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I am working on a circuit design that has multiple IC's on it. My MCU requires 2.5-3.3 VDC for it's power. Another IC that I am using requires 5V for it's logic power. So what is the cheapest and simplest solution for powering both of these devices. I figure my options are:

A) Use two voltage regulators a 3.3 and a 5V. B) Use just a 5V regulator and used diodes or something to drop the voltage down the an acceptable level for the PIC. C) Maybe there is a dual voltage regulator out there? D) Something else?

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5 Answers 5

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Microchip has a document called 3V Tips ‘n Tricks. It has some options you can chose that will help you.

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  • \$\begingroup\$ Looks like the link is dead. Try here: digikey.com/en/datasheets/… \$\endgroup\$
    – FredFury
    Commented Apr 28, 2020 at 19:55
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    \$\begingroup\$ @FredFury, it works for me. \$\endgroup\$ Commented Jul 9, 2020 at 2:29
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What you're asking for is pretty common since there are a lot of mixed-voltage IC applications out there with the same constraints (especially the two voltages you listed since most ICs operate off either of those two). Here are two options for you:

  • MIC5211-LXYM6 from Micrel should do what you want. It's a dual-output, +3.3V/+5V linear regulator but only can output 80mA.

  • TPS767D301 from Texas Instruments if you require more power than that. It has a +3.3V fixed output and an adjustable output which will go up to +5.5V. It can output up to 1A (per supply) and should be able to source most projects that I could think of.

If you require more power than either of these two, it would make sense to go to two separate supplies.

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There are plenty of dual regs, but it is often cheaper to use two seperate ones unless you have severe space constraints, as there are many more parts to choose from in single-output versions.

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  • \$\begingroup\$ Quite true, you often pay for integration. However some highly integrated parts are cheaper if you're using every single feature to it's full potential, e.g. door-zone ICs \$\endgroup\$
    – Nick T
    Commented Dec 2, 2010 at 19:01
  • \$\begingroup\$ what are door-zone ICs? \$\endgroup\$
    – cksa361
    Commented Dec 2, 2010 at 23:07
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Dual LDOs (even triple) are very common. Here is what Micrel offers, though ST, National, and Linear should have additional options.

If you're concerned about efficiency for battery life or heat dissipation's sake, it is also fairly common to use a SMPS to step some input voltage rail to 5 V, then use a LDO from there to 3.3 V. Providing standby power to the 3.3 V rail is somewhat difficult then...not knowing much else about your design I'll stop guessing though.

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You could use a 5v regulator and run a 3v from there by adding a resistor to the vout of the regulator. Alternatively, you could make it adjustable by using a potentiometer.

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    \$\begingroup\$ Among other things, this would be totally unregulated and is a bad idea. \$\endgroup\$
    – Nick T
    Commented Dec 2, 2010 at 17:44
  • \$\begingroup\$ but if the 5v is regulated, wouldn't the 3v be regulated since its on the vout? Others suggested this to me in questions I had asked in the past \$\endgroup\$ Commented Dec 2, 2010 at 18:21
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    \$\begingroup\$ In fact there is something similar in tip #13 of this document by microchip: docs.google.com/viewer?url=http://www.microchip.com/stellent/… \$\endgroup\$ Commented Dec 2, 2010 at 18:27
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    \$\begingroup\$ The 5V is regulated, but dropping it down to 3v with a resistor isn't. As you sink more current, your 3v will sag. I discovered this while trying to power an XBee pro using a voltage divider. The voltage sagged down to ~2.some V from 3.3ish due to the large current spikes from transmitting. \$\endgroup\$
    – pfyon
    Commented Dec 2, 2010 at 18:29
  • \$\begingroup\$ I wondered if putting a cap on the 3v side of the resistor would provide the necessary current during spikes to keep the voltage from sagging. Never tested it though. \$\endgroup\$
    – pfyon
    Commented Dec 2, 2010 at 18:40

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