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Hi readers, this is an AC to DC converter. I have been trying to build a power-saving rectifier by changing the capacitance for this specific load (20 ohms and it is a constant load) in order to obtain the best power_out/power_in ratio. However, the ,maximum power ration I obtained by LT spice (check it out, its free) simulation can get is only 0.5. Does anyone know how to get the maximum power_out/power_in (efficiency) while my load is a constant 20 ohms?

Thank you for reading

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2 Answers 2

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That 20 ohm load ideally draws about 500 mA continuously (at 10 V DC). Every time that the AC input is at its peak voltage (+10 V or -10 V), the diodes conduct and recharge the capacitor.

The problem is that the voltage is at its peak only a few percent of the time, so your peak current is like 4 A while a 1N4148 is only rated for 500 mA peak repetitive forward current. In other words, you are grossly overloading the diodes. Since it's a simulation, you apparently get crazy high forward voltage drops (around 2.5 V per diode) instead of just blowing them up.

If you want efficiency, get schottky diodes like 1N5818.

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  • \$\begingroup\$ Hi, thank you!! Is schottky diodes like 1N5818, the most efficient one among all diodes? \$\endgroup\$
    – Maximus Su
    Commented Aug 19, 2021 at 9:47
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    \$\begingroup\$ @MaximusSu Normal silicon diodes cause a voltage drop around 0.7 V when conducting. A full bridge rectifier has two diodes in series, so you lose 1.4 V just to the diodes. Might not matter at 230 V, but at 10 V that's only an efficiency of 86%. In schottky diodes that voltage drop is generally between 0.15 and 0.45 V (depending on the diode), so 2-4 times better. You can improve efficiency even further with active rectification (using transistors instead of diodes to rectify AC). \$\endgroup\$
    – jms
    Commented Aug 19, 2021 at 10:11
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Ideal diode bridges may has lower loss, at the cost of a higher circuit complexity.

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