On 6 mayo, 17:20, walt wrote:
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.

Walt, Maybe you should familiarize yourself with class E (and other
high efficient topologies). Set up a simulation (or measurement) and
try to apply your "conjugated match" thing. You will find out that you
can't make a class E PA that operates under conjugated match, unless
you are going to play with the power supply's internal resistance (so
the system becomes power supply limited). My current design outputs
500W into 4.5 Ohms, however the output impedance (load pulling) is 1
Ohm.

The reason for non-conjugated matched operation is that in class-E the
device is in voltage saturation for about 50% of the time. During that
time the device has no gain, so a device used in class-E has less gain
then the same device used in a non-saturated application. In other
words: tuning is designed for highest efficiency, not highest output
power.


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.

Quote from text above: "If the load impedance is the complex conjugate
of the source.....". This If-statement has a "false" result for many
PAs, try to broaden your view.


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?

Yes I agree with him, see my reaction to that statement

My class-E PA design that I am doing now is designed for a 4.5 Ohms
nominal load. If I change that load to 2.5 Ohms (VSWR=1.8), output
increases to 700W, but it will be destroyed due too non-favourite
combination of Vds and Id.

Walt

With kind regards,

Wim
PA3DJS
www.tetech.nl

On May 6, 11:43*am, Wimpie wrote:
You will find out that you
can't make a class E PA that operates under conjugated match, unless
you are going to play with the power supply's internal resistance ...

As I understand it, Walt's approach is to pick a point inside the
source at which the output becomes linear through filtering and call
the impedance at that point the linear "source impedance" (including
some boundary conditions). It is akin to the motion of a pendulum in a
clock being linear even though the sustaining energy comes in non-
linear pulses.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

On May 6, 12:43*pm, Wimpie wrote:
On 6 mayo, 17:20, walt wrote:









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.

Walt, Maybe you should familiarize yourself with class E (and other
high efficient topologies). Set up a simulation (or measurement) and
try to apply your "conjugated match" thing. You will find out that you
can't make a class E PA that operates under conjugated match, unless
you are going to play with the power supply's internal resistance (so
the system becomes power supply limited). *My current design outputs
500W into 4.5 Ohms, however the output impedance (load pulling) is 1
Ohm.

The reason for non-conjugated matched operation is that in class-E the
device is in voltage saturation for about 50% of the time. During that
time the device has no gain, so a device used in class-E has less gain
then the same device used in a non-saturated application. In other
words: tuning is designed for highest efficiency, not highest output
power.



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.

Quote from text above: "If the load impedance is the complex conjugate
of the source.....". This If-statement has a "false" result for many
PAs, try to broaden your view.

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?

Yes I agree with him, see my reaction to that statement

My class-E PA design that I am doing now *is designed for a 4.5 Ohms
nominal load. If I change that load to 2.5 Ohms (VSWR=1.8), output
increases to 700W, but it will be destroyed due too non-favourite
combination of Vds and Id.

Walt

With kind regards,

Wim
PA3DJSwww.tetech.nl


Well, Wim, I still don't see how a 2:1 mismatch can achieve 135w
forward and 15w reflected with a 100w source. How can a 2:1 mismatch
achieve 130w forward power, when that mismatch can only reflect 0.111
x 100w? Also, how can that mismatch achieve 15w of reflected power
when that mismatch can only reflect 0.111 x 100w? And how can 11.111
watts of forward power add to 100w to achieve 130w? I don't see how
the power reflected at a mismatched load can be affected by the nature
of the source.

But I do understand how with a 4.5-ohm load can deliver 500w and a 2.5-
ohm load can deliver 700w. I now understand what you mean by a
mismatch can increase the power delivery. (Actually, a change in
mismatch)

But Wim, you pulled a fast one one us!!! Until now you didn't tell us
that the source resistance of the source was 1 ohm!!!

You also didn't tell us what the power would be if the load was 1 ohm
(thus matching the load to the source) if the power supply was
sufficiently large so the power delivery would not be limited by the
power supply. So in reality, you really DON'T get an increase in power
delivery by mismatching, but actually a decrease. What you're really
doing is obtaining an increase in power delivery by decreasing the
amount of mismatch from 4.5:1 to 2.5:1.

