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
    Commented Mar 25, 2023 at 0:51
  • \$\begingroup\$ electronics.stackexchange.com/questions/449562/… \$\endgroup\$
    – DKNguyen
    Commented Mar 25, 2023 at 0:58
  • \$\begingroup\$ electronics.stackexchange.com/questions/624470/… \$\endgroup\$
    – DKNguyen
    Commented Mar 25, 2023 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
    Commented Mar 25, 2023 at 1:02
  • 1
    \$\begingroup\$ By "turn on" do you mean barely start to conduct? Or fully conduct? Look closer at the datasheet. You don't care about the Vgs to barely conduct when you intend to use the MOSFET as a switch. The datasheet I found meets my expectations and 6.4V and 7V are nowhere to be seen. \$\endgroup\$
    – DKNguyen
    Commented Mar 25, 2023 at 1:17

2 Answers 2


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


When using an N-channel MOSFET as the "high-side" switching element, that transistor is operating in "common-drain" mode, otherwise called "source-follower":


simulate this circuit – Schematic created using CircuitLab

As you can see, to bring the source potential (as shown on the volmeters) up to the full supply potential (+12V here), the gate must be held at a potential well above that.

In fact, source potential \$V_S\$ (not to be confused with the driver's use of \$V_S\$) is related to gate potential as follows:

$$ V_S \approx V_G - V_{GS(TH)} $$

where \$V_{GS(TH)}\$ is the MOSFET's threshold voltage. In this case it's about 4V, meaning that the source "follows" the gate, but always about 4V lower.

That is why the high-side gate driver supply voltage (\$V_S\$ on the datasheet, again, not to be confused with source potential \$V_S\$ here) needs to be several volts greater than the MOSFET-bridge's positive supply.

This can be achieved using a charge pump, which operates by charging a capacitor to +12V, and switching it into a position where it becomes in series with the +12V supply to obtain +24V. This is then used to drive the gate, and is sufficient to obtain the required +12V bridge output. This technique is employed by the IR2301 gate driver, and is called "bootstrapping".

For this technique to work, the driver needs to periodically re-charge the bootstrap capacitor, meaning that switching must occur frequently. This will not work well (if at all) if the driver state remains unchanged (output fixed high or low, as would happen for 0% or 100% PWM) for extended periods.

If you are using a P-channel MOSFET to switch at the high-side, then such bootstrapping with a charge pump is not required, since the transistor will be operating with "common-source", and behaviour is very different.


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