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The main goal of this circuit is to limit the current if the output side is shorted. LED1 Lights up when output is shorted. LED2 lights up with output isnt shorted. U1 is a 12V regulator with current limiting features.

Regulator Circuit

Problem: Why does LED1 light up when I apply 13.91V input voltage, even though it should not? Is this because the excess voltage is being dropped over LED1, given that the Vdrop is 12?

Can I added a resistor in parallel to the LED to bleed it out? Is that the best solution?

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2 Answers 2

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According to the LED's datasheet, its typical forward voltage at rated current (20mA) is 1.95V. When you apply 13.91V to the input, 13.91V - 12V = 1.91V drop across the LED (and series resistor). I'd be surprised if the LED didn't light up under these conditions.

Adding a resistor in parallel to the LED will work. It will form a voltage divider together with R1 to lower the voltage across the LED while the circuit is in operation. As a rough guesstimate, you could use 220 Ohms to raise the voltage differential at which the LED begins to light roughly threefold. If that prevents the LED from lighting when you want it to, increase the value.

Note that the resistor has to go directly across the LED's pins.

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  • \$\begingroup\$ Thank you for your help! Appreciate it. Ideally I can also change the LED which has a higher forward voltage yeah? \$\endgroup\$
    – S_D
    May 11, 2023 at 23:30
  • \$\begingroup\$ @S_D forward voltage roughly depends on emitted color and current through LED \$\endgroup\$ May 12, 2023 at 6:03
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It lights up because 2V is enough to get light from an LED, especially a red one.

You could add a green LED in series (doesn't have to be placed anyplace visible).

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