So now I understand that you haven't violated any laws of physics, but
IMHO, you have been misleading when you say you obtain a increase in
power by mismatching, because that statement isn't really true, is
it?

take care,

Walt

On 6 mayo, 22:18, walt wrote:
On May 6, 12:43*pm, Wimpie wrote:



On 6 mayo, 17:20, walt wrote:

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.

Walt, Maybe you should familiarize yourself with class E (and other
high efficient topologies). Set up a simulation (or measurement) and
try to apply your "conjugated match" thing. You will find out that you
can't make a class E PA that operates under conjugated match, unless
you are going to play with the power supply's internal resistance (so
the system becomes power supply limited). *My current design outputs
500W into 4.5 Ohms, however the output impedance (load pulling) is 1
Ohm.

The reason for non-conjugated matched operation is that in class-E the
device is in voltage saturation for about 50% of the time. During that
time the device has no gain, so a device used in class-E has less gain
then the same device used in a non-saturated application. In other
words: tuning is designed for highest efficiency, not highest output
power.

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.

Quote from text above: "If the load impedance is the complex conjugate
of the source.....". This If-statement has a "false" result for many
PAs, try to broaden your view.

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?

Yes I agree with him, see my reaction to that statement

My class-E PA design that I am doing now *is designed for a 4.5 Ohms
nominal load. If I change that load to 2.5 Ohms (VSWR=1.8), output
increases to 700W, but it will be destroyed due too non-favourite
combination of Vds and Id.

Walt

With kind regards,

Wim
PA3DJSwww.tetech.nl

Well, Wim, I still don't see how a 2:1 mismatch can achieve 135w
forward and 15w reflected with a 100w source. How can a 2:1 mismatch
achieve 130w forward power, when that mismatch can only reflect 0.111
x 100w? Also, how can that mismatch achieve 15w of reflected power
when that mismatch can only reflect 0.111 x 100w? And how can 11.111
watts of forward power add to 100w to achieve 130w? I don't see how
the power reflected at a mismatched load can be affected by the nature
of the source.

But I do understand how with a 4.5-ohm load can deliver 500w and a 2.5-
ohm load can deliver 700w. I now understand what you mean by a
mismatch can increase the power delivery. (Actually, a change in
mismatch)

But Wim, you pulled a fast one one us!!! Until now you didn't tell us
that the source resistance of the source was 1 ohm!!!

You also didn't tell us what the power would be if the load was 1 ohm
(thus matching the load to the source) if the power supply was
sufficiently large so the power delivery would not be limited by the
power supply. So in reality, you really DON'T get an increase in power
delivery by mismatching, but actually a decrease. What you're really
doing is obtaining an increase in power delivery by decreasing the
amount of mismatch from 4.5:1 to 2.5:1.

So now I understand that you haven't violated any laws of physics, but
IMHO, you have been misleading when you say *you obtain a increase in
power by mismatching, because that statement isn't really true, is
it?

take care,

Walt

Hello Walt,

Off course I agree with you that you can never have more power then
the conjugated matched power, but many PAs don't operate in this
regime (as my very extreme class-E case). It was the reason for
mentioning:

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

The problem is in whether you define mismatch based on the ohmic value
printed on the back of the PA, or on the actual output impedance of
the PA (that you mostly don't know in case of many solid state PA's).

As the actual output impedance of the PA may not be 50 Ohms (for
example 100 Ohms), a 50 Ohms load (as mentioned on the back of the PA)
will provide the stated power (for example 100W). By applying 100
Ohms (that is VSWR = 2 for a 50 Ohms reference), the net power will
increase to 112 W. Actually it wasn't me to experience this mismatch
isue first, but my father during the time that CB was very popular
overhere.

The above example assumes that the output impedance of the PA is
independent of load. Of course this is only true when the actual
voltage across the active device and current through it doesn't
saturate the active device too much.

