hello,

When considering maximum voltage and maximum current requirements should I worry about voltage or the current? I bought some LEDs that work on 3.2-4.0 volts and max 30mA so I connected them to the 3.3 volt supply from the arduino and they keep burning out. I thought I don’t have to worry about current because it would only draw what it needs. Is there something to “step-down” the current too?

Thanks

 

You need current limiting resistors.

Post your schematic.

Though I can't see how they could be burning out as I can't imagine the Arduino ports being able to supply much more than 30ma

If your forward voltage of the LEDs is 3.2V-4V then you do not have enough voltage at 3.3V to light them. You will need 5V at least.

 

You still want a current limiting resistor in series with your LED. Using the 3.3 Volt out or the 5.0 Volt out on your Arduino board make sure you know the max current available. Also know your working voltage and working forward current for your LED. I have no idea if your 30 mA is a maximum? If when connected to the 3.3 Vout of the Arduino board the LED is excessively bright some current limiting is in order. Place a resistor in series with your LED. I would start with about 5 to 10 Ohms. I also assume this is not connected to an Arduino pin but rether the Arduino board 3.3 Vout.

<EDIT> Spinaker beat me to it and yes a schematic would be nice and a link to the LED data sheet.

Ron

 

Current through the LED is what will burn it out, so you need to limit the current with a resistor. Remember that an LED is a diode (Light Emitting Diode), and once the voltage passes the forward voltage of a diode, it is essentially a short. So once the voltage is above the forward voltage, the diode cannot limit the current itself, so you need a resistor in series with it. This should be a good read:

https://learn.sparkfun.com/tutorials/diodes/real-diode-characteristics

And here's cheat sheet. Figure out the math yourself, then use this to check your answers:

https://www.digikey.com/en/resource...ors/conversion-calculator-led-series-resistor

 

Resistor calculators don’t work when the forward voltage is the same or higher than the supply voltage.

I would use a 10 ohm resistor.

 

You need current limiting resistors.

Post your schematic.

Though I can't see how they could be burning out as I can't imagine the Arduino ports being able to supply much more than 30ma

If your forward voltage of the LEDs is 3.2V-4V then you do not have enough voltage at 3.3V to light them. You will need 5V at least.

You still want a current limiting resistor in series with your LED. Using the 3.3 Volt out or the 5.0 Volt out on your Arduino board make sure you know the max current available. Also know your working voltage and working forward current for your LED. I have no idea if your 30 mA is a maximum? If when connected to the 3.3 Vout of the Arduino board the LED is excessively bright some current limiting is in order. Place a resistor in series with your LED. I would start with about 5 to 10 Ohms. I also assume this is not connected to an Arduino pin but rether the Arduino board 3.3 Vout.

<EDIT> Spinaker beat me to it and yes a schematic would be nice and a link to the LED data sheet.

Ron

Yes you are right, this is not connected to any digital output pins and instead it is connected to the +3.3v out.

Current through the LED is what will burn it out, so you need to limit the current with a resistor. Remember that an LED is a diode (Light Emitting Diode), and once the voltage passes the forward voltage of a diode, it is essentially a short. So once the voltage is above the forward voltage, the diode cannot limit the current itself, so you need a resistor in series with it. This should be a good read:

https://learn.sparkfun.com/tutorials/diodes/real-diode-characteristics

And here's cheat sheet. Figure out the math yourself, then use this to check your answers:

https://www.digikey.com/en/resource...ors/conversion-calculator-led-series-resistor

Thank you, I will start reading this now.

Resistor calculators don’t work when the forward voltage is the same or higher than the supply voltage.

I would use a 10 ohm resistor.

Excuse my novice-ness but that is my schematic. Before it did not have any resistors so I put a 560 ohm resistor with a 5% tolerance (Green, Blue, Brown, Gold). This is the lowest resistor I have at the moment. I am trying to power an Infrared LED which I can only detect using a camera with no IR filter. Before putting a resistor in, they kept getting really hot, smelling like burnt and eventually not lighting up. Someone gave these LEDs to me and they don't have a data sheet neither do they know the product reference. I have a feeling they are meant to be 1.6-1.8 volts just like some other IR LEDs I found online.

Using V = IR then V = 0.03*560 = 16.8 voltage drop, that means 3.3 - 16.8 = -13.5 volts? How does that work?

Thanks

 

With an LED it's all about the current. If you don't have the part number for the LED, then take a guess that it will be OK with 15mA and that it's forward voltage is 1.5V. So you take your supply voltage minus the forward voltage divided by the resistance to get the current. So if your supply is 5V, subtract the 1.5V forward voltage to get 3.5V. Now divide 3.5 by 560 ohm and you get 6.25mA. If you put two of those resistors in parallel then you'll have about 12.5mA, which should give you more light and is likely still safe for the LED. You can verify by measuring the actual current with your multimeter.

 

Using V = IR then V = 0.03*560 = 16.8 voltage drop, that means 3.3 - 16.8 = -13.5 volts? How does that work?

You've made calculations here based on the assumption that you need to calculate the voltage that results from your resistor choice and desired current. That's all turned around. You can apply the formula two ways right now:

1. Calculate expected current through LED based on existing voltage and resistor.
2. Calculate what resistor value you would need in order to achieve preferred current based on existing voltage and target current.

In both cases, the LED has a fairly stable voltage drop across it that changes very little with changes in current - it doesn't fit in neatly with simple ohms law calculations... although there's a simple method that gets close enough for most situations: just treat it as a fixed voltage drop which you subtract from the total supply voltage, then calculate current through the resistor. Here are two examples:

If you have 3.3V supply and assume 1.5V drop across LED, that leaves 1.8V. You're using a 560 ohm resistor right now. So rearrange V=IR into I=V/R and calculate. 1.8V/560=3.21mA.

Alternately, let's choose a new resistor, still assuming the same supply and LED, so still 1.8V across resistor. If your target current is 15mA, solve the formula for resistance: R=V/I, so 1.8V/0.015A=120ohms.

 

"140Ohms" is common for a "3.3VDC" power supply and a normal or ultra bright LED.