I want to charge a capacitor at 200V DC from a 220V AC power supply (peak voltage = 311 V). Also I want no power disipation once it reaches full charge. This goes inside an equipment that discharges the capacitor sporadically.

The first thing I came up with is a half-wave rectifier diode followed by a resistor voltage divider, but the drawback is the useless power loss in the divider once the capacitor is fully charged.

Another aproach is this: half-wave rectifier diode followed by a ~100 V Zener diode, a resistor and finally the capacitor. I should do this but I don't have the zener diode and I don't like this design because is not very flexible in case I need other voltage.

I suspect that there is another clever way of achieving this in a simpler and flexible way. Does anybody knows one?


  • Capacitor value is in the order of 1uF.
  • Charging time is not an issue as long as it is not > 1s.
  • Capacitive divisor is not an alternative (see my reply to one of the answers).
  • By "efficiently" I do not mean losing power on resistors DURING charging time; I mean it once the charging is done.
  • \$\begingroup\$ with 100V drop on the zener, you would be getting 1W of dissipation for every 10mA - so you'd need a big resistor to keep the zener from melting down, and that would make your charging time longer. btw, you didn't mention the size of the capacitor. \$\endgroup\$ – JustJeff Dec 6 '11 at 1:06
  • \$\begingroup\$ It is a 470nF capacitor. Maybe in the future I will have to change it to another similar value. Charging time is not an issue... I'm fine with < 1s. Thanks! \$\endgroup\$ – falbani Dec 6 '11 at 18:05
  • \$\begingroup\$ Remember that power line voltage varies over time and from place to place. 220 VAC is just an approximation. If you need it to be exactly 200 VDC, you're going to need some kind of regulation. \$\endgroup\$ – endolith Dec 12 '11 at 15:28

You had the right idea with a voltage divider, but as you say resistors will always dissipate power. The answer is therefore a capacitive voltage divider. It works just like a resistive voltage divider except that the impedance goes down with frequency. Your frequency is well known and controlled, so a capacitive voltage divider is quite appropriate. Adjust the capacitors so that the peak output voltage is the 200V you want to charge the output cap to, and put a diode to this output cap like you would from any other 200V AC source.

One thing to watch out for though. The power line will occasionally have large voltage spikes. These will make it thru the capacitive divider proportionally just like the intended voltage. This could possibly overcharge the output cap on rare occasions. It might be a good idea to put some kind of clamp circuit accross the cap that turns on when the voltage is a little above the normal value. This could save a few random mysterious field failures years after installation.


Here is what I was talking about in the comment:

You have 220V AC sine coming in, so the peak voltage is 220V * sqrt(2) = 311, and the peak to peak voltage is twice that or 622V. C1 and C2 must be sized to reduce that p-p to 200V, so C2 must have 2.11 times the capacitance of C1.

Think about the circuit with only C1 and C2. The node between these two now is at 200V p-p. Diode D1 pushes the DC bias on the capacitors so that the negative peak is 0V or more, and D2 pushes it so that the positive peak is 200V or less. When C3 is at less than 200V, then the circuit acts like a charge pump and each cycle will add a little charge onto C3. How much depends on the absolute value of the capacitors and the cycle frequency. Once C3 reaches 200V the AC peaks just hit the limits where the diodes conduct but don't actually transfer any current. In other words, once C3 is fully charged, there is no more power drain from the AC line.

Note that while there is no power drain when C3 is fully charged, there is current. However, this current is 90 degrees out of phase with the voltage, which is how no power is tranferred. Technically this presents a poor power factor. However, in general the AC power grid is more inductive than capacitive and power companies actually maintain banks for capacitors to try to even this out. In most cases you're actually helping the power company by offsetting some local inductive load. Also from what you say this is apparently quite low power. A few mA reactive current either way isn't a issue.

  • \$\begingroup\$ I even build this configuration. But before I tested it, I realized a "prohibiting feature": once my capacitor is discharged for its purpose, it will never be charged again, as the other one "expands" to 311 V. It is a "one shot" solution. I should have mention it. Thanks! \$\endgroup\$ – falbani Dec 6 '11 at 18:08
  • \$\begingroup\$ @Falbani: You can add a diode and turn it into a charge pump, but still use a additional capacitor to attenuate. The two capacitors bring the peak to peak swing down to the 200V you want. The didode to ground and the diode to the output cap will limit the swing to exactly that range. \$\endgroup\$ – Olin Lathrop Dec 6 '11 at 20:18

First, I'll assume you know how to work safely with line or mains voltages. If you don't, for your own safety, don't mess around with this. Remember that your charged capacitor is potentially dangerous even after line voltage is removed from your circuit.

If your line voltage is well-enough regulated that your zener solution would work, you could also just use a transformer to reduce your 311 V(peak) AC to 200 V(peak). Then use a half-rectifier and diode as you described to charge your capacitor.

