I have an MCU with VDD = 3V and a pFET where Vsource = 5V. Is there a simple way to gate the FET off and on with a 3V output pin?
The pin is configurable for either push/pull or open drain operation. So I am wondering if I can tie the pin/gate to 5V level with a resistor so that when the output pin is low (it can sink 25mA) the fet is on, and when the pin (a 5V tolerant pin) is 3V, can the resistor pull the gate up to at least 4V or so?
Here is a link to a picture of the IO circuit on the MPU:
The pin is on the right side of the picture. Gate of the pFET has a capacitance of 2900pF and it has a Vgs threshold of between -1 and -2V. pFET is going to drive a buck converter and I need to switch it fast enough to get 500kHz.
What exactly does 5v tolerant mean? In the picture it looks like the 5v voltage will drop through a diode and end up at the 3V rail and just get pulled down to 3.7V, but who knows?