Connect with us

MCU voltage regulator

Discussion in 'Microcontrollers, Programming and IoT' started by NMNeil, Feb 7, 2019.

  1. NMNeil

    NMNeil

    99
    9
    Oct 3, 2014
    Remember I'm a newbie for electronics and have only just started using PIC and Atmel microcontrollers in my simple learning projects, but a question.
    Why don't microcontrollers have a voltage regulator built into the chip?
    I have some old A3141 hall sensors that can take an input voltage of up to 24 volts because it has an internal voltage regulator, keeping the parts count down, But if I use a microcontroller in the same circuit I have to add the regulator and caps, so an MCU with a built in regulator would be very handy.
    There must be a valid reason why it's not so, it's just that I can't think what it could be.
     
  2. Harald Kapp

    Harald Kapp Moderator Moderator

    8,918
    1,760
    Nov 17, 2011
    2 reasons:
    1. The power dissipation of the voltage regulator will heat up the chip unnecessarily.
    2. The comparatively large structures required to handle the power within a voltage regulator are different from the small structures of the MCU itself.
    That said, this is not a rule set in stone. For example Intel's 4th generation microprocessors had integrated voltage regulators. However, those were removed in later versions.
     
  3. BobK

    BobK

    7,525
    1,604
    Jan 5, 2010
    Actually, many of the newer PICs do have a builtin regulator which takes the input voltage of 3.3 or 5V down to, I think 2.5V used internally. Any PIC that has a Vcap pin has this regulator.

    The reason for this is because the lower internal voltage allows higher speed at the same power but you need the higher supply voltage for I/O.

    There are also some PICs with an HV in the name that have a shunt regulator built in and can operate at up to 12V with only a series resistor to drop the extra voltage.

    In addition to the extra heat mentioned br Harald, having a more robust regulator would require a higher input voltage to overcome the dropout voltage, and this would waste power in a chip that is optimized for very low power.

    Bob
     
Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-