Hi All,

Following on from my previous thread http://forum.allaboutcircuits.com/showthread.php?t=9507

I was hoping someone would be able to explain what the bootstrap capacitors do for a high side gate driver?
If you look at page 9 of http://www.irf.com/product-info/datasheets/data/irams10up60a.pdf there is a typical application connection schematic.

Cheers
Shaun

 

The "bootstrap" capacitor provides a method of using N-channel MOSFETS on the high side of an H-bridge. There is a charge pump in the IC which charges the capacitor about 10v higher than Vdd.

In order to fully turn on an N-channel MOSFET, you generally need to get the gate in the vicinity of 10V higher than the source. Without the charge pump and bootstrap capacitor, when the gate turned on and Vss approached Vdd, the gate would begin to turn off; ie: partially conducting state.

 

Thanks once again SgtWookie!

So the bootstrap cap charges to vdd while the low side MOSFET is on, then when the high side gets turned on, the cap becomes in series with the source and so the driver gets drain+Vdd, which would maintain the 10v or so potential difference... presumably the value of the cap is chosen depending upon the length of time it needs to maintain this?

Have I understood this correctly?

Cheers
Shaun

 

Yes, you pretty well have it

Although the impedance of the gate is extremely high, it does require a certain amount of current, usually specified in the datasheet under "total gate charge" in nanocoulombs.
(1 coulomb (C) is the amount of electric charge transported by a current of 1 ampere in 1 second; 10-9 C = nC)

Bootstrap capacitors usually don't have to be very large. 0.1uF would likely fill the bill in all but unusual circumstances. Low-ESR ceramic or tantalum would be the kind to use. Electrolytics would not be a good choice due to comparatively higher ESR and inductance.