No it doesn't. If you input a square wave of 0 - 3.3 volts at point mpptB, you can expect an output at pins 1,4 of 0.6V - 2.7V. Now 2.7 volts may be enough to turn the MOSFET on fully, depends on the MOSFET.
It's now made sense why bucks offering 100% duty cycle (aka bypass) can't use bootstrapping so use a PFET on the upper.
I used the above circuit plus a similar one to drive the top. [dc] is limited to 16v. Both drivers are 600mA rated which should be plenty for the small fet's I intend to use.
dc=16V? -> yes, no more to protect the gate. May lower a bit.
Will the bottom MOFET turn on well at 4V on the gate? -> Yes, they lower dc helps to get fast low Vgs fets. Won't be 4 though, more like 2.5v!
What is the voltage of mpptB? 5V? -> 3.3v (the 2.5v is a bit of a worry but it'll still be "mostly on" and there's a schkotty just out of shot parallel to the fet)
Bottom MOSFET pin1 at gnd? -> correct
A number if companies make "gate drivers". This one needs a supply in the 5 to 30V range. (you have 16V)
You can choose inverting or not-inverting or one of each.
Input signal is "ttl levels" and the output is ground to Vcc.
Just a thought. RonS.
I've made another mistake, the PFET is swinging to ground.
I think a better solution would be to keep it as NFET and drive it with a higher voltage of say 15v. I know when directly connected to the load this would give a healthy 10Vgs but what about when there's an inductor in the path?