When powering a simple LED circuit (DC power source, LED, resistor), does the supply voltage matter, as long as the correctly calculated current limiting resistor value is used?

In other words, is there / could there be something inherently wrong by powering an LED with 12V or 24V as long as I used the correct resistor, knew the forward voltage of the LED, knew the maximum current, and calculated it using something like this, when I could have powered the same LED with a 3.5V supply knowing the same variables and using the same website?

I'm assuming there is a limit to the maximum amount of voltage to use for the LED here... when I look at the electrical characteristics chart for the CREE XP-G for example, it shows current as a function of voltage, with voltage starting at around 2.5V @ 0ma, maxing out at around 3.25V @ 1500ma (the maximum current the LED is rated at, as described in the Characteristics table in the same document.

After 3.25V, the chart depicts current quite rapidly approaching infinity.

I'm assuming this relates to my question, I'm just curious how its all related. I'm sure its all basic Ohm's law stuff, I'd just appreciate a clarification of the math at work.

  • 2
    \$\begingroup\$ Safety may come into play. LEDs are not built for their isolation performance. For example you must not connect them directly to mains power, no matter how good your series resistor is. \$\endgroup\$
    – jippie
    Jan 30, 2015 at 7:36
  • 8
    \$\begingroup\$ The LED only "knows" the voltage across itself; not the voltage across your resistor. \$\endgroup\$
    – user253751
    Jan 30, 2015 at 8:05
  • \$\begingroup\$ Diodes aren't ohmic (linear relationship between current and voltage). So no, it's not just Ohm's law at work here. \$\endgroup\$ Jan 31, 2015 at 3:58

6 Answers 6


There is no limit on the voltage, per se, that you use to power the circuit that drives the diode. The diode only cares about what the diode can see, and it can't see the voltage drop across the current limiting resistor.

That said, at some point what you're going to care about is the power dissipated across the resistor, which is \$ I^2R \$. If you want to keep the current to be constant in the case of growing required voltage drop, then R will eventually get big, and it will dissipate too much power. The power that run of the mill axial lead resistors can dissipate is 1/4 watt. For a 20mA current, that means to limit power across the resistor to 1/4 watt, you can't exceed 625 ohm, which means you can maximally drop 12.5 volts across it, and you're ceilinged out at a power supply of about 14.5V for a red LED. It's worse for small package SMD resistors, which are often 1/8 Watt or less. If you need more of a voltage drop, you would have to change to a higher power rated resistor, which can get physically big, as well as more expensive.

As to why the actual voltage across the LED doesn't change too dramatically given proper choice of current limiting resistor, one convenient way to look at this is with the "load line" technique. From https://i.stack.imgur.com/1cUKU.png, (Public domain image from Wikimedia):

enter image description here

The negative sloped line represents the resistor. If \$ V_D = 0 \$ there would be \$V_{DD}/R \$ of current through the resistor, and if \$ V_D = V_{DD} \$, then there is no current through the resistor (as there's no voltage drop across the resistor). The circuit "lives" at the equilibrium point where the resistor line and the diode curve intersect, as you MUST have the same current through the diode and the resistor. Note that changing R and \$ V_{DD} \$ less than dramatically won't move this point as much as you think it might in terms of the final voltage drop across the diode, because of how steep the diode curve gets.

  • 1
    \$\begingroup\$ TBH, that 20 mA/625 Ohm/14.5 Volt max is easily sidestepped by having 2 LEDs in series. The same light intensity now comes from 10 mA, which means you can now go up to 2.5 kOhm, a voltage drop of 25 Volt over the resistor and a total max voltage of 29 volt (twice as high). \$\endgroup\$
    – MSalters
    Jan 30, 2015 at 14:46
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    \$\begingroup\$ I'm not sure, @MSalters. I'd have to do a bit of data sheet peeking to figure out if light output is linear with current. I suspect the better, and cheaper, practice is just to use two parallel 1200 ohm resistors to give you a half watt capability. It might also be worth mentioning that its not a bad idea to derate the resistors. \$\endgroup\$ Jan 30, 2015 at 15:41
  • \$\begingroup\$ LED output is pretty darn close to linear with current (at least for typical 5-25 mA devices). To the point that I doubt you'll find a datasheet that would give you a spec to indicate the deviation. \$\endgroup\$
    – The Photon
    Jan 30, 2015 at 19:19

The purpose of the series resistor is to throttle the current through the LED. Forward voltage of the LED goes into the in calculation of the current limiting resistor.

$$ R = \frac{V_{cc} - V_f}{I_f} $$

There is nothing fundamentally wrong with using a higher voltage if you size the current limiting resistor appropriately for the voltage. At the same time, you will be dissipating more power on the current limiting resistance. So, you will need a resistor with sufficient power rating.

