Transmitter Output Impedance
 

On Apr 26, 7:20*am, Cecil Moore wrote:
On Apr 25, 7:35*pm, "Sal M. Onella" wrote:

Wisdom in any form would be appreciated. *Thanks.

Have you seen these?

http://www.w2du.com/QEXMayJun01.pdf

http://www.w2du.com/Appendix12.pdf

http://www.w2du.com/r3ch19a.pdf
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Well, I have _now_! 'Twill take just a bit of time to digest them.
Thanks.

"Sal"

Transmitter Output Impedance
 

On 26 abr, 02:59, "Ralph Mowery" wrote:
"Sal M. Onella" wrote in ...



This group has presented members with valuable lessons in antennas and
transmission lines, like how to measure, how to match, etc.

Something I haven't seen is a discussion of the source impedance of
the transmitter. *My curiosity was piqued today as I took some baby
steps into EZNEC. *A particular antenna had such-and-such VSWR if fed
with a 50-ohm cable and a different value if fed with a 75-ohm cable.

While this is hardly news, it got me wondering whether a 75-ohm cable
will load the transmitter the same. *Doesn't seem like it.

My point: *Using 75-ohm cable to improve the match at the antenna
won't help me *... IF ... I suffer a corresponding loss due to
mismatch at the back of the radio. *My HF radios, all solid state,
specify a 50 ohm load. As necessary, I routinely use an internal
autotuner and either of two external manual tuners. *(I'm aware of the
published 1/12 wavelength matching method.)

Wisdom in any form would be appreciated. *Thanks.

"Sal"
(KD6VKW)

A transmitter output impedance is designed for maximum power transfer at a
specific impedance. Most of the *the older tube transmitters impedance was
tunable within a *range.

In simple terms the impedance of the transmitter tube is the plate voltage
devided by the current. *This impedance is then transformed to the nominal
50 ohms of the antenna system. *If the transmitter has the usual tune and
load controls, the exect impedance will not mater as you adjust for maximum
transmitter output.

Most of the transistor transmitters are not adjustable so the output
impedance is usually fixed at 50 ohms for maximum power transfer. *If the
impedance of the antenna system is not 50 ohms, then the output power will
be less than the designed output. *You can use the antenna tuner to adjust
for a match.

Hello Ralph,

The actual output impedance can be anything, but is mostly not 50
Ohms. If you want it close to 50 Ohms, you have to spend money in
components and design time. As 50 Ohms isn't mostly required, one will
not design for that.

Just as an example, take a hard-driven totem pole or push pull stage
with only a series tuned circuit to suppress harmonics (so the LC
circuit shows zero ohms at the carrier frequency). As the active
devices are used a switches, the output impedance of this arrangement
is almost zero (at least far below 50 Ohms).

When you connect a 50 Ohms quarter-wave cable between the output and
the 50 Ohms load, the amplifier-cable combination has very high output
impedance (quarter wave transformer formula).

For power amplifiers, there is no relation between actual output
impedance and efficiency. When an amplifier is designed for 50 Ohms,
it only means that the amplifier will work correctly when terminated
with 50 Ohms. When you deviate from that, output power may decrease
or increase. This may result in more or less stress on the amplifier's
components.

With kind regards,

Wim
PA3DJS
www.tetech.nl

Transmitter Output Impedance
 

On Apr 26, 1:59*am, Richard Clark wrote:
On Mon, 25 Apr 2011 17:35:29 -0700 (PDT), "Sal M. Onella"

wrote:
Something I haven't seen is a discussion of the source impedance of
the transmitter.

sigh....

My point: *Using 75-ohm cable to improve the match at the antenna
won't help me *... IF ... I suffer a corresponding loss due to
mismatch at the back of the radio.

Hi OM,

Look at the prospective SWR and how much is
lost/reflected/absorbed/what-have-you? *More heat comes from a less
than optimal system efficiency than what your computation will reveal.
So much that it will swamp it.

But the trick here is that the reflected "power" (arguments turn on
this word) doesn't always mean heat and it could actually cool -
however hot or cold it may alter the situation, that same "power"
never got out into the air. *

Now, as for source impedance, that is a subject fraught with denial in
the face of the obvious: *Those fins in the back of your rig are to
help bookend your QSL cards into groups (the heat bears no relation to
efficiency nor match loss).

