L4931 3.3V Voltag3e regulator

Post date: Jan 12, 2016 3:48:58 AM

Features:

  • Very low dropout voltage (0.4 V)

  • Very low quiescent current

  • Typ. 50 µA in OFF mode, 600 µA in ON mode

  • Output current up to 250 mA • Logic controlled electronic shutdown

  • Output voltages: 2.7; 3.3; 3.5; 5; 8; 12 V

  • Automotive-grade product: 2.7 V, 3.3 V VOUT in SO-8 package only

  • Internal current and thermal limit • Only 2.2 µF for stability

  • Available in ± 1% (AB) or 2% (C) selection at 25 °C

  • Supply voltage rejection: 70 dB typ. for 5 V version

  • Temperature range: from -40 to 125 °C

Description:

Need a nice little 3.3V regulator? We rather like the very-low-dropout L4931 from ST! This little guy will help you get your 4-20V battery or wall adapter down to a nice clean 3.3V with 2% regulation. Perfect for just about all electronics! This is a TO-92 package version, with up to 250mA current capability, and has internal current limiting + thermal shut-down protection which makes it sturdy and pretty much indestructible - at least electronics-wise (we're pretty sure a hammer might work...)

This regulator has a very low 0.4V linear drop-out, way better than the 780X series' 2V. That means you must give it at least 3.7V to get a clean 3.3V out. This regulator is often used to get a 5V power supply or Lithium polymer/ion battery to a a clean 3.3V. There is a constant 'quiescent' current draw of 1mA (which increases up to 5mA as you draw 250mA) so it's good for portable and battery-powered projects

This regulator can provide up to 250 mA peak as long as you do not overheat the package. The higher your input voltage and output current, the more heat it will generate. Without an extra heatsink, you can burn off up to 0.6W. The wattage of your set up is = (InputVoltage - 3.3V) * AverageCurrentInAmps. E.g. a 9V battery and 0.2 Amp of average output current means the regulator is burning off (9 - 3.3)*0.2 = 1.14 Watts! That's way too much for the little package, it will overheat. Instead, reduce your average current to 0.1A (for 0.57W) or you could use a 5V power supply for (5-3.3)*0.2 = 0.34W, both of which would not require a heatsink.

Packaging and Pin:

Test Circuit: