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Why doesn't C1 get hot?

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throbscottle

Well-Known Member
I refer to the schematic on this page: **broken link removed** - I've also seen this done in a commercial product. The author provides a really good explanation of how it works, but I don't understand why the capacitor doesn't dissipate power, since when the circuit is loaded it has a voltage across it and passes significant current?????
 
I refer to the schematic on this page: **broken link removed** - I've also seen this done in a commercial product. The author provides a really good explanation of how it works, but I don't understand why the capacitor doesn't dissipate power, since when the circuit is loaded it has a voltage across it and passes significant current?????

Hello there,


Ideal capacitors do not dissipate power and therefore can not get hot, but any real life capacitor does dissipate some power and therefore could get so hot that it explodes. What matters is what the transitory current level is and how often it repeats. If the current level is high enough and the ESR high enough the capacitor gets hot the same way a resistor of the same value as the ESR would get hot because there is still an I^2 *R power loss. And this power loss accumulates over both positive and negative cycles, so even though there is no DC current there is still a transitory current every time the AC line cycle changes polarity. The current I is squared and so it dissipates real power for both half cycles.

We dont see this heating too much in light current circuits because the level of the current is low. But in power supplies it becomes more significant and that means sometimes the input filter caps have to be paralleled because one cap alone could get so hot it builds up pressure inside and the safety vent blows or the whole top blows off the cap (electrolytic caps).

For the circuit pictured the current is probably too low for any significant heating, but given the right circumstances and a cap will get very hot.
 
Capacitive reactance and resistance are seperate things, while the capacitor does in some ways act as a resistor it doesnt dissipate much heat, only the parasitics will generate heat, which for a good cap is very little.

Weird innit.
 
The current and voltage in an ideal capacitor are 90° out-of-phase. Thus the power factor is zero and no power is dissipated.
 
Don't forget that an applications like that has ripple voltage on the cap; hence there will be an AC ripple current. You do need to be careful that the capacitor is designed to handle the ripple. The problem may not be bad for 60 Hz, if the right cap is chosen, however switch mode supplies can operate at much higher frequency, which implies lower impedance, which implies higher current which implies more heating, which often implies ... BANG!

https://en.wikipedia.org/wiki/Capacitor

"Ripple current is the AC component of an applied source (often a switched-mode power supply) (whose frequency may be constant or varying). Ripple current causes heat to be generated within the capacitor due to the dielectric losses caused by the changing field strength together with the current flow across the slightly resistive supply lines or the electrolyte in the capacitor. The equivalent series resistance (ESR) is the amount of internal series resistance one would add to a perfect capacitor to model this. Some types of capacitors, primarily tantalum and aluminum electrolytic capacitors, as well as some film capacitors have a specified rating value for maximum ripple current."
 
I know this capacitor. I have not seen the data sheet recently but I believe it is good for amps. The circuit only has 15mA. The cap is good for 100x that. Don't worry about cap heating.

I have a simple way to think about capacitors. (not 100% correct but simple)
A resistor has no phase shift. Voltage X current = heat.
A capacitor has 90 degree phase shift. The voltage / current curve are shifted 90 degrees. When the voltage is at the peak the current is zero. When the current is at the peak the voltage is zero. Current X 0 = 0 watts. Voltage X 0 = 0 watts.

I know there are more complicated ways to explain; but on a simple level the current and voltage do not happen at the same time.
 
Take a look here Guys:

http://www.marcspages.co.uk/tech/6103.htm

I use a similar circuit but highly evolved from this Basic but very good explanation thereof...

A 2.2 uF X2 will happily give you around 120mA for ever off 220VAC. Without heating up even after you have fed it three times with 400VAC for three runs of 5 Minutes each. Unbustable no matter what your Grid or a runaway generator throws at you.

My little product will be launching latest end of the Year here in South Africa. Three Years of testing have not yielded one failed component or board so far.

No matter what gets thrown at it. Probably ready for the Market here in South Africa.

Regards,
tvtech
 
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Hi everyone

Thanks for the useful and informative responses (especially the article, TV)! My eyes have been opened to a whole world I only partially understood. I love you all :) Praise the gods for ETO...
 
Hi everyone

Thanks for the useful and informative responses (especially the article, TV)! My eyes have been opened to a whole world I only partially understood. I love you all :) Praise the gods for ETO...

