I understand how a current higher than what the LED is rated for can burn out an LED, but how does something equivalent happen with voltage?

If the correct current is on an LED but the voltage is too high, what causes it to burn out?

I just don't see what effect voltage has on the LED.

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    \$\begingroup\$ What do you mean when you say the voltage is too high? The supply voltage? The voltage across the LED itself? \$\endgroup\$
    – Null
    Jun 26, 2015 at 4:41
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    \$\begingroup\$ Impossible situation: current is fine but voltage is too large. \$\endgroup\$
    – Andy aka
    Jun 26, 2015 at 8:42
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    \$\begingroup\$ Maybe he means reverse voltage ? \$\endgroup\$ Jun 26, 2015 at 10:10
  • \$\begingroup\$ can you throw some rough figure of voltage you are applying and the resistor in series with the LED (assuming 1.5 V forward drop) \$\endgroup\$
    – User323693
    Jun 26, 2015 at 10:49
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    \$\begingroup\$ @Andy apply 50V forward bias- current will be just fine after a brief period. Lol. \$\endgroup\$ Jun 26, 2015 at 10:49

9 Answers 9


Voltage and current are intimately related. If you attempt to increase the voltage across an LED, the current will increase. Likewise, to increase the current through an LED, you must increase the voltage across it.

It is not possilbe to have the correct current through an LED, but too high a voltage across it.

  • \$\begingroup\$ What about something like a power line, with high voltage and low current after a transformer but on a smaller scale. Wouldn't that have an acceptable current but not voltage? \$\endgroup\$
    – Bob
    Jun 26, 2015 at 4:54
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    \$\begingroup\$ If you attempt to increase the voltage across an LED beyond its "natural" voltage for a given current, the current will increase. If the voltage source cannot supply the required current, the voltage will drop to a point where the supply can provide the current demanded by the LED. (or the power source will fail.) \$\endgroup\$ Jun 26, 2015 at 5:03
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    \$\begingroup\$ @Bob, look at the graph in Peter's answer: As long as the LED is not damaged, it's operating point will always be a point on that blue curve. Always. When you talk about a "high-voltage, low-current" power supply, the high voltage is the open circuit voltage. But even if that voltage is regulated, it is only regulated within some range of current. If your circuit (e.g., the LED) tries to pull more than the max current from the power supply, then the supply voltage will drop. \$\endgroup\$ Jun 27, 2015 at 21:48

As you can gather from the other answers, voltage (U) and current (I) are linked. In the case of a simple resistor:

U = R * I

where R is the constant resistance of the resistor. A diode is only slightly more complicated. Here we can use a graph to show the relationship. The graph uses i for current and V for voltage:

Enter image description here

The image is taken from Using larger resistor values.


First a diode(LEDs are diods) above a certain voltage is like a closed circuit. The problem is that like every wire, the diode has a critical point after which it will "burn", basically some irreversible transformations occur. So you can say the diode can sustain a certain power.

Now, power is related to voltage like this: P = I*V where I is the current, and V the voltage. Since it's a closed circuit, the current is ∞ over it. The source can't give ∞ current, and is limited to a maximum amount. So over a diode, it will use that maximum amount and thus, I becomes a constant. Since I is constant, this means power increases proportional with the variable left, which in our case is V(voltage).

  • \$\begingroup\$ Did you mean to say: "diode... above a certain voltage is like a closed circuit"? \$\endgroup\$ Sep 1, 2015 at 17:13
  • \$\begingroup\$ Yes, I corrected. Wrote it in a hurry and got confused in terms. Thanks for pointing out \$\endgroup\$
    – Luhter
    Sep 3, 2015 at 21:25

As mentioned in one of the comments, this:

If the correct current is on an LED but the voltage is too high

... is not possible.

If the current is "correct", then the voltage will be equal to the characteristic voltage of the diode.

For example:


simulate this circuit – Schematic created using CircuitLab

In the above schematic, Vdiode will be about 1.9V, because 10kV/1MΩ is about 10mA, and that's the voltage this particular LED arrives at if biased on by 10mA (datasheet PDF).

Were you to change the value of R1 to 1 ohm, then approximately 10kA would briefly flow through the LED, resulting it a burned-out LED.

A key concept to grok is the difference between Constant Current and Constant Voltage regulators. A typical "bench" power supply is Constant Voltage, meaning it puts out X volts at some current, and will regulate its output to remain at X volts whatever its load. Diodes approximate Constant Current regulators to a degree, because you can think of the voltage being dependent upon the current.

