Driving MOSFET from MCU with BJT

Question

1- What is the purpose of D1 in this circuit?

2- Can i connect the PWM output of MCU to Vi through a diode, to protect the MCU from 12V supply of the drive circuit? (e.g. if Vi got accidentally shorted to 12V)

3- Do i need to decrease R2 value if i connect the input through a diode, to compensate for diodes voltage drop, and maintain the base-emiter current? would decreasing R2 value to 560 ohms suffice?

(MCU's output high level is 5V, and i'm using TIP122 for BJTs)

Answer 1

1. D1 isolates the base drive of Q2 from the emitter output, so when Q1 turns on it pulls the base of Q2 to -0.7V relative to the emitter and turns it off before pulling the output down.
2. Yes, you could add a diode, however the drive will be reduced a bit by the lowered voltage. You can reduce the resistor value to compensate.
3. This totally depends on the drive voltage. Right now the base current is (Vi-0.7V)/1K mA. With a diode it will be more like (Vi-1.4V)/R2. So if Vi is 3.1V (say output voltage of an MCU with 3.3V supply) you have 2.4mA drive, and R2 should be more like 700 ohms (you could use 680, for example).

That's assuming you're using transistors, however you drop a bit of a bombshell at the end- TIP122 not a transistor- it is an NPN darlington with resistors, a really slow one- switching times in microseconds. It does not fit your schematic in the question nor is it really appropriate for this application.