I built a circuit powering an ESP8266 MCU from an MCP1811 LDO so I can use a 5V logic level. The MCP puts out 3.3V as desired by the ESP. The circuit looks as in the data sheet example: enter image description here

Assuming the ESP is constantly in deep sleep mode, it is supposed to use around 20uA. When I actually power the whole circuit with 5V through the MCP and monitor the current with my power analyzer device, my actual current draw is 18uA, which is great! I verified that the ESP is in proper working condition as well. Now I hook up a 3.3V supply directly to the VOUT of the MCP and let the VIN floating, by not hooking anything up to it, I measure a current draw of around 400uA! I am wondering how this happens, I would have assumed that the current draw would be the same and the MCP would just not operate... Any solution is appreciated!

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    \$\begingroup\$ You might be able to make that go away by adding a 1M resistor between Vin/nSHDN and Vout. \$\endgroup\$ Commented Sep 22, 2020 at 22:50

1 Answer 1


Now I hook up a 3.3V supply directly to the VOUT of the MCP and let the VIN floating

You are misusing the regulator outside of its intended regime of operation; the answer is not to do that; not only does it yield undesired results, it may over time cause damage.

One specific specification you are violating is the statement in the Absolute Maximum Ratings that the voltage on any pin may not exceed that on Vin by more than 0.3 volts. Naturally, with nothing on Vin this is violated.

Your diagram shows the /shutdown input wired to Vin, so if you have the "A" rather than "B" version of the chip, you have possibly also activated the regulator's "discharge transistor" which shunts the output pin to ground with the intent of draining downstream capacitors. I'd actually have expected a yet higher load from this, so maybe that isn't happening and you're only seeing a violation of the first concern.

Generally speaking, outputs are for output and should not be driven but something else. There do exist regulators which do allow backdriving the output, but it may take some searching to find one meeting your other needs. If may be simpler if you just use a 5v supply to the input on the bench; you may have headroom for diode voting on the input if you need to support multiple inputs.

  • \$\begingroup\$ Thanks for the answer, I actually verified with both versions of the LDO (A & B). Same result. I was just seeing another circuit featuring the MIC5353 instead of the MCP and when I tried powering the circuit as described above, I actually did not have that additional current draw but rather just the expected 20uA. For me with little experience the data sheets looked similar and I could not really figure out what to look for to pick the right part. Thanks a lot for answering! \$\endgroup\$
    – tlubes
    Commented Sep 24, 2020 at 0:06

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