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I have a reading lamp from IKEA. It has a bright LED. Too bright for my taste. So I inspected it and it is composed of a DC power supply unit that outputs 7V and a maximum of 3W. The LED unit is saying 1.5W.

My college memories tell me that this means that the LED is doing 214 mA at an equivalent of 32.7 ohms.

I am thinking to dim the LED using a potentiometer but I need to understand what resistance range should the right potentiometer have. If I choose a wide range the usable range within will be too short and I would have a hard time setting the desired light, choosing it too narrow it will not dim enough. I guess. Please correct any assumption that I make if needed.

Furthermore I remember that current vs light given by a LED is heavily non-linear. That means that if I manage to cut the current in half the light will not be half but probably 1/4 or even 1/5. If I'm not wrong again.

Considering that I would choose a potentiometer with a maximum resistance of the same equivalent resistance of the LED would that be a good choice for me?

I am aware this question might require editing and please provide comments that would help me pen the question better.

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    \$\begingroup\$ tape something on the lamp to restrict the light output \$\endgroup\$ – jsotola Jan 4 at 2:04
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    \$\begingroup\$ You understanding of the LED impedance is incorrect. Like zener diodes they have a threshold voltage and a roughly fixed resistance or voltage rise above the threshold voltage. A 1.5W 6V LED has a resistance around 1 Ohm and with another 4 to 5 Ohms added in series the voltage rises to 7V. So dropping the voltage by 0.5V with a Schottky Diode reduces the current by half. THis current is too much for most Pots except some. 10R WW pots \$\endgroup\$ – Sunnyskyguy EE75 Jan 4 at 2:12
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    \$\begingroup\$ Current is pretty linear with light brightness, but your eyes are logarithmic \$\endgroup\$ – Sunnyskyguy EE75 Jan 4 at 2:19
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    \$\begingroup\$ You can get logarithmic potentiometers. They're made for audio applications. \$\endgroup\$ – GDorn Jan 9 at 21:23
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Furthermore I remember that current vs light given by a LED is heavily non-linear. That means that if I manage to cut the current in half the light will not be half but probably 1/4 or even 1/5. If I'm not wrong again.

It is actually directly linear to the current (once you factor heat variation), but what is not linear is the eye response to light intensity.

The driver might not drive the LED in DC mode but perhaps in PWM mode. PWM mode allows to increase the power output of the LED, as it is flashing faster than the eye response, if you have a duty cycle of 20%, you can power the LED at 5 times higher for the same "load" (with abstraction to losses) and the eye will perceive a much brighter light as it only sees the "on" state.

If you want to DIM the IKEA LED, you need first to understand the driver schematics.

Adding a potentiometer before the driver won't work. Putting a potentiometer within the LED strip it will probably overheat as potentiometer are not designed to dissipate power. Also the driver will regulate the current, so increasing the impedance probably won't work.

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  • \$\begingroup\$ I tried to film the LED in 120fps and did not find signs of PWM mode and then I assumed (yeah, I know the word play with ass-u-me ;) ) that the LED is plain DC mode. I went on and bought a pot of 4W and 1KOhm (6$ equiv., I could reuse it etc) and put it in series with the actual LED. It dims the light nicely but, my eye at least, perceives it heavily non-linear. I am now letting it run for some time to see how much heat build-up the pot is getting. \$\endgroup\$ – Andrei Rinea Jan 5 at 1:20
  • \$\begingroup\$ In the end, after I settle on a certain resistance, I would add a simple resistor of enough power and a heat sink, if needed. \$\endgroup\$ – Andrei Rinea Jan 5 at 1:29
  • \$\begingroup\$ I guess I got lucky in this case for the LED driver is simple enough so I can hack this pathetic way of dimming the light.. \$\endgroup\$ – Andrei Rinea Jan 5 at 1:30
  • \$\begingroup\$ Well if it's work then it's great \$\endgroup\$ – Damien Jan 5 at 9:50
  • \$\begingroup\$ Thanks again! Yep, it works but I learned a lot placing this, honest, question. \$\endgroup\$ – Andrei Rinea Jan 5 at 22:14

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