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enter image description here

  • THE PROBLEM:

I used this LDO voltage regulator (SOT223 package) with a 4.5 V input from 3 AAA batteries to provide a steady 3.3 V as output for all the other components. The voltage regulator, C7 and C8 capacitor all get really hot to the touch. There doesn't seem to be any output from the voltage regulator at all because none of the LEDs are lighting up.

The temperature of the voltage regulator from my calculations is about 109.5 degrees Celsius. The total current from all of the load is 760.5 mA and the voltage drop across the regulator is 1.2 V. With P=VI, it is about 0.912 watts. Thermal resistance of the regulator is about 120 degrees per watt thus 0.912x120-109.5 degrees Celsius. I cannot guarantee if the calculation for 760.5 mA is correct but in any case, 109.5 degree is still lower than 150 which brings me to the next section.

I hope I can get the bottom of this heat problem with some help.

  • POTENTIAL PROBLEMS:

1. It is still too hot and heat is not properly dissipated

enter image description here

I know in electronics if something is too hot to touch then it is too hot, but in the case of this voltage regulator if it gets to 109.5 degrees, does it drive up the impedance to the point it blocks all flow of current? These two LEDS are supposed to act as indicators if there is current flow or if the board is on but neither of them light up. In this case, will putting a heatsink help at all? It still won't solve the heat of C7 and C8, though.

2. PCB layout

enter image description here

My power trace is 0.5 mm (19.7 mils) which I do believe is standard for power traces but is it somehow too small? Perhaps there is something wrong with my PCB layout that I am not taking into consideration?

3. My power source

enter image description here

Highly unlikely but there could be something wrong with the batteries? I put 3 of them in series to get a combined 4.5 V as mentioned early. 1.2 V drop shouldn't be a lot right?

4. Soldering issue

enter image description here

I used a stencil and relatively new soldering paste. I used a hot plate to melt the solder. I used 220 degrees Celsius for the hot plate which is well above the solder paste's melting point of 183 degrees Celsius. Should I have soldered this at a different temperature?

  • ADDITIONAL PICTURES

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  • 4
    \$\begingroup\$ As an aside, I'd look at driving the LEDs directly from the battery rather than wasting power in the LDO by powering them from the 3V3 line. \$\endgroup\$
    – Finbarr
    Sep 1, 2023 at 10:41
  • 1
    \$\begingroup\$ "I know in electronics if something is too hot to touch then its too hot". Actually, that is often not the case. 140 degrees F is generally hot enough for one to be unwilling to keep one's skin in contact. Many electronics components will continue to operate within spec at much higher temperatures. For personal projects, you may want to use the "too hot to touch" rule of thumb. But for commercial, one may need to do a cost-benefit analysis of allowing a component to run hot. \$\endgroup\$ Sep 1, 2023 at 11:54
  • 4
    \$\begingroup\$ Those poor AAAs won't live long when loaded by over 0.7 A. We're talking about tens of minutes of runtime at most. Are you sure you don't want to rethink your power supply? (And this is before we consider that basically one of your AAAs is used purely to heat the linear regulator.) \$\endgroup\$
    – TooTea
    Sep 1, 2023 at 13:51
  • 1
    \$\begingroup\$ You said you flowed the whole board using a hot plate? Or did you use the hot plate to heat the board up to near-soldering temperature, and then individually solder each pin with an iron? I see what I think are iron burn marks on your LED display. In any case, do any parts have pins that are not accessible with an iron? This looks like an easy board to solder everything by hand and you should be able to get much cleaner results than what's shown. Consider getting an old scrap PCB (dead mobo?) and practice soldering on that until you can get cleaner joints. \$\endgroup\$
    – SSilk
    Sep 2, 2023 at 13:29
  • 1
    \$\begingroup\$ And for joints that are already flowed and messy or shorted to nearby pins, a flux pen and good iron are your friends. Dab a bit of flux on the messy/shorted pins, touch them with a hot iron and they'll usually magically clean up by themselves and un-short in the process. If there's so much excess solder that they remain shorted, then solder wick + flux should be used to remove almost all of the solder, and then re-solder the joint from scratch. Avoid desoldering pumps. They can make a mess, splattering tiny tendrils or beads of solder across the PCB. \$\endgroup\$
    – SSilk
    Sep 2, 2023 at 13:32

3 Answers 3

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Looks like C1 is shorted, which is shorting out your 3.3V rail and causing the regulator to current-limit.

