It just smoked.
I’ve been operating and testing this PCB with a bench top power supply (13.8 VDC), and all has been well. When I hooked it up to my 18VDC battery, (20.1VDC), it smoked this 5.0 VDC regulator. According to THE DATA SHEET this regulator is made for input voltages from 2.7 to 30 VDC, so I thought I was fine. I see now the data sheet says,
Power dissipation depends on input voltage and load conditions. Power dissipation (PDISS) is equal to the product of the output current and the voltage drop across the output pass element, as shown in Equation 1:
PDISS = (VIN – VOUT) × IOUT
(1)
Ok, so I’ve got a 25 volt drop, and the current, well, I’m guess it’s less than 50ma. It’s just the Atmega328 chip, and a motor controller IC, which I think draws 5-8ma, and an LED drawing about 12ma. I couldn’t really find anything else in the data sheet that tell you how much power dissipation it can handle. It looks like it’s all based on thermal junction temp.
There’s also another identical regulator onboard, only it puts out 3.3VDC for an RF chip. It was not fried. It too is powered from the same 20VDC, but it lost power once the trace at the 5.0v regulator burnt up.
I’ll include a photo of the PCB trace/copper, and the board. If anyone can shed some light on this. Do I need 1) a different regulator? 2) different trace/copper/heat dissipation 3)are my high voltage traces too close to other? I’ve got 10mil clearance on all traces.