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I have an Adafruit Huzzah Feather with ESP8266 board, but I think my question will apply to most all boards (as most all have an onboard regulator). Most regulators have an Enable pin (EN) as does my Adafruit board. Apparently, I can shut down the regulator by pulling this pin low. I have a specific application where an external device will provide a high signal to me when I want the device to turn on, otherwise, I want it off at all times. So, what I have is a 1k resistor pulling the EN pin low, to ground. I tried other values 10-20k but they did not pull enough low to shut down the regulator. For example, 20k pulled me down from 3.3V to 2.6V but the regulator stayed on. On the other hand, 1k pulls me down to .3V on the EN pin and this shuts it down.

Then, when the external device gives me the high signal, which I have also tied to the EN pin, this will pull it back up and turn on the regulator.

This seems to be working just fine, but I wanted to ask if this sounds reasonable (I know this is a general question but I want to know if there is a better way). A specific, related question would be am I wasting too much power by using a 1k resistor when powered up or pulled high since some of the supply current will be dumped via the 1k resistor to ground.

UPDATE: In looking at the schematic on the Adafruit site, it appears that they have pulled EN to the input via a 10k resistor. So, I am fighting that which is probably why values over 10k do not pull it down enough to turn off. If I was doing the board design, I would have reversed this and pulled EN low by default but this is not the case when using the off the shelf board. So, I guess any value below 10k should work, like 4.7k? I guess I could also remove their pull up if I wanted to get really crazy. Thoughts on all of this?

Thanks!

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A better solution would be to use a PNP transistor to control the EN pin. A PNP is ON when the input to it is LOW (or lower than the collector minus 0.7V, that is).

Take this circuit, for instance:

schematic

simulate this circuit – Schematic created using CircuitLab

When the GPIO is HIGH the transistor Q1 is turned off (GPIO >= VCC-0.7). Also if the GPIO pin is disconnected it is turned off because R3 pulls it up to VCC.

Set the GPIO LOW and Q1 turns on since the base is now lower than VCC-0.7V. With Q1 turned on it connects the EN pin to ground, so it becomes LOW.

If you replace the PNP with an NPN (there's more chance you have one in your bits box) you can basically invert the logic:

schematic

simulate this circuit

Now when the GPIO is LOW or pulled down by R3 Q1 is off and EN is pulled HIGH. Set the GPIO HIGH and Q1 turns on and connects EN to ground setting it LOW.

In both these scenarios the EN pin defaults to HIGH. If you want it to default to LOW until you activate the GPIO pin (i.e., during bootloader operation when the GPIO is in INPUT mode) you can reverse R3 so it becomes a pull-down (for the PNP) or a pull-up (for the NPN).

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