The easiest solution is to use an integrated driver with high-side and low-side outputs. Here is a LM5109B as an example:
The low-side driver is a simple buffer with high current output. It drives the gate of the low side FET and is powered from the Vdd pin.
- Pick a driver with Vdd suply voltage compatible with your circuit and the gate drive voltage your FETs require. If your FETs require 10V gate drive you have to use a driver with 10-12V Vdd, not 5V.
When the low FET is ON, it shorts output node "HS" to GND so the voltage on node HS is close to 0V. The bootstrap cap (upper right corner) charges to near Vdd voltage through the bootstrap diode. Then, when the low FET turns OFF, the high-side driver uses this bootstrap capacitor as a power supply to drive the top FET.
This introduces one big gotcha: If the bootstrap cap is not charged with enough voltage, the top driver will not work.
This means the low FET must turn ON every millisecond or so to recharge the cap. These drivers cannot keep the top FET ON with 100% duty cycle, if you require 100% duty cycle you need a driver with an integrated charge pump, an extra supply, or use a PMOS as the top FET (with a suitable driver).
This also means that, if you make a buck converter and the HS node is connected to an inductor then to the output of a buck converter... At startup, if the output is 0V, the bootstrap cap will charge through the diode. However if the load is a battery or anything that has voltage on it, then the HS node might be at a high enough voltage that the bootstrap cap does not charge... and then the top driver won't work. So if you want to turn the top FET ON, you must first turn the bottom FET ON briefly to charge the cap, then you can turn the top FET ON.
Note that in some chips, BOTH drivers with turn off if the bootstrap cap is not charged, which makes them unable to start up when used in a buck converter if there is enough voltage present on the output. Since you are making a battery charger you have to check the datasheet for this.
Next gotcha, the chip above does not have anti cross-conduction logic. If you set both inputs to "Logic 1" it will do what you ask and turn both FETs ON. If they are connected across a capable enough power supply, expect fireworks.
Some chips will offer anti cross conduction but that could mean you can't choose the dead time. If you pick one which lets you choose the dead time, then you have to pick the correct value to make sure there is no cross conduction.
Finally some chips offer "diode emulation" with an internal analog comparator which is used to control the bottom FET. This is useful if you want discontinuous conduction in the bottom FET in a buck converter, and you don't like the FET internal diode characteristics or want to avoid losses.
Also you have to make sure the chip supports the frequency you will use (look at the timings), drive current according to your FETs, minimum on/off times, power dissipation, idle power if you care about low power, standby mode, suitable package, etc, etc.
Note FETs have a body diode, which is in the correct direction to let the voltage on the battery your buck converter is charging through, so when you connect the battery on the output, it will also power the input even if there is no power supply, which can cause problems if you don't design for it.
So make sure you're aware of all this... especially the gotchas.
Then, on the DigiKey search you can go to Gate Drivers and select:
High Side Voltage - Max (Bootstrap) : must be above Vin, so let's pick 55V to 600V
This one seems nice, it has anti cross conduction and adjustable dead time. You will have to check some datasheets and pick a favorite.
PS: Your FET requires at least 5V gate drive. Read this for more info.