I have used a AMS1117-3.3 voltage regulator to get 3.3V from a 12V power supply. The output capacitor used is 47uf (electrolytic) and there is a 22uf cap connected at the input. The device's heatsink is soldered to PCB copper.

The datasheet says the device can handle 1A. I connected a resistive load which draws a current of around .5A and the voltage drops to almost half. At a load current of a round 270mA however, the voltage seem to stay at 3.3V.

IS this the best one can expect from these regulators given that they were ordered from aliexpress and we super cheap?

  • 2
    \$\begingroup\$ If they come from AliExpress, they're commonly rejects or outright fakes. Anything goes, really. 🤷‍♀️ \$\endgroup\$
    – marcelm
    Commented Feb 5, 2023 at 13:43
  • \$\begingroup\$ well, after it was fed with 5V, it handles 500mA with no issues. Probably 1A too. \$\endgroup\$
    – peter
    Commented Feb 5, 2023 at 18:02

1 Answer 1


Linear regulators dissipate power proportional to the voltage drop from input to output and the current being supplied.

From 12V to 3.3V is a drop of 8.7V, which means at 0.5A you're dissipating \$8.7 \times 0.5 = 4.35W\$ of power as heat, which is way too much. The regulator is thermally overloaded and is probably oscillating in and out of thermal shutdown, which is why you're reading a much lower voltage.

The thermal resistance from junction to air, \$T_{JA}\$, is 80°C/W on the TO-252 package, 90°C/W on the SOT-223 package, and 160°C/W on the SO-8 package. The maximum junction temperature is 125°C. That means you're thermally limited to \$\frac {125 - 25} {80} = 1.25W\$ dissipation at 25°C ambient on the TO-252 package, and less on the other packages. Even at 270mA you'd be pushing the TO-252 into a temperature rise of 190°C above ambient, which is far beyond its maximum spec. It will eventually burn up.

Linear regulators are inefficient, and should only really be used for small voltage drops, e.g. from 5V to 3.3V, or when you only need a little bit of current.

For larger voltage drops you should consider a buck converter instead of a linear regulator. Linear regulators are fine with larger voltage drops if you only need a small amount of current, or if you only have a small voltage drop (e.g. 5V to 3.3V), but for both a large drop and large current at the same time you really want a switching regulator.

  • \$\begingroup\$ I put it through a 7805, so the input to AMS1117 is now 5V. It handles 500mA with very small drop now. I cannot check above 500mA as for some reason, the 7805 I use can only output 500mA max without a large drop, probably due to the same reason? The package of 7805 however is TO220. Thanks for the help. \$\endgroup\$
    – peter
    Commented Feb 5, 2023 at 7:03
  • 4
    \$\begingroup\$ @peter As you have a TO220 package on your 7805, you can bolt it to a proper heatsink to improve the output current \$\endgroup\$
    – Neil_UK
    Commented Feb 5, 2023 at 9:54
  • \$\begingroup\$ @peter [for completeness] aside from only needing a few mA where a 3-terminal linear regulator is much easier than even the simplest switching converter, linear regs generate far less noise. So you normally keep them for local (board-level) power with a decent smoothing filter: even a lot of digital devices will be unhappy if powered straight off a switching converter (sufficient passive filtering is perfectly possible, but the linear reg effectively operates as an active filter with defined loss to bring the voltage down), and anything analogue will probably be swamped by switching noise. \$\endgroup\$
    – 2e0byo
    Commented Feb 5, 2023 at 19:00
  • 2
    \$\begingroup\$ "even a lot of digital devices will be unhappy if powered straight off a switching converter" - citation needed on that one. Tons of stuff runs straight off a switching regulator; even really high speed stuff. As long as there's a low impedance power delivery path, with proper decoupling, it's fine. \$\endgroup\$
    – Polynomial
    Commented Feb 5, 2023 at 19:08

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