>
>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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