I have a quick questions: can I safely power a LED with 3.3v without a resistor? I have some LEDs that I typically power with 5v and a 220Ω resistor, but I can only supply 3.3v at the moment.

Additionally I have push-buttons that I typically pull-down with 10kΩ resistors when using 5v. What resistor should I use in this case? Is there a general rule on what resistor should be used given the voltage?

Thank you in advanced for your help!

  • \$\begingroup\$ It depends on the LED. Some LEDs won't even light at 3.3V. \$\endgroup\$
    – Hearth
    May 8 '18 at 16:37
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    \$\begingroup\$ My LED can light with 3.3v, and it's about as bright as when I use 5v with the resistor \$\endgroup\$
    – Microbob
    May 8 '18 at 16:39
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    \$\begingroup\$ All powers sources have a service or assistance if this was a lithium ion battery primary cell you can apply it directly if this was a computer power supply you may need a resistor of about 1 to 10 ohms So that the voltage drop across the resistor is your current and the resulting voltage matches the LED specifications. For example a 5 mm one maybe 3.1 V at 20 mA so it 200 mV drop with 200 mA is 10 ohms \$\endgroup\$ May 8 '18 at 16:40
  • \$\begingroup\$ The general rule depends on the LED specifications. \$\endgroup\$
    – Eugene Sh.
    May 8 '18 at 16:40
  • \$\begingroup\$ "can I safely power a LED with 3.3v without a resistor?" It depends. Define safely. "Is there a general rule on what resistor should be used given the voltage?" Not really. The resistance should be high enough to not use excessive current, and low enough that EMI and other disturbances won't trigger spurious button presses. \$\endgroup\$
    – Dampmaskin
    May 8 '18 at 16:44

Quick answer

If you were using to 220 ohms from 5 V and you drop the supply to 3.3 for a 2.1V RED then 5-2.1=2.9 or R reduces to 2.9/220*R=1.2V difference

and R=1.2 /2.9×220= answer

Very accurate method which I use to estimate R and determine the voltage threshold for dim at 10% Imax which is often 10% below the rated forward voltage . Then use the difference voltage between the supply and that threshold Vt to determine the total series resistance.

The LED resistance is I have found inverse to its power rating so a 1/16W is ~ 16 ohms , a 1 W is approximately 1 ohms or less. Thus the added Series R changes with power of the LED.

This may sound complicated but with practice it’s trivial. The total series resistance you would need to drop from 3.3 V just depends on the LED curve ESR plus the series resistance to get exact desired current with reasonable tolerances. If you search my answers in the window at the top of this page you will find I have written dozens of examples on this topic.

Also compute I^2R for high current LEDs

You may learn how to do this or not is your choice.

E.g. in search tab above type or paste user:"Tony Stewart" LED ESR R quote marks are needed due to space in "my name" . I see I have 63 hits for these keywords.

For users wanting to check self, enter "user:me" .... key words

  • \$\begingroup\$ I am a little confused as to where you got the 1.2V in 2.9/220*R=1.2V \$\endgroup\$
    – Microbob
    May 8 '18 at 17:38
  • \$\begingroup\$ Red Vf=2.1 Vcc=3.3 new difference = 1.2 old difference was 2.9 . I can put that into a ratio formula for scaling know R values or you can try to follow my general rule in answer for any situation. The result is the same with I=Vdrop/R where R can be on either side of the LED. Then in fancier high power methods current can be sensed on the low side and regulated with a comparator or difference amplifier. \$\endgroup\$ May 8 '18 at 18:01
  • \$\begingroup\$ Alright, so if I had 5v go to 2.2v after passing through an LED, I the resistance I should use on that same LED with 3.3v would be 1.1/2.8*220 ≈ 94Ω? \$\endgroup\$
    – Microbob
    May 8 '18 at 18:16
  • \$\begingroup\$ You got it. Vcc ≈ IR+ Vf = I*94Ω +2.1V= 3.3V so If≈13 mA, which is ok? but since ESR of LED ~ 15 Ω so can include that in the R value otherwise 1.2V drop If = 1.2V/(94Ω+ESR) Note that the ESR of 15 Ohms means the LED Vf drops 150mV for a drop of 10mA from rated 20mA. So minor corrections mean Vf might be 2.0V with same If=13mA, {R+ESR}=(3.3-2.0)V/13mA= 100 Ohms = R+ESR so R becomes 85 Ohms. or nearest value, This is why I use Vt in my formula. instead of Vf at rated current, where Vt is a dim thershold at 10% of rated current above which is fairly linear \$\endgroup\$ May 8 '18 at 18:48
  • \$\begingroup\$ as in my answer where Vt≈90%Vf and Vf = Vt + If * (ESR+R) ESR may also include battery ESR such as a 3.7V Li-Ion so optimum power LED's in flashlights match certain batteries to certain LEDs for cheap flashlights with no regulation, e.g. Some 10W LEDs are 2.85V at 350mA may be Vt=2.6V with ESR = 100 mΩ +0/-50% and Li-Ion ESR may be 50 mΩ +/- xx % per cell. The 90%Vf is an estimate but is better obtained from datasheet. \$\endgroup\$ May 8 '18 at 18:52

An LED has a point where the terminal voltage applied causes a significant rise in the current taken and if you increase that voltage by even as little as 0.1 volts, the current taken by the LED could be beyond its operational rating.

