Efficiency and maximum power transfer
Antonio Vernucci
impedances I have considered that the antenna newsgroup could be the most
appropriate one where to post it.
Let us regard a transmitter as an ideal RF generator with a resistance in
series. It is well known that, for maximum power transfer, the load resistance
must be equal to the generator resistance. Under such conditions efficiency is
50% (half power dissipated in the generator, half delivered to the load).
To achieve a higher efficiency, the load resistance should be made higher than
the generator resistance, although this would turn into a lower power delivered
to the load (the maximum power transfer condition is now no longer met). This
can be verified in practice: by decreasing the antenna coupling in a
transmitter, one obtains a higher efficiency though with a lower output power.
That said, now the question.
Usually, when a transmitter is tuned for maximum output power, efficiency
results to be higher than 50% (typically >60% for class-B, >70% for class-C).
This would seem to contradict the above cited fact that, under maximum power
transfer condition, efficiency is 50%.
Pertinent comments are welcome.
73
Tony I0JX - Rome, Italy
K7ITM
Simple: a transmitter is not an ideal voltage source with a resistor
in series.
I'm playing with a switching power supply design that delivers about a
kilowatt at 100 volts. The output is designed specifically to have a
negative resistance, so the output voltage increases as the current
drawn increases. The output dynamic impedance is about -1 ohms
(adjustable, actually). The linear model is a 100 volt battery in
series with -1 ohms. With an 11 ohm load, I get 10 amps load current,
with the battery thus delivering 1000 watts, the load dissipating 1100
watts, and the -1 ohm resistance dissipating -100 watts. Which shows
the absurdity of thinking of a dynamic output resistance being
anything like a real resistance. In my switching supply, I can adjust
the dynamic output resistance between a small negative value and a
rather larger positive value, with very little change in efficiency.
Although transmitters MAY have dynamic output resistances similar to
the recommended load resistance, that's not a necessary condition, and
has little to do directly with efficiency.
Cheers,
Tom
jgbo...@aol.com
In addition to Tom's comments, an RF Power Amplifier's efficiency is
defined as (Pout/Pin)X100%. Pout is RF and Pin is usually DC. So if
you pump 1000 watts DC in to a class B RF amp and get 600 watts rms
out you are 60% efficient. Your model of a Voltage Source in series
with an internal resistance does not apply here.
The maximum power theorem gives conditions where power in the load, is
equal to internal power in the generator. Not always a good idea. A
50HZ generator capable of Megawatts of power would dissiapate 1/2 in
the generator and 1/2 in our houses if they designed them to conform
to the MPT. The 50HZ generators would melt. Utilities design their
Generators to have nearly 0.0 ohms internal impedance.
Good question.
Gary N4AST
Cecil Moore
> Usually, when a transmitter is tuned for maximum output power,
> efficiency results to be higher than 50% (typically >60% for class-B,
> >70% for class-C). This would seem to contradict the above cited fact
> that, under maximum power transfer condition, efficiency is 50%.
The maximum power transfer theorem only applies to
linear sources. What is the linear source impedance
of a class-C amp?
--
73, Cecil http://www.w5dxp.com
J. Mc Laughlin
Thank you. Thank you. Thank you.
Let us put away forever the idea that a normal transmitter fits the
passive, linear model of an ideal voltage source in series with an actual
resistor.
The active device(s) in an amplifier (tube, BJT, FET, ...) needs to see
a certain impedance (at a given frequency) in order to have desirable
performance characteristics. Too many have extrapolated from that
information (found in data sheets for the active devices) the conclusion
that the active device has an internal Z that is the complex-conjugate of
the load Z.
The wording found in another string of messages communicates the right
idea. A normal transmitter "wants to see" a certain Z. That Z is most
often 50 ohms.
RIP 73, Mac N8TT
P.S. An IEEE paper explored this issue in the last year.
--
J. McLaughlin; Michigan, USA
Home: J...@power-net.net
"K7ITM" <k7...@msn.com> wrote in message
news:dba2b3fa-332b-4ff1...@x35g2000hsb.googlegroups.com...
Roy Lewallen
made in this case was to attempt to apply an unsuitable model (a voltage
source in series with a resistance) to a system to be modeled (a
transmitter). As the OP showed, the attempt leads to an impossible
result. The classic example of this is the "proof" that a bumblebee
can't fly, based on a flawed model and immediately shown to be false by
simply observing that they do, indeed, fly. Yet we see people falling
into this trap daily, not only in modeling electrical circuits, but also
in modeling such diverse processes as human behavior, economic systems,
and roulette wheel numbers.
Unfortunately, the bad results of applying unsuitable models aren't
always so obvious as they were here. So it's always wise to check to see
if the model fits before putting faith in the results.
Roy Lewallen, W7EL
Walter Maxwell
The source resistance appearing at the output of either a Class B or C amplifier is R = E/I, where E is the
peak voltage at the output terminals and I is the peak current at the output. Or RMS values can also be used.
Since E/I is simply a ratio, R is also a ratio. And we know that a ratio cannot dissipate power, or turn
electrical energy into heat, thus the output resistance R is non-dissipative. I have made many measurements
that prove this. It is also the reason why reflected power does not dissipate in the tubes, because it never
reaches the tubes. The reflected power simply causes a mismatch to the source, causing the source to deliver
less power than it would if there were no mismatch. The input at the pi-network in the xmtr is non-linear, but
the fly-wheel effect of the network tank isolates the input from the output, resulting in a linear condition
appearing at the output. Except for a very slight deviation from a sine wave due to a small amount of harmonic
content, the voltage E and current I at the output are essentially a sine wave, which one can easily prove
with a good oscilloscope, proving the output to be linear. I'm speaking for tube rigs with pi-network tanks,
not for solid-state rigs.
I nearly forgot. The only dissipation in the amp tube(s) is due to the filament-to-plate current as the
electrons bombard the plate. The efficiency is determined by the ratio of the DC input power to the RF output
power. The maximum power is delivered when the load resistance equals the output resistance R of the source.
But since resistance R is non-dissipative it is not a factor in determining efficiency. The only factors in
determining efficiency are the RF output power and the dissipation in the tube caused by the electrons
striking the plate. The non-dissipative output resistance is the reason Class B and C amps can have an
efficiency greater than 50 percent. If the output resistance were dissipative it would be the determining
factor in efficiency, which could never be greater than 50 percent if the load resistance was equal to the the
output resistance.
A report of my measurements will soon be available in Reflections 3, from measurements taken since those
reported in Chapter 19 of Reflections 2.
Cecil, if I send you a copy of the new chapter that has the report of my newer measurements do you have any
way to make it available to the guys on this thread?
Walt, W2DU
Owen Duffy
news:73353273-a079-499c...@z66g2000hsc.googlegroups.com:
...
> The maximum power theorem gives conditions where power in the load, is
> equal to internal power in the generator. Not always a good idea. A
> 50HZ generator capable of Megawatts of power would dissiapate 1/2 in
> the generator and 1/2 in our houses if they designed them to conform
> to the MPT. The 50HZ generators would melt. Utilities design their
> Generators to have nearly 0.0 ohms internal impedance.
Actually, the AC power distribution system from alternator down has a
manged substantial equivalent source impedance.
The source impedance serves to limit fault currents, which reduces the
demands on protection devices.
Sure, the network is not operated under Jacobi MPT conditions, but neither
does it have near zero source impedance.
Owen
Owen Duffy
news:4849a80e$0$18158$4faf...@reader3.news.tin.it:
> Someone may regard the following question a bit OT, but as it deals
> with impedances I have considered that the antenna newsgroup could be
> the most appropriate one where to post it.
>
> Let us regard a transmitter as an ideal RF generator with a resistance
> in series. It is well known that, for maximum power transfer, the load
> resistance must be equal to the generator resistance. Under such
> conditions efficiency is 50% (half power dissipated in the generator,
> half delivered to the load).
>
> To achieve a higher efficiency, the load resistance should be made
> higher than the generator resistance, although this would turn into a
> lower power delivered to the load (the maximum power transfer
> condition is now no longer met). This can be verified in practice: by
> decreasing the antenna coupling in a transmitter, one obtains a higher
> efficiency though with a lower output power.
>
> That said, now the question.
But your statements are not true. The model you propose for a transmitter
does not apply in general. Whilst it would be possible to build a
transmitter like that, most transmitters that hams use are not built like
that.
So... it is a loaded question of a type, a question premised on a
falsehood.
>
> Usually, when a transmitter is tuned for maximum output power,
> efficiency results to be higher than 50% (typically >60% for class-B,
> >70% for class-C). This would seem to contradict the above cited fact
> that, under maximum power transfer condition, efficiency is 50%.
>
If the equivalent source impedance is not important, ie it does not need
to be fixed by the design, there here is an analysis.
If you take the case of a grounded cathode triode in class C with a
steady signal, the conduction angle is usually somewhere around 120°. The
anode current waveform is a little like a truncated sine wave, but even
for the range of grid voltages where anode current is greater than zero,
the transfer characteristic is not exactly linear, and the wave will be
further distorted.
If the nature of the anode load is that it is some equivalent R at the
fundamental and zero impedance at all other frequencies, the power output
can be determined by finding the fundamental component of the anode
current waveform, squaring it, and multiplying it by R. The input power
is the average anode current multiplied by the DC supply voltage.
Efficiency is OutputPower/InputPower. By varying the grid bias, drive
voltage, load impedance and supply voltage for a given triode, different
efficiencies will be found, and the maximum could be well over 80%.
Nothing in this approach to design attempts to fix the equivalent source
impedance, the design is performed without regard to that characteristic.
Nevertheless, some argue that the output network performs magic and
achieves source matching naturally without designer intervention, and
does this irrespective of parameters like the dynamic anode resistance,
and the effects of feedback (such as cathode degeneration in grounded
grid amplifiers which in turns depends on the source impedance of the
exciter).
Owen
Owen Duffy
news:Xns9AB6BFE242...@61.9.191.5:
...
> voltage. Efficiency is OutputPower/InputPower. By varying the grid
> bias, drive voltage, load impedance and supply voltage for a given
> triode, different efficiencies will be found, and the maximum could be
> well over 80%.
If you want to explore this approach to design, I have implemented a
spreadsheet which in turn implements the methods described in CPI/Eimac's
"Care and Feeding of Power Tubes" it is described and can be downloaded at
http://www.vk1od.net/RFPATPC/index.htm . The spreadsheet is populated with
the example 4CX20000 in Class C from the above publication.
The spreadsheet calculates the anode current wave form from an
interpolation based on a number of points from the published
characteristics, and calculates the fundamental component of anode current
using an FFT. It then calculates anode efficiency and overall efficiency
given operating parameters.
Owen
Cecil Moore
> Cecil, if I send you a copy of the new chapter that has the report of my newer measurements do you have any
> way to make it available to the guys on this thread?
Walt, it is sure good to hear from you and I hope you are
doing well. I can certainly post your new chapter to my
web page thus making it available for downloading.
jgbo...@aol.com
Not really sure I agree. A multi-megawatt 60HZ generator by necessity
has near zero source impedance. The ones I am familar with require
forced air cooling on their output buses. If you are pumping out Mega-
watts, then any non -zero source impedance results in serious heat.
I^2R.
Gary N4AST
Owen Duffy
news:44e36c19-1767-403e...@c65g2000hsa.googlegroups.com:
> On Jun 7, 12:43�am, Owen Duffy <n...@no.where> wrote:
>> jgboy...@aol.com wrote
>> innews:73353273-a079-499c-89df-c11975b37c78@z66g200
Gary, you use the terms impedance and resistant as if they were
equivalent.
