>
>measures wrote:
>
>>>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?
>
>You really can't tune for max efficiency. The efficiency part of the
>equation is taken into account in the design of the amplifier,
>particularly the tank circuit. You really can't "tune" for efficiency.
? Sure you can. It takes a ton of calculation and a sheet of graph
paper.
>
>>>.......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.
>>
>OK, Rich. You're correct there. But so am I. Class AB is not purely
>linear, you state it yourself - it's only linear for over half of the 360
>degree cycle. The tube does act as a switch as well.
? inexpensive sausage
>What I meant by
>linear bias was linear biased over the entire 360 degree cycle (class A).
>>
? . The tank fills in the missing half provided that enough Q is
provided by the designer. Did I catch of whiff of herring?
>>>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. .
>>
>Agreed.
>
>.......
>>? When the tank is tuned for max out into a less than perfect load, one
>>is delivering all there is.
>
>Not necessarily, Rich. You tune the tank for the best obtainable match
>you are able to give.
? I instead tune the tank for max watts.
> But it isn't necessairly optimal.
? When the Tune and Load are peaked for max out, that is a optimal as
optimal gets.
>Unless you are
>able to conjugately match the output of the amplifier to the impedance
>seen looking down the feedline, you won't get 100% power transfer. Some
>power will be reflected back into the tank. The only way you'd really be
>able to do this is likely with a roller inductor for your L.
>
? A variable inductor with a variable Tune and Load C only allows you to
maintain a constant Q.
>> // Who tunes their amplifier for a
>>conjugate match?
>
>I would like to. Unfortunately, I don't have a roller inductor so I can
>only get as close as possible. Plus, most folks design their tank
>components for an output impedance near 50 Ohms. If you try to match a
>750 Ohm antenna into that kind of a design, you likely won't be able to
>get a conjugate match. So instead you tune for maximum power output. But
>maximum power delivered to the load is not the same as total available
>power. If your antenna is near 50 Ohms, your tank components can usually
>be matched to a conjugate match (or near enough for practical purposes).
>In this case tuning for maximum power out will likely give you a very
>close match to a conjugate match since maximum power transfer occurs with
>a conjugate match.
>
>If one were to use only a 750 Ohm antenna system with their amplifier,
>then their tank circuit should be designed to match from the load
>impedance of the tube to 750 Ohms, not 50 Ohms. This explains why Jim
>Reid needs more drive power to get 1500 Watts out of his amp with his
>mismatched antenna then when he matches the antenna with the tuner. The
>tank of the Henry (and all commercial amps for that matter) are designed
>to match to 50 Ohms, not the 23 Ohms that Jim is seeing.
>
>Now, sure, you can vary the impedance of the output of the amp by
>adjusting you tank components, but are their adjustment ranges enough so
>that you can tune to an impedance vastly different than 50 Ohms? And even
>if you are able to match to some impedance other than 50 Ohms, what is the
>Q then of the tank? Is the Q too high? Too low? Or just right? Who's
>to say. You are trying to tune to something for which the tank was not
>designed.
>
>In the real world, you have to many times sacrifice optimal performance
>for what you can really obtain practically.
>
>This is why I do advocate antenna tuners. Your antenna tuner is designed
>to match a wide variety of loads to 50 Ohms. And in the antenna tuner, Q
>is not a critical component like in your tank. So your tank matches the
>tube's load impedance to 50 Ohms, then your tuner matches the 50 Ohms to
>whatever it sees at your antenna. This way you can have the most
>efficient power transfer.
? if you are using a L-network tuner.
>
>
>>
>>>In other words, a mismatch between tube and tank decreases gain and
>>>efficiency.
>>>
>>? not if the tank is adjustable.
>
>Well, correct. But how adjustable is the tank, really? Most guys don't
>use roller inductors and their caps may not have the necessary range.
>Adjusting the load cap for maximum power out into a load, does not
>necessarily mean that the amp is matched to that load. Go through the
>mathematical equations and solve for the best network. It likely isn't
>your tank.
>
? "Best network" counts not a jot. The bottom-line is watts.
- Rich..., 805.386.3734, www.vcnet.com/measures.
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