How to calculate the capacitance of a capacitive dropper for a simple power supply

A capacitive-dropper power supply is the simplest way to power a low-power device.

For all its simplicity, it has two drawbacks:
1. It is galvanically connected to the mains! That’s why these power supplies are used only where there’s no chance of touching the contacts.
2. Its output current is fairly low. To increase it, you need a larger dropper capacitor, which quickly becomes quite bulky.

Warning: be very careful. Do not touch the contacts of this power supply while it is plugged in.

The simplest circuit for this power supply looks like this:


As you can see from the circuit, a capacitor is connected in series with the mains. It acts as a ballast, dropping part of the voltage.
A capacitor blocks DC, but the mains is AC, so the capacitor is constantly charging and discharging. As a result, current flows at the output. The current depends directly on the capacitor’s capacitance.

So, to calculate the capacitance, you need to know at least the output current of the power supply you’re planning to build. Be sure to include the regulator’s current consumption too, which is usually a few mA.

There are two formulas: a complex one and a simple one.
The complex formula works for any output voltage.
The simple formula works when the output voltage is no more than 10% of the input voltage.
I — output current of the power supply
Uin — mains voltage, for example, 220 V
Uout — power supply output voltage (or the voltage before the regulator, if there is one), for example, 12 V.
C — the capacitance we’re looking for.


For example, I want to build a power supply with an output current of up to 150 mA. The circuit above is one example; one possible application is a radio remote control powered at 5 V, plus a 12 V relay.
Plugging in 0.15 A gives a capacitance of 2.18 µF. We can use the nearest standard value, 2.2 µF, or, as they say, the “imported” notation: 225.


Everything seems fine—the circuit is simple—but there are a few drawbacks we need to address:
1. The inrush current when switching on can burn out the bridge rectifier.
2. If the capacitor fails, it can cause a short circuit.
3. If left as is, the input capacitor can give you an electric shock, since it may remain charged for a long time after the power supply is unplugged.
4. When the load is disconnected, the voltage across the capacitor before the regulator will rise to a fairly high level.


Solutions:
1. Resistor R1 in series with the capacitor.
2. A 0.5 A fuse.
3. Resistor R2 in parallel with the capacitor.
4. A 12 V TVS diode in parallel with the capacitor after the bridge rectifier. I don’t recommend using Zener diodes here: TVS diodes are designed to dissipate more power, so the circuit will be more reliable.

I’ve highlighted the new components in red in the circuit, and a small addition—a LED—in blue.


Dropper capacitors are also often used in cheap LED bulbs. That’s a bad thing, because these bulbs are less reliable and often have high light flicker.
Below is a simplified circuit for one of these bulbs.


Let’s calculate the capacitance for this application. Since the output voltage will clearly be more than 1/10 of the input voltage, we’ll use the first formula.
I’ve assumed an output voltage of 48 V: 16 LEDs at 3 V each. Of course, this is only an approximation, but it’s close to reality.
Current: 20 mA, a typical maximum current for most indicator LEDs.

I get a required capacitance of 0.298 µF. The nearest common values are 0.27 or 0.33 µF. The first is much harder to find, while the second would give us too much current. So we can make up the capacitance using two capacitors in parallel, for example, two 0.15 µF capacitors. In a parallel connection, capacitances add up.


That takes care of the capacitance. There are two more things to consider:
1. Capacitor voltage rating
2. Capacitor type.

The voltage rating is simple: you can use a 400 V capacitor, but 630 V is more reliable, even though it’s larger.

Choosing the type is a little more complicated. For this application, it’s best to use capacitors designed for it, such as K73-17, CL21, or X2.
The photo shows a CL21 capacitor.


This is a more reliable option. Don’t be put off by the 280 V marking: that’s its AC RMS voltage rating, and it will work more reliably than a K73-17 or CL21.


These capacitors can also look like this:


Now you can take another careful look at what it takes to build this “simple” power supply and decide whether you really need one.
In some situations, yes, it can help. But it has a whole host of drawbacks, so in my opinion it’s better to use a small switching power supply instead. It already has a regulated output voltage, galvanic isolation, and a higher output current.
As an example, here’s a link to a detailed review of four options, with tests, schematics, and teardown photos.


But there’s an even better option. Monolithic power supplies have become widely available. Essentially, they’re little cubes containing a miniature power supply.
For example, the HLK-PM01 from Hi-Link costs about two dollars apiece.


Or there’s the Chinese equivalent, the TSP-05 from Tenstar Robot. They’re a little cheaper, at $1.93 apiece.
In practice, their quality has proved comparable.


As I mentioned above, these are switching power supplies in a modular package. The power supply is housed in a plastic case and potted with epoxy resin.
They’re available with different output voltages and can maintain them at a fairly stable level.


Here’s a closer look inside; the photo shows the Hi-Link version.


That’s about it. I hope you found the article useful, and I’ll try to find more interesting topics in the future. I’d also love to hear what you’d like to see in the “For Beginners” section.