1
\$\begingroup\$

I need help.

I would like to dim LEDS of a Christmas tree with a potentiometer. Here is my electric diagram. (Please note I'm a beginner in electronics.)

enter image description here

I don't know how to choose a good value for the potentiometer. Is my diagram correct?

I noticed that by adding a resistance in series on the postive side of the out powering cable I manage to lower the brightness of the tree, the goal of the project is to be able to adjust the brightness with the potentiometer.

Here is a picture of the power supply.

enter image description here

The Christmas tree 31V P=3.6W

enter image description here

\$\endgroup\$
12
  • \$\begingroup\$ I think this isn't going to work properly, the power supply looks to me like it outputs a constant current. If you place a resistor in series with the output, the power supply will simply output a higher voltage to make the current the same again and your LEDs will burn as bright as they used to. If you do manage to dim the LEDs then you're hitting the maximum output voltage of the power adapter and in the long run, that might damage it because they're not designed to operate like that. \$\endgroup\$
    – Bimpelrekkie
    Commented Nov 27, 2020 at 16:57
  • \$\begingroup\$ @Bimpelrekkie Thank you for your message, Do you have any idea how I could reduce the brightness LEDs with an other safe way? \$\endgroup\$
    – Cypf
    Commented Nov 27, 2020 at 17:00
  • 1
    \$\begingroup\$ There is no "safe" way. These products are simply not designed to be dimmed. To make them dimmable a more complex power supply is needed. The one you show is a "generic and cheap" one that is used for many LED lights. \$\endgroup\$
    – Bimpelrekkie
    Commented Nov 27, 2020 at 17:04
  • \$\begingroup\$ @Bimpelrekkie What kind of power supply would it take to make these LEDs dimmable ? \$\endgroup\$
    – Cypf
    Commented Nov 27, 2020 at 17:24
  • \$\begingroup\$ would it work with pwm system with arduino with the original power supply or not at all? \$\endgroup\$
    – Cypf
    Commented Nov 27, 2020 at 17:26

1 Answer 1

Reset to default
1
\$\begingroup\$

Going out on a limb here because I don't have experience with these constant current contraptions, so please let me know if this is hare-brained or not:

The supply is rated at 3.6W, 31V, 116mA.

So it seems likely the circuit is a bunch of parallel LED strips each running drawing 5mA or whatever. enter image description here

So why couldn’t you just put a variable resistor across it to siphon off some of the current? enter image description here

True, some drain occurs even at max setting. Make the 470 smaller if not bright enough.

\$\endgroup\$
9
  • \$\begingroup\$ Hello, Thanks for your reply and your nice schematics ! I agree with you, I think they are parallel led strips. Yes indeed, i would like to make an variable resistor between power supply out and LEDS. But i can't put an potentiometer directly because it would be instantly damaged isn't it ? That's why i would like to do this with mosfet and pot if it's possible, i hope to be clear. \$\endgroup\$
    – Cypf
    Commented Nov 27, 2020 at 20:56
  • \$\begingroup\$ The fixed resistor is acting as a current limiter for the potentiometer. With values shown, the pot would have to dissipate as much as 500mW which, granted, is a little much for standard pots. Making the fixed resistor larger will spare the pot but limit the maximum brightness. You could optionally get full brightness back with a switch that disengages the resistor bleed. In any case, perhaps it’d be best to test this scheme with a fixed resistor of about 1K across the supply to verify it actually does work to dim the tree. \$\endgroup\$
    – td127
    Commented Nov 27, 2020 at 21:24
  • \$\begingroup\$ This might work but it will require some tweaking to make it work properly. I doubt that someone with limited experience in electronics can do this without burning up some components. Also, you would be "burning off" the excess power so when you dim the LEDs, the excess power is simply turned into heat. Since the total power is about 3.6 W it isn't that much but will require some attention. \$\endgroup\$
    – Bimpelrekkie
    Commented Nov 27, 2020 at 21:36
  • \$\begingroup\$ Agreed, it is certainly not an "eco-friendly" design in that the same power is just dissipated elsewhere. If you can find a supply with a variable current drive that would certainly be preferable! If not, maybe skip the problematic potentiometer and use a rotary switch to select a few different dimness settings by switching in different bleed resistors. Those resistors, as well as the fixed resistor in the drawn circuit should be power resistors, not little 1/4 Watt things. \$\endgroup\$
    – td127
    Commented Nov 28, 2020 at 0:09
  • \$\begingroup\$ @td127 Thanks you so much for your answer!I will try with a 1 K resistor and see how it looks like.What mean an power resistor? Do you have an example please? For the moment I only have the littles resistor. I have a question, If I would like to keep the actual power supply and have the possibility to variate the brightness thanks to a potentiometer. Would it be possible to create a circuit something like my first schematic to do that? And put the mosfet like a variable resistor so it will dissipate the excess heat? \$\endgroup\$
    – Cypf
    Commented Nov 29, 2020 at 0:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.