[Amps] Ceramic capacitor ratings

[Gary Schafer]

Hi Manfred,

Thanks for the comments. I understand what you are saying about the current
thru the capacitor by using the plate voltage to figure the current thru its
reactance but isn't there also circulating currents in the tank between coil
and capacitor? I read somewhere that an approximation of current thru the
capacitor was plate current times Q. I assumed that was loaded Q.
 I understand that when reducing L you must raise C to maintain resonance
and therefore the reactance of C goes down and current goes up.

But in my case where I installed a more efficient coil, I taped the coil so
that I had resonance with approximately the same amount of plate tune C as I
had before. The new coil has less resistive loss so the loaded Q should be
higher as evidence of more power output and also sharper tuning (less
bandwidth). So if the Q is raised keeping the same plate tune reactance then
circulating current has to go up and so doe's current thru the capacitor
does it not?

73
Gary  K4FMX


> -----Original Message-----
> From: amps-bounces at contesting.com [mailto:amps-bounces at contesting.com] On
> Behalf Of Manfred Mornhinweg
> Sent: Friday, April 13, 2007 3:31 PM
> To: amps at contesting.com
> Subject: Re: [Amps] Ceramic capacitor ratings
> 
> Hi ye all,
> 
> I feel you are worrying too much about Q and the like. Make it simple,
> guys! Start from the basics:
> 
> You know the operating frequency, and the total tuning capacitance
> value. Take the maximum value, 127pF in this case (200 in series with
> 347). Calculate the capacitive reactance from this: Xr = 1/(2*PI*f*C).
> Now take the maximum RF RMS voltage at the tube plate. It will be quite
> close to the DC plate voltage, so for simplicity you might even cheat
> and use the DC voltage. Now apply Ohm's Law, and you have the current
> through the tuning caps. Compare this current to the maximum current
> rated by the manufacturer of the caps (I hope you can find this value!),
>   not forgetting that you have two in parallel, and you are done.
> 
> If the results look very bad on the higher bands, it's because we are
> assuming maximum capacitance. Try doing the math again with the actual
> tuning capacitance used on each band.
> 
> For sure, I would add a healthy safety factor, but that's me. Many
> commercial manufacturers don't do it, as they have to watch the cost.
> 
> Now the practice: On 18MHz, 127pF has 70 ohms. With 3000V rms of RF,
> that would give 42A. Poof...!  But if the amp tunes to 18MHz with just
> 20pF total tuning capacitance, you get less than 7A, which seems fine.
> Since the amp anyway won't be able to run with a high tuning capacitance
> setting on this band, it's probably OK.
> 
> And assuming it needs the full 127pF at 1.6MHz, the reactance there
> would be 783 ohms, which gives you a current of 3.8A. Surely that's
> fine. But clearly this amp has a low loaded Q on the low bands, unless
> additional capacitance is switched in!
> 
> So, it seems you amp is designed with a sensible safety factor, but not
> going overboard.
> 
> The change of coil will not significantly change the current through the
> tuning caps, as long as you implemented it with the same inductances as
> the original one. If its unloaded Q is higher, as it should, the added
> power output will come from reduced dissipation in the coil, and the
> LOADING capacitor will be exposed to a slightly higher voltage and might
> need significantly different values than before.
> 
> Cheers,
> Manfred.
> 
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