Alternators have a designed value of leakage reactance, and they also
have resistance. The combination make the equivalent source impedance,
and it is sufficient to limit fault current to something typically in the
range of 20 to 50 times the rated output current.
Transmission lines and transformers in the transmission and distribution
networks are usually designed in the same way.
It is not zero, and it is not purely resistive. Most supply authorities
would not allow you to connect a capacitive load (a leading PF load), so
another concept, conjugate matching (in the Jacobit MPT sense) is also
not practiced.
Understanding the electricity network does not really give an insight
into a typical ham radio transmitter, they do not share the same design
objectives.
Owen
Antonio Vernucci
an unsuitable model for an RF transmitter.The easy part of the work is done!
Now the more difficult part. As, by the Thevenin theorem, any complex circuit
comprising resistors, voltage generators and current generators is equivalent to
a generator with a resistor in series, evidently the transmitter model must
comprise elements other than just resistors, voltage generators and current
generators.
Can one suggest how such a model looks like? (even a plain one, that does not
take into account second- or superior-order effects).
73
Tomy I0JX - Rome, Italy
Antonio Vernucci
> cannot dissipate power, or turn
> electrical energy into heat, thus the output resistance R is non-dissipative.
> I have made many measurements
> that prove this.
> Walt, W2DU
Hello Walter,
thanks for your explanation.
I remember having read your excellent articles on QST magazine many years ago,
in which you also explained, among many other things, why reflected power cannot
dissipate in the final stage of a transmitter.
I am not at all opposing your explanation of "non disspative resistance" (on
the other hand how may I contradict a person named Maxwell, hi), but I have some
difficulties to appreciate it.
In my understanding resistance just means that current and voltage are in phase.
There are two possibilities for this to occur:
1) dissipative resistance. Example is a pure resistor, in which power is
converted into heat.
2) non-dissipative resistance: Example is a DC motor, that converts electrical
power into mechanical power. An ideal motor would convert all absorbed power
into mechanical power, producing no heat
But I am unable to see how the second case could be fitted into the transmitter
model.
Thanks and 73
Antonio Vernucci
equal to internal power in the generator. Not always a good idea. A
50HZ generator capable of Megawatts of power would dissiapate 1/2 in
the generator and 1/2 in our houses if they designed them to conform
to the MPT. The 50HZ generators would melt. Utilities design their
Generators to have nearly 0.0 ohms internal impedance.
Good question.
Gary N4AST
Whay you write is perfectly true.
To maximize efficiency, the ratio between load resistance and generator
resistance must be as high as possible. That is the reason why the internal
resistance of the Megawatts generator you took as an example is always made
extremely low (very little power is so dissipated within the generator).
Those generators however do not operate in the maximum power transfer condition
(generator resistance = load resistance). As a matter of fact one cannot
decrease the load resistance below a certain threshold because of the generator
power delivery limitations.
In other words the generator can deliver the maximun power it is able to
deliver, but not under maximum power transfer conditions. Never mind, what is
important is that efficiency is good!
Walter Maxwell
Hi Tony,
The reason a Class B or C amplifier can have efficiencies greater than 50 percent is because the output source
resistance at the output of the pi-network is non-dissipative, as I said in an earlier post. I realize this
phenomenon is somewhat difficult to appreciate. However, I have explained it, and proved it with measurements
reported in Chapter 19 in Reflections 2. I have explained it further in an addition to Chapter 19 that will
appear in Reflections 3, which has additional proof from measurements made since the original Chapter 19 was
written.
This addition to Chapter 19 is Chapter 19A, which I asked Cecil to put on his web page for all to see. Just
scroll down to Cecil's post to see "Chapter 19A from Reflections III", and double click on the url. You will
see the entire Chapter 19A. Apparently you don't have a copy of Reflections 2, so I'm going to email you a
copy of the original Chapter 19. I hope these papers will help in understanding why the output resistance of
the pi-network is non-dissipative.
Walt, W2DU
-.-. --.-
Compliments, sincerely, to all you OM here. For sure in the posts here it is
the maximum efficiency to transfer the power of experience and knowledge to
anyone subrscribe this NG - and for sure to a novice as i am, with a bit of
losses due to things that i don't know when i'm reading something.
Anyway, true Radio Amateurs are here. Apologize for mistakes in my poor
english.
73,
-.-. --.- , Cristiano, Italy.
K7ITM
I believe you have mis-stated the Thevenin theorem. First, it applies
only to linear circuits. Fine -- over some narrow range at least, a
transmitter does indeed look like a linear circuit. But more
importantly, it describes ONLY what you observe at an external pair of
terminals, with no other connections, NOT what goes on inside the
"black box" containing those elements you mentioned.
A very simple example is a voltage source (a perfect battery) and two
resistors in series across the battery; the external terminals for
this example will be at opposite ends of one of the resistors. Let's
say the battery is 2 volts and each resistor is 2 ohms. That will
look like a Thevenin equivalent 1V in series with 1 ohm. Note that it
also looks like a one amp source in parallel with a one ohm resistor.
But it does not behave INTERNALLY like either of those. Consider also
the same internal circuit, except drop the voltage source to 1V and
add a 1/2A current source across the output terminals, polarity so
that there's no drop across the resistor between the voltage and the
current source (with no external load). Now figure the internal
dissipation for each of those two cases, with no load, with a 1 ohm
load, and with a short-circuit load.
Cheers,
Tom
Roy Lewallen
efficiency and/or power available from a voltage source occurs when the
source resistance equals the load resistance (or, more generally, when
they're complex conjugates). But this isn't universally true, as I'll
show with a simple example.
Suppose we have a 100 volt perfect voltage source in series with a
variable source resistance, and a fixed load resistance of 100 ohms. If
we make the source resistance 100 ohms, the source delivers 50 watts, 25
of which are dissipated in the source resistance and 25 watts in the
load. The efficiency, if you consider the source resistance dissipation
as wasted, is 50%. But what happens if we reduce the source resistance
to 50 ohms? Now the source delivers 66.7 watts, of which 22.2 is
dissipated in the source resistance and 44.4 in the load resistance. The
power to the load has increased, and the efficiency has increased from
50 to 66.7%. The efficiency and load power continue to increase as the
source resistance is made smaller and smaller, reaching a maximum when
the source resistance is zero. At that point, the source will deliver
100 watts, all of which is dissipated in the load, for an efficiency of
100%.
The well known and often misapplied rule about maximizing power transfer
by matching the source and load impedances applies only when you're
stuck with a fixed source resistance and can only modify the load.
Roy Lewallen, W7EL
Cecil Moore
> But what happens if we reduce the source resistance to 50 ohms?
> Now the source delivers 66.7 watts, of which 22.2 is dissipated
> in the source resistance and 44.4 in the load resistance.
> The well known and often misapplied rule about maximizing power transfer
> by matching the source and load impedances applies only when you're
> stuck with a fixed source resistance and can only modify the load.
What if we assume that we are stuck with the 50 ohm source
impedance above? How do we get greater than 50% power transfer?
Walter Maxwell
"Roy Lewallen" <w7...@eznec.com> wrote in message
news:JrCdnR01Yp20tNHVnZ2dnUVZ_sednZ2d@easystreetonline...
What seems to be overlooked here is that the source resistance at the output
terminals of the pi-nework in Class B and C amplifiers is non-dissipative, which
is the reason they can be loaded for delivering all available power for a given
grid drive, and still have efficiencies greater than 50 percent. One of the
myths circulated for years, and still prevelant, is that the reason for Class B
and C amps to have efficiencies greater than 50 percent is that the load
resistance must be greater than the source resistance. Tain't so.
I've proved the above to be true with extensive measurements using laboratory
grade instruments. Reports on those measurements are reported in Chapter 19 in
Reflections 2, and additional measurements taken after Reflections 2 was
published are reported in Chapter 19A, to be published soon in Reflections 3.
This additional chapter is listed here in the rraa for your information. If
anyone is interested in reading Chapter 19 in Reflections 2 it appears in my
website at www.w2du.com.
Walt, W2DU
jgbo...@aol.com
>
> - Show quoted text -
We are obviously talking about two different things, that as you say
have little to do with a Ham Radio transmitter.
Gary N4AST
Richard Fry
> What seems to be overlooked here is that the source resistance at the
> output
> terminals of the pi-nework in Class B and C amplifiers is non-dissipative,
> which
> is the reason they can be loaded for delivering all available power for a
> given
> grid drive, and still have efficiencies greater than 50 percent. One of
> the
> myths circulated for years, and still prevelant, is that the reason for
> Class B
> and C amps to have efficiencies greater than 50 percent is that the load
> resistance must be greater than the source resistance. Tain't so.
Walt - what is your thinking on the point that untuned, solid-state
amplifiers also can have PA DC-to-RF power conversion efficiencies of 70% or
more at the device level?
In fact the solid-state, analog FM broadcast transmitters supplied by Harris
Corporation and others need no tuning to produce their rated output power
into a 1.3:1 SWR or less, anywhere in the FM broadcast band 88-108 MHz.
Even the harmonic filter needs no changes, and maintains harmonics at -80
dBc or better. They are frequency agile, and can be reset from one carrier
frequency to another, anywhere in the FM band with a transition time of a
few seconds
The overall AC input to r-f output efficiency of these transmitters exceeds
60% (includes the exciter, control system, IPA, and cabinet fans).
RF
Roy Lewallen
I know this question was directed to Walt, but I'd like to mention that
I've designed and built solid state class C amplifiers at the 5 - 10
watt level which have measured efficiencies of greater than 85%.
Roy Lewallen, W7EL
Walter Maxwell
"Richard Fry" <rf...@adams.net> wrote in message news:g2jc1s$8hg$1...@aioe.org...
Hello Richard,
Sorry, Richard, I have no knowledge of solid-state untuned amps, so my thinking
on them is zero, nada. As you'll note, all of my discussion on the subject
concerns only tube amps with a pi-network output, and I've specifically stated
these conditions.
If you've read Chapter 19 and its addition as Chapter 19A, do you agree with my
position that the output resistance at the output terminals of the pi-network is
non-dissipative?
Walt, W2DU
Richard Fry
>The source resistance appearing at the output of either a Class B or C
>amplifier is R = E/I, where E is the peak voltage at the output terminals
>and I is the peak current at the output. Or RMS values can also be
>output resistance R is non-dissipative. I have made many measurements that
>prove this. It is also the reason why reflected power does not dissipate in
>the tubes, because it never reaches the tubes. The reflected power simply
>causes a mismatch to the source, causing the source to deliver less power
>than it would if there were no mismatch.
If the source resistance of a tuned r-f PA stage was truly non-dissipative,
and the tx simply supplied less power into poor matches, how would that
explain the catastrophic failures to the output circuit components often
seen when high power transmitters operate without suitable SWR protection
into highly mismatched loads?
Another reality is that r-f power from two co-sited, tuned transmitters on
two frequencies in the same band can be present in each others output stage
due to antenna coupling, which causes r-f intermodulation between them. The
non-linear (mixing) process occurs at the active PA stage. If reflected
(reverse) r-f energy never reaches the PA stage as you assert, then how
could this IM generation occur?
RF
Walter Maxwell
"Richard Fry" <rf...@adams.net> wrote in message news:g2ltq2$j1s$1...@aioe.org...
