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Re: [Amps] Ceramic capacitor ratings

To: garyschafer@comcast.net
Subject: Re: [Amps] Ceramic capacitor ratings
From: Manfred Mornhinweg <mmornhin@gmx.net>
Date: Fri, 13 Apr 2007 19:00:02 -0400
List-post: <mailto:amps@contesting.com>
Hi Gary,

> 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? 

Yes, and actually this current (to the coil) is much larger than the 
current coming through the the tube! But this doesn't affect the 
calculation at all, since the tube plate, tuning cap top end, and tank 
coil high impedance side are all the same point, RF-wise. So you don't 
need to do vectorial additions of individual currents. It's enough to 
take the single, well defined voltage at that point, and calculate the 
current through any of the circuit elements from there.

 > I read somewhere that an approximation of current thru the
> capacitor was plate current times Q.

Yes, that's a reasonable approximation.

 > I assumed that was loaded Q.

Of course.

>  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.

Yes. But it's not a good idea to do that.

> 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.

Good. That's the correct thing to do.

> 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?

No, it doesn't. You can have a higher loaded Q by having the same 
circulating current, but a lower resistance! You still have the same 
load resistance (50 ohms, probably), but the coil resistance is lower, 
and this simply reduces the total load on the tank circuit, raising the 
loaded Q.

I only wonder that the effect was so strong. I mean, if you get 300 more 
watts out, and really the conditions are the same other than the lower 
coil loss, it means that the original coil was loosing those 300 watts 
plus any power the present coil looses! That old coil must have gotten 
absolutely horribly hot! Otherwise there is some other variable in your 
case, such as changed loading on the tube.

Manfred.

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