Bypassing Sync Driver Bootstrap for 100% duty

[ACharnley]

Question largely regarding NCP81166 but likely to apply to others.

NCP81166MNTBG onsemi | Motor Driver ICs - LCSC Electronics

NCP81166MNTBG onsemi $0.311 - Half Bridge 2 MOSFET 16ns 11ns 4.5V~13.2V DFN-8(2x2) Gate Drivers ROHS datasheet, price, inventory C116731

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I require a driver which has a 100% duty (so no bootstrap driver). I have a 10v+ supply ready to use.

The output is maximum 5v and I'd ideally like 5v at the fet gate. The NCP81166 ticks a lot of boxes such as diode emulation mode but the bootstrap I'm unsure about. Reading through the documentation if I connect 10v to this chip, after the BST diode drop, that will give about 19.3v at the gate, close to destruction for 20v gate FET's.

What if I replaced the bootstrap capacitor with a resistor? It would waste a bit of power and possibly hinder the turn-on speed?

Anyone had the same issue?

 

[ACharnley]

The following diagram suggests that the turn on is getting the voltage through the BST diode plus the floating component. If I therefore just did away with the bootstrap capacitor then problem solved?

 

[rjenkinsgb]

The bootstrap cap charges through the diode when the lower FET is on and feeds the upper driver with a voltage somewhat above the DC supply level as the upper FET turns on and the output voltage switches high.

That could be close to double the DC supply.

The IC should have some internal limitation or regulation for the bootstrap pin input voltage, but I do not see any problem supplying it with a fixed DC voltage rather than using the cap.

 

[ACharnley]

Cool, just to double confirm, get rid of the cap and feed GVDD (in the above diagram) a high voltage?