# What will happen to an LED if the forward voltage (Vf) is greater than its supply voltage?

For example, if the supply voltage is 3.3V, and the forward voltage of the LED is 3.5V with a forward current of 20mA, what will happen to the LED?

I have already tested the circuit using a 220ohm resistor and the LED lights up just fine, but I have a feeling this is probably not a good idea for the long term.

Are there any alternative resistor value formulas I can use to continue using the 3.3V supply?

• Link to the LED's data sheet please. Jun 28, 2021 at 14:33
• but I have a feeling this is probably not a good idea for the long term Measure the voltage across the 220 ohm resistor and from that, calculate the current that is actually flowing. Is it (significantly) less that 20 mA? Then you have nothing to worry about, the LED should last a long time. If the current is close to 20 mA I would use a larger value resistor to make the current for example 5 mA. If that makes the LED light up brightly enough then I would use that. Jun 28, 2021 at 14:45
• What you MUST provide in your description is the source of current to the LED (5V.?) , the type of LED or link to spec (RGBW?) the required brightness , and why you need a separate supply for an LED if it is just an indicator. When any LED can be used with 3.3V with suitable design details. Jun 28, 2021 at 15:34

The thing about LEDs is that they are current driven devices, not voltage driven.

The formula you will see bouncing around the internet about how to calculate the series resistor for an LED ($$\R = \frac {V_{Supply}-V_{forward}}{I_{forward}}\$$) is an approximation that holds true as long as $$\V_{Supply}\$$ is higher than $$\V_{forward}\$$.

LEDs don't just "snap" to life as soon as you cross $$\V_{forward}\$$. There's a curve of current versus voltage, with $$\V_{forward}\$$ rising as $$\I_{forward}\$$ increases.

This is a current/voltage plot for a blue LED picked at random out of my junk box: Note that it is already conducting well before the typically quoted "3.5V" for blue LEDs. The forward voltage increases as the current is increased. At the end of the scale, $$\I_{forward}\$$ is about 2 milliampers and $$\V_{forward}\$$ is about 2.7V. What you can't see on the chart is that the LED was already eye-searingly bright with just those lousy 2 mA.

The expected 3.5V for a blue LED occurs somewhere near the maximum rated current of the LED (typically 20mA for indicator LEDs.) You don't need that much current just to indicate power on status - you can get away with far less.

What you should do is to use the formula ($$\R = \frac {V_{Supply}-V_{forward}}{I_{forward}}\$$) to calculate the worst case resistor value (never go under that value.) Use the full rated current and the lower end of the expected $$\V_{forward}\$$ to calculate a resistor value. That is the minimum resistance for use with a particular $$\V_{forward}\$$ and $$\V_{Supply}\$$. You can safely use a larger value resistor - the LED will be dimmer but nothing bad will happen.

In your case, you can't use the typical formula because you are outside of the conditions it is valid in.

You have a setup that lights up and doesn't immediately destroy the LED. Measure the voltage drop across the resistor, then use Ohm's law to figure the current through the resistor (this is also the current through the LED.) It'll most certainly be below 20mA. Any resistor value above that ought to be safe. Anything lower and you will need to measure the current again.

If you have a current/voltage plot of your LED, then you could pick a $$\I_{forward}\$$ from the chart and read off the $$\V_{forward}\$$ then use the above formula to determine a resistor value.

For example, for my LED, I picked $$\I_{forward} = 1mA\$$. That relates to $$\V_{forward} = 2.6V\$$ . I figure about 700 ohms in series with $$\V_{Supply}= 3.3V\$$ would work out about right (for $$\I_{forward} = 1mA\$$)

Some LED datasheets will have a current/voltage chart, some won't. With the typical mixed grab bag stuff hobbyists get from Amazon or e-Bay or wherever, you won't have a current/voltage chart.

You can make them yourself if you have (or make) a small I/V plotter.

• "You can safely use a larger value resistor - the LED will be dimmer but nothing bad will happen." Not sure why I though this, but I always thought each LED had an "optimal" resistor value (based on the supply voltage) which NEEDED to be used to maximize the lifetime of the LED. I guess there is no one optimal resistor value, but instead a "minimum" resistor value. Correct? So I can really just fine tune the brightness of an LED as I like as long as I don't go below a certain amount of resistance Jun 28, 2021 at 17:47
• Right. You don't have to give an LED its rated current. Less is fine, and will actually increase the life of the LED.
– JRE
Jun 28, 2021 at 18:02

You should be able to find a graph of the forward current vs. forward voltage in the datasheet for the LED. If the graph shows a significant current at 3.3 V then you will probably observe some amount of illumination from the LED. However, using a constant voltage source close to the specified $$\V_F\$$ is not a good idea...the actual forward current may be much larger than the rated current.

It doesn't hurt the LED to operate it below its rated current. However, using a constant voltage with an LED is tricky. It may work fine for one LED but burn our another one with the same part number.

What you will get is less current than the 20mA at its rated voltage, and less (or no) light. Look at a typical curve 