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I am trying to use the STCS05DR constant current LED driver IC by STMicro in my buck converter design to indicate input power with a 1206 red LED.

The input power is a variable DC battery source (Milwaukee M18 battery) up to 20V. The reason I went this route is obviously to keep the brightness of the LED relatively unchanged with the gradually falling input voltage over time.

I have had no success using this IC and am wondering what am I missing to utilize this correctly.

The first time I plugged the circuit in I ended up releasing the magic smoke from the IC, the LED lit up very briefly beforehand.

Upon troubleshooting, I noticed that pin 6 which is GND was disconnected, which is strange because I imported the KiCad symbol and footprint directly from SnapEDA, therefore I was surprised to find GND unconnected by default. I scraped away some of the solder mask on the ground plane in front of the pin and solder jumped the pin the plane to connect it.

After replacing the burnt up chip with a new one and also populating a new LED, the second time I plugged the circuit in the LED fried but the IC stayed working (I think).

When measuring the open circuit voltage across the LED I got 19ish volts, which seems like a lot that would make sense why it would fry the LED. I have the current set in my use case to 20mA with a 5 ohm resistor across FB and GND, which according to the datasheet is the correct way to set the current limit.

PWM I have left floating which I wouldn't think would have an affect on anything.

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Do I still need a resistor on the output to drop the voltage? Wouldn't this defeat the purpose of the IC? I also don't have a CDRAIN capacitor strapped across DRAIN and FB resistor such as the datasheet application suggests.

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  • \$\begingroup\$ C4 on the feedback pin looks like a mistake? That's probably going to prevent the converter from adjusting the output voltage. Also, show your layout. \$\endgroup\$ Mar 13, 2023 at 16:44
  • \$\begingroup\$ That circuit is the main buck conversion circuit and is irrelevant from the led driver issue, I just included that portion for reference. However you are kind of correct, I put a large value polarized capacitor there when it was supposed to be just a small 1nF unpolarized cap per datasheet recommendation for the LM2596-ADJ application. \$\endgroup\$ Mar 13, 2023 at 16:59
  • \$\begingroup\$ Measuring a high open-circuit LED voltage is not unexpected. The IC is trying to push 20mA through an open circuit and raising the voltage to try to achieve that. If you replace the LED with a 100R resistor then you should expect to measure 2V for 20mA. \$\endgroup\$
    – brhans
    Mar 13, 2023 at 17:57
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    \$\begingroup\$ OK, one thing I hadn't checked earlier was the pinout on the symbol. Looks completely wrong to me. Ground is pin 6, not 2. PWM is pin 2, not 3. Disc is 8, not 4. And you have two pins numbered 4. The PCB, I think looks about right though. What's going on there? Could the new LED have been put in backwards? \$\endgroup\$
    – DiBosco
    Mar 13, 2023 at 19:50
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    \$\begingroup\$ "PWM I have left floating which I wouldn't think would have an affect on anything" and how did you come to this conclusion? PWM is on/off input, so if you want your LED to be ON you have to pull it high. Same goes for bypass capacitors, they are in the datasheet for a reason. Ignoring the datasheet and then asking why it doesn't work will not take you far. \$\endgroup\$
    – Maple
    Mar 13, 2023 at 20:49

1 Answer 1

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I figured out the issue, when I went to scrape away at the solder mask on the ground plane to connect the previously disconnected pin 6 to GND, I actually ended up connecting pin 7 to GND. This was a very stupid mistake I can't believe I hadn't noticed before hand.

After connecting pin 6 to GND and populating a new LED, the circuit works fine and the LED lights up appropriately when power is applied.

The PWM pin is not required to be tied to GND nor pulled up to VCC for this IC to work properly, nor is CDRAIN strictly required.

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  • \$\begingroup\$ "The PWM pin is not required to be tied to GND nor pulled up to VCC for this IC to work properly" where in datasheet does it say this? because if not in datasheet then you just got lucky, or simply did not encounter intermittent problems yet \$\endgroup\$
    – Maple
    Mar 16, 2023 at 20:16

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