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I find it counter-intuitive that deliberately wasting some power makes the system perform more efficiently. (Another circuit where adding a little resistance makes it perform better: Ripple current in a linear power supply transformer ).

I find it counter-intuitive that deliberately wasting some power makes the system perform more efficiently. (Another circuit where adding a little resistance makes it perform better: Ripple current in a linear power supply transformer ).

Good eye -- the zener is the part that is the least efficient in this circuit. A linear regulator here would significantly improve the efficiency of this circuit.

If you assume ideal capacitors (which is a good assumption) and ideal diodes (not such a good assumption), no power is lost in those components. If you eliminate the safety protection resistors, then no power is lost in those components either. Since there's no where else for the power to go, such an idealized circuit would give you 100% efficiency. But it would also have some ripple. You may be able to follow this no-zener circuit with a linear voltage regulator to eliminate that ripple and still get a net efficiency over 75%.

The "law" that "a voltage regulator always has an efficiency of \$V_{out}/V_{in}\$" only applies to linear DC to DC regulators. That law doesn't apply to this circuit, because this circuit has AC input, and so this circuit can have much better efficiency than that "law" predicts.

Good eye -- the zener is the part that is part that wastes the most energy in this circuit. A linear regulator here would significantly improve the efficiency of this circuit.

If you assume ideal capacitors (which is a good assumption) and ideal diodes (not such a good assumption), no power is lost in those components. In normal operation, relatively little power is lost in the safety protection resistors. Since there's no where else for the power to go, such an idealized circuit would give you 100% efficiency. But it would also have some ripple. You may be able to follow this no-zener circuit with a linear voltage regulator to eliminate that ripple and still get a net efficiency over 75%.

The "law" that "a voltage regulator always has an efficiency of \$V_{out}/V_{in}\$" only applies to linear DC to DC regulators. That law doesn't apply to this circuit, because this circuit has AC input, and so this circuit can have much better efficiency than that "law" predicts.

EDIT: Dave Tweed points out that simply replacing the zener with a linear regulator actually makes this overall circuit less efficient.

I find it counter-intuitive that deliberately wasting some power makes the system perform more efficiently. (Another circuit where adding a little resistance makes it perform better: Ripple current in a linear power supply transformer ).

I wonder if there is some other way to improve the efficiency of this circuit, that is less complex than a 2-transistor switching regulator?

I wonder if further modifying the circuit by adding another capacitor across the AC legs of the bridge rectifier might result in something more efficient than the original zener circuit? (In other words, a capacitive divider circuit like this Falstad simulation ?)

Z = V/I.

If you assume ideal capacitors (which is a good assumption) and ideal diodes (not such an good assumption), no power is lost in those components. If you eliminate the safety protection resistors, then no power is lost in those components either. Since there's no where else for the power to go, such an idealized circuit would give you 100% efficiency. But it would also have some ripple. You may be able to follow this no-zener circuit with a linear voltage regulator to eliminate that ripple and still get a net efficiency over 75%.

The "law" that "a voltage regulator always has an efficiency of vout/vin" only applies to linear DC to DC regulators. That law doesn't apply to this circuit, because this circuit has AC input, and so this circuit can have much better efficiency than that "law" predicts.

$$Z = \frac{V}{I}$$

If you assume ideal capacitors (which is a good assumption) and ideal diodes (not such a good assumption), no power is lost in those components. If you eliminate the safety protection resistors, then no power is lost in those components either. Since there's no where else for the power to go, such an idealized circuit would give you 100% efficiency. But it would also have some ripple. You may be able to follow this no-zener circuit with a linear voltage regulator to eliminate that ripple and still get a net efficiency over 75%.

The "law" that "a voltage regulator always has an efficiency of \$V_{out}/V_{in}\$" only applies to linear DC to DC regulators. That law doesn't apply to this circuit, because this circuit has AC input, and so this circuit can have much better efficiency than that "law" predicts.