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I have this power supply which is part of an energy meter design already on PCB, and undergoing some tests. It passed all tests except the one that requires it to run at 60% or rated voltage, 240V, which gives us 144V enter image description here The AC supplies 240, the transformer output gives 6V, and the LDO gives me 3.3V. HOwever, at 144V, the input of the LDO is just about 3V, giving me no output.

I want to know the easiest way to make this work.

Thanks.

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The 60 % seems an unnecessarily severe requirement, given that mains voltage has a tolerance of only 10 %.

You can place the secondary windings in series instead of parallel, but the power supply will become very inefficient: 12.6 V AC rectified gives maximum 17.6 V DC, then the efficiency of a linear regulator is 3.3 V/ 17.6 V = 19 %. At 100 mA load the NCP1117 will dissipate 1.5 W.

If you do want to have 3.3 V in a relatively efficient way at a wide input voltage range you can use a switching power supply like this one by Recom. Input voltage range is 90 V to 277 V. This is for 1W or 2 W, but other modules with higher power outputs are available .

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You could also replace the linear regulator with a buck-boost regulator like the TI TPS63060. The part switches from buck mode to boost mode as needed to maintain regulation with an arbitrary input (higher or lower than the output voltage).

The entire circuit would be laid out on a small PCB and wired in to the existing layout as a rework operation (it would tie in where your three-terminal regulator goes).

(If the secondary voltage was capped at 5.5V, you could use an even simpler part like the TPS63001.)

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Your transformer's secondary windings are in parallel.

One way to hack this up is to cut some traces and solder some wires to make secondary transformer windings in series, this will give you twice the voltage (12V at 240V primary and ~7V at 144V), but less current.

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