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I want to limit current from an AC source using capacitors. As I am using dielectric (polar) capacitors, I thought I could use 2 capacitors with opposite polarity in parallel with a diode in front of each capacitor to protect it from reverse bias damage (see circuit diagram for clarification). My thinking was that every time the source's polarity switches, the current will go through a different capacitor-diode pair, always with the capacitor's reactance limiting the source's current.

When attempting to do this in practice, no current moves at all, so I must be misunderstanding something very important. Would any kind sage be interested in pointing out my error? I'm a physicist who likes to tinker, so if this is something embarrassingly stupid, that I should have learned in AC 101, I apologize. Circuit Diagram

I'm mostly interested in understanding the general idea of why this won't work, but if you'd like details, the AC source is mains (120VAC, 60Hz), the capacitors are 22uF, 400V, the diodes are 5A, 1000V rated.

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    \$\begingroup\$ How are the capacitors supposed to discharge? \$\endgroup\$ Sep 24, 2016 at 2:10
  • \$\begingroup\$ And what's the load? \$\endgroup\$ Sep 24, 2016 at 3:35
  • \$\begingroup\$ Ooops... So then would there be a configuration that would both protect the capacitor and discharge it? In this case the load is an AC heating element, but I was curious about general solutions. \$\endgroup\$
    – Chuck P
    Sep 24, 2016 at 3:40
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    \$\begingroup\$ Connect the capacitors in series, connecting both the positive wires together, or connecting both negatives together. Capacitance will half, voltage will increase some between 1.01 and 2 times depending on their quality and they will become unpolarised. \$\endgroup\$
    – Asmyldof
    Sep 24, 2016 at 7:00
  • \$\begingroup\$ @Asmyldof: That will result in a stippled ceiling. Your arrangement will have half mains voltage across each capacitor. You need reverse polarity prodection. See my answer. \$\endgroup\$
    – Transistor
    Sep 24, 2016 at 8:04

3 Answers 3

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I'm mostly interested in understanding the general idea of why this won't work ...

Because there is no discharge path once the capacitors are charged.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Rearranging the diodes provides a discharge path for the capacitors and reverse polarity protection.

How it works:

  • When L goes positive D1 is reverse biased and C1 charges. D2 prevents C2 being charged more than 0.7 to 1.0 V reverse.
  • When L goes negative the reverse happens.

The effective capacitance will be the value of one of the capacitors since only one is in use at any time.

I used this arrangement almost 40 years ago to replace a failed starting capacitor on a family friend's 230 V well pump. (It died on a Friday evening on a bank-holiday weekend. How do capacitors know what day it is?) I used a couple of valve amplifier capacitors and suitable diodes. It worked but I was concerned about the long-term reliability due to the small reverse bias on each cycle so I replaced it with an unpolarised capacitor as soon as possible.


Update after comments.

schematic

simulate this circuit

Figure 2. Original redrawn with GND applied between the two capacitors to assist in visualisation with the simulation.

enter image description here

Figure 3. Results of simulation of Figure 2 when V1 is set to 100 Vp-p.

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  • \$\begingroup\$ Thanks! I'm curious as to why the standard solution seems to not include the diodes, as your circuit looks like it would result in the capacitors operating nearly identically to their intended specs, likely increasing lifetime. Are the diodes left out just for bulk/cost reasons? \$\endgroup\$
    – Chuck P
    Sep 26, 2016 at 18:08
  • \$\begingroup\$ The following statement is wrong: "The effective capacitance will be the value of one of the capacitors since only one is in use at any time." While with the first cycle of the input voltage, the capacitors are uncharged and positive current only flows through C1 and D2 and negative current only flows through C2 and D1, both capacitors eventually will become charged and their + terminal will have a higher voltage than their - terminal. Thereby they reverse bias their parallel connected diodes and block current from flowing through them. \$\endgroup\$
    – GerdF
    Jun 25, 2019 at 7:13
  • \$\begingroup\$ ... All current now has to flow through both capacitors, making the effective capacitance the half of one capacitor (if both of them are of same size). \$\endgroup\$
    – GerdF
    Jun 25, 2019 at 7:13
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Here's what happens as time goes by...

enter image description here

For power applications, a capacitive dropper isn't always a good choice, but when it is it's far better to use a single film capacitor, as shown below, than to use a pair of electrolytics hooked in series opposition, as suggested earlier.

The reason for that is that the electrolytics will always dissipate more power in their electrolytes and dielectrics than the film caps will in their dielectrics, and the electrolytics will run hotter and vent their magic smoke sooner.

enter image description here

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  • \$\begingroup\$ Thanks! I was intending to change for 2 dielectric capacitors in series, but now I'll look into getting some film capacitors. \$\endgroup\$
    – Chuck P
    Sep 26, 2016 at 18:15
  • \$\begingroup\$ All capacitors depend on electrically insulating dielectrics to separate their plates and, as far as I know, there's no such thing as a "dielectric capacitor".\ \$\endgroup\$
    – EM Fields
    Sep 26, 2016 at 18:31
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Each C would conduct current during one half of the first cycle and then would be charged to the peak +/- voltage then stop conducting current.

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  • \$\begingroup\$ Ya, thanks. Ignacio pointed out the issue. Any suggestion for a different setup with polar capacitors in an AC circuit? \$\endgroup\$
    – Chuck P
    Sep 24, 2016 at 5:31

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