Hi throbscottle

You are always welcome. And I praise the Gods for ETO too. And I love you too ;) Without this Forum...well we would all learn very slowly.

And at least, tonight, I contributed a little too :D

All the best,
tvtech
 
ronsimpson,

A resistor has no phase shift. Voltage X current = heat.
A capacitor has 90 degree phase shift. The voltage / current curve are shifted 90 degrees. When the voltage is at the peak the current is zero. When the current is at the peak the voltage is zero. Current X 0 = 0 watts. Voltage X 0 = 0 watts.

I know there are more complicated ways to explain; but on a simple level the current and voltage do not happen at the same time.

Voltage and current do occur at the same time. True, at the maximum values for voltage, the current is zero and vice versa. But, in between those values, both voltage and current are nonzero. The charge from the current accumulation and depletion on the plates causes a energy storing electric field to form and collapse twice every cycle, thereby taking and giving back energy from the circuit twice each cycle also. This results in a net energy dissipation of an ideal cap of zero. However, the ESR and the dissipation factor of the dielectric will "steal" some energy causing heating to occur.

In the attachment, the color red is the sin (voltage), green is the cos (current), and blue is the product (instantaneous power). You can see the average of the instantaneous power is zero.

Ratch
 

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tvtech,

One gripe about the article is that the author says that current exists through the capacitor, which we both know does not happen. It does exist in the capacitor branch, however.

I have seen this "trick" used in other applications where is is desired to reduce AC down to a lower level. I wonder, if instead of capacitors, whether a miniature transformer could not do the same thing. I haven't priced out the cost and size, but it might be something to look into.

Ratch
 
tvtech,

One gripe about the article is that the author says that current exists through the capacitor, which we both know does not happen. It does exist in the capacitor branch, however.

That's not a good gripe. Displacement current is a valid form of current and it is the current the exists in the capacitor dielectric.

Man, you should have been happy that he didn't say the current flows through the capacitor. :)
 
steveB,

That's not a good gripe.

I think it is, and I will explain why below.

Displacement current is a valid form of current ...

Although it was called "displacement current" when it was first proposed, it has since been shown to be an inaccurate description.

...and it is the current the exists in the capacitor dielectric.

That is a rather vague description. Do you mean it is circular within the dielectric, or does a charge unit hop from one plate into the dielectric while another charge unit goes from the dielectric onto the opposite plate? I assume you mean the latter.

You can take it to the bank that no physical charge unit goes through a capacitor, unless there is leakage. If you don't believe that, then how about a vacuum capacitor? Even though no charge can flow, and thereby no current can exist through a capacitor, energy can flow and power can exist through a capacitor. If one divides the power existing in the capacitor by the voltage across it, the result is current. It should be called something like "equivalent current", because it is not a physical current. This equivalent current makes Maxwell's equations come out correctly, and is useful in other electromagnetic calculations.

Man, you should have been happy that he didn't say the current flows through the capacitor.

I am more relieved and grateful than happy.

Ratch
 
Ratch,

Even if you don't think the term displacement current is accurate, it is accepted (even if archaic) terminology. Hence, don't make a gripe against the author of the article. Instead make a gripe against the history of the terminology. ... Well, I guess you just did that. :)

Since we are talking about AC current, even the current in the wire does not flow through. It is charge back and forth. The distorted electron clouds in the dielectric is real charge moving back in forth (at least in some quantum mechanical sense).

The issue with vacuum is what it is: - one of the mysteries we still have not fully penetrated. Still, we say the vacuum is capable of displacement current. Terms are what they are defined to be. Am I being too pedantic? :)

Steve
 
steveB,

Even if you don't think the term displacement current is accurate, it is accepted (even if archaic) terminology.

The terminology may be accepted, but its meaning that you implied to it as being the same as ordinary drift current in a wire is not accepted.

Hence, don't make a gripe against the author of the article.

The author of the article did not mention displacement current, you did. The author implied that drift current existed through a capacitor.

Instead make a gripe against the history of the terminology. ... Well, I guess you just did that.

I only pointed out what others did long before me.

Since we are talking about AC current, even the current in the wire does not flow through. It is charge back and forth. The distorted electron clouds in the dielectric is real charge moving back in forth (at least in some quantum mechanical sense).

The atoms of every material above absolute zero move back and forth from thermal energy. Only when there is net movement in one direction is current defined. In AC there a definite movement in one direction for a time followed by an opposite movement for the same time.

