Using a buck converter with potentiometer for my battery-operated fairylights -- Blinking and no read-out on voltmeter

Question

I'm an extreme newbie when it comes to electronics, and decided to just dive right in.

I wired up my buck converter exactly like how it's done here: https://www.youtube.com/watch?v=fmdxSRaVPVg

Without a load, everything works fine. I was able to use the trimmer to set a max output of 6V, and the potentiometer (10k) allows me to vary output voltages from practically 0 to 6V. I'm using a 12V 2A DC wallwart as my input.

Now, I have some cheap fairy lights (The ones that are basically just parallel LEDs.) When I got them they were powered by 2x CR2302 batteries (in series.)

I was thinking everything would work out if I just connected my now liberated fairylight string to the output of the buck converter, as it's now outputing a clean 6V.

When I connect the fairylights to the converter, the lights will blink non-stop, and my buck converter stops displaying anything on it's built-in voltmeter. If I dial the potentiometer all the way down, I get a blinking red light on the buck converter for the 'IN'. The same issue occurs even when the potentiometer is removed.

Any ideas what's causing this and how to fix it?

MORE INFO:

1. I haven't had much luck finding the exact model of the buck converter, since whenever I find it, it's only labeled as 'LM2596ADJ'

However, I do know the following specs:

• Input Voltage 4.5~ 40V.
• Output Voltage range 1.25V ~ 37V
• Output Current max 3.0A (recommended 2.0A)

2. For the fairylights themselves, I really have no idea, but after using the multimeter, whenever I try to measure the current, it jumps around from 100mA to around 50mA.

EDIT, unfortunately, I don't know how to make schematics/diagrams, so hopefully some embedded photos will do.

Here's the completed board. Don't mind the poor solder joints, please.

Here's the potentiometer at the top. It's 10k.

How the pot-meter is joined to the board:

Here are some images/videos of what I've done: https://i.sstatic.net/7COEs.jpg

Answer 1

Your problem is pretty simple.

The fairy lights depended on the internal resistance of the CR2302 cells to limit the current. They only drew as much current as the cells could deliver - which is some few milliamperes.

Your power supply can deliver amperes of current at 6V. You are lucky that the power supply shuts itself off (that's the blinking,) else the LEDs would be destroyed. The LEDs get current at 6V, and try to draw as much current as the power supply can deliver. The power supply detects a short circuit and shuts off.

You need a series resistor for your fairy lights.

Assuming 3 milliamperes (about all you can reasonably draw from a CR2302) and that the lights are 3V LEDs all in parallel, you would need a resistor of \$R= \frac{V_{supply} - V_{forward}}{I_{forward}} = \frac{6V -3V}{3mA} = \frac{3V}{0.003A}= 1000 ohms\$.

1000 ohms is a (fairly) safe starting place. If it is too dim, use a smaller value resistor. If they don't light at all, try a 500 ohm resistor.