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I'm having problems with my design. I put the schematic here, maybe someone can find an error that I didn't find yet.

EDITED[I forgot the schematic, sorry...]

enter image description here

The microcontroller, dsPIC, has the fololwing current characteristics:
Typical ......................................................... 41 mA
Maximum ..................................................... 60 mA
Maximum current into VDD pin ...................... 300 mA

The FT232R' operating Supply Current is:
Icc1 Normal Operation ................................... 15 mA

Considering the Led current as:
I_led ........................................................... 10 mA

The total current would be at worse case, something aroud 325 mA, maybe a bit higher.

Below, the are 2 photos of my self-made PCB. The smd chip is the LM1117 SOT223. I put saliva on the Tab output pin and it instantaneously evaporated, I could hear and feel it, it was very hot. So, I think it was at much more than a 100°C. What could explain it, why it became so hot , and what can I do to fix it in another version ?

Image 1 Image 2

With my best regards.

(click for larger images)

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    \$\begingroup\$ I'm not seeing your schematic anywhere. \$\endgroup\$ – brhans Jul 12 '18 at 17:36
  • \$\begingroup\$ Lol at licking the board to see if it's hot \$\endgroup\$ – BeB00 Jul 12 '18 at 17:54
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    \$\begingroup\$ @BeB00 Sniffing the board is great way to check baud rate too :) \$\endgroup\$ – Maple Jul 12 '18 at 17:57
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    \$\begingroup\$ Please share the schematics.. Pictures of board files will be helpful but later \$\endgroup\$ – User323693 Jul 12 '18 at 18:55
  • \$\begingroup\$ @brhans Thank you, I edited the post and already added the schematics. Sorry, I really forgot it \$\endgroup\$ – Daniel Jul 13 '18 at 11:29
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Your regulator appears to be connected incorrectly.

enter image description here

The minus side of the two tantalum caps (presumably your ground) is connected to the "input" pin.

Putting reverse voltage on the chip is a good way to turn it permanently into a heater or NED (Noise Emitting Diode).


Your problem may be visible in the schematic, but it's probably more likely that you have a problem with the mapping of the schematic symbol pins to the footprint.

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    \$\begingroup\$ I agree he laid out the pads from top view but located on bottom side, so it is upside down. And he is using 2x as much solder as needed \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 13 '18 at 1:51
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    \$\begingroup\$ Yes, Spehro Pefhany was right, I inverrted the regulator and now it does not overheat anymore. But, is possible that the short period that the regulator became reversed could damage something ? For example, the output of LM1117 is about 1.64V and the FT232R appears to start overheating very slowly... \$\endgroup\$ – Daniel Jul 13 '18 at 14:30
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    \$\begingroup\$ If your input voltage was only 5V it seems unlikely the FT232R would be damaged (if it was connected correctly). \$\endgroup\$ – Spehro Pefhany Jul 13 '18 at 14:46
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I suggest looking at "10.1.1 Heatsink Requirements" in the datasheet. There is no heat dissipating copper on your PCB whatsoever. Also, check you power supply. Is it indeed 5V?

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    \$\begingroup\$ Second the power supply check. Your tantalum input cap looks burnt, although that could be from soldering - hard to say. \$\endgroup\$ – calcium3000 Jul 12 '18 at 17:56
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The voltage across the regulator is 5-3.3 or 1.7 volts. At 325 ma the power is 1.7*0.325 or 0.55W. The data sheet gives a temperature rise of 186°C per watt, so if you run 0.55 W the rise is 102°C. If room temperature is 25°C, this puts you about 127°C, so your part is probably working as it should.

enter image description here

That power (and heat) will always be there no matter which linear regulator you use, so the only way to decrease the local temperature is to add a heat sink. You could also use a switcher to increase the regulator efficiency.

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    \$\begingroup\$ 127 °C ought to begin thermal shutdown \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 12 '18 at 18:18
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    \$\begingroup\$ It will be close if there is really 325 ma. \$\endgroup\$ – John Birckhead Jul 12 '18 at 19:23
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    \$\begingroup\$ The termal resistance is 186°C/W, but for the TO package! For the SOT-223 package that the OP is using WITHOUT copper area it's about 130°C/W. With only about half a square inch of copper it would be less than 70°C/W. \$\endgroup\$ – Marco Zollinger Jul 12 '18 at 20:27
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EE's must use their mechanical and thermodynamic skills here.

\$R_{θJA}\$ = Junction-to-ambient thermal resistance 61.6 °C/W Assumes 1 sq in copper
but with almost none it's ~ 136 °C/W
\$R_{θJC}~\$= (top) Junction-to-case thermal resistance 42.5 °C/W

We can estimate the case to ambient Resistance (136-42.5)=94°C/W rise in temp or roughly 1 Watt for 120 °C with a voltage drop of correction) 5-3.3=1.7V or 590 mA/W out of 800 mA max or 125'C max

Next time read about thermal cooling requirements . I suggest 1~2 sq in of copper per Watt.

This is SOT-223 package with a thermal tab. enter image description here

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    \$\begingroup\$ But 5V - 3.3V is only 1.7V, which equals about 0.55W if his board really draws 325mA. So with your calculated 94°C/W, case temperature should only be about 52°C over ambient temperature. I don't think the microcontroller is even drawing 300mA, so that could imply there is a problem with his PCB that the regulator is becoming so hot. \$\endgroup\$ – Marco Zollinger Jul 12 '18 at 19:34
  • \$\begingroup\$ Yes Marco I agree \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 12 '18 at 21:51

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