The issue with vacuum is what it is: - one of the mysteries we still have not fully penetrated.

What mystery is that? Vacuum is free space, the very essence of Maxwell's equations.

Still, we say the vacuum is capable of displacement current.

And why not? Energy can be transferred across vacuum with magnetic and electric fields. If you have an energy transfer with a voltage, you have an equivalent current, even if no real current exists.

Terms are what they are defined to be. Am I being too pedantic?

No, your last statement was a simple circular definition, and not perspicacious enough to be pedantic.

Ratch
 
The terminology may be accepted, but its meaning that you implied to it as being the same as ordinary drift current in a wire is not accepted.

The author of the article did not mention displacement current, you did. The author implied that drift current existed through a capacitor.

Are you just making stuff up now? Who said the meaning implied the displacement current is the same as ordinary drift current? I didn't say that. I never saw a text or reference imply that. I didn't see the author of that article imply that. Also, I'm not implying that. Can you show me where that implication took place?

My interpretation is that the word "current" encompasses more than drift current and if someone is speaking generally, that is accepted terminology. You seem to be deliberately trying to find a fault that is not there, but maybe I missed it. If you can show me a place where the author said that drift current is through the dielectric of the cap, then I would understand your criticism. Did he really imply or say that?
 
There is a question here "Why doesn't C1 get hot?".
This thread is now about 'drift current'.
We need a way to explain current with out getting deep into drift current. I need a way to say 'current flows [through or in] a wire or capacitor, with out getting hung up on details. Same thing about 'the speed of electricity'. I want a simple way to explain electricity with out making a one hour talk about things that the questioner does not want to hear.

I think we got lost at ESR. Now drift current. Lets not talk about ESL, or dielectric heating (several kinds), or piezoelectric effect, etc. We need to answer the question, at the level of the questioner, with out writing a book. When talking face to face I can see when someone is tired of the answer. When I write I don't know when to stop.

Capacitors (and inductors) are not like resistors. A cap does not heat. [A 'perfect' cap does not heat.]
 
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steveB,

Are you just making stuff up now? Who said the meaning implied the displacement current is the same as ordinary drift current? I didn't say that. I never saw a text or reference imply that. I didn't see the author of that article imply that. Also, I'm not implying that. Can you show me where that implication took place?

In post #13 of this thread you said, "Displacement current is a valid form of current and it is the current the exists in the capacitor dielectric." Since "displacement current" is an equivalency and not a real physical current, and doesn't exist in the dielectric, I can be forgiven for assuming you are confusing the real current in a wire with the assumed current in the capacitor.

My interpretation is that the word "current" encompasses more than drift current and if someone is speaking generally, that is accepted terminology. You seem to be deliberately trying to find a fault that is not there, but maybe I missed it. If you can show me a place where the author said that drift current is through the dielectric of the cap, then I would understand your criticism. Did he really imply or say that?

The author says in the 5th paragraph "The current through the capacitor (red) will 'lead' the mains voltage (green) by 90º. " Since he says the current passes through the capacitor, he is also saying that current is passing through the dielectric, which it does not. He should have said something to the effect of (the branch containing the capacitor). Drift current is the only type current in a wire, so the author can only mean drift current.

Ratch
 
ronsimpson,

There is a question here "Why doesn't C1 get hot?".

And wasn't that answered by MrAl in post #3?

This thread is now about 'drift current'.

Onward to a new aspect of the problem. Specifically, that a capacitor does not support drift current.

We need a way to explain current with out getting deep into drift current. I need a way to say 'current flows [through or in] a wire or capacitor, with out getting hung up on details.
Do you want to define it or explain it? You define it by saying current is the net movement of charge. You explain it by saying that current in a wire moves by the drift of electrons, as I explained earlier using the marbles-in-a-hose analogy. The drift method is not a deep heavy concept that will derail anybody.

Same thing about 'the speed of electricity'. I want a simple way to explain electricity with out making a one hour talk about things that the questioner does not want to hear.

Electricity is a generic word. What aspect of electricity do you want to explain about? If you are referring to speed, the charge carriers travel at a slow drift speed in a wire, but the response moves at the speed of light.

Capacitors (and inductors) are not like resistors. A 'perfect' cap does not heat.

Yes, that is well known.

Ratch
 
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