> "Walter Maxwell" wrote
> >The source resistance appearing at the output of either a Class B or C
> >amplifier is R = E/I, where E is the peak voltage at the output terminals
> >and I is the peak current at the output. Or RMS values can also be
> >used.Since E/I is simply a ratio, R is also a ratio. And we know that a
> >ratio cannot dissipate power, or turn electrical energy into heat, thus the
> >output resistance R is non-dissipative. I have made many measurements that
> >prove this. It is also the reason why reflected power does not dissipate in
> >the tubes, because it never reaches the tubes. The reflected power simply
> >causes a mismatch to the source, causing the source to deliver less power
> >than it would if there were no mismatch.
> __________
>
> If the source resistance of a tuned r-f PA stage was truly non-dissipative,
> and the tx simply supplied less power into poor matches, how would that
> explain the catastrophic failures to the output circuit components often
> seen when high power transmitters operate without suitable SWR protection
> into highly mismatched loads?
You must be talking about solid-state tx when you mention SWR protection. I
don't know of any tube tx that have such protection. Any catastrophic failures
in tube tx with pi-network output circuits due to poor impedance match at the
output without retuning to match the the network to the load simply leaves the
tx detuned away from resonance. The result is excessive plate current that would
be reduced to normal by resonating the tank circuit. No operator in his right
mind would allow the tx to be operated with the tank not tuned for the resonant
dip in plate current.
> Another reality is that r-f power from two co-sited, tuned transmitters on
> two frequencies in the same band can be present in each others output stage
> due to antenna coupling, which causes r-f intermodulation between them. The
> non-linear (mixing) process occurs at the active PA stage. If reflected
> (reverse) r-f energy never reaches the PA stage as you assert, then how
> could this IM generation occur?
>
> RF
>
:You said two tx operating on two different frequencies, which means the RF
signals from the two tx are not phase coherent. In this condition the signal
from each tx does enter the other. On the other hand, the wave reflected from a
mismatched termination is phase coherent with the source wave, resulting in the
addition of the reflected wave to the source wave when either the antenna tuner
or the pi-network is adjusted to deliver all available power at the desired grid
drive.
W2DU
Dave
>>The source resistance appearing at the output of either a Class B or C
>>amplifier is R = E/I, where E is the peak voltage at the output terminals
>>and I is the peak current at the output. Or RMS values can also be
>>used.Since E/I is simply a ratio, R is also a ratio. And we know that a
>>ratio cannot dissipate power, or turn electrical energy into heat, thus
>>the output resistance R is non-dissipative. I have made many measurements
>>that prove this. It is also the reason why reflected power does not
>>dissipate in the tubes, because it never reaches the tubes. The reflected
>>power simply causes a mismatch to the source, causing the source to
>>deliver less power than it would if there were no mismatch.
> __________
>
> If the source resistance of a tuned r-f PA stage was truly
> non-dissipative, and the tx simply supplied less power into poor matches,
> how would that explain the catastrophic failures to the output circuit
> components often seen when high power transmitters operate without
> suitable SWR protection into highly mismatched loads?
if the load impedance seen at the transmitter terminals is outside the range
that it was designed for you end up with either arcing from excessive
voltage or meltdown from high currents. remember, the pa output as long as
it is connected to a linear load and you consider only the sinusoidal steady
state condition can be completely replaced by a lumped load impedance, hence
no reflections necessary to figure it out... same load, same result, hence
the reflections have no effect on the internals of the pa.
>
> Another reality is that r-f power from two co-sited, tuned transmitters on
> two frequencies in the same band can be present in each others output
> stage due to antenna coupling, which causes r-f intermodulation between
> them. The non-linear (mixing) process occurs at the active PA stage. If
> reflected (reverse) r-f energy never reaches the PA stage as you assert,
> then how could this IM generation occur?
non-coherent input would of course pass through the matching network going
into a pa, there is nothing that precludes that. that current would cause
mixing in the non-linear tube or transistor and therefore im generation.
but that incoming rf is not reflected, it is the incident wave, with nothing
coherent to interfer with why would it not pass into the pa?
Now, have fun with this one... two transmitters on exactly the same
frequency feeding a common load through equal lengths of coax... what
happens if you change the phase relationship between them? make it
simpler, remove the coax and connect both pa outputs togther with a single
load, now no reflections to worry about... does it matter? do the two pa's
feed all their power into each other? where do the reflections go??
Richard Harrison
"If the source resistance of a tuned r-f PA stage was truly
non-dissipative, and the tx simply supplied less power into poor
circuit components often seen when high power transmitters operate
without suitable SWR protection into highly mismatched loads?"
The PA is a switch. Almost no voltage across it when it is closed and no
current through it when it is open. Some of its impedance is dissipative
and some is non-dissipative.
A conjugate match to its total impedance is the way to deliver maximum
power from the transmitter to its load. Alexander H. Wing wrote on page
43 of "Thansmission Lines, Antennas, and Wave Guides":
"If a dissipationless network is insrted between a constant voltage
generator of internal impedance Zg1 and a load ZR such that maximum
power is delivered to the load, at every pair of terminals the
impedances looking in opposite directions are conjugates of each other."
An operating transmitter is normally adjusted for conjugate match with
its load.
Normal plate dissipation occurs when electrons strike the anode and
there is little damage to the tube when the current and cooling are
within limits.
Let an arc strike across the transmission line and it may effectively
become a short circuit which may impose an enormous mismatch in an
instant to the transmitter. That`s why a d-c supply is often connected
in series with a relay coil across the transmission line. The arc
completes the d-c circuit energizing the relay which breaks the
interlock circuit. The transmitter instantly is shut down until it is
manually restarted.
Tubes are often destroyed by internal arcs if overloads don`t act in
time.
Best regards, Richard Harrison, KB5WZI
Jim Lux
> "Walter Maxwell" wrote
>> The source resistance appearing at the output of either a Class B or C
>> amplifier is R = E/I, where E is the peak voltage at the output
>> terminals and I is the peak current at the output. Or RMS values can
>> also be used.Since E/I is simply a ratio, R is also a ratio. And we
>> know that a ratio cannot dissipate power, or turn electrical energy
>> into heat, thus the output resistance R is non-dissipative. I have
>> made many measurements that prove this. It is also the reason why
>> reflected power does not dissipate in the tubes, because it never
>> reaches the tubes. The reflected power simply causes a mismatch to the
>> source, causing the source to deliver less power than it would if
>> there were no mismatch.
> __________
>
> If the source resistance of a tuned r-f PA stage was truly
> non-dissipative, and the tx simply supplied less power into poor
> matches, how would that explain the catastrophic failures to the output
> circuit components often seen when high power transmitters operate
> without suitable SWR protection into highly mismatched loads?
the output network might be non-dissipative, but that doesn't mean that
the voltages and/or currents won't exceed the limitations of the components.
Consider a gedanken.. you have an active device acting as a shunt DC
regulator that puts out 1 amp at 1000 volts into a matching network, the
other side of which is a 50 ohm load. The power supply is a 5000 volt
power supply with a 2000 ohm dropping resistor, and 1 amp is also
flowing through the active device. The 2 amps total flowing through the
dropping resistor results in the 1000 volt output into the network. The
implication is that the matching network, with the 50 ohm output, is
presenting a 1000 ohm load.
Now, say you change the load impedance on the network so that the
impedance on the input side is zero ohms. Now, the voltage across the
active device is zero volts, the voltage across that dropping resistor
is 5000 volts, instead of 4000 volts, and it will get a bit hotter, and
possibly fail.
Or, say you change the load impedance so that the input of the network
presents an infinite impedance. The same 1 amp is flowing through the
active device (say it's a controlled current source), but, since there's
no load current, the voltage drop across the resistor is now 2000 volts
instead of 4000 volts, and the voltage across the active device is now
3000 volts instead of 1000. If the device can only take 2000 volts,
it's cooked.
Owen Duffy
2...@storefull-3258.bay.webtv.net:
...
> The PA is a switch. Almost no voltage across it when it is closed and no
> current through it when it is open. ...
The OP's question related to Class B and Class C.
Have you hopped on another tram?
Your statement is definitely not true for Class B RF PAs. In fact it is
difficult achieving theoretical Class B efficiency with tetrodes running on
relatively low HT (eg 750V as in many transceivers).
If what you state was true, you would expect conversion efficiencies in
excess of 95%... and that just doesn't happen with practical electron
devices.
Practical Class C electron devices cannot achieve those efficiencies at
maximum continuous output power. The anode voltage waveform in practical
continuous Class C amplifiers with resonant loads is hardly the rectangular
pulse waveform that you suggest. In fact, if you observed it with a CRO you
would be hard put to argue that it wasn't an almost pure sine wave.
Owen
Walter Maxwell
"Jim Lux" <james...@jpl.nasa.gov> wrote in message
news:g2n4sd$pc3$1...@news.jpl.nasa.gov...
> Richard Fry wrote:
> > "Walter Maxwell" wrote
> >> The source resistance appearing at the output of either a Class B or C
> >> amplifier is R = E/I, where E is the peak voltage at the output
> >> terminals and I is the peak current at the output. Or RMS values can
> >> also be used.Since E/I is simply a ratio, R is also a ratio. And we
> >> know that a ratio cannot dissipate power, or turn electrical energy
> >> into heat, thus the output resistance R is non-dissipative. I have
> >> made many measurements that prove this. It is also the reason why
> >> reflected power does not dissipate in the tubes, because it never
> >> reaches the tubes. The reflected power simply causes a mismatch to the
> >> source, causing the source to deliver less power than it would if
> >> there were no mismatch.
> > __________
> >
> > If the source resistance of a tuned r-f PA stage was truly
> > non-dissipative, and the tx simply supplied less power into poor
> > matches, how would that explain the catastrophic failures to the output
> > circuit components often seen when high power transmitters operate
> > without suitable SWR protection into highly mismatched loads?
>
> the output network might be non-dissipative, but that doesn't mean that
> the voltages and/or currents won't exceed the limitations of the components.
I fail to understand how the limitations of the components have anything to do
with the output resistanceof the network. Can you please explain how this is
relevant?
> Consider a gedanken.. you have an active device acting as a shunt DC
> regulator that puts out 1 amp at 1000 volts into a matching network, the
> other side of which is a 50 ohm load. The power supply is a 5000 volt
> power supply with a 2000 ohm dropping resistor, and 1 amp is also
> flowing through the active device. The 2 amps total flowing through the
> dropping resistor results in the 1000 volt output into the network. The
> implication is that the matching network, with the 50 ohm output, is
> presenting a 1000 ohm load.
>
> Now, say you change the load impedance on the network so that the
> impedance on the input side is zero ohms. Now, the voltage across the
> active device is zero volts, the voltage across that dropping resistor
> is 5000 volts, instead of 4000 volts, and it will get a bit hotter, and
> possibly fail.
>
> Or, say you change the load impedance so that the input of the network
> presents an infinite impedance. The same 1 amp is flowing through the
> active device (say it's a controlled current source), but, since there's
> no load current, the voltage drop across the resistor is now 2000 volts
> instead of 4000 volts, and the voltage across the active device is now
> 3000 volts instead of 1000. If the device can only take 2000 volts,
> it's cooked.
Sorry Jim, I'm not familiar with a 'gedanken'. Is it distant cousin of a Class B
or C amplifier? Your discussion above doesn't appear to have any relation
whatsoever to those types of RF amplifiers. Are you using your discussion in an
attempt to prove that the output resistance of those types of amps is not
non-dissipative? Or what are you trying to prove? I don't get it.
Walt, W2DU
Richard Harrison
"Have you hopped on another tram?"
Maybe. I agree. Radio power amplifiers have clean sinusoidal outputs,
otherwise they would generate unacceptable harmonics. All except Class A
amplifiers which have continuous plate current flows and a maximum
efficiency of 50%, have plate currents which flow in pulses.
