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I'm using the IR2104 gate driver in a DC motor driver design, and I noticed something in functional block diagram from the datasheet that I didn't understand.

IR2104 Functional Block Diagram

The low side schematic is as expected, but the high side has lots of extra parts, including a pulse generator, pulse filter and SR Latch. I would have thought that the high side could be connected similar to the low side, but with an inverted output signal so that it has the opposite logical output from the low side.

From what I've read it seems that this extra circuitry is for bootstrapping and giving high-side bias, but I'm not sure what is meant by high-side bias and don't know what the goal of this is and why it is necessary.

An explanation of what high-side bias is and its purpose here would be really appreciated since I'm a bit lost.

For reference: IR2104 Datasheet

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  • \$\begingroup\$ I suggest that you find a more complete circuit using such a gate driver which also shows the N-channel MOSFETs that such a gate driver IC controls. Then think about the \$V_{GS}\$ that would be needed to turn those MOSFETs fully on. The proper conclusion should be that you need a voltage that is higher than the supply voltage. \$\endgroup\$
    – Bimpelrekkie
    Commented Sep 17, 2021 at 17:52

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When driving the N MOSFET, most drivers put out a voltage 12V or more than the source, which is typically enough to drive even the most demanding MOSFETS.

The problem is when your Vcc is not high enough, you cannot depend on that Vcc+12 being enough, so the driver has to generate a gate voltage that is Vs+12, and this is where the pulse generator and extra circuitry requirements come from.

If you look at most MOSFET drivers, even if not for a half-bridge, they will either provide a voltage that is +12 to +15 over Vcc or +12 to +15 over the MOSFET source. The former type of driver does not need a bias, but the latter requires access to the source pin so it can superimpose the voltage.

Hope that helps.

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  • \$\begingroup\$ That explains it, thanks! I didn't realise it was for boosting the gate drive voltage. Much appreciated! Does that mean I should feed the same voltage that I'm putting into the H-bridge into the Vcc pin of the IR2104? That way regardless of the H-bridge input voltage the gate drive would always be higher \$\endgroup\$
    – RobotGuy
    Commented Sep 17, 2021 at 18:17
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    \$\begingroup\$ Yes, I learned the hard way, I used a non-boosting driver in a circuit that would only work with Vcc < 12.2V or so; burned a few MOSFETS before realizing the big difference between the two types of drivers. Most of the time, the non-Vs referenced drivers are for low-side applications. \$\endgroup\$
    – Yiannis
    Commented Sep 17, 2021 at 18:17

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