There definitely are PN half bridge drivers, and a pmos for the high side is fine nowadays for not very high power applications, but you need to get the correct driver for it (some have the PN fets integrated, like the FAN7093).
It used to be that they had much lower performance than their counterpart n-type mosfets, so the drivers were made to use n-type for the high and low sides. N-type still have better performance because of their nature of using electrons instead of holes for conduction (electrons have higher mobility), but not as severely as before. On the other hand, P-type mosfets don't need a charge pump.
Take a look at this application note from Vishay:
P-Channel MOSFETs, the Best Choice for High-Side Switching
An EE-Times article also talks about this:
A primer on high-side FET load switches
simulate this circuit – Schematic created using CircuitLab
Note 1: The input voltages are only \$V_{cc}\$ and \$V_\text{High Voltage}\$. You don't apply anything at the \$V_{BS}\$ node. It is only for representation.
Note 2: Notice that there are two different type of grounds. Those grounds must not be directly connected to each other.
You must drive the MOSFET between its gate and source terminals. Since the source terminal voltage of a high side MOSFET will be floating, you need a separate voltage supply (VBS: \$V_\text{Boot Strap}\$) for the gate drive circuit.
In the schematic below, VCC is the voltage source of the rest of the circuit. When the MOSFET is off, ground of the boot strap circuit is connected to the circuit ground, thus C1 and C2 charge up to the level of Vcc. When the input signal arrives to turn the MOSFET on, ground of the gate drive circuit rises up to the drain voltage of the MOSFET. The D1 diode will block this high voltage, so the C1 and C2 will supply the driving circuit during the on-time. Once the MOSFET is off again, C1 and C2 replenish their lost charges from VCC.
Design criteria:
- RB must be chosen as low as possible that will not damage D1.
- Capacity of C2 must be chosen enough to supply the driving circuit during the longest on-time.
- Reverse voltage rating of D1 must be above \$V_\text{High Voltage} - V_\text{CC}\$.
The input signal must be isolated from the boot-strap circuit. Some possible isolaters are:
Optocoupler
Optocoupler is the most basic method for isolation. They are very cheap compared to other methods. The cheap ones have propagation delay times down to 3\$\mu\$s. The ones with less than 1\$\mu\$s propagation delay are as expensive as isolated gate drivers though.
Pulse Transformer
Pulse transformer is a spacial type of transformer for transferring rectangular pulses. They have less number of turns in order to avoid parasitic capacitance and inductance and larger cores for compensating loss of inductance due to reduced number of turns. They are much faster than optocouplers. Delay times are less than 100ns in general. The image above is for illustration only. In practice, the current they can provide is not enough for driving a MOSFET fast; so they need additional circuitry in practice.
Isolated Gate Driver
Isolated gate driving is a relatively new technology. All the complexity of gate driving is encapsulated in one single chip. They are as fast as pulse transformers, yet they can provide a few amperes of peak gate current. Some products also contain on-chip isolated DC-DC converters, so they don't even need boot-strapping. However, all these super features come with a cost.
Best Answer
In normal circumstances the voltage on the boot pin will be about 12 to 15 volts higher than the voltage on the source pin and this is due to the charge on the boot capacitor. As this cap connects between boot and source this is the voltage seen across it normally.
However, if your circuit shuts down with a load connected to the high voltage and source the cap will need to be rated at the full power rail because one side will be taken to the high power rail by the load and the other will decay to 0 volts.
The reverse voltage rating on the diode needs to be at the full power rail by the way.