In Class B, the plate current in individual tubes flows in pulses of
approximately a half cycle. Actual efficiency is about 60%.
In Class C, the plate current pulses last less than a half cycle.
Practical efficiencies are in the range of 60 to 80 per cent.
Jim Lux
I was attempting to discuss the comment above:
"how would that explain the catastrophic failures to the output
circuit components often seen when high power transmitters operate
without suitable SWR protection into highly mismatched loads?"
"gedanken" refers to an intellectual exercise describing an experiment
with (usually) contrived or idealized circumstances so that the
underlying concepts can be understood. Classics are Schroedinger's cat,
Maxwell's demon, etc.
http://en.wikipedia.org/wiki/Thought_experiment for more details and
history than anyone could want.
>
Roy Lewallen
>
> If the source resistance of a tuned r-f PA stage was truly
> non-dissipative, and the tx simply supplied less power into poor
> matches, how would that explain the catastrophic failures to the output
> circuit components often seen when high power transmitters operate
> without suitable SWR protection into highly mismatched loads?
This question has been asked many times (several times by you, as I
recall) and answered (several times at least by me) on this forum. The
answer is that when a transmitter is terminated in an impedance too far
from the one it's designed to see, voltages and/or currents in the
transmitter and tank circuit rise to unacceptably high values. This
causes the damage or destruction.
I don't think this is difficult to understand. So the reason for the
need to keep asking is that it's apparently not being believed,
presumably because of the deeply compelling need to see waves of
"reflected power" entering the transmitter to wreak havoc. But the
simple answer is true. This is very much like trying to convince an
astrology believer that he stubbed his toe because he didn't watch where
he was going rather than because Mars was lined up with Jupiter.
> Another reality is that r-f power from two co-sited, tuned transmitters
> on two frequencies in the same band can be present in each others output
> stage due to antenna coupling, which causes r-f intermodulation between
> them. The non-linear (mixing) process occurs at the active PA stage.
> If reflected (reverse) r-f energy never reaches the PA stage as you
> assert, then how could this IM generation occur?
Power from one antenna is coupled to the other antenna by a process
known as "mutual coupling". This power enters the other transmitter via
normal transmission along the transmission line, where it reaches the
final stage for mixing. No mysterious "reverse r-f energy" is needed to
explain this well-known and well-understood phenomenon.
Roy Lewallen, W7EL
Cecil Moore
> This question has been asked many times (several times by you, as I
> recall) and answered (several times at least by me) on this forum. The
> answer is that when a transmitter is terminated in an impedance too far
> from the one it's designed to see, voltages and/or currents in the
> transmitter and tank circuit rise to unacceptably high values. This
> causes the damage or destruction.
But the offending impedance equals (Vfor+Vref)/(Ifor+Iref), a
virtual impedance caused by the superposition of the reflected
wave and the forward wave.
> I don't think this is difficult to understand.
What is difficult to understand is the denial of the role that
the energy content of the reflected wave plays in causing the
offending impedance. If it were not for the reflected waves,
that offending impedance would never happen.
Owen Duffy
> Owen Duffy wrote:
> "Have you hopped on another tram?"
>
> Maybe. I agree. Radio power amplifiers have clean sinusoidal outputs,
> otherwise they would generate unacceptable harmonics. All except Class
> A amplifiers which have continuous plate current flows and a maximum
> efficiency of 50%, have plate currents which flow in pulses.
>
> In Class B, the plate current in individual tubes flows in pulses of
> approximately a half cycle. Actual efficiency is about 60%.
Pulses is not very descriptive, it is definititely not rectangular
pulses as implied by your earlier posting.
For a single ended RF common cathode valve PA with resonant load, an
active device that is perfectly linear and zero saturation voltage,
theoretical efficiency is 79% in Class B.
If saturation is only 20% of the supply voltage, that figure drops to
63%.
For HV tubes, saturation of about 10% of the supply voltage is
reasonable, and efficiency is that case is 71%.
Then the output circuit loss need to be factored in, and 90% is not an
unreasonable output circuit efficiency.
The comination of 10% saturation and practical output circuit arrives at
64% efficiency overall (ie plate and output circuit) ... but tetrodes on
low voltage can't usually deliver low saturation voltage and won't reach
that figure.
>
> In Class C, the plate current pulses last less than a half cycle.
> Practical efficiencies are in the range of 60 to 80 per cent.
Again, not rectangular pulses, not nearly.
For most practical valves in continous mode, they cannot develop their
rated maximum power at very small conduction angles without exceeding
rated cathode current, so there is often little benefit in operating at
conduction angle much below 120 degrees.
Owen
Richard Harrison
"In Class C, the plate current pulses last less than a half cycle.
Practical efficiencies are in the range of 60 to 80 per cent."
Owen Duffy wrote:
"Again, not rectangular pulses, not nearly."
A pulse does not need to be rectangular. According to my electronics
dictionary:
"Pulse - 1. The variation of a quantity having a normally constant
value. This variation is characterized by a rise and decay of finite
duration. 2. An abrupt change in voltage, either positive or negative,
which conveys information to a circuit. (See also Impulse.)"
A rectified sine wave could properly be called a string of pulses being
constantly off between pulses then rising and falling between pulses for
a short finite duration.
J. Mc Laughlin
Owen says:
>
> For most practical valves in continous mode, they cannot develop their
> rated maximum power at very small conduction angles without exceeding
> rated cathode current, so there is often little benefit in operating at
> conduction angle much below 120 degrees.
>
Recall that valve/tube diodes in rectifier service were not used with
capacitive filters because the rated peak to average current ratio was
something like 1.5:1. Modern, solid-state diodes can have ratios of 40:1
and can be happy with capacitive filters in rectifier service.
Peak current limits of high-voltage, high-power tubes is a real physical
limit. The cathode can only emit so much.
73, Mac N8TT
--
J. McLaughlin; Michigan, USA
Home: J...@power-net.net
Richard Harrison
"What is the linear source impedance of a class-C amp?"
A conjugate match is necessary for maximum power transfer.
A Class C amplifier is not inherently linear. That is disasterous for an
already modulated AM signal but it is of no importance to an FM signal.
As Richard Fry points out, no tank circuit is required. A low-pass
filter to suppress all harmonics is all that is needed for a clean
signal. No tuning is required of a tank at the operating frequency.
A tank circuit is not selective enough to prevent intermod anyway. When
I worked the morning shift at 790 KHz in Houston, and I fired up the
transmitter, I would hear 740 KHz`audio coming out of the monitor
speaker. They started programming earlier than we did. They were 15
miles away. Our transmitting antenna made a dandy receiving antenna. The
received 740 KHz modulated our final amplifier and we rebroadcast it
although at a level much lower than our own modulation. Any one
listening to 790 KHz who turned up the volume to hear the 740 KHz audio
got their ears knocked off when our modulation started. We had high
level plate modulation of the final amplifier for our own signal. For
the 740 KHz signal the level of modulation was much lower, millivolts
not kilovolts. Either program was cleanly modulated on our carrier. The
only difference was the enormous difference in modulation levels.
Richard Harrison
"Most supply authorities would not allow you to connect a capacitive
Incorrect. Overexcited synchronous machines are commonly used to correct
the power factor causing reduced line current and lower power loss on
the a-c power transmission line.
Owen Duffy
> Owen Duffy wrote:
> "Most supply authorities would not allow you to connect a capacitive
> load (a leading PF load),-----."
>
> Incorrect. Overexcited synchronous machines are commonly used to
> correct the power factor causing reduced line current and lower power
> loss on the a-c power transmission line.
In this part of the world, connection of an overall leading PF load
usually requires approval of the supply authority. Most electricians
would be aware of the restriction.
The reason being is that it can cause an increase in supply voltage
beyond spec in the local area.
Yes, large factories for example may use static or rotary power factor
correction to improve their PF for tariff reaons, but if an installation
reaches an overall leading PF it usually breaches rules in this part of
the world, and commonly permission is required to attach individual
'devices' with capability for leading PF. Some tariffs are based on
maximum demand, and often maximum apparent power (ie VA), not maxiumum
real power (ie W), hence the economic interest in PF correction.
Today, active static power factor correction that can dynamically track
a varying load at varying PF is a lot more attractive than rotating
machines.
Owen
Owen Duffy
1...@storefull-3253.bay.webtv.net:
> I wrote:
> "In Class C, the plate current pulses last less than a half cycle.
> Practical efficiencies are in the range of 60 to 80 per cent."
>
> Owen Duffy wrote:
> "Again, not rectangular pulses, not nearly."
>
> A pulse does not need to be rectangular. According to my electronics
> dictionary: ...
Yes Richard, pulse can mean all things to all people... but you did say
"The PA is a switch. Almost no voltage across it when it is closed and no
current through it when it is open." That implies a rectangular current
pulse... and I think we are now agree that is not the case for Class B or
Class C (which were the stated scope of the OP's question).
Owen
Richard Fry
> No mysterious "reverse r-f energy" is needed to explain this well-known
> and well-understood phenomenon.
The quote above reads as though the existence of reverse (reflected)
r-f energy is being denied.
For skeptics, the link below leads to a field report showing a measurement
of the reflection of a narrowband r-f pulse by an analog broadcast TV
antenna, back toward the source. The H.A.D. of the sin² pulse used
represents the shortest transition time that can be accommodated in a ~ 4
MHz transmission channel.
Note that the reflected pulse appears some 6.2 µs after the incident pulse,
which corresponds to a round trip through the ~1,525 feet of 6" OD, 75 ohm,
air pressurized transmission line leading to the antenna in this system.
The return pulse amplitude indicates a far-end match (elbow complex +
antenna) of about 1.05 VSWR (2.3% reflection).
http://i62.photobucket.com/albums/h85/rfry-100/RFPulseMeasurement.gif
RF (RCA Broadcast Field Engineer 1965-1980)
Cecil Moore
> Cecil, W5DXP wrote:
> "What is the linear source impedance of a class-C amp?"
>
> A conjugate match is necessary for maximum power transfer.
Is the class-C amp conjugately matched during
the 75% of the cycle when it is off? Is there
any such thing as an instantaneous conjugate
match? Don't we have to move downstream from
non-linear sources for our linear math models
to start working?
Cecil Moore
> "Roy Lewallen" wrote
>> No mysterious "reverse r-f energy" is needed to explain this
>> well-known and well-understood phenomenon.
> __________
>
> The quote above reads as though the existence of reverse (reflected)
> r-f energy is being denied.
Especially strange since the "reverse r-f energy"
is the cause of the impedance responsible for the
mismatch problem. If there were no "reverse r-f energy",
the impedance causing the mismatch wouldn't even exist.
Why is the cause of the offending impedance of no importance?
Walter Maxwell
"Cecil Moore" <nos...@w5dxp.com> wrote in message
news:XM74k.11586$Ri....@flpi146.ffdc.sbc.com...
Richard, its a common myth that Class C amps are non-linear, but the truth of
the matter is that although the condition at the input of the pi-network is
decidedly non-linear, the energy storage in the pi-network tank circuit isolates
the input from the output and the result is a totally linear condition at the
output of the pi-network. Evidence proving this is true is that the output of an
unmodulated signal at the output of the network is an almost pure sine wave.
With a Q of at least 12 the difference between a pure sine wave from a signal
generator and that from the pi-network output can not be seen on a dual trace
scope with the traces overlapping.
I don't know about the energy storage in the filters you mention, but I would
assume that if the filter output is a sine wave then the energy storage required
to produce a linear output is sufficient.
Walt, W2DU
Jim Lux
> Cecil, W5DXP wrote:
> "What is the linear source impedance of a class-C amp?"
