On May 6, 6:30*am, Wimpie wrote:
On 6 mayo, 03:19, walt wrote:









On May 5, 5:26*pm, Wimpie wrote:

On 5 mayo, 19:23, Jim Lux wrote:

John KD5YI wrote:
On 5/4/2011 7:54 PM, Jim Lux wrote:
John KD5YI wrote:

Acceptable is what the manufacturer recommends for his gear. What does
this have to do with the device's output impedance?

Absolutely nothing.. which is the point.

Are we arguing the same point?

but the summary is,

That it is a bag of worms? I'm waiting with baited breath...

Exactly..

In fact, as interesting as it would be to measure the output impedance
of my radio, I started to think about what it would buy me, and came to
the conclusion, almost nothing (other than satisfying curiosity).

It *might* be interesting to look at (and write an article for QST/QEX
or something) "optimizing radiated power". *Answering the question: do
you really want a 50 ohm match on your antenna analyzer, or do you want
maximum net power at the antenna feedpoint, and what that might mean for
typical 100W solid state rigs, antennas, etc.

(as a practical matter, this is what automatic antenna tuners actually
adjust for: either minimum reflected power, or maximum fwd-ref)

but it's possible that deliberately running a mismatch (as shown on your
rig's SWR meter) might actually result in more radiated power. *e..g. if
at 1:1 you have 100W fwd and 0 rev, but at 2:1 you have 150W fwd and 15W
reflected, so you're actually net 135W vs 100W; assuming your rig
doesn't otherwise have any problems.

Hello Jim,

I agree with the PA load mismatch issue. It is possible to get (some)
more net power by applying mismatch to a PA stage (from experience).
But frequently it comes with too much increase in input power (so very
hot heatsink).

In case of high efficiency designs, the active devices may communicate
to you by means of smoke or ejection of (hot) particles.

Wim
PA3DJSwww.tetech.nl
In case of PM, tell the pigeon that abc is not in the address.

Hi Wimpie and KD5YI,

Will you please explain how it is possible to get more power delivered
by applying a mismatch to the output of a PA?

And for KD5YI specifically, I believe you have presented some
inaccurate math calculations. You begin with delivering 100w into a
matched load. Then you say you mismatch to 2:1 and get 135w forward
and 15w reflected, leaving 120w delivered. You must be kidding!!

No he is not kidding. I observed the same. At that time I was lucky to
have full access to HP, Advantest and Rohde & Schwarz equipment to
double-check everything (I first blamed my own gear). * You should
leave the idea that all PA's have 50 Ohms output impedance, then it is
easy to explain yourself.

A certain load that has mismatch referenced to 50 Ohms may have a nice
match to a system with non-50 Ohms output impedance.



First, with a 2:1 mismatch and 100w delivered by the PA, the reflected
power is 11.111w, which when added to 100w from the source, the
forward power is 111.11 watts. When the reflected power that returns
to the load is subtracted from the forward power, the result is 100w.
You've heard the expression 'there is no free lunch'?

So please explain to me, if you can, how you can deliver 120w when the
source is 100w.

Walt

The extreme cases where you can get significantly more net output
power by applying mismatch, are PA's with high efficiency (class-E, -
D, -DE, etc). * I am currently designing a balanced class-E 500W
stage. It can deliver 1 kW, but within very short time the mosfet's
will explode (if the supervisory circuit doesn't act).

Wim
PA3DJSwww.tetech.nl


Wim, are you saying that by using Class E amps you are able to violate
the Laws of Physics pertaining to the Conjugate Matching Theorem and
the Maximum Power Delivery Theorem? I cannot agree.

A load doesn't care what the source is. If the load impedance is the
complex conjugate of the source, all available power will be delivered
to the load. Then, if the load impedance is either increased or
decreased, the power delivered will decrease. Are you now saying that
the concept I just stated above is no longer true? If you are, please
explain in detail why this is so. How does 'high efficiency' overcome
the requirement for impedance matching in the delivery of power?

And are you agreeing with an earlier poster that with a 100w source
and a mismatch of 2:1 the forward power will be 135w and 15w
reflected, the power delivered to the load will be 130w? If so, will
you please explain in detail how this can occur?

Walt