I think the saturation issue is one of the advantages of a linear PA
with accessible R an X tuning. You just tune for maximum output given
a certain load and drive (you may use plate current as a guide also).
As long as your SSB signal's PEP stays below that output power, your
active device will not go into voltage saturation and IMD will likely
be acceptable.

In case of a wide band push-pull PA, VSWR = 2 (with inconvenient
phase) may provide a load where the active devices (mosfet or BJT),
will go into voltage saturation at a net output power below the rated
power of the PA. For constant envelope modulation this isn't problem,
but for SSB/AM it isn't good.

With kind regards,


Wim
PA3DJS
www.tetech.nl

On Fri, 6 May 2011 13:18:16 -0700 (PDT), walt wrote:

Well, Wim, I still don't see how a 2:1 mismatch can achieve 135w
forward and 15w reflected with a 100w source.

Hi Walt,

Irrespective of who the author is of this statement (I've lost track),
I too would like to see how this is reconciled.

But seeing that you've asked several times....

73's
Richard Clark, KB7QHC

On May 6, 5:50*pm, Wimpie wrote:
On 6 mayo, 22:18, walt wrote:

On May 6, 12:43*pm, Wimpie wrote:

On 6 mayo, 17:20, walt wrote:

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.

Walt, Maybe you should familiarize yourself with class E (and other
high efficient topologies). Set up a simulation (or measurement) and
try to apply your "conjugated match" thing. You will find out that you
can't make a class E PA that operates under conjugated match, unless
you are going to play with the power supply's internal resistance (so
the system becomes power supply limited). *My current design outputs
500W into 4.5 Ohms, however the output impedance (load pulling) is 1
Ohm.

The reason for non-conjugated matched operation is that in class-E the
device is in voltage saturation for about 50% of the time. During that
time the device has no gain, so a device used in class-E has less gain
then the same device used in a non-saturated application. In other
words: tuning is designed for highest efficiency, not highest output
power.

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.

Quote from text above: "If the load impedance is the complex conjugate
of the source.....". This If-statement has a "false" result for many
PAs, try to broaden your view.

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?

Yes I agree with him, see my reaction to that statement

My class-E PA design that I am doing now *is designed for a 4.5 Ohms
nominal load. If I change that load to 2.5 Ohms (VSWR=1.8), output
increases to 700W, but it will be destroyed due too non-favourite
combination of Vds and Id.

Walt

With kind regards,

Wim
PA3DJSwww.tetech.nl

Well, Wim, I still don't see how a 2:1 mismatch can achieve 135w
forward and 15w reflected with a 100w source. How can a 2:1 mismatch
achieve 130w forward power, when that mismatch can only reflect 0.111
x 100w? Also, how can that mismatch achieve 15w of reflected power
when that mismatch can only reflect 0.111 x 100w? And how can 11.111
watts of forward power add to 100w to achieve 130w? I don't see how
the power reflected at a mismatched load can be affected by the nature
of the source.

But I do understand how with a 4.5-ohm load can deliver 500w and a 2.5-
ohm load can deliver 700w. I now understand what you mean by a
mismatch can increase the power delivery. (Actually, a change in
mismatch)

But Wim, you pulled a fast one one us!!! Until now you didn't tell us
that the source resistance of the source was 1 ohm!!!

You also didn't tell us what the power would be if the load was 1 ohm
(thus matching the load to the source) if the power supply was
sufficiently large so the power delivery would not be limited by the
power supply. So in reality, you really DON'T get an increase in power
delivery by mismatching, but actually a decrease. What you're really
doing is obtaining an increase in power delivery by decreasing the
amount of mismatch from 4.5:1 to 2.5:1.

So now I understand that you haven't violated any laws of physics, but
IMHO, you have been misleading when you say *you obtain a increase in
power by mismatching, because that statement isn't really true, is
it?

take care,

Walt

Hello Walt,

Off course I agree with you that you can never have more power then
the conjugated matched power, but many PAs don't operate in this
regime (as my very extreme class-E case). *It was the reason for
mentioning:

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

The problem is in whether you define mismatch based on the ohmic value
printed on the back of the PA, or on the actual output impedance of
the PA (that you mostly don't know in case of many solid state PA's).