>
> A conjugate match is necessary for maximum power transfer.
*in a linear system*
>
Richard Harrison
"in a linear system"
It produces no significant harmonics, so the system is linear.
Richard Harrison
"Is the class-C amp conjugately matched during the 75% of the cycle it
is off?"
We must consider the complete cycle.
Working with spark ignition systems (Hettering) you may have encountered
a "dwell meter". It indicates the % of the time ignition points are
closed. When the points are closed, impedance between the meter and the
battery is insignificant. The meter if left continuously connected
through the points would indicate full-scale. When the points open,
their impedance is infinite. Left continuously open, the meter indicates
zero on the dwell scale.
Dwell is measured while the engine is rotating and the meter is being
connected intermittently to the battery through the ignition points.
Intermittent opening and closing of the points causes the same scale
reading that would be caused by replacing the points with some
particular value of fixed resistance (a resistor).
The main difference is that no dissipation occurs in the open ignition
points and precious little energy is lost in the closed points. Voila!
We have produced a dissipationless resistance.
The Class C amplifier is a switch which operates in the same manner. The
Kettering ignition points have a low-resistance ignition coil primary in
series, and the Class-C amplifier has a tuned plate circuit in series,
but both are being switched on and off repeatedly.
Richard Harrison
"I don`t know about the energy storage in the filter you mention, but I
would assume that if the filter output is a sine wave then the energy
storage required to produce a linear output is sufficient."
As a former FCC official, Walt knows their regulations sharply limit
harmonic content of FM broadcast carrier frequencies. A sine wave must
be pure, otherwise it has harmonic content (not allowed).
Jim Lux
> Cecil, W5DXP wrote:
> "Is the class-C amp conjugately matched during the 75% of the cycle it
> is off?"
Matched across what boundary? from output pi network to load?
From active device to pi network?
I'd venture that between active device and pi network, there isn't a
conjugate match, at any given instant, and perhaps not even considered
over the entire cycle (without resorting to some things like "apparent
impedance" which doesn't have a real clean definition).
In order to provide a true "conjugate match" to a device that is
changing state, the load must also be changing state: i.e. the match is
single valued, for any Zload, there is a single Zmatch that maximizes
power transfer; except perhaps for some trivial cases, like Z=zero or
infinity, but in such cases, the load doesn't dissipate ANY power, so
what is there to maximize.
Furthermore, if one looks at situations where you have, for instance, a
very low source impedance (a stiff voltage bus) or a very high source
impedance (a constant current source), power transfer is maximized to a
given load impedance when the reactive components are conjugate. In such
a case, the source and load resistances are not equal.
One might look at
http://p1k.arrl.org/~ehare/temp/conjugate_match/conjugate_match_theorum.pdf
http://mysite.orange.co.uk/g3uur/index.html
>
> We must consider the complete cycle.
>
> Working with spark ignition systems (Hettering) you may have encountered
> a "dwell meter". It indicates the % of the time ignition points are
> closed. When the points are closed, impedance between the meter and the
> battery is insignificant. The meter if left continuously connected
> through the points would indicate full-scale. When the points open,
> their impedance is infinite. Left continuously open, the meter indicates
> zero on the dwell scale.
>
> Dwell is measured while the engine is rotating and the meter is being
> connected intermittently to the battery through the ignition points.
>
> Intermittent opening and closing of the points causes the same scale
> reading that would be caused by replacing the points with some
> particular value of fixed resistance (a resistor).
> The main difference is that no dissipation occurs in the open ignition
> points and precious little energy is lost in the closed points. Voila!
> We have produced a dissipationless resistance.
I would say "apparent resistance".. the "conjugate match" and any other
linear circuit analysis can't necessarily be "averaged". Something like
Kirchoff's current law or voltage law (or Ohm's law, for that matter)
has to be true at any instant.
The challenge faced by folks faced with analyzing "real" circuits is
that you have to be careful about how you turn a circuit that is likely
time-varying AND nonlinear into a linearized approximation. For
instance programs like SPICE's transient analysis uses linear circuit
theory (via matrix analysis) in combination with an iterative
differential equation solver, and tries to treat the circuit as linear
at a given instant. (granted, newer versions of SPICE and its ilk are a
bit more sophisticated, since they can handle nonlinear terms in the
matrix).
> The Class C amplifier is a switch which operates in the same manner. The
> Kettering ignition points have a low-resistance ignition coil primary in
> series, and the Class-C amplifier has a tuned plate circuit in series,
> but both are being switched on and off repeatedly.
Complicated substantially by the fact that the active device in a RF
amplifier generally doesn't act as an ideal switch. So the piecewise
linearization you describe isn't totally applicable. For instance, a BJT
acts like a constant current source if base drive is fixed, but in RF
circuits, the base drive isn't fixed. In FET circuits you worry about
the gate capacitance.
This is why there are all sorts of variants of SPICE modified for
switching power supplies.
Owen Duffy
1...@storefull-3252.bay.webtv.net:
...
> The Class C amplifier is a switch ...
If you say it enough times, will it become true?
Owen
Cecil Moore
> Jim Lux wrote:
> "in a linear system"
>
> It produces no significant harmonics, so the system is linear.
I'm not trying to be difficult, but I just resurrected the
Fourier equation for the output of a class-B amplifier. It is:
i = 0.318Im + 0.500Im*sin(A) - 0.212Im*cos(2A) - 0.0424Im*cos(4A) - ...
The second harmonic is 42% of the amplitude of the
fundamental frequency at the plate of the tube amp.
Walter Maxwell
OK Cecil, if the 2nd harmonic is 42% of the amplitude of the fundamental at the
plate of the tube, can you determine the amplitude of the 2nd harmonic at the
output of a pi-network having a Q of 12?
Walt, W2DU
Walter Maxwell
OK Cecil, if the 2nd harmonic is 42% of the amplitude of the fundamental at the
Cecil Moore
> OK Cecil, if the 2nd harmonic is 42% of the amplitude of the fundamental at the
> plate of the tube, can you determine the amplitude of the 2nd harmonic at the
> output of a pi-network having a Q of 12?
I think I understand ideal class-A operation pretty well.
What I am trying to understand is: In class-B operation,
if the frequency of interest is made up of less than 1/2
of the current through the amplifier tube, is more than
1/2 of the current not conjugately matched?
Richard Fry
> Note that the reflected pulse appears some 6.2 盜 after the incident
> pulse, ...
_________
Correction: the reflected pulse appears 3.1 盜 after the incident pulse (the
time base was 0.5 盜/cm).
The rest of the statements there remain valid.
RF
Richard Harrison
"Matched across what boundary?"
Where there is a conjugate match in the transmitter-antenna system, it
exists at every pair of terminals. That is, the impedances looking in
opposite directions are conjugates. The resistive parts of the impedance
are equals and the reactances looking in opposite directions are
opposites of each other.
Cecil Moore
> Where there is a conjugate match in the transmitter-antenna system, it
> exists at every pair of terminals.
Quoting w2du's web page:
“The Conjugate Theorem also shows that in a sequence of
matching networks it is necessary to match at only one
junction *if the networks are non-dissipative*. In actual
practice, since there is usually some dissipation, it
is frequently desirable to adjust at more than one point.”
Richard Harrison
"The Class C amplifier is a switch...
If you say it enough times, will it become true?"
True is true no matter what anyone says. I`ve never seen Terman
misspeak.
On page 255 of his 1955 opus Terman wrote:
"---the Class C amplifier is adjusted so the plate current flows in
pulses that last less than half a cycle."
On page 450 he wrote:
"The high efficiency of the Class C amplifier is a result of the fact
that plate current is not allowed to flow except when the instantaneous
voltage drop across the tube is low; i.e. Eb supplies energy to the
amplifier only when he largest portion of the energy will be absorbed by
the tuned circuit."
Sounds like a switch to me. When switched on, voltage drop across the
tube is low. When switched off, voltage drop across the tube is Eb, but
since current is zero, no power is lost at that instant in the tube.
Richard Harrison
"Quoting W2DU`s web page:"
Sure hope Walt finds a publisher soon for his latest edition of
"Reflections". Web TV (a Microsoft company) doesn`t allow me to read
pdf.
Jim, K7JEB
Richard, try this: Let Google translate it from PDF to
HTML for you.
Do a Google search for "w2du reflections
chapter 19A".
One of the first entries will be a hit
at http://w2du.com/r3ch19a.pdf .
Under the Google entry for this, carefully look for
the words:
"File Format: PDF/Adobe Acrobat - View as HTML"
The words 'View as HTML' will be linked to Google's
HTML translation of the PDF.
The translation isn't pretty, but the essentials are
there. You will have to fill in some of the blanks
yourself.
Posting this as well for anyone else experiencing this
problem.
Jim, K7JEB
Cecil Moore
> Cecil, W5DXP wrote:
> "Quoting W2DU`s web page:"
>
> Sure hope Walt finds a publisher soon for his latest edition of
> "Reflections". Web TV (a Microsoft company) doesn`t allow me to read
> pdf.
Do a Google search for <NIST definition of "conjugate match">
One of the first entries will be:
[PDF]
Chapter 01
File Format: PDF/Adobe Acrobat - *View as HTML*
Quoting from Robert W. Beatty, NIST, Microwave Mismatch Analysis, (Ref
120):. 1) Conjugate match—The condition for maximum power absorption by
a load, in ...
w2du.com/Appendix09.pdf - Similar pages
Click on <View as HTML> and the section I quoted should appear
in HTML format.
Owen Duffy
> Owen Duffy wrote:
> "The Class C amplifier is a switch...
> If you say it enough times, will it become true?"
>
> True is true no matter what anyone says. I`ve never seen Terman
> misspeak.
>
> On page 255 of his 1955 opus Terman wrote:
> "---the Class C amplifier is adjusted so the plate current flows in
> pulses that last less than half a cycle."
>
> On page 450 he wrote:
> "The high efficiency of the Class C amplifier is a result of the fact
> that plate current is not allowed to flow except when the
> instantaneous voltage drop across the tube is low; i.e. Eb supplies
> energy to the amplifier only when he largest portion of the energy
> will be absorbed by the tuned circuit."
>
> Sounds like a switch to me. When switched on, voltage drop across the
> tube is low.
Lets plug some real world numbers in...
Take a DC supply of 1000V, and a valve that saturates at 200V, the RF
approximately sinusoidal voltage swing on the anode is from 200V to
1800V. (For avoidance of doubt, whilst the RF voltage on the anode is
approximately sinusoidal, the anode current waveform is not.)
If the conduction angle is 120 degrees (typical for Class C amplifiers),
the valve starts conducting at about 1000-800*sin((180-120)/2) or 600V
instantaneous anode voltage... and continues conducting as the
instantaneous anode voltage passes through the minimum and rises again,
cutting off when then instantaneous anode voltage again reaches 600V. In
this case, the anode voltage during conduction varies between 400 and
600V, 40% to 60% of the supply voltage.
The switch analogy is not a good one.
>When switched off, voltage drop across the tube is Eb, ...
That is wrong. The anode voltage is approximately a sinusoidal voltage
swing of almost (70% to 90%) Eb zero to peak superimposed on the DC
supply voltage (Eb).
You have taken a quote from Terman and weaved your own flawed extensions
(being the switch analogy and the statement about instantaneous anode
voltage).
Owen
Richard Harrison
"When switched-off, voltage drop across the tube is Eb"
Owen Duffy wrote:
"That is wrong."
Yes as a general statement, that is wrong. The instantaneous drop across
the tube is the sum of Eb and EL, the signal voltage across the load.
Terman shows this in Fig. 13-1(b) on page 449 of his 1955 opus.
