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I have a small amplifier with a bright blue LED (dome-shaped with a height of 6mm and a diameter of 3mm). I am building a new case and I want to replace the LED with (three) smaller ones, preferably white. How can I calculate what the required specifics of the new LEDs need to be to create an equivalent circuit to the one blue LED? Is this even possible?

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    \$\begingroup\$ Check what resistor is in series with the LED, then check from what supply voltage it is powered Then you'll know how much voltage you have available, and you can decide to wire your new LEDs in series or in parallel with resistors. \$\endgroup\$
    – bobflux
    Oct 17, 2021 at 10:12
  • \$\begingroup\$ Thank you! There is a 10kOhm resistor in series with the LED. The voltage between the power supply and just before the resistor is 11.5V. The voltage drop over the resistor is 8.2V. This means there is 3.3V available for the LEDs? Can I replace the resistor with a smaller one to make more voltage available so I can connect three in series? \$\endgroup\$ Oct 18, 2021 at 16:48
  • \$\begingroup\$ It all depends on how the led is driven… if it's a simple resistor it's easy (already answered) if it's something more complex you need to study the circuit \$\endgroup\$ Oct 21, 2021 at 7:05

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There is a 10kOhm resistor in series with the LED. The voltage between the power supply and just before the resistor is 11.5V. The voltage drop over the resistor is 8.2V. This means there is 3.3V available for the LEDs?

So there is 3V3 across the LED, which is normal for a blue LED.

Power supply is 11.5V and there is 8.2V across the 1k resistor, so current through the LED is 8.2mA.

If you use 3 white LEDs, these will have Vf somewhere around 3V-3.4V, so 9-10V total. You can wire them all three in series with a 470 ohm resistor, so there will be 1.5-2.5V on the resistor you'll get about 5mA maximum. Three LEDs at 5mA should be a bit brighter than one LED at 8.2mA, but that depends on the LEDs, so adjust resistor value accordingly.

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  • \$\begingroup\$ Allright, I think I get it now. I don't think it matters much but the resistor is in fact a 10kOhm, but this will result in all currents divided by 10 and therefore a dimmer light (than expected with 1k resistor)? \$\endgroup\$ Oct 18, 2021 at 17:22
  • \$\begingroup\$ Yes, correct. You don't have to replace the LED/resistor, if that's easier leave them in, just disconnect a pin, and connect the new LEDs where it is most convenient. Also, super bright power buttons are annoying, especially on an amp if you use it while watching a movie at night, so maybe 10x lower current is a good idea. \$\endgroup\$
    – bobflux
    Oct 18, 2021 at 18:44
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Similar size and construction white and blue LEDs are typically interchangeable electrically (they both use a blue die).

Replacing one with three is more involved to do it "right" and you'd need to figure out the circuit and measure the current (since the specs for the original LED are unlikely to be available).

Just putting three white LEDs in parallel in place of the original blue one may work well enough and you might want to try that first. Worst case is the LEDs are too uneven brightness or they don't last long, it should not hurt the rest of the amplifier if you do it correctly.

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