Electronic – Capacitive power supply – Bleeding resistor reliability considerations

This circuit is one of a category of circuits called a "Transformerless AC to DC Powersupply" or a "CR dropper circuit". For other examples, see "Massmind: Transformerless AC to DC Powersupply" or "Massmind: Transformer-less capacitive bleed power conversion" or "ST AN1476: Low-cost power supply for home appliances".

Such a device has a power factor near 0, making it questionable whether it meets EU-mandated power factor laws, such as EN61000-3-2. Even worse, when such a device is plugged into a "square wave" or "modified sine wave" UPS, it has much higher power dissipation (worse efficiency) than when plugged into mains power -- if the person who builds this circuit does not choose safety resistors and zener big enough to handle this additional power, they may overheat and fail. These two drawbacks may be why some engineers consider the "CR dropper" technique "dodgy and dangerous".

How does the capacitor step down the voltage?

There are several ways of explaining this. One way (perhaps not the most intuitive):

One leg of the capacitor is attached (through a safety resistor) to the "hot" mains which oscillates at over 100 VAC. The other leg of the capacitor is connected to something which is always within a few volts of ground. If the input were DC, then the capacitor would completely block any current from flowing through it. But since the input is AC, the capacitor lets a small amount of current flow through it (proportional to its capacitance). Whenever we have a voltage across a component and current flowing through the component, we electronics people can't resist calculating the effective impedance using Ohm's law:

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

(Normally we say R = V/I, but we like to use Z when talking about the impedance of capacitors and inductors. It's tradition, OK?)

If you replace that capacitor with a "equivalent resistor" with a real impedance R equal to the absolute impedance Z of that capacitor, "the same" (RMS AC) current would flow through that resistor as through your original capacitor, and the power supply would work about the same (see ST AN1476 for an example of such a "resistor dropper" power supply).

Does the capacitor waste power as heat?

An ideal capacitor never converts any power to heat -- all of the electrical energy that flows into an ideal capacitor eventually flows out of the capacitor as electrical energy.

A real capacitor has small amounts of parasitic series resistance (ESR) and parasitic parallel resistance, so a small amount of the input power is converted to heat. But any real capacitor dissipates far less power (far more efficient) than a "equivalent resistor" would dissipate. A real capacitor dissipates much less power than the safety resistors or a real diode bridge.

If the zener were gone and the output was let to float around 50V ...

If you can tweak the resistance of your load, or swap out the dropping cap for one with a different capacitance of your choice, you can force the output to float at close to whatever voltage you choose. But you will inevitably have some ripple.

If the zener were gone and the output was let to float ... would it approach 100% efficiency?

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 ?)

Use a relay driven by the rectified/filtered power. Connect the NC terminal to the bleeder resistor.

^{simulate this circuit – Schematic created using CircuitLab}

Find a relay with suitable coil voltage to not need separate regulation/resistance for your output voltage.

Btw: Your diagram is DANGEROUSLY WRONG. With 120V AC input, the load/device you're showing marked as "64V" will see up to 180V, as the 120V is RMS-to-zero, not peak-to-peak, and if there is no load, the capacitor will keep the voltage at the peak value rather than the RMS value.

If you really need 64V unregulated, I'd recommend a 2.5:1 transformer added to your circuit. Also, the resistor in the diagram (please label your components!) seems like a waste. It will just burn off power if you draw power, and do nothing if you don't. If it's there to lower the voltage to 64V, then that will only happen if the load is very even and well-characterized. And the resistor will still get very warm -- power equals voltage squared divided by resistance. If it has to drop 100 Volts at 60 mA, it's about 1.7 kOhm, and 60 mA through 1.7 kOhm is over six watts!

I'm tempted to ask what probably you're actually trying to solve here. Working with mains power is pretty dangerous, and extra so if you're going to hook something into power permanently, or if anyone other than you will ever go near the device.