[AMPS] Impedance of amplifiers

[Tom Rauch]

> >  > since 1995, the Handbook has had an inaccurate method of calculating
> >  > RF
> >> current in a DC blocking cap.  My guess is that it will remain there
> >> until sometime around 2040.  .  
> >
> >What is the correct method?
> 
> € Calculate  the difference between instantaneous anode potential at max.
> conduction and anode supply potential.  Divide by 2^0.5 to convert peak v
> to RMS v. .

OK, you found the RMS anode voltage for a sinewave at the anode.

  Disconnect the C1 side of the blocking cap.  Measure anode-C.
> with tube in socket  Calculate capacitive-reactance in ohms. 
> Current=volts/ohms.    

I'm sorry Rich, something appears to be seriously missing in your 
"improved" method.

You have determined anode RMS voltage, and then assumed that 
voltage drives the anode capacitance of the tube and that is the 
current through the blocking cap.

Let's try the "improved" method with a 3CX1200.

Let's assume the RMS voltage is 3000 volts. The anode 
capacitance is about 14 pF. 

On 2 MHz, the anode capacitive reactance is 5.68 k ohms.

3000/5680 is .528 amperes. 

Now let's see if that works. The tube is delivering 1800 watts. 

We have .528 amperes flowing in the blocking cap with 3000 v 
RMS, for a power of 1584 watts. Hmmm. We lost some power.

Let's move the PA to 200 kHz and check the blocking cap current.

We now have .0528 amperes at 3000 volts. Our amplifier is still 
putting out 1800 watts and only  .0528 amperes at 3000 volts is 
driving the tank. That's neat!!

If we use your formula at audio, we can have 1800 watts output 
with nearly zero anode current. If we rectify the output and feed it 
back to the power supplies of similar stages, we can run an infinite 
number of PA's without need for power from the power mains.

Where does tank circuit Q and power enter the formula? Did you 
leave it out on purpose?

Something seems very very wrong with your formula. Maybe that's 
why they didn't use it?



  
73, Tom W8JI
w8ji@contesting.com

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