Hi all,

I've been doing some research on high side switching and I've come across a question that some of you may know the answer to (and it actually may sound silly):

If I'm using an N-channel FET as a high side switch, why would I need to use a high side gate driver? From looking at various datasheets it seems like using a low side driver nets the same results. Why wouldn't a low side driver IC work?

 

Because the voltage on gate needs to be higher than the voltage on source to turn it on, maikng it effectively higher then you supply voltage. Thus you need a charge pump to provide voltage relative to source and not relative to supply voltage.

 

Because the voltage on gate needs to be higher than the voltage on source to turn it on, maikng it effectively higher then you supply voltage. Thus you need a charge pump to provide voltage relative to source and not relative to supply voltage.

Do you mean higher than the voltage on the drain? Similar to what is shown in Figure 1a here? How is it not relative to supply voltage?

 

I mean higher than source. Vgs is very similar to base current in bipolar transistors. To turn an NPN fully on on the high side you need 0.7V higher voltage than the power supply, or it will act like voltage follower with 0.7V difference between base and collector. Mosfet will have a little more than Vth between gate and source in follower configuration.

edit: see this gate driver with output transistors. http://s.eeweb.com/resized/images/r...e-Gate-Driver-GaN-FETs-1324309807_600_361.png