Mosfet driving circuit(IR2110)

[rokai1990]

i look forward to making a buck-boost converter ,the problem now is the firing of the MOSFET,i tried to fire it using IR2110 ,but i failed to fire one side only of the two sides ,so if any one has an idea about that,please help me or give me an easier way to fire the mosfet by the pwm coming from the uc.
thanks in advance

 

[kak111]

More info is needed to answer your questions.............

Show us the circuit shematic of your converter.

or for new design , all needed parameters. ( Vinmin, Vinmax, Vout, Iout, etc..... )

BUCK AND BOOST CIRCUITs
http://www.intersil.com/engineeringtools/tools/buckandboostcircuit.pdf

App.note for IR2110
http://www.irf.com/technical-info/appnotes/an-978.pdf

 

[rokai1990]

this is the cct schematic
i want a method to fire the mosfet used as aswitching device in i this cct

 

[Hz2020]

Hello rokai1990

u can not use high side driver of the IR2110 as the source of the MOSFET does not reach zero voltage (Vin when MOSFET is on and Vx when MOSFET off), So, the bootstrap capacitor will not charged, So, the MOSFET will not turn on.

u can use the low side driver of the IR2110, but u will use another supply for firing process(12v supply), the GND of that supply will connected to the source leg of the MOSFET and the 12v enters to the MOSFET gate according to the firing signals on the LIN pin.

Also u can use other firing ICs such as TLP250 opto.

Note that the MOSFET that i talking about is N channel MOSFET

 

[mtwieg]

Yeah, you can't use a bootstrapping driver for this, since the FET source voltage will be going up to Vin and down to Vout.

What you'll want is an optoisolated gate driver IC. But those do not provide the supply on the FET side needed. For that you'll either need a small isolated power supply, or some kind of more sophisticated bootstrapping circuit. This is one of the reasons you don't see boost buck DC-DC converters often.