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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.

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  • \$\begingroup\$ extra 6ma won't kill LED's, and you can just find the right resistor value with a multimeter. Or you can use a current source (2 resistors, 2 diodes and a transistor), and measure voltage drop on it and find out the equivalent resistance. \$\endgroup\$ – ilkhd Aug 5 '15 at 7:59
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    \$\begingroup\$ There is something wrong with the simulation. Try to run the same schematic with LT spice. \$\endgroup\$ – Sanjeev Kumar Aug 5 '15 at 8:33
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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.

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    \$\begingroup\$ That makes sense. If the datasheet only has a Vf at 20mA, is there a method to calculate an accurate or approximate value to enter into the simulation for Vf at 1A? \$\endgroup\$ – parrowdice Aug 5 '15 at 7:57
  • \$\begingroup\$ @parrowdice I think you have a mistake in you comment... \$\endgroup\$ – ilkhd Aug 5 '15 at 8:01
  • \$\begingroup\$ @ilkhd Sorry, I'm a newb. Can you please explain? \$\endgroup\$ – parrowdice Aug 5 '15 at 8:14
  • \$\begingroup\$ @parrowdice You want Vf at 20 ma in your post but Vf at 1a in your comment \$\endgroup\$ – ilkhd Aug 5 '15 at 8:16
  • \$\begingroup\$ @ilkhd Yes, I know Vf at 20mA, but as Nicolas said, the simulation wants Vf at 1A. I want to know if there is a way to calculate that. It's there something wrong with that statement? \$\endgroup\$ – parrowdice Aug 5 '15 at 9:01
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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.

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  • \$\begingroup\$ Amazing, thank you. Great information there. You mention I should make the circuit robust enough for variations, and I completely agree, but I don't know how. What techniques are there to account for this? Many thanks \$\endgroup\$ – parrowdice Aug 7 '15 at 9:09

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