Fig. 13-1(e) shows the plate current pulse which is less than 180
degrees in duration as the tube is biased beyond cut-off.
My switch analogy is imperfect but good enough to exemplify
dissipationless resistance as a part of the output impedance of a Class
C amplifier.
Cecil Moore
> My switch analogy is imperfect but good enough to exemplify
> dissipationless resistance as a part of the output impedance of a Class
> C amplifier.
Well, there are mechanical switches, digital switches,
and analog switches. I suspect a class-C amp falls
under the heading of an analog switch.
jimkel...@gmail.com
> Since E/I is simply a ratio, R is also a ratio. And we know that a ratio cannot dissipate power, or turn
> electrical energy into heat, thus the output resistance R is non-dissipative. I have made many measurements
> that prove this.
Hi Walt,
R is by definition a physical "property of conductors which depends on
dimensions, material, and temperature". So if we multiply both sides
of our "ratio" equation by I^2 to convert to power we get V*I =
I^2*R. Given that V, I, and R are all non-zero, why would you ask us
to believe that I^2*R and V*I could be zero? It's true that V^2/R is
a ratio. And I guess it's probably also true that the equation itself
doesn't dissipate power. But what would you have us believe that that
is supposed to prove?
73, Jim AC6XG
Owen Duffy
485181...@storefull-3257.bay.webtv.net:
> Jim Lux wrote:
> "in a linear system"
>
> It produces no significant harmonics, so the system is linear.
That is a new / unconventional definition of 'linear'.
The term is usually used in this context to mean a linear transfer
characteristic, ie PowerOut vs PowerIn is linear.
Considering a typical valve Class C RF amplifier with a resonant load:
Conduction angle will typically be around 120°, and to achieve that, the
grid bias would be around twice the cutoff voltage.
If you attempted to pass a signal such as SSB though a Class C amplifier
that was biased to twice the cutoff value, there would be no output
signal when the peak input was less than about 50% max drive voltage, or
about 25% power, and for greater drive voltage there would be output. How
could such a transfer characteristic be argued to be linear?
Owen
Alan Peake
I always believed that a ratio was a comparative measure between like
units - e.g. forward voltage to reverse voltage, output power to input
power etc. Voltage to current is not a ratio. V/I has dimensions of
resistance - ratios are dimensionless.
Alan
Cecil Moore
> R is by definition a physical "property of conductors which depends on
> dimensions, material, and temperature".
That's only one definition. From "The IEEE Dictionary",
the above is definition (A). Definition (B) is simply
"the real part of impedance" with the following Note:
"Definitions (A) and (B) are not equivalent but are
supplementary. In any case where confusion may arise,
specify definition being used."
Definition (B) covers Walt's non-dissipative resistance.
A common example is the characteristic impedance of
transmission line. In an ideal matched system V^2/Z0, I^2*Z0,
or V*I is the power being transferred under non-dissipative
conditions.
Walter Maxwell
<jimkel...@gmail.com> wrote in message
news:17e24476-2725-4317...@t54g2000hsg.googlegroups.com...
Hi Walt,
73, Jim AC6XG
Hello Jim,
I don't understand how my statement in the email above indicates that I^2*R and
V*R could be zero. The simple ratio of E/I is not zero, yet it defines a
resistance that is non-dissipative because a ratio cannot dissipate power.
Walt
Cecil Moore
> I don't understand how my statement in the email above indicates that I^2*R and
> V*R could be zero. The simple ratio of E/I is not zero, yet it defines a
> resistance that is non-dissipative because a ratio cannot dissipate power.
"The IEEE Dictionary" is careful to differentiate between
an E/I ratio equaling an impedance vs an "impedor" consisting
of physical components.
jimkel...@gmail.com
Good point. You may be right.
73 de jk
jimkel...@gmail.com
> I don't understand how my statement in the email above indicates that I^2*R and
> V*R could be zero. The simple ratio of E/I is not zero, yet it defines a
> resistance that is non-dissipative because a ratio cannot dissipate power.
> Walt
Hi Walt -
If E and I are not zero, then E*I is not zero. But you are correct
that the equations themselves do not dissipate power. :-) Resistors
do, however. If there isn't an actual resistor located where you make
your measurement, then of course there's no power being dissipated
there.
73, ac6xg
Richard Harrison
"V/I has dimensions of resistance - ratios are dimensionless."
Yes, until we name them. Cycles / seconds is now called Hertz.
My electronics dictionary defines ratio -
"The value obtained by dividing one number by another."
Simple and no qualifications.
From Newton:
Acceleration = force / mass
You must pick the right units or use constants to make the numbers work.
Some people are persuaded that resistance = loss. Not so at all.
Resistance is just a name given to the ratio of voltage to current. A
perfect reactance produces a voltage drop but no power is lost. A
transmission line can be lossless enough to qualify and have a Zo = sq
rt of L/C. Reg Edwards used to say that if your perfect line were long
enough you could measure its Zo with an ohmmeter. Reg was right because
no reflection would ever return to change the current supplied by the
ohmmeter.
Free-space has a lossless Zo of 120 pi (or 377 ohms) according to page
326 of Saveskie`s "Radio Propagation Handbook". This is a ratio which is
related to volts and amps but is actually the ratio of the electric
field strength to the magnetic field strength in an EM wave. The volts
and amps are in phase so it has the units of a pure resistance.
Walter Maxwell
"Owen Duffy" <no...@no.where> wrote in message
news:Xns9ABD9B0E61...@61.9.191.5...
Owen, 'linear transfer characteristic' isn't the only context for the use of the
word 'linear'. Even though the input circuit of a Class C amplifier is
non-linear, the output is linear due to the energy storage of the tank circuit
that isolates the input from the output, therefore, the output is linear. Proof
of this is that the output signal is a sine wave. In addition, the voltage and
current at the output terminals of the pi-network are in phase. Furthermore, the
ratio E/I = R appearing at the network output indicates that the output source
resistance R is non-dissipative, because a ratio cannot dissipate power. This
resistance R is not a resistor.
Walt
Walter Maxwell
<jimkel...@gmail.com> wrote in message
news:2b72179f-67e7-4035...@59g2000hsb.googlegroups.com...
> Walt
Hi Walt -
73, ac6xg
Jim, have you reviewed the new section Chapter 19A that appears on Cecil's
website that he uploaded on June 7? It appears there posted as 'Chapter 19A from
Reflections 3'. If you haven't reviewed it I urge you to do so, especially the
last portion where I report the measurements I made with a complex impedance
loading the amplifier. These measurements prove two things: 1) that the output
resistance of the amp is non-dissipative, and 2) that no reflected energy
reaches the amp tube, and in fact the measurements show that the tube doesn't
even see the reflected power. When you see the numbers and understand the
procedure I used in obtaining them you will be hard pressed to disagree with the
results.
Walt, W2DU
Walter Maxwell
"Alan Peake" <adp...@nosspam.activ8.net.au> wrote in message
news:48535A79...@nosspam.activ8.net.au...
Alan, I disagree with you when you say that 'voltage to current' is not a ratio.
IMHO, your are definine 'ratio' to narrowly. Below is a quote from Google:
Ratio
From Wikipedia, the free encyclopedia
. Learn more about using Wikipedia for research .Jump to: navigation, search
This article is about the mathematical concept. For the Swedish institute, see
Ratio Institute. For the academic journal, see Ratio (journal).
This article or section is in need of attention from an expert on the
subject.
Please help recruit one or improve this article yourself. See the talk
page for details.
Please consider using {{Expert-subject}} to associate this request with a
WikiProject
The ratio of width to height of typical computer displays
A ratio is a quantity that denotes the proportional[citation needed] amount or
magnitude of one quantity relative to another.
Ratios are unitless when they relate quantities of the same dimension. When the
two quantities being compared are of different types, the units are the first
quantity "per" unit of the second - for example, a speed or velocity can be
expressed in "miles per hour". If the second unit is a measure of time, we call
this type of ratio a rate.
Fractions and percentages are both specific applications of ratios. Fractions
relate the part (the numerator) to the whole (the denominator) while percentages
indicate parts per 100.
Note, Alan, the expression "When the two quantities being compared are
different types......
Walt, W2DU
Richard Clark
wrote:
Hi All,
It has taken considerable restraint not to ask some pointed questions:
1. Is a metal wire wound resistor NOT a resistance because it is not
carbon?
2. Is a carbon resistor NOT a resistance because it is not metal
wire-wound?
3. Is a Tube NOT a resistance because it contains no metal?
4. Is a Tube NOT a resistance because it contains no carbon?
5. Is a cathode resistor NOT a resistance when the tube conduction is
zero?
6. Is that same cathode resistor NOT a resistance because it conducts
non-linearly for some speciously constrained (and myopically chosen)
incomplete cycle of time?
finally, and possibly the only compelling logic that seems to flow
from this thread:
7. Is a Tube NOT a resistance simply because it lacks the familiar
shape of an axial lead resistor? (Or, rather, that a familiar axial
lead resistor cannot be found soldered between cathode and plate
within the vacuum?)
I have offered a spectrum of questions guaranteed to be accessible to
the buffet style of responding to cosmetic issues instead of
substance.
As this is all Rhetoric, I will take the author's prerogative to
short-cut the anticipated sputterings of denial, condemnation,
damnation, and outrage.
1. 2. 3. and 4. Carbon is a metal.
3. and 4. Plate dissipation bears scant relation to Ohmic Loss. And
yet there is heat there that is correlatably and causally related to
match, loss, and drive - from any "source."
4. there are many power tubes with Graphite (carbon) plates. The
original 813B comes to mind. Some power tubes have their screen grids
graphite (carbon) coated too! (Ohm's law still does not appreciably
account for plate dissipation.)
5. and 6. are sucker bait for those who would prove the world is
non-linear because of the discontinuity at the time of the big-bang
(or creation, take your pick).
7. Is, as I intimated, the implicit populist choice (masked as a
question) for those who cannot say what the source resistance IS, but
are over fulsome by half to say what it is NOT.
Describing what source resistance is NOT is like moving the stacks of
brass disks between the Towers of Hanoi. You can do that forever
without really coming to any conclusion. Given the length of many
threads that imitate this behavior, its popularity marks its less than
stupendous insights. But lest I interrupt the modern interpretation
of that game as it is played here, I would point out that parsing
"ratio" is even more funny. It sure beats Brett dragging the cesspool
for newspaper reports of cures for cancer.
-Phew-
73's
Richard Clark, KB7QHC
Owen Duffy
news:ziU4k.62$JK5...@newsfe07.lga:
Hi Walt,
A few issues....
Yes, I understand the context in which you mean linear (though I have
issues with your proposition)... but my comment was referring to the
assertion that 'no harmonics' relates to linear operation which seems to
me to refer to the transfer characteristic linearity context.
I do have issue with your stated 'proof'. Firstly, I must qualify that we
are talking steady state... the mention of resonant loads means we are in
the frequency domain. Whilst it might seem that the tank circuit / pi
coupler / whatever is just a network of passive parts and they are all
linear, the energy that is supplied to that circuit in each cycle depends
on the resonant load impedance and traditional PA design methods suggest
that that Eout/Iout relationship is not linear for changes in load Z,
although it might be approximately linear over a small range.
I recognise a distinction between resistance (the ratio of E/I) and a
resistor (one type of component that exhibits resistance)... but I would
not claim that resistance is just a 'ratio' because it implies it is a
dimensionless ratio.
Owen
Alan Peake
Walter Maxwell wrote:
> "Alan Peake" <adp...@nosspam.activ8.net.au> wrote in message
> Alan, I disagree with you when you say that 'voltage to current' is not a ratio.
