First of all, it can be unnatural using a translator. I'm sorry. I'm trying to drive the high side switch through gate driver that doesn't use bootstrap method. I think we can use optocoupler type gateriver. I think the high side can use isolated DC/DC converter to power it. I thought it was possible by referring to the following data. Is it the right data? Additionally, can I use this type of gate driver to get circuit diagram information that drives half or full bridges? Thank you. enter image description here In addition, it's a circuit diagram that I'm thinking about. enter image description here

I referred to the following link. Optocoupler Reference

  • \$\begingroup\$ kogyu - Thank you for adding that link to the source of the copied image. However Stack Exchange has specific requirements when copying material from their site (even into another question, as you have done) as part of the CC BY-SA license. Just adding a link is not enough. See here for details. || If no-one else corrects this insufficient reference, I'll do it in a few hours. || In future, please note that copying from a site can require you to do additional things to comply with that license. This is all your responsibility. \$\endgroup\$
    – SamGibson
    Feb 23 at 5:24
  • 2
    \$\begingroup\$ You can still use a bootstrap gate driver IC. Just remove the bootstrap cap an diode and connect the output of an isolated supply where the boostrap capacitor would normally go. That's basically what using optocouplers is where the high-side opto output is powered by an isolated supply, except that it doesn't have optical isolation. Both approaches work with half-bridges, H-bridges, and 3-phase bridges. \$\endgroup\$
    – DKNguyen
    Feb 23 at 5:48
  • \$\begingroup\$ Do you mean that we only need to supply the power to the high side gate as an insulated power source? \$\endgroup\$
    – kogyu
    Feb 23 at 6:28
  • \$\begingroup\$ @kogyu Exactly. Half-bridge drivers are always designed to run the high side driver on a floating supply. But very often, this floating supply is created by simply adding a diode. Think of the diode as the most simple means to generate this floating supply, but being so simple, it has the drawback of an upper duty cycle limit. \$\endgroup\$
    – tobalt
    Feb 23 at 7:39
  • \$\begingroup\$ @kogyu Not sure what you mean when you say "using a translator is unnatral". It's by far the most common way to drive the high-side switches. Way simpler than using separate DCDC convertors and cheaper. Also, the level translation complexity is already handled internally by pretty much all gate drivers you can buy. Is it the isolation that's required causing you trouble? \$\endgroup\$
    – MOSFET
    Feb 23 at 7:48

1 Answer 1


Looks good, but you will have to connect your -5V Iso to the switch node of the first half-bridge leg. That way, the isolated power will be always 10 V above the source of the high-side MOSFET. You don't need a bipolar isolated supply. A single voltage isolated supply is enough.

And also take note of DKNguyen's comment. Using a half-bridge driver instead of the two optos, will make the signal translation much easier, faster and probably less skewed.

  • \$\begingroup\$ Then, is 5V.Iso enough power to be produced by the insulated power source? The gate driver must send out a gate signal of 10V voltage. \$\endgroup\$
    – kogyu
    Feb 23 at 10:14
  • \$\begingroup\$ @kogyu This will happen, if you tie the "negative 5V" rail to the switch node. (+5V) - (-5V) = 10V \$\endgroup\$
    – tobalt
    Feb 23 at 10:43
  • \$\begingroup\$ I understood connecting to a high side switch source instead of -5V.iso. But I don't understand why a positive insulation power source is not required. Is it wrong to need a floating power source of 10V to make a gate signal of 10V? Also I'm wondering if I need an isolated power source of 5V for the 5Viso position or an isolated power source of 10V. \$\endgroup\$
    – kogyu
    Feb 23 at 11:52

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