  • 1
    \$\begingroup\$ +1, Just to add: Sufficient power rating OR high enough resistance so the current gives a power dissipation (P = I^2*R) low enough for a given power rating. \$\endgroup\$
    – Doombot
    Jan 30, 2015 at 16:17
  • \$\begingroup\$ @Doombot How can you increase the resistance without changing the current and voltage supplied to the LED? \$\endgroup\$
    – apraetor
    Dec 11, 2016 at 17:40

In the general, the diode current increases exponential with the voltage:

$$I = e^{cU}$$

where c is a constant depending on the geometry, doping, temperature etc.

This is the reason why a high power LED should always be driven by a constant current, and not a constant voltage source. Tiny variations to c (eg. temperature change) or U would cause a massive change in current.

The series resistor works, because its resistance is usually a lot higher than the differential resistance of an LED. From the perspective of the LED, the voltage source plus the resistor behaves like a current source.

  • \$\begingroup\$ I think this answers the second part of the question well: the transfer function goes "to infinity" because the exponential grows extremely fast when the voltage is high enough. \$\endgroup\$
    – Greg d'Eon
    Jan 30, 2015 at 13:01
  • \$\begingroup\$ This only works for small enough voltages, i.e. smaller than opening voltage. For higher voltages the current will grow linearly. \$\endgroup\$
    – Ruslan
    Jan 31, 2015 at 16:29

does the supply voltage matter, as long as the correctly calculated current limiting resistor value is used?

No. Diodes are current devices. They have a voltage drop that you should take into account in your circuit, but they are current driven and as long as you limit the current appropriately, and cool the diode if needed (for high power LEDs), then there is no supply voltage limit.

The voltage at the LED itself will be the diode's voltage drop, which will depend a little on the current through the diode, but mostly on the composition of the diode. Applying too large a voltage at the diode terminals (ie, without the current limiting) will result in current above the diode's limit, and will damage the LED.

With the appropriate current limiting, though, you can use a million volt power supply to power an LED. Although at that point you'll have to check for adequate insulation between the terminals of the various parts...


An LED has a "maximum voltage" because its resistance decreases dramatically - just like in any other diode - as its forward voltage is increased past its knee, and this increase in voltage across the LED coupled with the increase of current through it (because of the decrease in its forward resistance) increases the power the LED must dissipate and, thus, its operating temperature. Then, if the the current through the LED's junction is allowed to rise past its absolute maximum rating, its lifetime will be shortened and the magic smoke will escape, sooner or later.

In the case of the CREE XP-G you referred to, I've taken the Forward Voltage VS Forward current plot from the data sheet and overlaid it with the derived Forward voltage VS Forward Resistance plot as shown below. Rather crude because I didn't do any curve fitting, but it's easy to see the huge change in forward resistance for the smallish 250 millivolt change in forward voltage from 2.5 to 2.75 volts.

enter image description here

Because of this extreme sensitivity to voltage and because the location of a diode's knee can't be predicted with great certainty, LEDs aren't generally driven by raw voltage sources, but by constant current or current limited voltage sources designed to never allow the product of the current through the LED and the voltage dropped by the LED to exceed the power rating of the LED.

For high power, not inexpensive LEDs like the XP-G, a constant- current supply can be used to good advantage because it'll keep the current through the LED fixed regardless of the variations in LED Vf or voltage input to the constant-current supply. Most commonly, though, a resistor is used in series with a voltage source in order to limit the current through the LED.

The value of the resistor is determined by subtracting the LED's specified minimum Vf from the source's maximum output voltage, and then dividing that difference by the desired LED current. That resistance will assure that the LED's "maximum voltage" will never be exceeded, and you can see that there is no limit (well...) to the source voltage allowed since the resistor will get rid of everything the LED doesn't need.


I am going to cover the Peak Inverse Voltage (sometimes seen as the reverse voltage) of the diode. The PIV is the voltage at which the diode junction begins to break down when it is reverse biased (ie voltage is backwards). For most LEDs it is relatively low (5V is typical - I did a quick search and found 3 different manufacturers that all had 5V). Depending on the power source it may not matter (a low voltage battery makes it a relatively moot point). Other power sources such as AC/DC convertors can have a high voltage with the opposite polarity of design for a brief time when the source or controlled devices such as relays are turned on and off.
Therefore any application that has a power source greater than 5V should have reverse protection for the LED. This can be a reversed biased diode across the LED for simple protection or other more advanced techniques.


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