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We are designing an isolated power supply for our IGBT Driver. We took the reference design from this & this

This is reference design 1:

enter image description here

This is reference design 2:

enter image description here

We are using a zener diode (MMSZ4702T1G) and a resistor (10 kΩ) to split +24 V into +15 V and -8 V, but when we connect the load +15V decreases to +12V and -8V decreases to -12V.

Here is our schematic:

enter image description here

How can we improve this design, and how much current can be drawn from this setup?

Are there any alternatives for generating an isolated +15 V and -8 V from a single supply?

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  • \$\begingroup\$ What voltage does the IGBT switch? What dv/dt? What isolation capacitance is needed? \$\endgroup\$
    – Andy aka
    Commented Sep 6, 2022 at 8:11

1 Answer 1

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What is the resistance of the load? The 10k resistor supplies less than 1 mA to the zener, so your load on the 15V supply can't draw much more than half that current. If that is a half watt 15V zener, you should have at least 10 mA, which would be about 800 ohms.

Another approach might be a negative VR like 7915 for the high side supply, or a 7808 for low side supply.

It would help to see details of the components connected to the gate driver output.

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  • \$\begingroup\$ +15V and -8V is connected to IGBT Driver (HCPL-316J) \$\endgroup\$ Commented Sep 6, 2022 at 6:19
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    \$\begingroup\$ If you read the datasheet media.digikey.com/pdf/Data%20Sheets/Avago%20PDFs/HCPL-316J.pdf, you will see that the output supply current Icc2 is 5-10 mA. So I would use about 500 ohms for R70. \$\endgroup\$
    – PStechPaul
    Commented Sep 6, 2022 at 6:34
  • \$\begingroup\$ Icc2 flows from Vcc2 to Vee though, not Ve. Ve is rated for about -0.4mA -- which is relatively large compared to the 0.8mA in the schematic, but shouldn't be the actual problem if this is the case. That said, they only give Ieh/Iel at Ve = Vee, and do not seem to mention what it is for intermediate voltages(!). \$\endgroup\$ Commented Sep 6, 2022 at 9:00

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