Because that voltage (for your LED) might be OK set at 3.3 volts, it doesn't mean you won't blow the back off it at 3.4 volts. This is why we put a current limit resistor in series. If your LED takes 100 mA at 3.3 volts (for instance) you would need to drop 1.7 volts from 5 volts with a resistor.

That means a resistor of value 1.7 volts / 100 mA = 17 ohm.

Now, if your 5 volt rose to 5.1 volts, the 100 mA would rise to no more than 1.8 volts / 17 ohm = 106 mA. If your LED is rated at a maximum current of 120 mA, you could in fact allow the 5 volts to rise to 5.34 volts before being on the cusp of exceeding its ratings.

Do you see how adding a resistor protects your LED from over-current?

For pull-down resistors on switches you need to ensure that if there is a wetting current requirement to be met, the resistor is low enough to allow that current to flow.

  • \$\begingroup\$ Let me say thanks, i like your answers and tonight i'm lucky to see you twice during my nightly researches \$\endgroup\$
    – payam_sbr
    Dec 26 '18 at 20:23
  • \$\begingroup\$ So if led is rated [2.8v - 3.1v] is it ok to connect it to an ideal 3v supply without resistor? \$\endgroup\$
    – payam_sbr
    Dec 26 '18 at 20:27
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    \$\begingroup\$ No it isn’t. You need to control the LED current. Try googling LED voltage current characteristic. \$\endgroup\$
    – Andy aka
    Dec 27 '18 at 10:51

You cannot safely power most LEDs with a 3.3V source when the source can supply more current than the LED's capacity.

Two characteristics of the LED need to be known to calculate the resistor.

  1. The Forward Voltage
  2. The maximum current capacity

If the current output of the 3.3V source is less than the LED's maximum, then it is safe.

If you need a resistor you select your desired current which must be equal to or less than the LED's maximum current.

Today most new white LEDs driven at less than maximum (e.g. test current) will have a forward voltage of about 2.85V. At maximum current about 3.1V. Red LEDs are about 2.1V.

A very common current for LEDs used in lighting is 65mA. Indicator LEDs are 20-50 mA.
For an example let's use 60 mA as the target current and an LED with a 2.85V Vf

You then use an LED resistor calculator the find the value of the resistor.

The resistor value is 7.5Ω for 60 mA and a 2.85 Vf powered by a 3.3V source,

enter image description here
Source: Hobby Hour LED Series Resistor Calculator

A red indicator LED with a 220Ω resistor draws about 13 mA at 5V. 13 mA at 3.3V would use a 91.3Ω (1%) or 100Ω (5% 12mA) resistor

enter image description here

enter image description here

  • \$\begingroup\$ These calculators have innacuracies based on user input of 2.1 @ 13mA which is incorrect, It should be reduced 7mA *ESR. but dont take my advice. \$\endgroup\$ May 9 '18 at 1:46
  • \$\begingroup\$ @TonyStewartolderthandirt OP said 5V with 220Ω resistor. The only assumption is a 2.1V red led as stated. That works out to 13mA when powered with 5V. Then I changed the 5V supply voltage to 3.3 leaving the 13 mA. The calculated resistance value is then 92.3Ω. Prior to that I explained step by step how to calculate the resistor for any LED. ESR? I do not need to calculate ESR. I only need desired current and forward voltage at that current. ESR gives me nothing. ESR will remain exactly the same at 13mA no matter what the supply voltage is. Thanks for the down vote. \$\endgroup\$ May 9 '18 at 3:32
  • \$\begingroup\$ The assumption is wrong and I indicated why. If you put blind faith into a calculator you get errors unless you understand the impact ESR has on Vf. Capiche? \$\endgroup\$ May 9 '18 at 4:10
  • \$\begingroup\$ If you care to understand why I down voted , try reading my other answers on LED ESR in my links posted in my answer. When you understand let me know and I'll reverse the -1 \$\endgroup\$ May 9 '18 at 4:20
  • \$\begingroup\$ @TonyStewartolderthandirt It's not "blind faith". The math is not that difficult. ESR is forward voltage divided by current. LEDs have a dynamic ESR based on the current. If the current through the LED remains the same so does the ESR. Notice in the calculator for both 5V and 3.3V supply voltage the "power consumed by LED" remains the same. Why? The ESR remains the same so therefore forward voltage remains the same because the current remained the same as will as the power consumed. Where did you get lost?? It not that complicated. \$\endgroup\$ May 9 '18 at 4:29

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