Drawer light circuit burns out

Question

I made my own drawer light, but suspect I've done something wrong. After working for a few weeks (lights are only on for a few seconds per day), it stopped working.

Here's the gist of what I did:

• U1 is a buck converter with 3.3v out
• U2 is an IR proximity sensor
• It's not just 3 LED's, it's a 12V strip

Without the LED's connected, the indicator light on the proximity sensor works like normal. When the LED's are connected, it's always on, but the LED's turn on/off in response to proximity... at least for a while.

Is there some flaw with this design?

Answer 1

Yes, well there could be many things wrong.

That's no way to connect an SFH900 to a transistor or a power supply. But what you drew as SFH900 is not even an SFH900 but some sort of module.

So there is no need to believe anything else in the drawn schematics either. What is the 12V supply might not be a 12V and the specs of the LED strip is unknown too, it might not be simply LEDs in series with a resistor but some LEDs driven by an electronic driver module, such as CC buck regulator.

The sensor module does have schematics on the website, but the modules you get from these e-commerce sites do not necessarily match the item advertized. For example the sensor is advertized with buzzwords like "Good conductivity and corrosion resistance", which is of course, pure nonsense.

This is proven by the fact that for example the regulator you bought is advertized as a DC/DC step down buck converter. What a lie, it isn't, as AMS1117 is simply a linear regulator with low drop-out.

It is also unknown how much current your LED light strip takes and what is the 12V supply really.

For example the AMS1117 regulator can't drive out much current before it overheats due to the input being 12V. For example even a small load like 100mA will make it dissipate 0.87W and could raise the IC tenperature by 75C, so in 25C room temperature it could boil water. But there does not seem much load on 3.3V, so likely not an issue, just something to watch out.

Another thing is that the 12V is awfully near the absolute maximum limir of 15V for the AMS1117 input voltage. If the supply voltage ever shoots up beyond 15V, the regulator will damage.

The IR module output is via 10k resistance and via LED and 1k resistance. This means the base current is only about 2mA. It could drive 4A collector current, but at a VCE drop of 1.4V. So as the base current is so small, there might be huge losses on the transistor and it might dissipate too much and damage from the heat. But it depends on how much voltag and current your LED strip requires.

So when buying these modules from online e-commerce shops, you must be prepared to live with the fact that the modules may not be exactly what was advertized or you may not even get any documentation how they work and how to use them, so you may need to reverse-engineer how they work yourself.

Also try measuring voltages and currents in the circuit, it helps to pinpoint any problems.

Answer 2

The IR proximity sensor isn't designed to drive a Darlington transistor directly. You can try a 330 ohm resistor from the base to +3.3V to get more base current.

Assuming that the online schematic is correct, the IR proximity sensor doesn't have any hysteresis. Hysteresis is positive feedback to prevent oscillation around the set point. Without hysteresis, there will be a point where the output will oscillate and stress the transistor.

To properly add hysteresis, the circuit will need significant modifications. But you can hack some hysteresis as shown in the image. The value is a guess. If I knew the pot output voltage, I could guess better.

https://www.amazon.com/dp/B08DR1W3BK