MOSFET driver

Hello,
I would like to drive a synchronous two mosfets which work as buck converter. The pwm source is a PIC mcu. How can I drive properly without a IC.

Regards

If the MOSFET has a 5 volt or less gate saturation, PIC pin to gate and either pull down or up (NPN, PNP) the gate / PIC pin with a 100K resistor as sometimes the MOSFET will not "shut off".

Double check the the MOSFET will switch as fast as you need, if not you can use an IGBT the same way.

I have uploaded my circuit. When i use a diode instead of Q2 it works but i see little voltage about 13V with %98 duty cycle. How can I add Q2 to the circuit and drive them synchroniously. I use IR2111 and all of them are dead.

Input voltage is 17V.
Pwm Freq: 100Khz.

you need to make your mind up on which vesrion you are using PBP or PDS ?

I see several problems here. First you are using N-Channel FETs for both devices. An N-Channel will only turn on when the gate is positive with respect to the source. You need to use a P-Channel device for Q1. You then hook the Source to your positive supply and pull the gate negative to turn it on. But you have another problem there - the gate-to-source voltage of most FETs is limited to 20V, so if you have a 17V supply, you should not pull the gate all the way to GND if you want to be safe.

I would use a transistor as a driver for Q1. A 2N7000 FET would be a good choice. Connect the drain of the 2N7000 through a 1 K resistor to the gate of Q1 and connect a 4.7K resistor from Q1's gate to your positive supply. Connect the source of the 2N7000 to GND and the gate of the 2N7000 to your PIC.

The 1K between the 2N7000's drain and Q1's gate, along with the 4.7K resistor acts as a voltage divider that prevents more than a 14V swing on Q1's gate.

Q2 can be driven directly from the PIC, but it needs to be driven out-of-phase compared to the gate of the 2N7000. You will also need a little "dead-band" between the two outputs.
That is, you do not want to drive the gate of the 2N7000 positive at the same instant you drive the gate of Q2 negative. You will need a little delay so that there is no time that Q1 and Q2 are turned on at the same time.

I would probably replace Q2 with a Schottky diode and eliminate that hassle. The circuit's efficiency will be slightly less, but it is much easier in the long run.