A standard definition (courtesy of Wikipedia) for Return Loss is:
* * * * where Zs is the impedance toward the source and
* * * * Zl is the impedance toward the load.
and we find from the values you supply that it is
* * * * 0.20

Of course, such a definition is utterly useless when the concept of Zs
is replaced with (in most cases) "it ain't 50 Ohms, thet's fur
shure").

If, perchance, some brave soul steps into the breach of NOT 50 Ohms to
suggest what Zs is, then we can give it the acid test of engineering
(an act that I am usually reminded is beyond the understanding of
readers and the province of discussion here). *Let's be gentle and go
only by an order of two (which is reasonably available and can be
coaxed out of my TS-440). *Return loss for a rig exhibiting an Zs of
25 Ohms into the 75 Ohm line (presuming it is infinite in length)
would give us:
* * * * 0.50
that doesn't look good, so let's try Zs of 100 Ohms:
* * * * 0.14
that looks better all 'round. *Even intuition agrees.

Let's press intuition to the proximal limit and say that Zs is 74 Ohms
(yes, my thumb is on the scale):
* * * * 0.01

What does intuition affirm? *What is preferable? * * * *

73's
Richard Clark, KB7QHC

Thanks Richard. Intuition is that the Zs is near 50 ohms for as many
frequencies as the designer can manage. I am on record (including in
this group) of letting intuition lead me down the path to ruin.

I get from you that there's a presupposition that I know the source
impedance or can easily establish it. Hm-m-m ... not so. One big
problem I see is the need to try to measure power delivered in a
non-50-ohm system with my existing instruments that depend on a 50-ohm
system. I don't have a nominal 75-ohm power meter. Won't putting a
50-ohm meter into a 75-ohm circuit not only read wrong but introduce
reflection losses, too?

I think I'd need a collection of non-inductive load resistors and an
accurate rf ammeter. I'd need to connect them and calculate power at
a few points in every band.

Maybe the papers that Cecil cited for me will fill in the gaps or
suggest other approaches.

"Sal"

Transmitter Output Impedance
 

On Tue, 26 Apr 2011 13:12:23 -0700, Jim Lux
wrote:

(If I had a very efficient op amp, I could simulate any arbitrary output
impedance, without dissipating any power in the source)

I can see why this is parenthetical, because it covers a lot of sins
of omission.

First we bang up against the wall of Gain Bandwidth Product. If you
are talking about resistive loads at low power DC, then your statement
is trivially valid.

Second, the ability to "simulate" any arbitrary output (or input for
that matter) impedance for an OP AMP is well defined in the closed
loop gain (which robs from the open loop gain for frequency by
proportion to GBP).

Taking the conventional RF Power Deck of any consumer (Ham) product,
the similarity to an OP AMP is wholly foreign, and for good and
commercial reason. If one were motivated to engineer in the necessary
noise amplifier (a term coined by H.W. Bode who defined this topic of
source Z and applied it to the negative gain or feedback path); then
we would find that the exact same loss is exhibited in the exact same
component(s).

However, by virtue of OP AMP characteristics we would benefit to
vastly better distortion figures, far less spurious content, and
virtually no need for either the conventional impedance transformer,
nor the bandwidth filter that follows the same power deck (provided,
of course, that the drive input is sinusoidal - which it never is,
unfortunately, for this scenario). This novel OP-AMP/Power-Deck
redesign would also confer considerable power supply rejection (that
voltage could sag or rise without appreciable effect) and noise
rejection (the internal noise from other circuitry would not migrate
into the signal output). ALL such benefits are strictly derived from
the amount of negative feedback (not to be confused, as are many
readers to this topic, with the rather ordinary compensation cap in
the last stage).

Why isn't this done as a service to the customer?

Cost.

Again, OP AMP design merits are paid for in lost gain and bandwidth.
The price is found in the amount of negative feedback that goes to
lower the overall amplification. Would you pay for this improved cool
performance to run 10W in the 80M band from a formerly crackly and hot
100W 10-80M band capable source?

73's
Richard Clark, KB7QHC

Transmitter Output Impedance
 

On 26 abr, 22:21, Jim Lux wrote:
Measuring the output impedance (for relative small change in load) is
possible, but is not a simple task. Very likely other people will
comment on this.

ON9CVD's website I linked to has a very simple technique.. 50 ohm dummy
load and a 220 ohm resistor you can switch in.

At 100W (into 50 ohms), the 220 ohms would only dissipate 22W. *You
could get some of those non-inductive resistors from Caddock and series
them up to do something like this.