> IMHO, your are definine 'ratio' to narrowly. Below is a quote from Google:
>.............
Well, someone has redefined "ratio" since I went to school. My old maths
text book says "The term ratio is used when we wish to compare the size
ofr magnitude of two quantities (or numbers) of the same kind, i.e.,
expressed in the same units, and is measured by a fraction"
All my dictionaries say much the same thing. There is no mention of
comparing quantities of different units. That to me would be like
comparing apples with oranges.
Alan
Alan Peake
Richard Harrison wrote:
> Some people are persuaded that resistance = loss. Not so at all.
> Resistance is just a name given to the ratio of voltage to current.
If you define resistance as simply V/I with no regard to phase, then
what you say is true but if V and I aren't in phase then you have
impedance consisting of real and imaginary components - resistance AND
reactance.
> Free-space has a lossless Zo of 120 pi (or 377 ohms) according to page
> 326 of Saveskie`s "Radio Propagation Handbook". This is a ratio which is
> related to volts and amps but is actually the ratio of the electric
> field strength to the magnetic field strength in an EM wave. The volts
> and amps are in phase so it has the units of a pure resistance.
I suppose you could also say that a real resistor is also lossless as
the heat due to I*I*R is radiated into space and thus is not lost :)
Alan
Richard Harrison
lost in cyberspace somewhere. I was responding to Owen Duffy.
Owen wrote:
"How could such a transfer characteristic be argued to be linear?"
I responded:
Conditioning.
Class C amplifiers are used lawfully in great abundance. That is proof
enough that they are relatively free from distortion. Pulses in plate
current don`t prevent the output of the Class C amplifier from becoming
a pure sinusoid. Just as an internal combustion engine uses an almost
endless string of exlosions in its cylinders to produce a smooth uniform
rotation of its crankshaft and flywheel, the Class C amplifier uses an
almost endless series of pulses to produce a smooth sinusoid.
I will quote B. Whitfield Griffith, Jr., Principal Engineer (retired) at
Continental Electronics, Dallas Texas, builder of many of the world`s
most powerful radio transmitters. Griffith says on page 500 of
"Radio-Electronic Transmission Fundamentals", that it is important where
you couple the load to the Class C amplifier:
"Figure 56-2 shows how the class C amplifier might look in a typical
arrangement. Many refinements of the circuit, which are necessary for
practical reasons, are omitted here, since we are concerned only with
the fundamental principles of its operation at this time. The plate load
impedance consists of a tank circuit of a type similar ro that of Fig.
15-5; the difference is that the load resistor is in series with the
inductance rather than the capacitance. This is the preferred
arrangement, because the harmonic components of the plate current all
have frequencies higher than the fundamental and quite naturally tend to
follow the capacitive branch of the circuit. By extracting power from
the inductive branch, therefore we can expect to find less harmonic
energy in the output than would be present if we loaded the capacitive
branch. This load resistance may be an actual resistor, if we wish to
feed the output of this amplifier into a dummy load for measurement
purposes, or it may be the input resistance presented by some type of
impedance-matching network so arranged that the loading of the amplifier
can readily be varied. Another common method is to couple resistance
effectively into the tank inductance by means of the mutual inductance
between the tank and a secondary coil which is coupled to it
magnetically, where resistive loads appear in the secondary circuit.
There is also shown in Fig. 56-2 the r-f waveform of voltage and current
which we would expect to find at various points in the amplifier
circuit. No allowance is made in these illustrations for the differences
in potentials of various portions of the circuit; these diagrams are
merely representative of the behavior of the r-f potentials and
currents. Notice particularly that the r-f plate voltage is 180 degrees
out of phase with the r-f grid voltage. The reason for this is easily
understood. When the grid is its at its most positive potential, the
plate current is at its maximum. As the plate current is drawn through
the load impedance, the increase in plate current causes a corresponding
reduction in plate voltage. The plate voltage therefore swings downward
at the moment the grid voltage swings upward. We also see that the
current in the load resistor is lagging the r-f plate voltage by an
angle of a little less than 90 degrees. Correct operation of the tank
circuit requires that the resistance of this load resistor be much
smaller than the reactance of the coil."
Walter Maxwell
news:26406-485...@storefull-3256.bay.webtv.net...
action of the energy storage of the tank circuit and that of a automobile
engine, so I'd like you to read a portion of Chapter 19 from Reflections 2 to
see how I approached the same analogy for the book that I quote below:
Therefore, the pi-network must be designed to provide the equivalent
optimum resistance RL looking into the input for whatever load terminates the
output. The current pulses flowing into the network deliver bursts of electrical
energy to the network periodically, in the same manner as the spring-loaded
escapement mechanism in the pendulum clock delivers mechanical energy
periodically to the swing of the pendulum. In a similar manner, after each plate
current pulse enters the pi-network tank curcuit, the flywheel effect of the
resonant tank circuit stores the electromagnetic energy delivered by the current
pulse, and thus maintains a continuous sinusoidal flow of current throughout the
tank, in the same manner as the pendulum swings continuously and periodically
after each thrust from the escapement mechanism. The continuous swing of the
pendulum results from the inertia of the weight at the end of the pendulum, due
to the energy stored in the weight. The path inscribed by the motion of the
pendulum is a sine wave, the same as at the output of the amplifier. We will
continue the discussion of the flywheel effect in the tank circuit with a more
in-depth examination later.
....
We now return to conduct a close examination of the vitally important
flywheel effect of the tank circuit. The energy storage (Q) in the tank produces
the flywheel effect that isolates the nonlinear pulsed energy entering the tank
at the input from the smoothed energy delivered at the output. As a result of
this isolation the energy delivered at the output is a smooth sine wave, with
linear voltage/current characteristics that support the theorems generally
restricted to linear operation. We know that the widely varying voltage/current
relationship at the tank input results in widely varying impedances, which
precludes the possibility of a conjugate match at the input of the tank circuit.
However, the energy stored in the tank provides constant impedance at the output
that supports both the Conjugate Matching and the Maximum Power-transfer
Theorems.1
The acceptance by many engineers and amateurs of the notion that the output
of the RF tank is nonlinear is a reason some readers will have difficulty in
appreciating that the output of the RF tank circuit is linear, and can thus
support the conjugate match. Valid analogies between different disciplines are
often helpful in clarifying difficulties in appreciating certain aspects of a
particular discipline. Fortunately, energy storage in the mechanical discipline
has a valid and rigorous analogous relationship with energy storage in LC
circuitry that makes it appropriate to draw upon a mechanical example to clarify
the effect of energy storage in the RF tank circuit. (A further convincing
analogy involving water appears later in the Chapter, in which the origin of the
term 'tank circuit' is revealed.)
The smoothing action of the RF energy stored in the tank circuit is
rigorously analogous to the smoothing action of the energy stored in the
flywheel in the automobile engine. In the automobile engine the flywheel smooths
the pulses of energy delivered to the crankshaft by the thrust of the pistons.
As in the tank circuit of the amplifier, the automobile flywheel is an energy
storage device, and the smoothing of the energy pulses from the pistons is
achieved by the energy stored in the flywheel. In effect, it is the flywheel
that delivers the energy to the transmission. The energy storage capacity
required of the flywheel to deliver smooth energy to the transmission is
determined by the number of piston pulses per revolution of the crankshaft. The
greater the number of pistons, the less storage capacity is required to achieve
a specified level of smoothness in the energy delivered by the flywheel. The
storage capacity of the flywheel is determined by its moment of inertia, and the
storage capacity of the tank circuit in the RF amplifier is determined by its Q.
Owen Duffy
48547F...@storefull-3256.bay.webtv.net:
...
> Class C amplifiers are used lawfully in great abundance. That is proof
> enough that they are relatively free from distortion. Pulses in plate
> current don`t prevent the output of the Class C amplifier from becoming
> a pure sinusoid.
... a very long dissertation on Class C amplifiers snipped.
Richard, analysis of the Class C amplifier excited with a constant
amplitude single frequency sine wave is revealing about their transfer
linearity.
I do not disagree that a Class C amplifier excited with a constant
amplitude single frequency sine wave driving a resonant load produces a
low distortion constant amplitude single frequency sine wave output.
But the absence of harmonic distortion in such an amplifier is not
evidence that the amplifier transfer characteristic is linear. You may be
able to use harmonic distortion to detect non-linearity in, for example,
audio amplifiers... but not in RF amplifiers with a resonant load... for
the reasons set out in your quotation.
A Class C amplifier is unsuited to amplfying SSB telephony because it is
manifestly non-linear. In fact, a Class C amplifier is so non-linear that
it is well suited to use as a relatively efficient harmonic multiplier.
Class B and AB RF amplifiers are extremely sensitive to non-linearity in
the region near cut-off and must have sufficient idle current in every
active device (which means conduction ange is > 180°) so that distortion
products are sufficiently low. This means that the theoretical conduction
angle of 180° for Class B is just not realisable because of distortion,
much less 120°.
Owen
Walter Maxwell
"Owen Duffy" <no...@no.where> wrote in message
appropriate to include it in view of Richard's similar discussion on the
automotive engine analogy to the RF tank circuit. I'll try to keep my comments
shorter from now on.
Walt, W2DU
Walter Maxwell
"Owen Duffy" <no...@no.where> wrote in message
Owen Duffy
> Sorry about the 'long dissertation on Class C amps', Owen, but I
> thought it appropriate to include it in view of Richard's similar
> discussion on the automotive engine analogy to the RF tank circuit.
> I'll try to keep my comments shorter from now on.
Walt, it wasn't so much that it was long, but it was long and for all
that was said, it didn't address the linearity issue.
I understand your position to be that the behaviour of the tank circuit
is independent of the transfer linearity of the active device... but
asserting that 'things' are linear because there are no harmonics is
wrong and being so, is no support for your argument.
I am wary of analogies, the switch analogy that was raised is not a good
approximation and I haven't even thought about the car engine.
I am genuinely insterested in your argument. I don't accept it (yet?) as
you know, and I have spent some time over the last 18 months or so
exploring the concept you describe.
Fundamentally, I am trying to reconcile what you say with the techniques
commonly accepted for designing such a PA. Those design techniques give
us a method of predicting power output at different load impedances, and
the E/I characteristic for different loads is not always a straight line
(as it would be if a Thevenin equivalent circuit exists), though it might
appear fairly straight over a narrow domain. Since working from
characteristic curves is so prone to error, my modelling has been based
on an idealised triode transfer characteristic, but with similar
behaviour to an 811A. The analysis is waiting for me to build the
analytical equations for the negative feedback due to cathode
degeneration in a grounded grid configuration. I need to apply more time
to it, and the revived discussion might focus me for a bit!
Owen
Richard Clark
>The analysis is waiting for me to build the
>analytical equations for the negative feedback due to cathode
>degeneration in a grounded grid configuration.
Hi Owen,
Consult the work of H.W. Bode taken from his lectures at Bell Labs ca.
1939, and then rendered into text as:
Network Analysis and Feedback Amplifier Design,
Chapter IV Mathematical Definition of Feedback
4.2 Return Voltage and Reduction in Effect of Tube Variations p 46
It attends specifically (grounded grid triode) what you call out
above.
In a nutshell, Output (or Input) Z can be tailored by what is called
the "noise gain" of an amplifier. In today's parlance, that is that
portion of Open loop gain that is fed back to the input to create what
is called closed loop gain ("noise gain" is simply the difference when
all gains are expressed as dB). The higher the "noise gain" the lower
the output Z (or higher the input Z) compared to the native (open
loop) Z. There are a host of other characteristics improvements that
flow from this same boon offered by "noise gain" (dynamic range, noise
rejection, linearity, CMRR, PSRR, and so on).
