Transmitter Output Impedance
 

On May 5, 12:23*pm, Jim Lux wrote:
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, ...

Strangely enough, for any fixed configuration with reflections, one
way to maximize radiated power is to maximize reflected power at the
antenna feedpoint.:)

Transmitter Output Impedance
 

On Thu, 05 May 2011 10:38:27 -0700, Jim Lux
wrote:

That doesn't mean that we actually want to use a 50 ohm load in real
life. Maybe I have an amplifier designed to drive a loop antenna with a
resistive 10 ohm impedance.

Interesting. For a community that is so tight-fisted with cash, and
so brag-hearty with power claims, absolutely no one has ever tossed
their hat into the ring that with lower than 50Ohm termination on
their rig (and I'm not talking about 45Ohms) or with a more than 50Ohm
termination on their rig (and this is certainly achievable with a
72Ohm Dipole and 70 Ohm coax) that they have then proclaimed they
substantially exceeded 100W radiated (or lost to heat for that matter)
by the same degree of offset from 50.


Maybe that's because of tradition.. it's not only hams who do that.

Well, that and the rest doesn't respond to the observation that NO ONE
makes a claim of more power from a rig that DOES NOT have a source
resistance of 50 Ohm.

in the professional world, there's a strong tendency to make the
"interfaces" between subsystems 50 ohms, even if overall system
efficiency might be improved by, say, a 25 ohm impedance.

Retail engineering is ruled by a market economy. You have offered a
reductio-ad-absurdum of the native components offering ~10 Ohms
without the cost of intervening transformers. Such an "improved
efficiency" at lower cost option has been available for decades and
has yet to appear as a retail product for the tight-fisted Ham market.
Market logic shows by the force of economy that such a solution is not
viable.

a far cry from your supposition. Off by 1000%?

Just an example.

An incredibly poor one in the context of 100W transmitters when garden
variety examples abound.

I know that modern parts are MUCH lower.

That deliver 97W into 50 Ohms at 30 MHZ with a nominal 12VDC supply? I
smell another poor example.

It might happen that you could hook up a 40 ohm load, or a 20 ohm load,
and it will work just fine, and may even put out more than 100W.

"May" is lazy, "Does" is more authoritative. Any reports of "Does?"

The link from several days ago for the gentleman who measured the output
Z being around 40 ohms would be a "does"..

"Does" with how much more power? When I observe the contents of the
page at that link = NOT ONE WATT MORE!

In fact, if we follow the logic of lowered Rs yielding more power, his
own data flips this! He reports the same 100W available at higher Rs.

However, all that aside, when you observe your comment above, the link
"Does not" offer any support to it. In fact on9cvd explicitly offers:
"This is an ambiguous message"

Which is curious when the manufacturer of the RD70HHF1 power
transistors gives them a "No destroy" rating into a mismatch of 20:1.
Such is the inertia of 1950s design-think with modern components.

Yes, but the design limit might not be the transistor mismatch. It
might be a thermal dissipation limit, or some other component that's the
limiting factor.

Consult the original spec. It is quite specific. The characteristic
of
Load VSWR tolerance: Load VSWR=20:1(All Phase): No destroy
is not a dissipation reference.

73's
Richard Clark, KB7QHC

Transmitter Output Impedance
 

On 5 mayo, 17:06, "J. C. Mc Laughlin" wrote:
Dear Wimpie: *The content of the paragraph below may well rise above the
noise of this thread.
I expect to learn something of value from your comments about why twice the
(apparent) output Z of an amplifier was of importance and what you did to
have the amplifiers conform.

73, * *Mac * N8TT
-------------------"Wimpie" *wrote in message

...

Hello Dave and John,

snip

Regarding the "academic discussion" I also agree. In my professional
career where I designed several RF PA's, only 2 times the output
impedance of the amplifier was of importance. *In one of these cases I
couldn't meet the specs and had to insert attenuation (some waste of
power…).

snip

With kind regards,

Wim
PA3DJSwww.tetech.nl
Remove abc first before setting free the pigeon.
--------------

J. C. Mc Laughlin
Michigan U.S.A.
Home:

Hello,

Regarding the defined output impedance.

The first time was during my thesis (third harmonic peaking PA). The
stability margin was not very large (expected), and it could be
improved by keeping the impedance seen from the base within certain
limits. My teacher said, be careful, you only have two devices
(BLW76). I tried to make an exciter (based on 2SC1307 BJT) with
defined output impedance, but without success. So in the end I
increased the output power of the exciter and inserted a 3 dB
attenuator, not elegant, but it did the job.