BTW, this is a simplified version of what's called a "load pull" test...
which makes me wonder if one could cobble up a quick test set that could
be controlled by a computer to do automated output Z measurements of an
HF transceiver over a reasonably wide range... One approach would be to
use a RS-232 controlled antenna tuner and, maybe, a antenna relay box
with several different load resistances).

The challenge (having actually looked at doing this with a LDG AT200PC)
is that the Z of the tuner isn't very well defined. *It's a pretty big
calibration project in itself.

Maybe, though, one could build a few test dummy loads.. say a 25 ohm and
a 75 or 100 ohm, along with your vanilla 50 ohm, and the antenna switch
(like an RCS-8V). *Basically, you're building a "high power resistor
substitution box"

You'd want some sort of nice inline watt meter (like an LP100) to make
the measurements.

Hello Jim,

Other method is injecting a slightly off-carrier frequency signal into
the amplifier (this emulates a constant small VSWR shown to the PA
(wtih 50 Ohms load), but with continuous varying phase). Because of
the difference in frequency, one can measure the forward (towards the
PA) and reverse (reflected by PA) signal with a two channel VSA.
This will give you the PA's complex output impedance.

Tom (K7ITM from my head) did this with a HP89410 with couplers.

With kind regards,


Wim
PA3DJS
www.tetech.nl

Transmitter Output Impedance
 

On Tue, 26 Apr 2011 14:37:58 -0700 (PDT), "Sal M. Onella"
wrote:

Thanks Richard. Intuition is that the Zs is near 50 ohms for as many
frequencies as the designer can manage. I am on record (including in
this group) of letting intuition lead me down the path to ruin.

Well, if you miss the path, you are certain to be reminded where it
is.

I get from you that there's a presupposition that I know the source
impedance or can easily establish it. Hm-m-m ... not so.

It is printed in the specifications. There are other ways to derive
it, of course, and they would merely confirm that number.

One big
problem I see is the need to try to measure power delivered in a
non-50-ohm system with my existing instruments that depend on a 50-ohm
system. I don't have a nominal 75-ohm power meter. Won't putting a
50-ohm meter into a 75-ohm circuit not only read wrong but introduce
reflection losses, too?

It is easier to measure voltage and current and use your calculator.
Now having said that, measuring voltage and current is damned hard at
HF. It is achievable with care, but now we are back into your same
question with many hands pointing at that path to ruin.

As both are difficult (power or voltage times current), you could
trust authority (which confirms what is intuitive), or you could
listen to argument (which at the distance of time and recall becomes
murky and opaque).

I think I'd need a collection of non-inductive load resistors and an
accurate rf ammeter. I'd need to connect them and calculate power at
a few points in every band.

Bravo! This reduces complexity because RF Power resistors of high
accuracy and bandwidth are commercially available. You will need to
practice your skill at mounting to a heat sink, however (another
non-trivial achievement).

Maybe the papers that Cecil cited for me will fill in the gaps or
suggest other approaches.

Walt and I have been corresponding over these matters just these past
two weeks. Skimming that content will once again confirm intuition.

73's
Richard Clark, KB7QHC

Transmitter Output Impedance
 

Richard Clark wrote:
On Tue, 26 Apr 2011 13:12:23 -0700, Jim Lux
wrote:

(If I had a very efficient op amp, I could simulate any arbitrary output
impedance, without dissipating any power in the source)

I can see why this is parenthetical, because it covers a lot of sins
of omission.

yep.. not possible to build such a thing, anymore than one can build a
zero ohm output impedance RF source with any signficant power.
Suggested more as an example that the power dissipation in the source
doesn't necessarily correlate with match, load Z, or anything else in
general.

(You can get pretty darn close at powers less than a watt and HF, though..)


However, by virtue of OP AMP characteristics we would benefit to
vastly better distortion figures, far less spurious content, and
virtually no need for either the conventional impedance transformer,
nor the bandwidth filter that follows the same power deck (provided,
of course, that the drive input is sinusoidal - which it never is,
unfortunately, for this scenario). This novel OP-AMP/Power-Deck
redesign would also confer considerable power supply rejection (that
voltage could sag or rise without appreciable effect) and noise
rejection (the internal noise from other circuitry would not migrate
into the signal output). ALL such benefits are strictly derived from
the amount of negative feedback (not to be confused, as are many
readers to this topic, with the rather ordinary compensation cap in
the last stage).

