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I am evaluating IR2125 and IRS20752LPBF and IR2184 high side gate drivers to drive a MOSFET that is to be used as a battery switch for a 48V battery, the continuous current from the battery is around 25A. I have selected the IRF100B201 MOSFET for the same. If I pull up the IN pin to 3.3V or 5V will my MOSFET turn on, and if I give a logic 0 signal to IN pin will the MOSFET turn off. I am not talking about a PWM signal, I am talking about a fixed DC voltage level. Can someone please help in clarifying my doubt regarding these gate drivers?

Datasheet Links-

IR2125 https://www.infineon.com/dgdl/Infineon-IR2125-DS-v01_00-EN.pdf?fileId=5546d462533600a4015355c85ba31694

IRS20752LPBF https://www.infineon.com/dgdl/Infineon-IRS20752L-DS-v01_02-EN.pdf?fileId=5546d462533600a401535675e709278e

IR2184 https://www.infineon.com/dgdl/Infineon-IR2184(4)(S)-DataSheet-v01_00-EN.pdf?fileId=5546d462533600a4015355c955e616d4

In my application I want the switch to always remain in on-state, i.e drive all of my electronics during normal operation, and when there's a fault microcontroller should turn it off. This switch acts as an input to different buck converters and DC motor drivers.

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3 Answers 3

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As already answered by other you need a gate voltage above the source. Depending on the mosfet type it could be 5 to up 15V (usually, the more, the better, unless you break the Vgs maximum). Since you are driving in continuous your losses will be almost only from Rds, so a large gate drive will help (there are curves or specs for that in your mosfet datasheet). These days mosfets are usually characterised at least 4.5V and 10V.

Now, for driving the gate. You don't need a big current even if it's a big gate since you are essentially making a load switch. In fact a small gate current helps softening the load transient turning on the mosfet gradually.

Most commercial gate drivers are designed for PWM: they have large current capabilities to reduce switching losses and they generate the gate voltage by bootstrap (a kind of charge pump). In short, they don't work in high side without at least some turn off interval (specified in the driver datasheet).

For switching your 48V load you'll need something like 55-58V (or something like 10V referred to the source). The first option is simply using an high side gate driver designed for 100% duty cycle. They have an embedded charge pump, you solder them in and it's all done. Another option (mentioned in another answer) is a photovoltaic gate driver. They are somewhat expensive but work well; they also give you galvanic insulation which is handy with IGBTs, for example.

If you have many of these load switches you could consider to generate your own gate voltage with a converter. A boost of 25% for some mA is really easy to do. Maybe you can also tap some similar voltage from somewhere else. In any case remember to protect the gates with zeners if there is a risk of running over the Vgs maximum, especially since probably your 48V battery can presumably go low (to 36V, maybe? depends on the application)

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  • \$\begingroup\$ I'm using an Isolated DC-DC converter from Mornsun to power the high side MOSFET gate. I have tested it, and it works. Your answer was insightful and gave me many ideas. I am using A1212S-1WR2. I am turning off the 12V input to the isolated converter using a logic-level MOSFET. mornsun-power.com/public/uploads/pdf/B_LS-1WR2.pdf \$\endgroup\$ Mar 8, 2021 at 15:26
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to turn the mosfet on you need to put the gate voltage to be 3.3V above the source voltage. This means that if you want the battery voltage on the source you need more voltage than the battery has on the gate.

these drivers all use capacitors to make the high voltage by adding VCC to it, but it only makes a pulse of charge, so if you're not occasionally turning it off there's a possibity that it'll run out of charge and the MOSFET will start to turn off and possibly overheat.

I'd consider using a P channel mosfet instead.

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  • \$\begingroup\$ But to turn off a PMOS, I'll also need a Vgs voltage of 10 V for most of the MOSFETs out there. If my source is at a battery voltage of 48V, to turn it completely off, I'll need a voltage of 58V on the gate w.r.t ground. Here's a 24V battery switch app note from infineon infineon.com/dgdl/… Sadly it's for 24V, I need something like this for 48V. the gate driver used in this app note is rated for 36V max. \$\endgroup\$ Mar 3, 2021 at 8:26
  • \$\begingroup\$ Perhaps you can use two of this in parallel: (1) IRF9Z34N P-Channel Power MOSFET (TO220, Vdss = -55V, Rds(on) = 0.01R, Id = -19A, Vgs(th) = -2 < -4V) irf.com/product-info/datasheets/data/irf9z34n.pdf. Just brainstorming. There should be better choices. Cheers. \$\endgroup\$
    – tlfong01
    Mar 3, 2021 at 8:48
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If I pull up the IN pin to 3.3V or 5V will my MOSFET turn on, and if I give a logic 0 signal to IN pin will the MOSFET turn off.

Yes. Applies to all three gate drivers mentioned. None of them have inverted inputs or outputs. You can look for the timing diagrams in the datasheets to see for yourself.

That said, there’s a catch. None of the gate drivers you linked will actually work in your application without adding a step up converter to provide a supply voltage for the output stage of the driver.

There are specialized gate drivers that can provide the gate drive voltage without relying on an external supplementary power supply. Look for photovoltaic gate drivers, they are made for pretty much exactly what you’re asking for. Example chip: https://www.vishay.com/docs/83469/vom1271.pdf

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  • \$\begingroup\$ Thank you for the help. The FDA217 has a higher open-circuit voltage of 12V. So it should be a better option to completely turn on the IRF100B201. I haven't tried this solution but have ordered FDA217 for testing from Digikey. \$\endgroup\$ Mar 8, 2021 at 15:17

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