Another solution might work if you are only discharging the capacitor infrequently (every few minutes?). This will give a more carefully regulated charging voltage (say 5% instead of the 10 or 20% that your line voltage might vary) than the transformer approach. Use a half-rectifier, followed by a resistor, followed by a relay, followed by a diode to charge the capacitor. Size the resistor to give a fairly long time constant (10's of cycles of the input line) in combination with the capacitor. Use a voltage divider to sense the capacitor voltage and only close the relay when the capacitor voltage drops below your target value. The divider circuit can have a series resistance in the range of 10 MOhms to minimize leakage when the cap is charged.

  • \$\begingroup\$ Thanks for the advices! I'm a "near graduate" EE working in a laboratory where this things are done frequently enough for making everybody in here very careful. The capacitor is ideally used only once at start time of the equipment, but on rare occasions it will need to be used again. Your proposed solution (and variations based on sensing) are more complex than what I expect for this simple job. Besides I need this working, it is also an interesting intellectual challenge that I feel it has to have a simple solution. \$\endgroup\$ – falbani Dec 6 '11 at 18:16
  • \$\begingroup\$ I think the transformer solution is probably conceptually the simplest, although its also likely to be bulky and expensive and not flexible if you want to change the output voltage. Olin's capacitive divider circuit is probably the next simplest and probably the best solution overall. My second solution is meant for the case when you really need to accurately set 200 V on the capacitor (not 195 or 205 V). If you're okay with the capacitor voltage ranging from maybe 180 to 22o V, then the simpler solutions are definitely better for you. \$\endgroup\$ – The Photon Dec 6 '11 at 21:11

NB Uninsulated mains kills.
The whole circuit below is potentially at mains voltage.

If discharges are only occasional a transformer based 200V power supply makes sense.

There is NO way to charge a capacitor efficiently using resistive drop or wires that dissipate energy (= low ohm resistors) from a fixed supply voltage (Note "trick" below) .

To gain efficiency with a fixed source supply you MUST "boost" the voltage or "buck" the voltage using either capacitors standing on each other's shoulders techniques (usually not practical) or inductive energy transfer. Charging the capacitor at a voltage just in excess of its current voltage can achieve this. This can be from the output of a boost or buck converter BUT ...

A trick: Your input AC waveform assumes all voltage levels between 0 & 200 V as it heads towards its 311 V peak. If the capacitor started off at 0 V and was then attached to mains at 0 V (via a diode or switch) then as the mains rose the capacitor would follow it. As main impedance will be lower than capacitor impedance this would be quite effective AS LONG AS your capacitor can tolerate the current required to charge in under a 1/4 cycle of mains. ie about 5 mS. What size is you capacitor?

You could do this over multiple cycles a burst at a time. Efficiency will drop

The capacitor may be kept charged efficiency" by connecting it to main occasionally as the mains voltage transits the capacitor voltage. This can be exciting if you get it wrong , and takes some head scratching but can probably be done with a few suitable voltage rated transistors and due thought.

This circuit works on a similar principle and you can get ICs that do this.
Here the FET is turned on by gate pulses from the mains BUT when Vmains rises too high the 2n2222 clamps the gate and keeps the FET off. This circuit would notionally solve yur whole task subject to suitably rated components. The gate would need to be zener clamped when you use such a high voltage. Needs more than usually careful thought.

What it does that's good:

  • When the cap is fully discharged the FET gate ispulled high via the 47k turning the FET on

  • Cout (why are there 2?) :-) - charges via FET. Cap voltage tracks mains volotage so efficincy is high. Cap charges in under 10 mS.

  • When V= raches 200V so Vcap ~= 200V then transistor base is at 200 x 1/331 ~= 0.6V
    Transistor turns on clamping FET gate turning off FET so that cap stays at 200V.

  • If Vcap falls below 200V it will be "topped up.

  • As seen here loss in 47k at 200 V =~ 0.85 Watt = too high. A far more power effective gate circuit is possible with minimal power loss.

Hacked about it looks something like this:

enter image description here

Having started from this:

enter image description here

Here's a really fun buck converter solutin using an LNK304 IC about$1 /1 at Digikey

Data sheet above or application comment here

enter image description here

  • \$\begingroup\$ I'm very grateful for all the time you took, but I'm afraid this is way complex than I had in mind. I only hope this will be useful for another user reading this post. I should clarify this: by "efficiently" I do not mean losing power on resistors DURING charging time; I mean it once the charging is done. \$\endgroup\$ – falbani Dec 6 '11 at 18:27
  • \$\begingroup\$ @falbani - No problem. If only power on standby then my 1st anseer applies = "If discharges are only occasional a transformer based 200V power supply makes sense." This could have a simple "pump up when low then turn off regulator". \$\endgroup\$ – Russell McMahon Dec 6 '11 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.