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I'm trying to step down my source voltage of 4.5V to a 3.3V suitable for the MPR121. The chip has at the best settings a typical current of 393μA. The 3.3V has to be quite accurate as to not damage the keypad. When researching, I've come across two possible solutions that could achieve what I want to do:

  • Voltage Divider
    In this case, can the output be stable and accurate enough to be able to step down and voltage and if so, what value resistors would I use?

  • Voltage Regulator
    This is then another IC just to step down the voltage (footprint wise), any suggestions on which regulator I would use (linear, switching)?

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    \$\begingroup\$ You basically cannot use a fixed-resistor voltage divider for a variable load. A linear voltage regulator is effectively a variable-resistor voltage divider with a feedback circuit which adjusts the resistor to maintain the output voltage as load or input voltage varies. \$\endgroup\$ Jun 26 '17 at 22:25
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Since the voltage drop ratio is relatively low and the output current requirements also low, use a linear regulator. 3.3 V is a common voltage, so there are many fixed linear regulators available at that voltage.

These things have only three pins and are very simple to use. The pins are the input voltage, ground, and the output voltage. You will also need a 1 µF or so ceramic cap between input and ground, and between output and ground.

You are dropping (4.5 V) - (3.3 V) = 1.2 V. You have to be careful to choose a regulator that can work with that headroom. These are often called LDOs (Low DropOut).

The efficiency from the voltage drop will be 73%, plus a little more loss for the quiescient current. At only 400 µA output, the overall wasted power will be very small.

Also take a look at the quiescient current spec. For some linear regulators, that would add significantly to your 400 µA figure. Others work with only a few µA.

Take a look at the MCP1700 series, but there are many many others that would be fine too.

Some older LDOs are not "0 ESR output cap stable". Simply stay away from them. They were designed before the era of small and cheap ceramic capacitors that could do a few µF.

The MCP1700 series I mentioned is 0 ESR output stable, requires a maximum of 350 mV headroom, has only 4 µA quiescient current, and can deliver up to 250 mA. These are my "jellybean" LDOs, meaning that's what I use unless there is a good reason not to. I don't see one in this case.

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    \$\begingroup\$ Is a LDO really the most efficient option for this sort of load, or is it just "cheap and not too inefficient"? This kind of design strikes me as the reason all sorts of "turned off" devices are constantly wasting significant (total) power. I have little practical experience but my impression is that the answer should essentially always be "buck/boost". \$\endgroup\$ Jun 27 '17 at 2:48
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    \$\begingroup\$ @R.. Bucks are great for high load applications, but suck at low power ones. To increase efficiency, they're even paralleled with LDOs which take over in the close-to-no-load conditions when buck is shut down to conserve power. e2e.ti.com/blogs_/b/powerhouse/archive/2014/08/26/… \$\endgroup\$
    – Agent_L
    Jun 27 '17 at 8:47
  • \$\begingroup\$ @R..: We do engineering here, which means numbers always win over impressions. A MCP1700 at 400 uA out and 4 uA operating current will be 73% efficient. It will waste 500 uW. Yes, microWatts. The control electronics of a buck switcher would likely use up more than that, so it would lose even if the power train were 100% efficient. I can't even guess why you think a buck/boost makes sense here instead of a buck, but in any case, a LDO is the clear choice unless something very unusual is going on the OP hasn't told us about. \$\endgroup\$ Jun 27 '17 at 10:46
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Do not use a voltage divider. The problem here is that as the current in your load changes the divider voltage will change in proportion. Sometimes this can be worked around to a suitable level of accuracy by making the current through the divider resistors be twenty or forty times the average load current but this solution is very wasteful of energy.

For your solution an appropriate solution will be to use Low Drop Out linear voltage regulator. A normal linear voltage regulator will not work for this because the input to output voltage differential for those needs to be on the order of 3.5 volts or more. A properly selected LDO should work for a input to output differential at the 1.2V as needed by the OP.

A small switcher may also be used but to convert the voltage levels needed here at the very low current requirement the LDO regulator will be the easiest and lowest cost to implement.

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For this current (less than 1mA) and this low voltage drop (about a volt) your best choice for higher efficiency is linear regulator.

There are LDO's with low Input Quiescent Current like this for example - MCP1701A - only 5uA!

You can also make a shunt regulator with TL431 - a bit cheaper, but power consumption will be more (at least 1-2mA).

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    \$\begingroup\$ The TI ATL431 is much more suitable: ti.com/lit/ds/symlink/atl431.pdf ..This has much lower current than the TL431 and is ideal for a solution down to 100's of uA. \$\endgroup\$ Jun 26 '17 at 22:40
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    \$\begingroup\$ The MCP1701 is not 0 ESR output stable if I remember right. The MCP1700, however, is. It also has slightly less quiescient current, also both are well below the OP's output current, so that doesn't matter. \$\endgroup\$ Jun 26 '17 at 22:40

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