Disclaimer: Super newbie and first time poster here.

The MCP1700 LDO regulator datasheet states on p3 that the Droptout voltage is 350mV for Vout > 2.5V @ I=250mA. I thought great I can power my ESP32 from an 18650 battery. 3.7V - .35V = 3.35V. Works out perfectly! Well it doesn't. The regulator's Vout kept dipping below 2.5V and the ESP32 kept browning out.

I couldn't understand this until I went back and read the datasheet, carefully this time, and saw this note 5: Dropout voltage is defined as the input to output differential at which the output voltage drops 2% below its measured value with a VR+ 1V differential applied.

My question: If I need a whole extra 1V for the regulator to properly work, then why tell me that the drop out is 350mV? What's the significance of this information? I came across the same thing with multiple regulators so I must be missing something.


  • \$\begingroup\$ Do you know the actual load current? Have you decoupled the input and output as required? \$\endgroup\$ Commented Apr 29, 2020 at 9:13
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    \$\begingroup\$ The 1 V extra is just a reference point to measure the dropout voltage. With a 3.3 V regulator, then 3.3 V + 1 V = 4.3 V is applied. Then the output voltage is measured, it will be around 3.3 V. Let's say it is 3.305 V. Now we lower the input voltage slowly until Vout has dropped 2%, that is when Vout = 3.305 V - 2% = 3.239 V. Then Vdropout = Vin - Vout. \$\endgroup\$ Commented Apr 29, 2020 at 9:14
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    \$\begingroup\$ The LDO output is stable when using only 1µF output capacitance. What input/output capacitors did you use? .. as @PeterSmith also asked. An oscillating output may have fried your ESP32. \$\endgroup\$
    – Huisman
    Commented Apr 29, 2020 at 9:14
  • 3
    \$\begingroup\$ The ESP draws current in a spikey manner- have you confirmed the actual input voltage dips under load with an oscilloscope? \$\endgroup\$ Commented Apr 29, 2020 at 9:32
  • \$\begingroup\$ Yep I used capacitors just like described in the datasheet. 1uF Cin and 1uF Cout. I also tried adding an electrolytic capacitor 47uF. \$\endgroup\$
    – MisterM
    Commented Apr 29, 2020 at 19:04

1 Answer 1


Alright so I think I found a clear explanation for that statement. According to this paper one test method is to:

measure the output voltage when VIN is 1 V above the nominal output voltage and then load the LDO with the specified test current. The input voltage is then reduced until VOUT drops by a specified number of millivolts. For example, 4.3 V is applied to a 3.3-V LDO. The output voltage is measured (assuming at 3.3 V), and then the input voltage is reduced until the output voltage measures 3.2 V. Dropout is defined as the input voltage minus the output voltage at this point. If VIN = 3.5 V when VOUT = 3.2 V, the LDO has a 300-mV dropout voltage. An LDO tested in this method will have a note similar to that shown below.

“Dropout voltage is defined as the differential voltage between VOUT and VIN when VOUT drops 100 mV below the value measured with VIN = VOUT + 1 V “

In the datasheet I was reading the manufacturer opted to use 2% or 66mv in the case of 3.3V as opposed to the 100mv in the example above.

As for my specific application, I replaced the regulator with one rated for 500mA. The new one is working perfectly. I will wait for my current probe to see using my scope how much exactly is the ESP spiking. It seems like such a quick spike over 250mA that my MM's Max function can't catch it.


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