The second time was for H-field generation where wider bandwidth was
achieved by adding a second resonator. When driving from a 50 Ohms
source, it had a nice Chebyshev type pass band. Because of the ripple,
it shows reflection in the pass band (this happens with Chebyshev
response). However when driving from a PA (that was flat within the
pass band when loaded with 50 Ohms), everything went wrong.

I redesigned the filter/coil combination, designed a switching PA
(half bridge in class DE operation) and skipped the 50 Ohms (I
designed around 8 Ohms). The PA drives the filter directly in such
away that most of the time the PA sees a nice (mismatched) load (that
is inductive for harmonics). This resulted in the desired pass band
with significantly increased efficiency. The strength of the H-field
is controlled by varying the supply voltage (PWM circuit).

Leaving out the 50 Ohms in between, saved several capacitors and
inductors.

With kind regards,


Wim
PA3DJS
www.tetech.nl

Transmitter Output Impedance
 

On May 5, 4:48*pm, Wimpie wrote:
On 5 mayo, 17:06, "J. C. Mc Laughlin" wrote:









Dear Wimpie: *The content of the paragraph below may well rise above the
noise of this thread.
I expect to learn something of value from your comments about why twice the
(apparent) output Z of an amplifier was of importance and what you did to
have the amplifiers conform.

73, * *Mac * N8TT
-------------------"Wimpie" *wrote in message

....

Hello Dave and John,

snip

Regarding the "academic discussion" I also agree. In my professional
career where I designed several RF PA's, only 2 times the output
impedance of the amplifier was of importance. *In one of these cases I
couldn't meet the specs and had to insert attenuation (some waste of
power…).

snip

With kind regards,

Wim
PA3DJSwww.tetech.nl
Remove abc first before setting free the pigeon.
--------------

J. C. Mc Laughlin
Michigan U.S.A.
Home:

Hello,

Regarding the defined output impedance.

The first time was during my thesis (third harmonic peaking PA). The
stability margin was not very large (expected), and it could be
improved by keeping the impedance seen from the base within certain
limits. My teacher said, be careful, you only have two devices
(BLW76). I tried to make an exciter (based on 2SC1307 BJT) with
defined output impedance, but without success. *So in the end I
increased the output power of the exciter and inserted a 3 dB
attenuator, not elegant, but it did the job.

The second time was for H-field generation where wider bandwidth was
achieved by adding a second resonator. When driving from a 50 Ohms
source, it had a nice Chebyshev type pass band. Because of the ripple,
it shows reflection in the pass band (this happens with Chebyshev
response). *However when driving from a PA (that was flat within the
pass band when loaded with 50 Ohms), everything went wrong.

I redesigned the filter/coil combination, designed a switching PA
(half bridge in class DE operation) and skipped the 50 Ohms (I
designed around 8 Ohms). The PA drives the filter directly in such
away that most of the time the PA sees a nice (mismatched) load (that
is inductive for harmonics). This resulted in the desired pass band
with significantly increased efficiency. *The strength of the H-field
is controlled by varying the supply voltage (PWM circuit).

Leaving out the 50 Ohms in between, saved several capacitors and
inductors.

With kind regards,

Wim
PA3DJSwww.tetech.nl


Hey Mac, are you planning on attending the Chapter 10 QCWA meeting in
Cadillac Saturday? I'm planning on attending, so hope to see you
there!!!

Walt, W2DU

Transmitter Output Impedance
 

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
PA3DJS
www.tetech.nl
In case of PM, tell the pigeon that abc is not in the address.

Transmitter Output Impedance
 

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!!

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

Transmitter Output Impedance
 

On May 5, 9:19*pm, 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!!

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

Ooops, my error, you said you're getting 130w delivered, not 120w as I
said incorrectly.

Walt

Transmitter Output Impedance
 

On 5/5/2011 8:24 PM, walt wrote:
On May 5, 9:19 pm, wrote:
On May 5, 5:26 pm, wrote:









On 5 mayo, 19:23, Jim 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!!

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

Ooops, my error, you said you're getting 130w delivered, not 120w as I
said incorrectly.

Walt


I did not present any numbers, calculations, or assumed conditions,
inaccurate or otherwise. You might want to reply to Jim Lux.

John - KD5YI

Transmitter Output Impedance
 

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
PA3DJS
www.tetech.nl

Transmitter Output Impedance
 

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