  • \$\begingroup\$ Not "whatever its load", but for some range of current. Exceed the maximum of that current range, and the voltage will drop (one way or another). \$\endgroup\$ Jun 27, 2015 at 21:49

I understand why you are having a tough time here. An LED is not like a resistor/heat lamp per se. An LED is like any other diode, except in the forward conducting mode as the electrons flow through the junction they cause the atoms to shake at a specific frequency, and not just randomly like a normal conductor. This shaking causes light.

Think of them as a whistle. One note, one amplitude. (Much like a blade of grass held between your thumbs.) --- and that takes energy. If you force too much air, because of a higher pressure (voltage), you will blow out the reed that is making the vibrations.

  • \$\begingroup\$ ...they cause the atoms to shake at a specific frequency, and [...] This shaking causes light. Um, No. en.wikipedia.org/wiki/Light-emitting_diode#Physics \$\endgroup\$ Jun 27, 2015 at 21:52
  • \$\begingroup\$ ok its the electrons. \$\endgroup\$
    – SkipBerne
    Jun 29, 2015 at 12:22
  • \$\begingroup\$ Yep, electrons, that's good, but I've never yet heard a physicist say, "shake" when describing photon emission. \$\endgroup\$ Jun 29, 2015 at 13:30
  • \$\begingroup\$ what else do you find wrong with the analogy? \$\endgroup\$
    – SkipBerne
    Jun 30, 2015 at 14:06

The above response is correct in that voltage and current are intimately related. If you think of a regular resistor that follows Ohm's law, then you can see the relationship V = I*R. With a diode, this relationship still exists, but it is not linear, which is why in data sheets of LED's you will see plots of the Voltage and Current. So if you increase the voltage across the LED above a certain threshold, the current will also increase, burning out the LED.

The reason why a power-line has high voltage and low current is that power-lines are super long, which increases its resistance. Big Voltage = Big Resistance * smallerCurrent. It still indicates that voltage and current are immediately related.


Diodes inhibit Avalanche in Reverse Bias. LED are not exception they being diodes. LEDs have a tolerance of a specific amount of reverse voltage that it can withstand but once this is exceeded, the LED can be damaged. So any excessive reverse voltage (VR) applied can cause avalanche breakdown.

The easiest you can do is adding a simple PN-junction diode in series with the LED (if you wish to save your LED from getting damaged).

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    \$\begingroup\$ LEDs only work with forward bias, so suggestions of reverse bias are irrelevant (but excessive reverse bias will damage an LED, but that's not what the OP was asking). \$\endgroup\$ Jun 26, 2015 at 4:59
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    \$\begingroup\$ @PeterBennett: While it is True that LEDs only work in Forward Bias, I answered w.r.t Reverse Bias Voltage considering the statement "I just don't see what effect voltage has on the LED." OP said. So I thought the term voltage would mean forward as well as reverse. \$\endgroup\$
    – WedaPashi
    Jun 26, 2015 at 5:14
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    \$\begingroup\$ I think Peter Bennett's point was the OP clearly has a limited understanding of how LEDs work at all. It is likely he doesn't understand terms like avalanche, reverse bias, and PN-junction, so going into details like those may only serve to confuse him further. Your answer is factually correct though. \$\endgroup\$
    – Dan Laks
    Jun 26, 2015 at 6:52
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    \$\begingroup\$ -1 ...the last sentence isn't correct though. Two diodes in series won't stop the LED from being damaged by over voltage or excessive reverse voltage. \$\endgroup\$ Jun 26, 2015 at 20:34
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    \$\begingroup\$ @DrFriedParts: The second diode will limit the total current to its reverse saturation current, which is quite low. This prevents that the LED enters the breakdown. So WedaPashi is correct. It does not help against forward overvoltage, but that he also does not claim. \$\endgroup\$
    – Andreas H.
    Jun 26, 2015 at 22:37

To burn a LED it's not the voltage alone! If you apply enough voltage to al LED it will sink a current; when the current grows over the absolute maximum current, the LED burns. This is a common effect of any kind of device. Also a copper wire will burn if Voltage * Current grows too much.


http://led.linear1.org/vf-help.php looks like voltage is constant or varies a little with current in led [Not sure for all diodes but works with leds with me] the diode is designed to take specific voltage not less assume Ex:3.3v for white led and max current 20mA and R=V/I so if u have 10 volts u should have 3.3 on led and the 7.7 should be dissipated by a resistor [which has Volt=7.7v and current 20mA] so R=7.7/0.020 = 385 ohm

so u got 3.3V and 20mA to light a led Now if u +voltage Ex: 15 volts[volt on led is same]
=> 3.3v is for led 11.7v for 385 ohm resistor since I=v/r= 11.7/385= 30.3mA
==> Current exceed max (20 mA ) so As supply voltage increase current on led increase & led voltage is constant so as u increase voltage u need to increase the resistance to keep current in safe range [nearly 20mA]


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