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I would like to turn off and on a circuit by another circuit. I believe that could be easily achieved by a transistor but when there is a voltage regulator and an ESP8266 with wide range of input current I'm lost.

I have the ESP8266 programmed to send a notification when it's turned on. The V1 supply has unfortunately too low current to power the ESP so I have to power it by the V2 (4xAA 1.5V batteries should be probably enough to feed the voltage regulator 3.3V + 1.1V it's dropout and some transistor dropout)

The circuit below is my amateurish attempt that can be probably all wrong.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ My advice is using a voltage regulator with a shutdown input. There are plenty of those available. Simpler, cheaper, known to work. \$\endgroup\$ – Janka Oct 23 '18 at 15:47
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    \$\begingroup\$ You are going to have a hard time turning on that transistor using high-side switching. The reason for this is that it will be difficult to get the gate voltage high enough above the source voltage. \$\endgroup\$ – evildemonic Oct 23 '18 at 16:00
  • \$\begingroup\$ A high side switch typically needs to be a P-FET or a PNP. To use an "N" device you need a voltage booster. Something like a Richtek USB power switch chip has that built in, though there are better alternatives. Since you need a regulator anyway, picking one with a shutdown input and low current in shutdown mode would be a lot simpler. \$\endgroup\$ – Chris Stratton Oct 23 '18 at 16:07
  • \$\begingroup\$ Follow Janka's advice. \$\endgroup\$ – StainlessSteelRat Oct 23 '18 at 16:14
  • \$\begingroup\$ Would you like to turn your ESP "on or off" remotely ? Because what you are describing sounds kind of like a "sensor" when someone opens a switch, in practical use for example when somebody opens a door. Then it would be much easier to define one pin as an input pin, so that when the loop ( switch ) is closed, you get a positive or negative input, and therefore the notification. \$\endgroup\$ – Bruce John Speedster Oct 23 '18 at 16:34
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As suggested in one of the comments, there is a simpler and more efficient way to do this using a DC-DC converter rather than a linear regulator. Here's one way to do this:

schematic

simulate this circuit – Schematic created using CircuitLab

This uses a Microchip MIC23050 high-efficiency buck regulator so you can power it directly from your 4 AA battery supply. The EN input can be driven directly from a logic output. The choice of the diode D1 is not critical -- we just need a small (~0.5V) voltage drop from the nominal +6V of the supply.

Not only is this simpler and has fewer parts, but it's also pretty inexpensive - around US $1 for everything.

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    \$\begingroup\$ Beware that using a DC-DC converter in a radio circuit may require taking extra care to ensure that it does not introduce a source of interference or distort the output signal. It definitely can be done, and it is more efficient in most cases, but it requires more care than a linear regulator would. Naturally both are available with enables. \$\endgroup\$ – Chris Stratton Oct 23 '18 at 18:00
  • \$\begingroup\$ Good point. Also, circuit layout is typically more critical with this kind of supply vs. a linear regulator, but it's much more efficient which may be useful for a battery-powered application. \$\endgroup\$ – Edward Oct 23 '18 at 18:02
  • \$\begingroup\$ Thank you for the provided link but unfortunately I can't get one around or find it on ebay. I have found LM2576T-3.3 that seems to be a good alternative!? \$\endgroup\$ – pejey Oct 26 '18 at 13:29
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    \$\begingroup\$ The LM2576 is a good part, if a little more expensive and a little less efficient, but very solid part and available from multiple vendors. My recollection is that it needs a relatively higher input voltage (5.4V?) so you will want to make sure that it meets your needs. \$\endgroup\$ – Edward Oct 26 '18 at 13:50

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