[AMPS] Conjugate matching and amplifiers

[measures]

>
>Roger D. Johnson wrote:
>
>>Poor Jon is going to think I'm picking on him but I must disagree on his
>>statements about conjugate matching of amplifiers. In reality the PA tube is 
only
>>matched to the extent that it produces the desired power output.
>
>Nope, not picking on me!  It's a good discussion!
>
>First of all, I do agree with your statement.  In reality your right, we 
>match the tube for our desired power output.  But is "desired" power 
>output really the most efficient or is it the maximum available power 
>output?
>
€   Who tunes for max efficiency?  

>>We can see this by the design
>>steps required. First, the output power is selected. Then we pick a tube and 
a
>>reasonable anode voltage to produce that power. Then we compute the necessary
>>current and multiply that by a fudge
>>factor for the class of operation. The anode voltage is divided by the
>>previous value and that gives the impedance the tube must work into to 
produce the
>>desiredpower output.
>
>OK, You are correct here.  The tube needs a specific load impedance to 
>work into.  For my 4-1K if I remember correctly it is on the order of 
>4700 Ohms real resistance.
> 
>>A matching network is then designed to transform that impedance to 50 ohms 
(usually).
>
>Correct.
>
>>Notice that at no time was the actual output impedance or load
>>resistance of the tube used!
>
>Huh?  

€  it's gotta be a typo.  

>.......because the tube is not linearly biased.  It acts more as a switch 
......

€  Undoubtedly unsliced bologna.  You outta know better, Jon.  The tube 
is typically biased linearly for over half of the 360-degree cycle.  


>and therefore that impedance varies over the drive cycle.  

€  The critical moment is the instant of max. peak emission when 
instantaneous anode volts reach a minimum.  .  

> This is the reason 
>also that the tank circuit needs to have a fairly decent Q so that it can 
>constantly deliver energy to the load over the drive cycle of the tube.  
>I wouldn't want to stake my life on it, but that specified load impedance 
>is probably pretty close to the average output impedance of the tube over 
>the drive cycle.  That's the way the laws of physics require it to be in 
>order to maximize power transfer.
>
>>The fact that the tube is not "matched" to the load also explains why
>>most of the reflected power is again reflected at the amplifier and goes 
back towards
>>the antenna.
>
>No, that's not correct.  If the output of the amplifier tank circuit is 
>conjugately matched to the impedance of the feed line, you won't get any 
>power reflected back to the antenna.  The fact that the tube might not be 
>matched completely to the tank doesn't affect that.  The tube not being 
>completely matched to the tank means that you won't be able to deliver 
>all of the available power to the tank in the first place.  

€  When the tank is tuned for max out into a less than perfect load, one 
is delivering all there is.  //  Who tunes their amplifier for a 
conjugate match?

>In other words, a mismatch between tube and tank decreases gain and 
efficiency.
>
€  not if the tank is adjustable.  
>........

-  Rich..., 805.386.3734, www.vcnet.com/measures.  


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