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Re: How could a pulse cap operate in TC?
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- Subject: Re: How could a pulse cap operate in TC?
- From: "Tesla list" <tesla@xxxxxxxxxx>
- Date: Thu, 21 Jul 2005 12:08:08 -0600
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- Resent-date: Thu, 21 Jul 2005 12:15:31 -0600 (MDT)
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Original poster: "Gerry Reynolds" <gerryreynolds@xxxxxxxxxxxxx>
Hello father,
Original poster: "father dest" <dest@xxxxxxxxxxx>
> >ESR = dissipation factor / (2 * PI * F *C)
>
> ??? This doesn't sound right. CDE 942C20P15K caps (0.15uf) specifies
> a typical ESR of 5 milliohms
what? have you even try to calculate the numbers? i see that you don`t :-P
so go (for example) here:
http://hot-streamer.com/TeslaCoils/MMCInfo/942C.pdf
I went there, and didn't see anything surprising. It shows ERS for
CDE caps in the range I said. So I guess I'm missing your point.
"ESR - Maximum value measured at 25oC, 100 KHz"
at 100 khz (2 * PI * F *C)=0,0942
DF~0.0005 for PP, so ESR=0,0053 ohm
what did you say - "This doesn't sound right"? :-P
I dont know where your equation comes from. It isn't in my text
books and not at the link you point to. The text books say the DF is
fairly constant over a wide frequency range and is very close to the
PF for DF's <0.15. If DF is almost constant then PF is almost
constant suggesting that the ERS is almost constant. Your equation
says the ERS goes DOWN with INCREASES in frequency and this is
opposite of what I would think. It still sounds like something is not right.
> I believe Steve said that the dielectric loss was insignificant at our
> frequencies.
the question is not in "our frequencies", but in value of current!
it`s seems that no one here can explain it simply and clearly to me, so -
ok, just forget it :-)
For a circuit where the current is determined by the capacitance
value, sure the rms current will go up as the capacitance goes
up. This doesn't mean the ERS goes up. It does mean the losses goes
up (I^2 * ERS). The ERS number of a component tells me how to
compute the heating effect within the component. Purely reactive
components can not heat. It take an Effective Series Resistance to
heat the component. So I'm still at a loss as to your point.
Gerry R.
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