As the actual output impedance of the PA may not be 50 Ohms (for
example 100 Ohms), a 50 Ohms load (as mentioned on the back of the PA)
will provide the stated power (for example 100W). *By applying 100
Ohms (that is VSWR = 2 for a 50 Ohms reference), the net power will
increase to 112 W. * Actually it wasn't me to experience this mismatch
isue first, but my father during the time that CB was very popular
overhere.

The above example assumes that the output impedance of the PA is
independent of load. *Of course this is only true when the actual
voltage across the active device and current through it doesn't
saturate the active device too much.

I think the saturation issue is one of the advantages of a linear PA
with accessible R an X tuning. You just tune for maximum output given
a certain load and drive (you may use plate current as a guide also).
As long as your SSB signal's PEP stays below that output power, your
active device will not go into voltage saturation and IMD will likely
be acceptable.

In case of a wide band push-pull PA, VSWR = 2 (with inconvenient
phase) may provide a load where the active devices (mosfet or BJT),
will go into voltage saturation at a net output power below the rated
power of the PA. For constant envelope modulation this isn't problem,
but for SSB/AM it isn't good.

With kind regards,

Wim
PA3DJSwww.tetech.nl


Thanks for the reply, Wim, but you seem to be evading my questions and
concerns by what appears to be going off on tangents unrelated to my
questions.

Perhaps I need to refresh Electronics 101, because I simply can't
understand the example you provided.

You have a 100-ohm source terminated with a 50-ohm load that provides
100w. But you say with the VSWR = 2 the net power will increase to
112w. How does this happen?

Wim, you are making statements that seem to disagree with known
principles, yet you give no explanation of how these statements can be
justified in relation to the known principles.

And please tell me, Wim, what RF amps do you know of that have the
source impedance indicated on the back? And you refer to solid-state
amps, when you know my discussion (and experience) is with tube amps
with pi-network outputs. I am totally ignorant on the operation of
solid-state amps.

I would like to be learning something from my discussions with you,
but I'm sorry, Wim, you're making me feel more stupid the more we
continue.

Walt

On 5/6/2011 5:04 PM, Richard Clark wrote:
On Fri, 6 May 2011 13:18:16 -0700 (PDT), wrote:

Well, Wim, I still don't see how a 2:1 mismatch can achieve 135w
forward and 15w reflected with a 100w source.

Hi Walt,

Irrespective of who the author is of this statement (I've lost track),
I too would like to see how this is reconciled.

But seeing that you've asked several times....

73's
Richard Clark, KB7QHC

Here is the header from that message:

From: Jim Lux
Newsgroups: rec.radio.amateur.antenna
Subject: Transmitter Output Impedance
Date: Thu, 05 May 2011 10:23:10 -0700
Organization: JPL Information Services, InterNetNews
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In-Reply-To:

On May 6, 5:24*pm, walt wrote:
You have a 100-ohm source terminated with a 50-ohm load that provides
100w. But you say with the VSWR = 2 the net power will increase to
112w. How does this happen?

Walt, here is probably what Wim means by that.

Source------1/2WL 50 ohm lossless------RLoad

Vsource = 212v, Rsource=100 ohms

If Rload = 50 ohms, PLoad = 100w

If Rload = 100 ohms, PLoad=112.5w

Wim must be assuming a 50 ohm SWR in both cases.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Dear Wimpie: Many thanks. I am always on the look-out for applied
electronic problems for a colleague who is particularly skilled at crafting
practice problems for the professional engineering exam (chartered engineer
in Europe).

Your comments about filter response reminds me of the (analog) FM receivers
that had wonderful suppression of the next channel over, but horrible
performance in the presence of noise from ignition - while other receivers
had great performance in both respects. As you know, it had to do with
phase response. Chebyshev is not always desirable.

Warm regards, Mac N8TT

"Wimpie" wrote in message
...

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


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

Dear Walt: I am giving a lecture on firearm law on Saturday and so can not
attend. Sorry not to see you and your bride.
Warm regards, Mac N8TT

"walt" wrote in message
...


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


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