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Somehow LEDs with a lower forward voltage than the source with a 100 Ω resistor tend to blow up every time the switch is turned on. How can I understand forward voltage better?

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  • \$\begingroup\$ It's unclear what you're asking. Could you elaborate on the problem? \$\endgroup\$
    – Hearth
    Commented Jan 31, 2023 at 3:59
  • \$\begingroup\$ LEDs, too, may breakdown biased in reverse (just reminding). \$\endgroup\$
    – greybeard
    Commented Jan 31, 2023 at 5:06

4 Answers 4

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It's not just forward voltage you need to know, you also need to know the rated current. From these, and the supply voltage you can calculate the resistance needed.

Example, you have an LED with a \$V_f\$ of 1.8 V, and a current rating of 20 mA. You are going to connect it to a 5 V supply. To find the resistor needed you have use Ohm's Law which tells us that resistance is voltage divided by current.

So we need to know how much voltage there will be across the resistor, this will be the supply voltage minus \$V_f\$, so 5 V - 1.8 V = 3.2 V.

Now you have the voltage and current, so the resistance needed to limit the current to 20 mA will be 3.2 V / 20 mA = 160 \$\Omega\$. If you had used a 100 \$\Omega\$ resistor the current would be 32 mA and the LED would probably have burned up. To be on the safe side you would go up to the next standard value resistor, 180 \$\Omega\$ and the current would be a little under 18 mA, leaving a little wiggle room.

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  • \$\begingroup\$ Apparently, the higher current is what's making the LED's in the simulator to blow up. I will consider applying this solution to the simulator. \$\endgroup\$ Commented Jan 31, 2023 at 5:29
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LEDs do not have a well specified 'forward voltage', only a range, that varies a bit with temperature, roughly as the logarithm of the current, and a lot from device to device. That means if we design to a 'voltage drop', perhaps we think we want 3.0 V across a white LED, then we will get huge variations in current, perhaps over orders of magnitude, taking us from dim to blown, as the device temperature varies, or we swap in one device for another.

LEDs do however have a well specified current, which will get rated brightness without overheating, or blowing up.

If we have as much or more voltage across a dropper resistor than we have across the LED, then it's possible to choose a resistor and drive voltage to give us a suitable current. The LED forward voltage variations will be swamped by the extra voltage across the resistor.

However, that's wasteful of voltage drop, and what we invariably do when we can is to current-drive LEDs. This is most efficient when done from a switch-mode constant current power supply.

It's best to think of 'LED forward voltage' as the voltage that happens to develop across a LED when driven by its rated current, which is the minimum output voltage a constant current LED driver should produce.

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  • \$\begingroup\$ Forward voltage is an approximation, but it does exist. \$\endgroup\$ Commented Jan 31, 2023 at 15:18
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The problem stems from assuming a current-limiting resistor value of 100 Ω instead of calculating the required value.

The following schematic will be of help.

enter image description here

Vs - Supply voltage

Vf - LED forward voltage

If - LED forward current

R Ω - Value of current-limiting resistor

Vr - Voltage drop across the resistor.

The resistance value is to be calculated.

Vs = Vr + Vf

Vr = Vs - Vf

Vr = If x R

R = Vr / If

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Heat sink!

If these are illumination LEDs, they may require a heat sink. Make sure that is dealt with.

Any lighting LED will fry if it's not appropriately heat sinked.

Don't guess at resistor value

Others have explained the math. If you can't be bothered to crunch the numbers, then start with about ten 100 ohm resistors in series with the LED and an ammeter. Bypass resistors one at a time until the LED gets bright and the ammeter shows an appropriate value for current.

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