simulate this circuit – Schematic created using CircuitLab
I did add the NPN 2N3904 because I want to put on sleep the Arduino. So I need to put LOW the GPIO when goes in sleep.
This is a prototype. I will port on an Arduino Mini 3V3.
I did forget the pullup from gate, but cannot edit the schema 🙁
Load is 5V display, with MAX current usage of 67mA. So per OHM Law, I got ~75 Ohm.
Arduino is 5V and his GPIO drive ON/OFF the 2N3904.
I did study that it is common to assume that base current is 1/10 of the collector current. But, what the collector current in my case? That 67mA?
If yes, I need to select a resistor for
5V (Arduino) – 0.9 (Vbe per datasheet) = 4.1/ (0.067 /10 ) = ~600 ohm?
For R2, Do I need to do same absumption?
Best Answer
Your circuit has some problems, particularly that there is nothing pulling the gate of the FET high when the transistor is off.
However, you don't need Q1 at all. If M1 is indeed a P channel FET that can work well enough with 5 V base drive (I didn't look it up, but that's how you are trying to use it), then just control it directly from the 0 to 5 V digital output. Driving the digital output low turns on the FET, which powers the load. Driving the digital output high turns off the FET, which removes power from the load. Yes, it's that simple.
Added
No, you don't. Any competent microcontroller can set it's pins high, low, or sometimes to high impedance when sleeping. Pin state can't be changed during sleep, but the polarity is arbitrary.
A single P channel FET that works with 5 V gate drive is still all you need.
Added 2
I'm not familiar with the Atmega series, but this doesn't sound right. Unless you can show me the datasheet snippet where this is stated, I simply don't believe you and conclude that you misunderstood something.