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Yes, I suppose for the amateur builder, applying power to your design in and finger testing is an option.? If it doesn’t burst into flames, I suppose its ok?? (In commercial designing we don’t often have that option so we’ll have to stick to our calculus).? It might be wise to invest in one of those IR temperature guns from Harbor Freight Tools for $20 to save your fingertips though!

For LP design work at the approx.. 10 watt level… this just isn’t gonna be an issue unless you are using the wrong filter for the band and the filter is dissipating all of the power.? Once you get to the kW+ region… best use your math.

Like I said in my original note, if you are uncomfortable with the math, there are many calculators on the internet that work.? Just google them.

To be clear about the specification… it says that you would be operating at 0.1% of the full/ capable dissipation properties of the capacitor if it were used in a 50 VRMS circuit at 1 MHz? Using this one data point, you can translate that to any voltage/ frequency point and ratio it to this point to determine what % of dissipation you are operating (designing) at.

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From: [email protected] [mailto:[email protected]] On Behalf Of Jerry Gaffke via Groups.Io
Sent: Sunday, February 18, 2018 2:06 AM
To: [email protected]
Subject: Re: [BITX20] New ideas for the Audio TX/RX pop and PA output stage improvements on ubitx.net

I find that train of thought rather confusing.
Not sure where that 0.3amps average comes from, I'd think you could just find
the impedance from 1/(2*pi*Hz*Farads) and divide that into the 50 volts RMS to find current
rather than bother with the calculus.
Not at all clear how a 0.1% dissipation spec somehow means a max of 0.2 Amps at 30mhz.
A few more details would be appreciated.

I think most builders just try some np0's and if they get warm then replace with a bunch of smaller value caps in parallel.
I'm not aware of any Bitx40 LPF caps blowing out, and I doubt they are anything special.

Jerry

On Sat, Feb 17, 2018 at 02:11 pm, K9HZ wrote:

The current through a capacitor is given by I=C*dV/dt so the average current over time is I=C*int[dV/dt]/t… For your example: let’s use a 100pf capacitor in an 30 MHz circuit at 50 VRMS which is 140V P-P.? Iavg = 0.42 * 0.707 = 0.3 amps average.? Now the spec from that capacitor in your finding is 0.1% at 1 VRMS at 1MHZ which is 0.00028A * 0.707RMS *1000% = 0.2 Amps continuous at 30 MHz.? Yikes we are over by 50%!? Note that this DOES vary with capacitance… that’s why the larger the capacitor, the more current capacity is needed.
?

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