flowing generates a small quantity of heat like any other material do

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy