Not understanding Forward Voltage and Voltage Drop

Question

I have a simple circuit:

+12V -- R1 -- LED1 -- LED2 -- LED3 -- ground

Falstad simulation link

If the Forward Voltage of an LED is 3V, and the Forward Current is 20mA, I can (I believe) calculate the required resistance of the resistor as (12V - (3 * 3V)) / 0.02A = 150Ω.

From what I understand, that should give me a Voltage Drop of 3V over the resistor and each LED respectively, and a current of 20mA through the circuit - perfect.

In the simulation of this circuit, I get a Voltage Drop of 4.01V, 2.66V, 2.66V, 2.66V respectively, and a current of 26.74mA through the circuit, which is too high for the LEDs.

This makes me think that I don't understand the relationship between Forward Voltage and Voltage Drop, and therefore, how am I supposed to calculate a correct resistor value that won't burn out the LEDs?

Apologies if this is asked a lot or is really simple, but I've been searching for ages and haven't come up with anything.

Answer 1

For your simulation, you specify the led's forward voltage as "3V at 1A". This means the forward voltage of the leds at about 20mA will be much lower.

Everything is right there, you just need to read the leds datasheet to find the forward voltage at arount 20mA.

Answer 2

In the simulation of this circuit, I get a Voltage Drop of 4.01V, 2.66V, 2.66V, 2.66V respectively, and a current of 26.74mA through the circuit, which is too high for the LEDs.

The problem is using a very simplistic simulator. Falstad is not well suited to answering this kind of question.

You want to know what the exact forward voltage will be, but Falstad is only really going to tell you whether the LED is "on" or "off".

A SPICE-like simulator, on the other hand, would model the diode using the Shockley diode model:

$$I = I_s\left(\exp\left(\frac{qV}{nRT}\right)-1\right)$$

This would allow you to specify \$I_s\$ and \$n\$ to get an IV curve very close to what a typical part will produce. With Falstad's model that fixes the I-V curve at only one point, you're not likely to get a result that accurately predicts the voltage over a wide current range.

That said, even if you use a better model, you should account for temperature changes and variation between parts. Not all parts (even of the same part number) will have the same \$V_f\$, even at the same temeprature, and your circuit design needs to be robust enough to account for that, not carefully tuned to the exact nominal behavior of a part type.