# Very broad input voltage range, low-current power supply

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.

• 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. Jul 19, 2022 at 15:00
• 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.
– Uwe
Jul 19, 2022 at 18:26
• @Uwe I think that's the circuit being measured, and the question is asking about the measurement device. Jul 19, 2022 at 18:40
• 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. Jul 19, 2022 at 18:42
• @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. Jul 19, 2022 at 21:26