With reference to Mosfet driver ir2301s, why is VS actually powered from a capacitor? The datasheet reads "High side floating supply offset voltage" for pin "VS", not that I really understand what that means, but is that not the high-side voltage?

A comment in the linked thread claims:

In normal operation, where the gate driver controls a MOSFET half bridge, turning the low side transistor on pulls VS low, charging C2 through the diode D1. If you keep the low side transistor off for an extended time, the voltage at VS is likely higher than VCC, causing C2 to slowly discharge and the high side to turn off, even when HIN is high.

I don't know much about high-side drivers. Currently I am using a low-side TC4420CPA driving a HY3912W to disconnect a 57 V, 20 A supply, but I am trying to to work out if I can disconnect the high-side line rather than the low-side in order to keep the ground plane.

I dont really know where to start; currently I am just reading datasheets on high-side driver ICs.

Ideally I would like to find a high-side driver to replace my low-side TC4420CPA that will hold a 57 V, 20 A load continuously on when given a 5 V signal (ideally using HY3912W MOSFETs as I have loads).

I do not want an accompanying low-side MOSFET switch, when in the off state I am expecting an open circuit (output NOT ground), like a mechanical switch really would be open circuit when no power is supplied to the coil. the MOSFET will be HIGH for hours at a time (100% duty cycle) and all the high-side driver ICs I look at do not seem to hold the line high permanently.

  • 1
    \$\begingroup\$ That is a bootstrap capacitor to act as a cheap temporary floating voltage supply to supply a source-referenced gate voltage. Do you understand you cannot drive the high side NMOS gate from a ground referenced voltage? Because the NMOS cares about the gate-source voltage difference? Notice how the source pin on the high side NMOS is not fixed can change based on how it is conducting, reaching as high as Vcc. \$\endgroup\$
    – DKNguyen
    Mar 25 at 0:51
  • \$\begingroup\$ electronics.stackexchange.com/questions/449562/… \$\endgroup\$
    – DKNguyen
    Mar 25 at 0:58
  • \$\begingroup\$ electronics.stackexchange.com/questions/624470/… \$\endgroup\$
    – DKNguyen
    Mar 25 at 0:59
  • \$\begingroup\$ emmm, i understand that the voltage difference of the gate is relative to the line that it is disconnecting, so if you plan to cut the high side, that is a floating ground plane of its own NOT ground of the entire circuit??? that's what i think i make of it anyway, EG stack 2 pencil batterys, ground of the top battery is 1.5v making its positive 3v. a MOSFET would have its own ground plane relative to BASE, thus the requirement for a MOSFET driver IC to make the potential difference? but then i may just be talking trash as i am no expert \$\endgroup\$
    – Jay Dee
    Mar 25 at 1:02
  • \$\begingroup\$ "so if you plan to cut the high side, that is a floating ground plane of its own NOT ground of the entire circuit??" Correct. Note the thing about 100% duty cycle and refreshing the bootstrap cap in the links. Just turning off the high side NMOS won't refresh the cap. Something else needs to be done which can be natural for some circuits (half-bridges, inverters) but not others (simple high-side switched loads). \$\endgroup\$
    – DKNguyen
    Mar 25 at 1:03

1 Answer 1


If you do not also a low-side NMOS to drive, there is no point to a typical high-side gate driver IC. These are intended for things like half-bridges. If you only have a high-side NMOS to drive, the only thing you would really be using from the IC is the level shifter to control floating circuitry from a ground referenced signal. You would still need to add floating regulators and such since there is no low-side switch to refresh a bootstrap cap. –

Discrete solutions are by pairing an isolated gate driver or optocoupler with an isolated regulator floating with the high-side NMOS on the other side. However, there are ICs that contain integrated charge pumps meant for to drive NMOS load switches but these usually only float under 12V. If you do use something like optocouplers, remember that if an optocoupler does not have a push-pull output (most don't...they are just a transistor that conducts or blocks) then they cannot drain the gate-source capacitance to turn off the MOSFET. You need something else to do that (like a pull resistor).

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.