Reading an LED data sheet, voltage drop and supply voltage in an LED ring

Question

I am looking to make an LED ring using these white LEDS.

I've drawn this schematic:

I was planning to use 5V supply so I followed Ohm's law and did 5/0.060 to get 83.3 so I rounded it up and used a 100 ohm resistor.

When looking at the data sheet there are 2 values:

• Forward voltage
• Reverse voltage

The forward voltage is 3.4V max, and reverse is 5V.

My understanding is that the LED with a resistor in series will act as a voltage divider, and the LED then can be operated from 5V.

• Why is the forward voltage 3.4V?
• I correct in assuming that 5V will be okay for this application with 100 ohms resistor?

Answer 1

Ignore the reverse voltage, I don't think it matters here

The forward voltage means when you apply + and - correctly, the diode will start to conduct. But however, unlike a resistor, it will conduct at a single (sorta) voltage. Yours are 3.4V But really, that's 3.4V at the designed current (is it 20mA??) It means when conducting, the LED will have 3.4V drop across it.

SO you do this...

5V - 3.4 = 1.6V --- That's the voltage the resistor has to drop

Then 1.6V / 0.020A = 80 ohm, because ohms law works ;)

So you use an 80 ohm resistor in series with the 3.4V VF diode, and you'll end up with 20mA per diode.

Make sense???

If the LED's are 60mA target, same math...

5 - 3.4 = 1.6

1.6 / .060 = 27 ohms

Here's a fun little trick.... The LED almost certainly will NOT be 3.4V. That's nominal (as other answer mentions). So if you want REALLY accurate math, you can measure the Vf of the LED. If you have a constant-current power supply available, Set the voltage for about 5V. Short circuit the terminals and set the current limit for 20mA (or 60mA or whatever your target is). Remove the short. Then power the LED. The constant-current will prevent the LED from melting down, you know exactly how much current is flowing (i.e. 20mA), and you can directly read the voltage because the power supply will be putting out exactly that diodes Vf.

Answer 2

The 2.8 to 3.4V is a typical operating voltage range for a white LED. It will have this rated voltage over it when running at the rated current.

The resistor value is calculated incorrectly. The false assumption is that all the 5V will be over the resistor, but it is not, since there will be 2.8V to 3.4V over the LED, so only the remaining voltage is over the resistor.

And you should not simply use the assumption that LEDs have always the maximum value of 3.4V over them, and calculate the resistance based on that, since the LEDs could just as well have the minimum value and then the current would exceed the nominal 60mA value and it can be over 80mA.

Answer 3

Forward voltage is the voltage drop across the LED when driven by the rated current. You must connect the anode to + and the cathode to - for the LED to be forward biased.

LEDs are current operated devices, not voltage. The forward voltage rating says that when the current flows through the LED, then that voltage will appear across the LED.

You have incorrectly calculated the resistor.

You need to use \$R = \frac {V_{supply}-V_f}{I}\$ to calculate the resistor. That's \$R = \frac {5 - 2.8}{0.060} = \frac {1.6}{0.060} = 36 ohms\$ for your case.

Calculate the resistor at the lower forward voltage because that's where you'll get the higher value resistor and the lowest current. You want to stay below the rated forward current. (0.06 A for your LED.)

Note that this is only approximate. The forward voltage of the LED could be anywhere from 2.8 to 3.4 volts. Using the 37 ohm resistor, that would be 60 milliamperes at 2.8 volts or 43 milliamperes at 3.4 volts.

Reverse voltage has nothing to do with the supply voltage at all.

Reverse voltage tells you that if you hook up the LED backwards then it will be OK unless you connect it to more than 5V backwards. More than 5V (cathode to +, anode to -) and the LED will most likely be damaged.

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It is purely coincidence that the reverse voltage rating of 5V matches your power supply voltage of 5V. You could use 12V with an appropriate resistor (143 ohms) and it would work just as well.