This shorthand can be found expanded in discussion in
5.5 Effect of Feedback on Input and Output Impedances of
Amplifiers
bullets 1. through 4. but it serves the reader to really stick with
the first 4 chapters to gain the proficiency to tackle the remaining
15 as the text is heavily cross-referential.
The general formula can be found at:
5.11 Exact Formula for External Gain with Feedback (5-30)
Bode was not simply a chalk-and-talk theorist wholly ignorant of the
practical realities as are evidenced in several chapter headings:
Chapter VII Stability and Physical Realizability
Chapter IX Physical Representation of Driving Point Impedance
Functions
Chapter XI Physical Representation of transfer Impedance Functions
Chapter XIII General Restrictions on Physical Network
Characterizations
Ultimately, it takes very little reading applied to the conventional
designs found in Amateur class amplifiers to discover there is really
very, very little modification of amplifier characteristics offered
through negative feedback design (it costs too much). In fact, I
would say none whatever - hence the heavy filtering at the outputs and
the customers' universal acceptance of barely mediocre performance. It
might be said that every transmitter owned by hams is a museum of
1930s performance. And for those who mistake the feedback of
stabilization (barely found in those same cheap designs) - this is not
negative feedback, it is compensation. It too has scant effect on
tailoring (reducing/increasing) impedances.
As I am undoubtedly the only copy holder of this book in this group,
access can be obtained through:
http://books.google.com/books?client=firefox-a&um=1&q=H.W.+Bode&btnG=Search+Books
which will provide a surplus of leads, if not the exact title. Some
links might provide a pdf, others full access, yet others limited
access, and most have links to copies in the market place.
Given the usual confusion over what constitutes a Conjugate match
(when most argue an Impedance match in its place) says discussion of
"Efficiency and maximum power transfer" without more rigorous
resources fails to even reach the level of tepid conjecture.
Bottom line is the source presents a real resistance and no appeal to
ratios, linearity, load lines, fly-wheels, or partial cycles is
necessary to arrive at a definitive value (which, to this point has
been notably absent in the face of obviously localized heat and loss).
There is plenty of discussion of what it is NOT, but none seem to know
what it IS. That the typical Amateur amplifier source Z is
demonstrable is embarrassment enough to this shortfall of expertise.
(The pile of theories, books and formulas merely support the obvious,
not replace it.)
Tom Donaly
Hi Richard,
A more modern treatment is _High Linearity RF Amplifier
Design_ by Peter B. Kenington. ISBN 1-58053-143-1. I think Amazon
still carries it.
73,
Tom Donaly, KA6RUH
Jim Lux
>
> Ultimately, it takes very little reading applied to the conventional
> designs found in Amateur class amplifiers to discover there is really
> very, very little modification of amplifier characteristics offered
> through negative feedback design (it costs too much). In fact, I
> would say none whatever - hence the heavy filtering at the outputs and
> the customers' universal acceptance of barely mediocre performance. It
> might be said that every transmitter owned by hams is a museum of
> 1930s performance. And for those who mistake the feedback of
> stabilization (barely found in those same cheap designs) - this is not
> negative feedback, it is compensation. It too has scant effect on
> tailoring (reducing/increasing) impedances.
probably not "every transmitter", but certainly the vast majority of
designs, particularly those for HF based on tubes in the ARRL handbook
(and by extension, those sold to readers of the handbook).
Cost *is* a factor. The Harris PWM modular transmitters are very cool,
but beyond the means of most hams as a commercially manufactured item
(in that, the NRE for a consumer mfr to get there would be prohibitively
high)
One should also not neglect that the hobby aspect of ham radio provides
an incentive (for some) to preserve fine (or not so fine) examples of
past radio art. No more unusual than steam train fans or classic auto
collectors. There is a visceral satisfaction of seeing those glowing
tubes with the plates changing color, notwithstanding that the RF
performance, in objective terms, is horrid.
>
> As I am undoubtedly the only copy holder of this book in this group,
> access can be obtained through:
>
I'll bet not..<grin>
Jim Lux
> richard...@webtv.net (Richard Harrison) wrote in news:23000-
> 485181...@storefull-3257.bay.webtv.net:
>
>> Jim Lux wrote:
>> "in a linear system"
>>
>> It produces no significant harmonics, so the system is linear.
>
> That is a new / unconventional definition of 'linear'.
>
> The term is usually used in this context to mean a linear transfer
> characteristic, ie PowerOut vs PowerIn is linear.
Or, as I used it, that superposition holds.
One can build an amplifier or other device where the Pout(Pin) =straight
line, but is not linear in the formal sense. Say you built a widget that
measured the input frequency and amplitude, then drove a synthesizer at
that frequency and amplitude = 2*input amplitude.
>
> Considering a typical valve Class C RF amplifier with a resonant load:
>
> Conduction angle will typically be around 120°, and to achieve that, the
> grid bias would be around twice the cutoff voltage.
>
> If you attempted to pass a signal such as SSB though a Class C amplifier
> that was biased to twice the cutoff value, there would be no output
> signal when the peak input was less than about 50% max drive voltage, or
> about 25% power, and for greater drive voltage there would be output. How
> could such a transfer characteristic be argued to be linear?
It would not be.You're right
The active device isn't linear.
neither is the whole assembly.
I think, though, that sometimes we take a more casual view of linear
(e.g. people talk about the linearity of a log detector.. referring to
the deviation from a Voltage out=dBm in straight line.)
And, some confusion about nonlinear devices in a building block that is,
by and large, linear (e.g. a power op amp with an AB2 output stage and a
fair amount of negative feedback) with some constraints on frequency and
amplitude.
>
> Owen
Richard Harrison
"... but asserting that things are linear because there are no harmonics
No one is arguing that an amplitude modulated wave can be amplified by a
Class C amplifier stage unimpaired by amplitude distortion.
Terman wrote on page 525 0f his 1955 opus:
"Amplitude distortion exists when the modulation envelope contains
frequency components not present in the modulating signal. Thus if the
modulating signal is a sine wave, then amplitude distortion will cause
the envelope to contain harmonics of the modulating signal, which in
turn denotes the presence of high-order sideband components that differ
from the carrier frequency by harmonics of the modulating frequency."
I`ve used microwave system performance monitors which alarmed on this
principle.
If there are no harmonics there is no distortion no matter how lousy the
transfer function. It is legal to filter out noise and distortion.
Michael Coslo
> On Mon, 16 Jun 2008 03:32:23 GMT, Owen Duffy <no...@no.where> wrote:
>
>> The analysis is waiting for me to build the
>> analytical equations for the negative feedback due to cathode
>> degeneration in a grounded grid configuration.
>
> Hi Owen,
>
> Consult the work of H.W. Bode taken from his lectures at Bell Labs ca.
> 1939, and then rendered into text as:
Hi Richard,
In this group, would not the work of Vaughn Bode be more appropriate?
- 73 de Mike N3LI -
Walter Maxwell
"Jim Lux" <james...@jpl.nasa.gov> wrote in message
news:g3611p$s19$1...@news.jpl.nasa.gov...
Owen, I didn't realize that this thread was specific to 'linear transfer
characteristic'. I thought the thread topic was sufficiently broad so as to
include the subject of linearity of the output of the tank circuit that permits
the use of theorems that require the output to be linear. Richard H's and my
posts were simply reminders that the energy storage in the tank circuit is the
reason for the linear relation between voltage and current at the output of both
Class B and C amplifiers that results in a sine wave. From that perspective I
believed our posts were legitimate to the thread topic. Apparently we were
wrong.
And Owen, I'm somewhat surprised that you don't agree with the flywheel analogy
with respect to the smoothing effect of the energy storage in the tank circuit.
This analogy has been around for decades--it's not my invention. IMHO, the
periodic energy spurts from the pistons entering the flywheel is precisely an
analog of the energy spurts of the periodic current pulses entering the tank
citcuit. Why do you not agree? Even the pendulum swing is appropriate, because
if you trace the position of the pendulum with respect to time you'll discover
the trace is a perfect sine wave, while the short spurt of energy supplied by
the spring at the beginning of each cycle is just sufficient to overcome the
energy dissipated due to friction at the axis plus the aerodynamic resistance.
How could this not be an appropriate analogy? Sorry to have forced you away from
the thread topic with questions not pertaining to the thread.
I am also curious as to why the subject of 'linear transfer characteristic' with
respect to Class C amps was even considered, because the Class C amp has always
been known to have a distorted output relative to its input. I would agree that
the subject is appropriate when considering Class AB and B amplifiers, but not
C.
Walt, W2DU
Richard Clark
<dtdo...@sbcglobal.net> wrote:
>Hi Richard,
> A more modern treatment is _High Linearity RF Amplifier
>Design_ by Peter B. Kenington. ISBN 1-58053-143-1. I think Amazon
>still carries it.
>73,
>Tom Donaly, KA6RUH
Thanx Tom.
Owen Duffy
...
>> That is a new / unconventional definition of 'linear'.
>>
>> The term is usually used in this context to mean a linear transfer
>> characteristic, ie PowerOut vs PowerIn is linear.
>
> Or, as I used it, that superposition holds.
> One can build an amplifier or other device where the Pout(Pin)
> =straight line, but is not linear in the formal sense. Say you built a
> widget that measured the input frequency and amplitude, then drove a
> synthesizer at that frequency and amplitude = 2*input amplitude.
>
Yes Jim, I should have written Vout/Vin is linear, that Vout(Vin) has no
significant terms higher than first order.
Noting that a single ended Class B or AB amplifier can only be linear
when a resonant load or suitable filter is included as part of the
system.
Elsewhere it was suggested that I do not accept the 'flywheel'
explanation of the tank circuit. That is not true, but it is a limited
explanation, simple, and appealing, but limited.
Another explanation is to view the anode current waveform as containing a
DC component, a fundamental component and harmonic components and a
filter that adequately reduces the undesired components provides the
solution to a single ended Class B or AB linear amplifier. The filter is
not restricted to a resonant 'tank' circuit.
I have modelled the operating characteristics of my HF linear using 4
572B in AB2. An FFT of the anode current reveals the spectral content, it
is plotted at http://www.vk1od.net/lost/572BIaSpectrum.png . Of course,
the output filter must only select the fundamental component for linear
operation, selection of a harmonic would not be acceptable for a complex
input waveform because it would destroy the absolute relationship between
different frequency components of the input.
Owen
Owen Duffy
Richard stated "It produces no significant harmonics, so the system is
linear." It is that with which I disagree.
...
> And Owen, I'm somewhat surprised that you don't agree with the
> flywheel analogy with respect to the smoothing effect of the energy
> storage in the tank circuit. ...
I have not disagreed with that in anything that I wrote.
> ...
> I am also curious as to why the subject of 'linear transfer
> characteristic' with respect to Class C amps was even considered,
> because the Class C amp has always been known to have a distorted
> output relative to its input. I would agree that the subject is
> appropriate when considering Class AB and B amplifiers, but not C.
>
Because Richards statement quoted above (which must be about transfer
linearity) is being used to support your assertion that the PA is linear
in its terminal V/I response with changing load.
Walt, the thread has become muddled with helpers muddying the water. Your
proposition needs to be argued with a single logically developed sound
argument. Your Chapter 19 tries to do that.
I have already stated that (as yet?) I am unconvinced, and I make the
observation that I am not alone. I will work through resolving the
apparent inconsistencies in my own time and without the confusion of
whether or not harmonics exist or more correctly the extent to which they
exist, and what that might mean.
Owen