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I might be searching for the jack of all trades with this one, but I'm still curious if it is possible.

I want to measure current (<32 A) in a voltage range of 5 V to 400 V in both AC & DC circuits, which is not a problem.

However, the remaining circuit (Hall-effect sensor SoC, MCU, WiFi, etc.) needs a power supply of 3.3 V and 200 mA max.

I plan on using the same two conductors I measure the current on to supply the device with power.

My question is: Which one is the most efficient way (in terms of energy efficiency - cost does not matter) to step down such a broad voltage range to a voltage of either 3.3 V or 5 V DC? Would it be a jelly bean buck/boost or SEPIC, a flyback converter, an S3R, or something completely different?

For further clarification - the power supply does not need to be isolated.

I'd appreciate your help!

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    \$\begingroup\$ Personally speaking, I don't think that an input voltage range of 5 volts DC to 566 volts (peak) AC is going to be easy to deal with. \$\endgroup\$
    – Andy aka
    Jul 19, 2022 at 15:00
  • \$\begingroup\$ You should think about a power supply with water cooling. To remove up to 12.8 kW with air only would not be easy at all. \$\endgroup\$
    – Uwe
    Jul 19, 2022 at 18:26
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    \$\begingroup\$ @Uwe I think that's the circuit being measured, and the question is asking about the measurement device. \$\endgroup\$
    – user253751
    Jul 19, 2022 at 18:40
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    \$\begingroup\$ A current transformer outputs a huge voltage if disconnected, right? (a common problem that is warned about) What if we connect a current transformer to a Zener diode? Then it should develop the Zener diode voltage, but no more. The current will be the current you are measuring, so if you add a resistor as well, and measure the voltage across the resistor, you can also have a current measurement. Only works for AC. \$\endgroup\$
    – user253751
    Jul 19, 2022 at 18:42
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    \$\begingroup\$ @user253751 I have a couple of inexpensive current-sensing relays which almost certainly operate very much as you've described. They have no power terminals of their own, but they do sense current through an internal CT and use that to power an internal relay. \$\endgroup\$
    – brhans
    Jul 19, 2022 at 21:26

1 Answer 1

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Which one is the most efficient way ...

Dynamic input or output range is a huge tradeoff for efficiency and stability.

So 2:1 max:min ratio is common while 100:1 is a contradiction in efficiency and stability for step load overshoot.

But if cost AND efficiency is no matter which is the cost of energy, it would be any buck-boost high-efficiency design for the pre-regulator and a post-regulator for stability.

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  • \$\begingroup\$ That is true. I am well aware of the inefficiency of such a broad range on voltage, however, i was wondering what the most efficient feasible way would be. \$\endgroup\$
    – Jonas
    Jul 19, 2022 at 17:10

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