ESP8266 reduce power consumption caused by extra ICs/components on board during deep sleep

Question

I need to reduce the power consumed by a circuit based on ESP8266 and extra circuitry (I2C sensors, light resistor based sensor, etc.). By now when I set ESP8266 in deep sleep I still have a relevant power consumption caused by extra components. Not all ICs have a 'deep sleep' mode that can be set to reduce power, so I tried to disconnect VCC line that power all extra components on the breadboard and the current drop enough to allow running the entire circuit by battery.

So my idea is to manage the sensors VCC line with a mosfet driven by the ESP8266, something like:

• on ESP8266 power up (Arduino setup() method), as first thing I switch ON the mosfet to power all sensors
• small pause to allow all ICs to init and stabilize
• perform all operation (read sensor, calculation, etc.)
• turn mosfet OFF
• turn ESP8266 to deepsleep

So I would ask:

1. is a feasible way or can led to any kind of issue?
2. if I use an N-channel mosfet the result is a low-side-switch, that leave GND floating. This seems bad to me, correct?
3. if I use a high-side-switch (P-channel mosfet) is a pull-up resistor enough to assure mosfet is full-close when micro going to deep sleep?

The entire circuit VCC is 3.3v

Answer 1

Removing power from the sensors is the best way to reduce power consumption when you don't need them. You can do a low-side switch because your sensors don't use a lot of current so that ground will be so close to real ground that it won't matter. The pins of the sensors however may be at VCC level so you have to watch where they connect, for example the I2C lines to your MCU, the inputs there have to be floating. On the other hand if you use the high-side switch then of course your I2C lines need to be low or otherwise the pull-up resistor on those will send power into your sensors (something to keep in mind if you are really trying to minimize power consumption). When you talk about a mosfet and you say "full-close" that means it is "ON" because you have to think of it as a switch in a circuit, for example think of a knife switch. It's not a valve in a water line so that can be confusing, but to answer your question a pull-up on the p-channel is enough to keep it completely off (aka open) (not accounting for leakage, check datasheet when in doubt).