# Efficiency of capacitor as AC current limiter

Discussion in 'General Electronics Chat' started by rudyauction8, Jun 15, 2014.

1. ### rudyauction8 Thread Starter Member

Jan 27, 2012
252
2
I've recently been using old induction motor run capacitors to limit AC current mainly because of how simple it is. Say I have a 24v 5 amp transformer and need to charge a 12 volt battery. Directly connecting it through a rectifier gives me a burned transformer. A switching current regulator is ~70% efficient and somewhat hard to build, but adding a couple capacitors between the transformer and rectifier is simple and works great.My question is how efficient are they? There's a large voltage drop yet they stay cool to the touch unlike a linear regulator, so I'm assuming they are as efficient or better than the switching regulator but I'd like to be sure.

2. ### crutschow Expert

Mar 14, 2008
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3,353
A capacitor dissipates no power is it is basically 100% efficient at reducing voltage/current in an AC circuit. But note that a capacitor can only be used in series with a full-wave bridge rectifier, not a single half-wave rectifier.

3. ### Alec_t AAC Fanatic!

Sep 17, 2013
5,949
1,129
That is true only if the cap has no resistance. In practice a cap has an Equivalent Series Resistance (ESR). Caps carrying high ripple current in power supplies can get hot enough to destroy them if they develop a high ESR.

4. ### #12 Expert

Nov 30, 2010
16,655
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and to the contrary...One of my teaching demonstrations is to connect a motor run capacitor directly across the power line, teach the class for an hour and then show them that they have been sitting next to a capacitor "shorting" the power line for an hour, and it's not even warm.

A good motor run capacitor is a marvelous thing. It can pass a few amps of current all day without getting hot. Just don't try a low voltage capacitance meter on them. They don't read right at 2 or 3 volts when they are designed for 200V to 440V.

5. ### wayneh Expert

Sep 9, 2010
12,362
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It's not really to the contrary, it's just a special case where the frequency is low enough and the capacitor capacity small enough, and the package big enough, that the current and heat dissipation is not a problem. High capacity in small package, plus high frequency, equals poof.

Makes me wonder where the boundaries are - what combination of parameters is OK and what combination is bad news.

Last edited: Jun 16, 2014
6. ### #12 Expert

Nov 30, 2010
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I guess the best you can do is run the math on a case by case basis.

7. ### Bernard AAC Fanatic!

Aug 7, 2008
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Voltage/ current limiting with a C is load dependent, transformers, small motors works well but if load is V or I sensetive [ LED ], trash on line can feed thru & cause trouble.

8. ### AnalogKid Distinguished Member

Aug 1, 2013
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At any particular frequency, a capacitor has an impedance that can be calculated. But note that even though it is stated in Ohms, this is an impedance, not a resistance. For a sine wave, a capacitor introduces a 90 degree phase shift between the phase angle of the voltage across it and the current through it. It is this phase shift that leads to the relatively low device heating. When the voltage is at its peak, the current is zero, and when the current is at its peak the voltage is zero.

Other factors such as materials, internal construction and size also affect heating, but the biggie is the phase shift. The math behind this uses complex numbers and the imaginary operator i. This is why capacitors and inductors are sometimes said to consume or dissipate "imaginary power."

Transformers step voltages up and down with very high efficiency, and switching power supplies do essentially the same thing, relying on the action of a transformer to do the real power conversion. A series capacitor is a very different animal, and uses different physics and math to achieve what might appear to be the same end result. For this reason, using the word "efficiency" to relate the two can lead to a misunderstanding of what is going on

ak

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9. ### #12 Expert

Nov 30, 2010
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@rudy:
The fact that your capacitors run cool is a reasonable indicator that they aren't using up much power. (Your instincts are correct.) The how and why were covered by other people. If you want to get a better measurement, you might tape a thermocouple to one of the caps and wrap it in insulation. Change in temperature/change in time will give you a clue that you can use math on.

10. ### Austin Clark Active Member

Dec 28, 2011
409
44
I've done the math before and was shocked to learn that charging a capacitor is only 50% efficient if it's done through a resistor. I imagine you can get greater efficiencies with inductors combined, but I haven't looked into it yet myself.

11. ### crutschow Expert

Mar 14, 2008
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True. A ideal inductor will charge an ideal capacitor with 100% efficiency. That's the basis for the high efficiencies of boost/buck regulators.

12. ### Austin Clark Active Member

Dec 28, 2011
409
44
How would they be configured to do this? Do you have to somehow match the Inductance and the Capacitance?

I may need to start a new thread on the topic. Sounds interesting.

13. ### rudyauction8 Thread Starter Member

Jan 27, 2012
252
2
Thanks, that's what I thought, I'll keep using them in my battery charger circuits.

A side question, if I was to, for example (I don't plan to do this), short hot to ground of an electrical outlet with a capacitor, it'll limit current and power, but where is the power it lets through dissipated? Like if it lets through 1 amp that's ~120 watts, where does that power go? Will it heat up in this situation, will the wires in the wall heat up? Will the power somehow just circulate back through the grid? I'm just trying to figure out how exactly they work.

EDIT: replied before refreshing page after my computer sat a couple days, I'll have to read through the other posts now!

14. ### inwo Well-Known Member

Nov 7, 2013
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It's 120 VA not 120 watts.
The voltage and current are out of phase. ie.
Yes, current is circulated. The only power consumed is in any resistance it might encounter. In the wire, capacitor plates, and transformer windings.

Last edited: Jun 18, 2014
15. ### AnalogKid Distinguished Member

Aug 1, 2013
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If you size the capacitor for 1 amp of current, that current goes to where the current for a 120 W light bulb goes - back to the generator in a complete circuit. The capacitor is not a short circuit, it is a relatively low power load just like a light bulb, so no, the wires in the wall will not heat up. The capacitor will, though not nearly as much as a light bulb, for the reasons in the previous posts.

ak

16. ### crutschow Expert

Mar 14, 2008
13,462
3,353
Here's some info to start.

The capacitor is not "matched" to the inductor (except for Resonant
Switching designs). The capacitor is just a storage filter that smooths the pulses from the switched current through the inductor.

Austin Clark likes this.
17. ### rudyauction8 Thread Starter Member

Jan 27, 2012
252
2
OK, but since there's still 120 watts of current going through where is the rest of the power dissipated, or am I missing something? Would my electric meter say 120 watts with the capacitor?

18. ### #12 Expert

Nov 30, 2010
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P = IE cos theta. Hint: What's the cosine of 90 degrees?

19. ### inwo Well-Known Member

Nov 7, 2013
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See post 14..................

It's not 120 watts. No your meter would not be registering 120 watts.

Amps times volts does not = power in ac circuits. As #12 explains mathematically. volts X amps X zero

I see this mistake used in power saving scams.
Current is measured at an appliance.
It is then plugged into "magic device".
Current is less.

Does less current prove lower light bill?
Not necessarily.

20. ### rudyauction8 Thread Starter Member

Jan 27, 2012
252
2
OK, that explains a lot. So if after I use a capacitor through a rectifier my multimeter says 4 amps DC to battery is that the correct reading, and since the battery is at 12v I'm charging at about 50 watts, depending on the exact voltage of the battery? Just making sure my charger hasn't been lying to me this whole time.