# Forward voltage drop across an LED

## Main Question or Discussion Point

Hi guys,

I'm trying to design circuits that include mostly ohmic devices with the exception of some LEDs. In particular, I'm having trouble coming up with an effective voltage drop across an LED. This is the datasheet for the particular LED that I'm trying to include in the circuit. It has a maximum power dissipation and a forward voltage range given a minimum supply current of 20mA.

I'm just not sure how to read a datasheet for a diode.

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The thing about LED's is that you cannot predict a precise voltage drop. And that's it. One thing that is normally done is to test how many amperes you want to run through it, before it gives the light-intensity that you want. This is can be done using a current-limited power supply for example. At the light intensity you want, you can measure a voltage drop across the LED. This voltage drop you can use in your circuit design then.

So lets say you figure out that you want 15 mA to run through the diode when its on, and this specific LED shows a 1.5V drop at this current. Depending on the voltage you have available in your circuit, you would use a resistor in series with the LED to produce the correct voltage and current.

For example, lets say that you have a 3V available, and you need 1.5V across the LED, you would need 1.5V to be dropped across the resistor.

Since you measured that at 15mA the light intensity was correct and this corresponded to 1.5V across the LED, you can calculate the resistance that is needed. In this specific case we have that

1.5V = 15mA * R <=> R = 100 Ohm

Putting this resistor in series with the LED will provide exactly 15mA through the resistor.

IF YOU DO not want to do it this way, a quicker idea is just to look at the min and max values of the LED and pick one at random (not to high and not to low) and then do the same with the resistor. This way you dont have to do the measurement of the LED.

Baluncore
2019 Award
You can model a LED as a voltage step in series with a small resistance. The series resistance is a manufacturing parameter. The voltage step is determined by the LED wavelength based on photon energy being Plank's constant times the radiation frequency. Photon wavelength is determined by the semiconductor band gap chemistry.

That becomes simply; Forward voltage = 1240 / wavelength in nm.

you want your led voltage at around 1.2 volts with 20milliamps of current. so with a 9volt battery you would use a 390 ohm resistor. 9v - 1.2v = 7.8v * .02a = 390 ohm resistor.

Baluncore
2019 Award
According to the data sheet the forward voltage could be anything from 1.5V to 2.6V. You need to specify material or colour of the LED you intend to use, then look up forward voltage.

You have a supply voltage of Vs and a LED forward voltage of Vfwd. The current limiting resistor will drop (Vs – Vfwd) volts. For a current of i amps the resistor should have a value of;
R = (Vs – Vfwd) / i

I think nevere actually meant / not * in the non-equation so; 9v - 1.2v = 7.8v / .02a = 390 ohm resistor.

Thanks for the replies, guys. I'm guessing that generally LEDs, if they're not a major component of a device, are not a big deal to factor into any design. The main concern, I guess, is not to blow the LED.
I just wanted to get a general idea of how I could get an effective voltage if I ever wanted to.

You can model a LED as a voltage step in series with a small resistance. The series resistance is a manufacturing parameter. The voltage step is determined by the LED wavelength based on photon energy being Plank's constant times the radiation frequency. Photon wavelength is determined by the semiconductor band gap chemistry.

That becomes simply; Forward voltage = 1240 / wavelength in nm.
Did you derive this from the general diode equation?

Baluncore
2019 Award
Not from general diode equation.
You select a chemistry. That decides a bandgap voltage, and fixes the colour.
When an electron falls across that bandgap voltage it releases energy as a photon.
The change in electron energy when it crosses a one volt field is one “electron volt” = eV.

A photon has an energy of plank's constant times frequency. E = h * u
Frequency = speed of light divided by wavelength. u = c * w
Therefore E = ( h * c ) / w, where ( h * c ) is a constant.
h = 4.13566733 x10^-15 (eV.sec) and c = 299792458. (metre/sec)
The constant is therefore 1239.84187 (nm.eV)

The LED forward voltage is therefore 1240 / (photon wavelength in nm)

Last edited:
• 1 person
Not from general diode equation.
You select a chemistry. That decides a bandgap voltage, and fixes the colour.
When an electron falls across that bandgap voltage it releases energy as a photon.
The change in electron energy when it crosses a one volt field is one “electron volt” = eV.

A photon has an energy of plank's constant times frequency. E = h * u
Frequency = speed of light divided by wavelength. u = c * w
Therefore E = ( h * c ) / w, where ( h * c ) is a constant.
h = 4.13566733 x10^-15 (eV.sec) and c = 299792458. (metre/sec)
The constant is therefore 1239.84187 (nm.eV)

The LED forward voltage is therefore 1240 / (photon wavelength in nm)
Ah, I understand. I have another related question.
In the provided datasheet for the LED I want to use (LG3330), why is there a minimum and maximum Vfwd for a given constant current of 20mA? Even though diodes are non-linear, they should still give a constant forward voltage given a constant current, right?

AlephZero
Homework Helper
The properties of semiconductor devices can be very variable. For example the small signal gain (beta) of transistors can vary by a factor of 2 or more between nominally "identical" components with the same part number. The bottom line is, you have to design your circuits so this variability doesn't matter.

From your data sheet, if the forward voltage drop is given as 1.7V to 2.6V and you have a 9V power supply, the voltage drop across a series resistor is 7.3V to 6 4V, so the resistance to give 20mA current is 365 to 320 ohms. But in practice you would probably use a 330 ohm resistor since that is a commonly available value, and the current will be near enough to 20mA. Remember the battery won't give "exactly" 9V all the time, as it discharges!

If you want a more precise and stable value of the LED current for some reason, you can use a constant current source circuit like http://www.onsemi.com/pub_link/Collateral/AND8109-D.PDF

The properties of semiconductor devices can be very variable. For example the small signal gain (beta) of transistors can vary by a factor of 2 or more between nominally "identical" components with the same part number. The bottom line is, you have to design your circuits so this variability doesn't matter.

From your data sheet, if the forward voltage drop is given as 1.7V to 2.6V and you have a 9V power supply, the voltage drop across a series resistor is 7.3V to 6 4V, so the resistance to give 20mA current is 365 to 320 ohms. But in practice you would probably use a 330 ohm resistor since that is a commonly available value, and the current will be near enough to 20mA. Remember the battery won't give "exactly" 9V all the time, as it discharges!

If you want a more precise and stable value of the LED current for some reason, you can use a constant current source circuit like http://www.onsemi.com/pub_link/Collateral/AND8109-D.PDF
This is the way I was initially going about it, but I just wanted to make sure it was logical. So, using the proper calculations, as long as I use a resistor that allows about 20mA, the Vfwd will be somewhere between 1.7V and 2.6V. This is basically what the table means?

Baluncore