Dim a 158V DC Led

Question

I bought a led lamp and it's too bright. When I saw the converter, it's not a standard 12 or 24 VDC one. it's a 120-215 VDC; Picture attached. I didn't trust that. Borrowed a multimeter and checked, it's 158VDC. I could not find dimmers hopefully a smart one like this for this 158 VDC adapter.

I tried to use 12V adapter. No light.
I tried a wall dimmer. Expectedly, LED is flickering but no dimming.

How do I solve this? What do I need?

I have some basics in electronics and comfortable in soldering.

Answer 1

That looks like it's a constant current supply. It says 240 mA \$\pm\$ 5 %.

What it would do if that's what it is, is supply a constant 240 mA current to a series string of LEDs by varying the output voltage between 120 and 215 VDC. This is done to keep all of the LEDs at the same brightness and allow for some variation in the number of LEDs. The lamp you have might have something like 75 individual LEDs in it all connected in series where each one drops roughly 2.1 V giving you the 158 V reading. If this supply was used with a a lamp with 100 LEDs it would output 210 V at 240 mA, 50 LEDs would be 105 V. (These values are just an example of the concept and would depend on the voltage drop of the LEDs used)

Dimming it is probably not going to be possible, changing the input voltage will make no difference over its rated range of 175 - 265 V, and below that it will likely not work at all. Adding resistance in the output will make it adjust its output voltage to keep a constant 240 mA within the range it's rated for, outside of that range it's likely to not work and possibly cause damage to the supply or lamp. Trying to pulse width modulate the output would probably not work either, as the supply would be trying to vary the output voltage to account for what it would see as an alternating open circuit, again possibly damaging the equipment.

One possible way of dimming a constant current fed LED array would be to bypass some of the current through a shunt load.

For example, if the current is 240 mA and you wanted to reduce it by 100 mA you could add a load resistor across the output. The resistance would be: $$ R_{shunt} = \frac{V_{lamp}}{I_{shunt}} = \frac{158 V}{100mA} = 1580\Omega $$ So a standard value of 1500\$\Omega\$ would be close. The problem then is that the shunt would dissipate $$ P = \frac{158 V^2}{1500\Omega} = 16.64 W $$ So a pretty good amount of waste heat, you'd need a rather large power resistor on a heat sink. If you want it variable you'd probably need to make it an electronic shunt. With a fixed resistor you would be pretty much guessing at the right amount of current to shunt to get the brightness you want. And you have to make sure it's safe, 158 V is nothing to toy with.

One thing you might look into is using an optical filter.

Answer 2

The converter is supplying a fixed 240mA current to the LED strings. There doesn’t seem to be a way to modify that directly at the controller, at least based on what you’ve shown.

In case you were thinking about it, adding a series resistor to the LED output won’t work: the converter will compensate for it by increasing the output voltage.

Your wall dimmer probably didn’t work because it is a type for incandescent, which uses ‘leading edge’ control. That doesn’t work so well with LED.

What you can try is an LED-compatible dimmer, specifically, one that does ‘trailing edge’ dimming. This will chop the input waveform to reduce the average output power. Below is a photo of one that I use for dimming LED lights in my kitchen.

More about dimmer types here: https://www.lamps-on-line.com/leading-trailing-edge-led-dimmers