enter image description here

Other pins on the chip appear to be shorted as well. Please inspect the board carefully for possible additional issues:

enter image description here

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  • \$\begingroup\$ I just looked at some of the previous pictures of the board after C1 was soldered and it most probably shorted during the soldering process. But I cant help wonder if this could be the product of poor PCB layout. Any thoughts on if was shorted by excess current as a result? \$\endgroup\$ Sep 1, 2023 at 6:27
  • 7
    \$\begingroup\$ Just looks like bad soldering to me. I don't know what process you used so can't comment on what was bad. Layout looks okay at first glance. Shorts are generally not caused by excess current, but burned parts, vaporized traces and so on are. \$\endgroup\$ Sep 1, 2023 at 6:30
  • 3
    \$\begingroup\$ This is the correct answer. \$\endgroup\$ Sep 1, 2023 at 16:53
  • \$\begingroup\$ @SpehroPefhany OP states he used a hot plate to reflow the whole board. Don't know if that fact was included initially or added later. \$\endgroup\$
    – SSilk
    Sep 2, 2023 at 13:26
  • 2
    \$\begingroup\$ Too much paste, too little flux, too little heat, or any combination of the three I would say. \$\endgroup\$ Sep 4, 2023 at 6:10
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@Spehro Pefhany probably found the problem.

I just wanted to complete on your thermal calculations. You say (I haven't checked the datasheet myself) :

  • R_th = 120°C/W
  • P = 0.912W

However, in your calculations, you forgot to taken into account the ambient temperature : T_silicium = T_ambient + R_th * P = 25°C + 120*0.912 = 134.4°C

NB : I just did a guess about the ambient temperature, you can adjust according to the real one if you want.

Anyway, if you are testing some else than in the freezer or in Antarctica, you are very close to the thermal limit. In addition, your layout isn't optimized at all for thermal dissipation (no copper plane on the components side, single via to ground plane, thin tracks), so if (as very often) the thermal resistance is given for a standard layout/PCB (or even an optimized one), your thermal resistance is probably worse (ie higher) than the one specified in the datasheet.

So it is quite likely that you are running your regulator at or even beyond it's specified max temperature (don't forget, it's the temperature inside the IC that counts, which is already quite a bit hotter than the outside of the package.

I think @Sphehro's explanation is more likely, but in my opinion you are pushing your LDO beyond what's reasonable. If you can't redo the board with another (switching?) regulator, I would consider adding a fan if your power budget allows it, or at least a passive heatsink

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In addition to what the others said, I'll use another LDO (LDL1117) as an example.

enter image description here

Thermal resistance junction to ambient is very high, this is the value to use for a layout like yours. However thermal resistance junction to case, which means to the tab and middle pin, is quite low. This means with a copper plane as a heat sink and several vias, this package can dissipate surprisingly high power, depending on copper area connected to the tab and center pin. Here's an example:

enter image description here

So if you use a layout like this, with enough thermal vias:

enter image description here

The resulting thermal resistance allows one watt of dissipation.

You can salvage the board by soldering some copper wire or foil to the tab to act as a heat sink. Not pretty but effective.

Note your LDO has two pinout options, only one will work on your board. The part number on the schematic seems to be the correct one.

enter image description here

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5
  • \$\begingroup\$ Yes! As-is, OPs layout is abysmal in thermal terms. \$\endgroup\$ Sep 2, 2023 at 0:16
  • \$\begingroup\$ If the OP would like to make the board work as is, a possibility would be to solder a Schottky diode in line with the batteries, to drop some of the voltage and reduce the thermal load on the LDO. This would in effect increase the dropout voltage of the voltage regulation circuit, so the board would lose voltage regulation earlier, while there was still some life left in the batteries. \$\endgroup\$
    – swineone
    Sep 2, 2023 at 4:46
  • \$\begingroup\$ I'd take a bit of 1mm² copper wire, bend it into a T, and solder that on the LDO tab. Makeshift heat sink! \$\endgroup\$
    – bobflux
    Sep 2, 2023 at 10:01
  • \$\begingroup\$ Based of figure 27, it seems to have a dedicated polygon as a ground plane for the voltage regulator. Im guessing I should do the same for a new design. Will this perhaps help with my capacitors (C7 and C8) overheating? Maybe the capacitors are heating up because of the heat that spreads out from the voltage regulator? Also thank you for this suggestion. Ill be sure leave out more space for my voltage regulator. \$\endgroup\$ Sep 4, 2023 at 3:07
  • 2
    \$\begingroup\$ Both the tab and center pin help conduct heat so it makes sense to use a larger pad for the center pin too. Ceramic capacitors have very low ESR so then normally don't heat up unless ripple current is very high which is not the case here. Maybe your caps are cracked (internal short) or heated up by the LDO... \$\endgroup\$
    – bobflux
    Sep 4, 2023 at 6:17

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