Posted by
Don Wilhelm-4 on
Nov 25, 2011; 3:44am
URL: http://elecraft.85.s1.nabble.com/wattage-on-k3-vs-kpa500-tp7026865p7030041.html
If I may, let me reinforce and paraphrase Ron's comments.
Today's transceivers are designed to work into a 50 ohm resistive load.
That means they will develop full power into a load which measures an
SWR = 1.0.
Now, most transmitters are designed to work properly into a SWR = 2.0
load - that means that they will develop full power output - however,
the efficiency may suffer, so the transmitter current draw may increase
when the load differs from a 1-1 SWR.
The power output will remain the same if the SWR is not 1-1, but there
will be more heat produced by the PA because the efficiency has been
reduced.
The reflected power is not dissipated in the finals (read the antenna
article on my website www.w3fpr.com), but the fact that the finals are
not operating at the full efficiency point, there will be increased heat
- that is just the way the laws of physics work.
73,
Don W3FPR
On 11/24/2011 9:06 PM, Ron D'Eau Claire wrote:
> I believe an ATU or other matching network provides a conjugate match. That
> results in optimum power transfer because the RF current and voltage are in
> phase and the SWR meter will read 1:1.
>
> When there is a mismatch, as indicated a high SWR, the finals in the rig do
> not see the optimum load impedance for efficient power transfer so they do
> not operate at optimum efficiency. Lower efficiency means that more of the
> d-c power applied to the finals is consumed in heat instead of RF, and the
> amount of heat the finals must dissipate goes up. I believe that's what led
> many Hams to mistakenly believe that the "reflected" power was being
> absorbed in the finals.
>
> The solution is, of course, to provide a proper termination for the finals.
> When the antenna or other load doesn't present the right impedance, a
> matching network (a.k.a. 'tuner' or ATU) can be inserted to provide one.
>
> Ron AC7AC
>
>
>
> -----Original Message-----
> Most likely the K3 is putting out 60 watts, real watts. The wattmeter is
> also probably correct reading 80 watts forward, with 20 reflected, equal to
> a net 60 watts. And 60 watts is radiated by the antenna. No power get
> reabsorbeds by the amp because it is presenting a congugate match. i.e. any
> power coming back from the antenna sees a perfect mismatch at the
> transmitter and gets re-reflected back to the antenna. Explained 100 times
> better than I can by Walt Maxwell in his books "Reflections" or maybe the
> current one is "Reflections III" something like that. Anyway the point is,
> none of this rf power heats up the finals.
>
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