One can also do a lot of this with various clever schemes if the input
to your PA is coming out of some signal processing. Generically,
predistortion, but it can be so much more.
People have literally spent their lives working out ever more
sophisticated approaches



Why isn't this done as a service to the customer?

Cost.

Like race cars... how fast do you want to go..just bring money

Transmitter Output Impedance
 

Wimpie wrote:
On 26 abr, 22:21, Jim Lux wrote:
Measuring the output impedance (for relative small change in load) is
possible, but is not a simple task. Very likely other people will
comment on this.
ON9CVD's website I linked to has a very simple technique.. 50 ohm dummy
load and a 220 ohm resistor you can switch in.

At 100W (into 50 ohms), the 220 ohms would only dissipate 22W. You
could get some of those non-inductive resistors from Caddock and series
them up to do something like this.

BTW, this is a simplified version of what's called a "load pull" test...
which makes me wonder if one could cobble up a quick test set that could
be controlled by a computer to do automated output Z measurements of an
HF transceiver over a reasonably wide range... One approach would be to
use a RS-232 controlled antenna tuner and, maybe, a antenna relay box
with several different load resistances).

The challenge (having actually looked at doing this with a LDG AT200PC)
is that the Z of the tuner isn't very well defined. It's a pretty big
calibration project in itself.

Maybe, though, one could build a few test dummy loads.. say a 25 ohm and
a 75 or 100 ohm, along with your vanilla 50 ohm, and the antenna switch
(like an RCS-8V). Basically, you're building a "high power resistor
substitution box"

You'd want some sort of nice inline watt meter (like an LP100) to make
the measurements.

Hello Jim,

Other method is injecting a slightly off-carrier frequency signal into
the amplifier (this emulates a constant small VSWR shown to the PA
(wtih 50 Ohms load), but with continuous varying phase). Because of
the difference in frequency, one can measure the forward (towards the
PA) and reverse (reflected by PA) signal with a two channel VSA.
This will give you the PA's complex output impedance.



Yes.. that's another way to do it.

I think one would want variable load impedances for the testing in any
case, because I'll bet that most ham rigs have a load dependent Z.

I've occasionally kicked around the idea of what would it take to do it
simply, especially with nifty devices becoming available to help with
the measurements.. $500 VNAs, $400 power meters that directly read
current and voltage, etc.

What I'm not sure about is whether it is "useful" to know. Consider a
ham with a manual or auto tuner. They'll adjust for minimum reflected
power, which is probably as good as anything else. Hams, as a class,
don't much care about "DC power to RF radiated" efficiency (because the
regulatory requirements are imposed at the "RF output" measurement plane..

Power dissipation in the PA is only a second order concern.. can I plug
it into a standard outlet? Will it get too hot in my car?

backpack QRPers are concerned about DC power consumption, but I'm not
sure they're worried about whether their PA is 30% or 35% efficient.

The people designing battery powered tracking transmitters ARE concerned
about this, as are high power broadcasters (since they're both directly
paying for the supply power and have a "radiated RF power" requirement
to meet their need).

Transmitter Output Impedance
 

On Tue, 26 Apr 2011 15:27:28 -0700, Jim Lux
wrote:

(If I had a very efficient op amp, I could simulate any arbitrary output
impedance, without dissipating any power in the source)

I can see why this is parenthetical, because it covers a lot of sins
of omission.

yep.. not possible to build such a thing, anymore than one can build a
zero ohm output impedance RF source with any signficant power.
Suggested more as an example that the power dissipation in the source
doesn't necessarily correlate with match, load Z, or anything else in
general.

(You can get pretty darn close at powers less than a watt and HF, though..)

OP AMPs are a constant of my admiration in the possibilities offered.
That and the signal processing you suggest (plus digital oscillators)
"could" change the playing field - if conventional design weren't so
universally fallen back upon.

73's
Richard Clark, KB7QHC

Transmitter Output Impedance
 

On Apr 26, 4:49*pm, Wimpie wrote:
Other method is injecting a slightly off-carrier frequency signal into
the amplifier (this emulates a constant small VSWR shown to the PA
(wtih 50 Ohms load), but with continuous varying phase). Because of
the difference in frequency, one can measure the forward (towards the
PA) and reverse (reflected by PA) signal with a two channel VSA.
This will give you the PA's complex output impedance.

Unfortunately, the impedance encountered by the off-carrier frequency
signal is probably not the same as the impedance encountered by the
carrier frequency so the results don't correlate and are not very
useful. The carrier frequency has interference components that the off-
carrier signal doesn't encounter.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK