On 1/24/06 12:57 PM, in article 6xwBf.11951$bF.2404@dukeread07, "Uncle
Peter" wrote:


"Straydog" wrote in message
My understanding of AM transmitter technology would estimate that a 32v3,
with ~120 DC input (two 6146s, or were they still using one 4D32?) would
have at most (class C, plate modulated) 70% X 120 = 80 watts of CW carrier
output. 60 watts of audio on that final tube (as a non-linear high level
mixer) will at best, double the _instantaneous_ (peak) input voltage,
therefore power to 240 watts (plate current will _not_ double even if the
plate voltage doubles on peak audio cycle [look at your tube curves again
of iP vs vP at constant biases]) which you could only attempt to measure
with an oscilloscope. Peak output? Could it be more than 240 x 0.7 = 168
watts? I doubt it (unless he's got something like "super-modulation" in
the rig).


Without delving into the limitations of the 32V3, according to the info
from an ARRL publication:

"..since the amplitude at the peak of the upswing is twice the unmodulated
amplitude, the power at this instant is four times the unmodulated, or 400
watts."

Average power, on the other hand, will be 1.5 times carrier. A Class C
amplifier with high level modulation should produce an instaneous PEP
of 4x carrier power.

Pete






Getting back to basics: A 120W (input) power, class C stage, will require
60W of audio (using a high-level, e.g. plate, modulator) for 100%
modulation. If we assume 85% efficiency, then the output will consist of a
Carrier of 102W and two sidebands of 25.5W each.

In my opinion, any other explanation is useless. Do remember that the
carrier amplitude does NOT vary with modulation.

Don

On Tue, 24 Jan 2006 13:40:54 -0800, Don Bowey
wrote:

On 1/24/06 12:57 PM, in article 6xwBf.11951$bF.2404@dukeread07, "Uncle
Peter" wrote:


"Straydog" wrote in message
My understanding of AM transmitter technology would estimate that a 32v3,
with ~120 DC input (two 6146s, or were they still using one 4D32?) would
have at most (class C, plate modulated) 70% X 120 = 80 watts of CW carrier
output. 60 watts of audio on that final tube (as a non-linear high level
mixer) will at best, double the _instantaneous_ (peak) input voltage,
therefore power to 240 watts (plate current will _not_ double even if the
plate voltage doubles on peak audio cycle [look at your tube curves again
of iP vs vP at constant biases]) which you could only attempt to measure
with an oscilloscope. Peak output? Could it be more than 240 x 0.7 = 168
watts? I doubt it (unless he's got something like "super-modulation" in
the rig).


Without delving into the limitations of the 32V3, according to the info
from an ARRL publication:

"..since the amplitude at the peak of the upswing is twice the unmodulated
amplitude, the power at this instant is four times the unmodulated, or 400
watts."

Average power, on the other hand, will be 1.5 times carrier. A Class C
amplifier with high level modulation should produce an instaneous PEP
of 4x carrier power.

Pete






Getting back to basics: A 120W (input) power, class C stage, will require
60W of audio (using a high-level, e.g. plate, modulator) for 100%
modulation. If we assume 85% efficiency, then the output will consist of a
Carrier of 102W and two sidebands of 25.5W each.

In my opinion, any other explanation is useless. Do remember that the
carrier amplitude does NOT vary with modulation.

Don


I don't remember the 32v3 specs but a pair of 6146B's is rated for
120 watts carrier output on AM. 6146A's are rated for 100 watts output
on AM.

Assuming the 120 watts carrier output, when modulated 100% the voltage
doubles and the current also doubles on modulation peaks. Doubling the
voltage and doubling the current works out to 4 times the power. This
is of course Peak Envelope Power of the signal which would be 480
watts.

You can not just add the audio power to the carrier power to find PEP.
You must first add the voltages together.

Peak envelope power is what the FCC is concerned with for maximum
allowable power of 1500 watts.

Although when advertising an AM transmitter it is common to state the
carrier power and not try to confuse people by stating the PEP power
and not stating that is what is being speced.

73
Gary K4FMX

"Gary Schafer" wrote in message
...
Although when advertising an AM transmitter it is common to state the
carrier power and not try to confuse people by stating the PEP power
and not stating that is what is being speced.

73
Gary K4FMX


Agreed.. I was just giving an explanation of the seller's somewhat
cryptic sales pitch.

Pete

Uncle Peter wrote:
"Gary Schafer" wrote in message
...

Although when advertising an AM transmitter it is common to state the
carrier power and not try to confuse people by stating the PEP power
and not stating that is what is being speced.

73
Gary K4FMX



Agreed.. I was just giving an explanation of the seller's somewhat
cryptic sales pitch.

Pete



How much would it be in P.M.P.O. watts? Coupla hundred kilowatts?



-Bill

On 1/24/06 2:54 PM, in article , "- exray
-" wrote:

Uncle Peter wrote:
"Gary Schafer" wrote in message
...

Although when advertising an AM transmitter it is common to state the
carrier power and not try to confuse people by stating the PEP power
and not stating that is what is being speced.

73
Gary K4FMX



Agreed.. I was just giving an explanation of the seller's somewhat
cryptic sales pitch.

Pete



How much would it be in P.M.P.O. watts? Coupla hundred kilowatts?



-Bill

No, that's eBay Watts.

On 1/24/06 2:31 PM, in article ,
"Gary Schafer" wrote:

On Tue, 24 Jan 2006 13:40:54 -0800, Don Bowey
wrote:

On 1/24/06 12:57 PM, in article 6xwBf.11951$bF.2404@dukeread07, "Uncle
Peter" wrote:


"Straydog" wrote in message
My understanding of AM transmitter technology would estimate that a 32v3,
with ~120 DC input (two 6146s, or were they still using one 4D32?) would
have at most (class C, plate modulated) 70% X 120 = 80 watts of CW carrier
output. 60 watts of audio on that final tube (as a non-linear high level
mixer) will at best, double the _instantaneous_ (peak) input voltage,
therefore power to 240 watts (plate current will _not_ double even if the
plate voltage doubles on peak audio cycle [look at your tube curves again
of iP vs vP at constant biases]) which you could only attempt to measure
with an oscilloscope. Peak output? Could it be more than 240 x 0.7 = 168
watts? I doubt it (unless he's got something like "super-modulation" in
the rig).


Without delving into the limitations of the 32V3, according to the info
from an ARRL publication:

"..since the amplitude at the peak of the upswing is twice the unmodulated
amplitude, the power at this instant is four times the unmodulated, or 400
watts."

Average power, on the other hand, will be 1.5 times carrier. A Class C
amplifier with high level modulation should produce an instaneous PEP
of 4x carrier power.

Pete






Getting back to basics: A 120W (input) power, class C stage, will require
60W of audio (using a high-level, e.g. plate, modulator) for 100%
modulation. If we assume 85% efficiency, then the output will consist of a
Carrier of 102W and two sidebands of 25.5W each.

In my opinion, any other explanation is useless. Do remember that the
carrier amplitude does NOT vary with modulation.

Don


I don't remember the 32v3 specs but a pair of 6146B's is rated for
120 watts carrier output on AM. 6146A's are rated for 100 watts output
on AM.

My Viking had a 4D32 final and it would load to well over 100W.


Assuming the 120 watts carrier output, when modulated 100% the voltage
doubles and the current also doubles on modulation peaks. Doubling the
voltage and doubling the current works out to 4 times the power. This
is of course Peak Envelope Power of the signal which would be 480
watts.

Where does the double voltage come from at 100% modulation? I can only
account for a 50% rise in voltage.


You can not just add the audio power to the carrier power to find PEP.
You must first add the voltages together.

Good idea, if one knows the voltages......


Peak envelope power is what the FCC is concerned with for maximum
allowable power of 1500 watts.

Although when advertising an AM transmitter it is common to state the
carrier power and not try to confuse people by stating the PEP power
and not stating that is what is being speced.

73
Gary K4FMX


My point is that listing the PEP capability of an AM transmitter isn't as
useful as stating it can output about 100 watts.

Don

On Tue, 24 Jan 2006 16:07:30 -0800, Don Bowey
wrote:

On 1/24/06 2:31 PM, in article ,
"Gary Schafer" wrote:

On Tue, 24 Jan 2006 13:40:54 -0800, Don Bowey
wrote:

On 1/24/06 12:57 PM, in article 6xwBf.11951$bF.2404@dukeread07, "Uncle
Peter" wrote:


"Straydog" wrote in message
My understanding of AM transmitter technology would estimate that a 32v3,
with ~120 DC input (two 6146s, or were they still using one 4D32?) would
have at most (class C, plate modulated) 70% X 120 = 80 watts of CW carrier
output. 60 watts of audio on that final tube (as a non-linear high level
mixer) will at best, double the _instantaneous_ (peak) input voltage,
therefore power to 240 watts (plate current will _not_ double even if the
plate voltage doubles on peak audio cycle [look at your tube curves again
of iP vs vP at constant biases]) which you could only attempt to measure
with an oscilloscope. Peak output? Could it be more than 240 x 0.7 = 168
watts? I doubt it (unless he's got something like "super-modulation" in
the rig).


Without delving into the limitations of the 32V3, according to the info
from an ARRL publication:

"..since the amplitude at the peak of the upswing is twice the unmodulated
amplitude, the power at this instant is four times the unmodulated, or 400
watts."

Average power, on the other hand, will be 1.5 times carrier. A Class C
amplifier with high level modulation should produce an instaneous PEP
of 4x carrier power.

Pete






Getting back to basics: A 120W (input) power, class C stage, will require
60W of audio (using a high-level, e.g. plate, modulator) for 100%
modulation. If we assume 85% efficiency, then the output will consist of a
Carrier of 102W and two sidebands of 25.5W each.

In my opinion, any other explanation is useless. Do remember that the
carrier amplitude does NOT vary with modulation.

Don


I don't remember the 32v3 specs but a pair of 6146B's is rated for
120 watts carrier output on AM. 6146A's are rated for 100 watts output
on AM.

My Viking had a 4D32 final and it would load to well over 100W.


Assuming the 120 watts carrier output, when modulated 100% the voltage
doubles and the current also doubles on modulation peaks. Doubling the
voltage and doubling the current works out to 4 times the power. This
is of course Peak Envelope Power of the signal which would be 480
watts.

Where does the double voltage come from at 100% modulation? I can only
account for a 50% rise in voltage.


You can not just add the audio power to the carrier power to find PEP.
You must first add the voltages together.

Good idea, if one knows the voltages......


Peak envelope power is what the FCC is concerned with for maximum
allowable power of 1500 watts.

Although when advertising an AM transmitter it is common to state the
carrier power and not try to confuse people by stating the PEP power
and not stating that is what is being speced.

73
Gary K4FMX


My point is that listing the PEP capability of an AM transmitter isn't as
useful as stating it can output about 100 watts.

Don


I agree stating PEP output of an AM transmitter does little. But a
properly operating transmitter should be able to give pep at 4 times
the carrier power. Some transmitters do not have that capability
because of a poor modulator or too small finals, or power supply etc.
I, like Peter, was trying to dispel the somewhat misleading add of the
original poster.

As far as the voltage doubling with modulation, you only need to look
at the output on an oscilloscope at the composite signal and you will
easily see that it does. Set the scope to show the carrier level at
say 2 divisions on the screen. With modulation you will see the
positive peaks reach 4 divisions on the scope. The negative peaks will
reach zero on the scope.

Another way to look at this is when modulating the final the peak
audio voltage must equal the plate voltage for 100% modulation. In
order for the modulation to go to negative 100% the audio voltage must
cause the plate voltage to swing down to zero. By the same note in
order to reach 100% positive modulation the audio voltage must cause
the plate voltage to go to twice the dc voltage.

It may seem confusing because if you add the average output power up a
100 watt transmitter is only 150 watts. 100 watts carrier and 25 watts
in each side band. However if you add the voltage of the carrier plus
the voltage of each audio side band and then calculate the power you
will see that it is 4 times the carrier power.

100 watts into 50 ohms = 70.7 volts
25 watts into 50 ohms = 35.35 volts
25 watts into 50 ohms = 35.35 volts
Total voltage = 141.4 volts (which is 2 x carrier
voltage)

P= E squared / R

141.4 x 141.4 = 19994
19994 / 40 = 400 watts

73
Gary K4FMX

I think Gary Schafer's analysis (below) is basically correct but I have
minor comments to add (in addition to my earlier post, also quoted below,
claiming the 400+ peak output just could not be possible, but I think I
was wrong about that). See below.

On Tue, 24 Jan 2006, Gary Schafer wrote:

On Tue, 24 Jan 2006 16:07:30 -0800, Don Bowey
wrote:

On 1/24/06 2:31 PM, in article ,
"Gary Schafer" wrote:

On Tue, 24 Jan 2006 13:40:54 -0800, Don Bowey
wrote:

On 1/24/06 12:57 PM, in article 6xwBf.11951$bF.2404@dukeread07, "Uncle
Peter" wrote:


"Straydog" wrote in message
My understanding of AM transmitter technology would estimate that a 32v3,
with ~120 DC input (two 6146s, or were they still using one 4D32?) would
have at most (class C, plate modulated) 70% X 120 = 80 watts of CW carrier
output. 60 watts of audio on that final tube (as a non-linear high level
mixer) will at best, double the _instantaneous_ (peak) input voltage,
therefore power to 240 watts (plate current will _not_ double even if the
plate voltage doubles on peak audio cycle [look at your tube curves again
of iP vs vP at constant biases]) which you could only attempt to measure
with an oscilloscope. Peak output? Could it be more than 240 x 0.7 = 168
watts? I doubt it (unless he's got something like "super-modulation" in
the rig).


Without delving into the limitations of the 32V3, according to the info
from an ARRL publication:

"..since the amplitude at the peak of the upswing is twice the unmodulated
amplitude, the power at this instant is four times the unmodulated, or 400
watts."

Average power, on the other hand, will be 1.5 times carrier. A Class C
amplifier with high level modulation should produce an instaneous PEP
of 4x carrier power.

Pete






Getting back to basics: A 120W (input) power, class C stage, will require
60W of audio (using a high-level, e.g. plate, modulator) for 100%
modulation. If we assume 85% efficiency, then the output will consist of a
Carrier of 102W and two sidebands of 25.5W each.

In my opinion, any other explanation is useless. Do remember that the
carrier amplitude does NOT vary with modulation.

Don


I don't remember the 32v3 specs but a pair of 6146B's is rated for
120 watts carrier output on AM. 6146A's are rated for 100 watts output
on AM.

My Viking had a 4D32 final and it would load to well over 100W.


Assuming the 120 watts carrier output, when modulated 100% the voltage
doubles and the current also doubles on modulation peaks. Doubling the
voltage and doubling the current works out to 4 times the power. This
is of course Peak Envelope Power of the signal which would be 480
watts.

Where does the double voltage come from at 100% modulation? I can only
account for a 50% rise in voltage.


You can not just add the audio power to the carrier power to find PEP.
You must first add the voltages together.

Good idea, if one knows the voltages......


Peak envelope power is what the FCC is concerned with for maximum
allowable power of 1500 watts.

Although when advertising an AM transmitter it is common to state the
carrier power and not try to confuse people by stating the PEP power
and not stating that is what is being speced.

73
Gary K4FMX


My point is that listing the PEP capability of an AM transmitter isn't as
useful as stating it can output about 100 watts.

Don


I agree stating PEP output of an AM transmitter does little.

I also think it doesn't mean much even for an SSB signal (which is
difficult to compare with AM coming from the same station) because the
S-meter damping makes it difficult to measure signal strength. Also, for
the power in two sidebands (only one of which is needed) and the waste in
the carrier, the usual efficiency of a linear amp is about half of that
for a (non-linear) AM final amp.

But a
properly operating transmitter should be able to give pep at 4 times
the carrier power.

I think this contradicts something you said below, and contradicts what I
said in my post, above. See more below.

Some transmitters do not have that capability
because of a poor modulator or too small finals, or power supply etc.
I, like Peter, was trying to dispel the somewhat misleading add of the
original poster.

As far as the voltage doubling with modulation, you only need to look
at the output on an oscilloscope at the composite signal and you will
easily see that it does. Set the scope to show the carrier level at
say 2 divisions on the screen. With modulation you will see the
positive peaks reach 4 divisions on the scope. The negative peaks will
reach zero on the scope.

Yes, and I have done this, myself and seen a carrier "band" on my scope,
and when speaking into the microphone (on a Johnson Ranger), seen the high
peaks go up to about double the height of the carrier and the valleys go
down to about zero (below zero would take the carrier away thus leading
to splatter).

Another way to look at this is when modulating the final the peak
audio voltage must equal the plate voltage for 100% modulation. In
order for the modulation to go to negative 100% the audio voltage must
cause the plate voltage to swing down to zero. By the same note in
order to reach 100% positive modulation the audio voltage must cause
the plate voltage to go to twice the dc voltage.

Yes, but none of this explains where the "4X pep" statement comes from. In
fact, even at the instantaneous double the plate voltage, there is no
plate current increase. The (non-linear) tube is not a (linear) resistor
where you double the voltage accross the resistor and cause the current
to double, thus a quadrupling of power. Look at the curves in your tube
manuals for any given control (triodes if no other grids are present) or
screen grid (tetrodes or pentodes) bias. Above some threshold plate voltage
the plate current is independent of plate voltage. Plate current is only
affected by grid voltages.

It may seem confusing because if you add the average output power up a
100 watt transmitter is only 150 watts. 100 watts carrier and 25 watts
in each side band.

I think this is right.

However if you add the voltage of the carrier plus
the voltage of each audio side band and then calculate the power you
will see that it is 4 times the carrier power.

I don't think this is quite right and, after thinking about all of this,
part of the reason I already gave above is also not quite right.
Here is another way I think we can look at this question (see below):

100 watts into 50 ohms = 70.7 volts
25 watts into 50 ohms = 35.35 volts
25 watts into 50 ohms = 35.35 volts
Total voltage = 141.4 volts (which is 2 x carrier
voltage)

I think this is a bit of a mistake and it would be better to calculate
peak power in the following manner:

First, there is no peak output power from the carrier, the carrier is
always there and at the same strength no matter if there is modulation or
not.

So, power contribution FROM THE CARRIER (whether modulated or not) is
still only your 100 watts, period. The carrier contributes NO extra power
to the peak power we're all interested in.

Now, lets look at the power in the sidebands. I'll accept that 25 watts as
converting to 35 volts, but that 35 volts is not added to carrier
power because it is in the 2-3 kc spectrum above or below the carrier.
So, use your formula below and get (35 X 35)/50 = 1225/50 = 24 watts.
And, for the second (other) sideband, there will be another 24 watts.
Total: 48 watts of audio translated to RF in addition to the 100 watt
(constant) carrier. Thus peak power is 148 watts.

Where is the conflict between my analysis and yours?

Its in the way we think about modulator power output (usually stated as
audio power must be about half of final DC input, thus a 120 w DC input
class-C final needs about 60 watts of audio). So, when you look up, for
example the specs on a pair of 6146s in modulator service (go look in the
back of your ARRL handbooks, any of them) and see them talk about 110-130
watts for the pair in either AB1 or AB2! They don't tell you that is _peak_
audio power! For the same tube in class C, they are showing 50-70 watts
(continuous RF) out for one tube. Not much difference in power specs per
tube, but you won't get anywhere near those 110-130 watts of audio in
continuous (i.e. average) power because the heat dissipation will melt
the plates since class AB is much less efficient than class C.

So, your example of 25 watts per sideband is more like an _average_ power
specification and what we should be looking at is what is the peak
audio power (or voltage) coming out of the modulator. That peak audio
voltage out of the modulator has to be equal to the plate voltage and in
the same direction to double the final amp plate voltage, and valley
bottom audio equal to the final amp plate voltage but in the opposite
direction to reduce plate voltage to zero or near zero. Final B+ voltage
plus the audio peak thus shows up on the scope, transiently, as RF output
voltage at double the height of carrier alone, and final B+ minus the
negative audio peak, at the negative peak, causes the height of the scope
trace to go, transiently, to zero. So, when THEY talk about 60 watts of
audio power to modulate a 120 w DC input final amplifier, they are talking
about more like 60 watts average power which really means something like
120 watts of peak power (and is in all of the tube manuals where specs
for all of the amplifier classes are shown next to each other! [this is
not the case for receiving tube manuals which talk about average signal
output for tubes like 6L6s in classes A and maybe AB]). This gets us into
the audiophiles' endless arguing about what audio power means and under
what conditions and specifications (eg. the distortions) need to be made
for considering peak audio power, especially in music audio (with eg. drum
beat transients) rather than voice audio (more or less steady).

My general feeling is that in the AM transmitter situation (and the SSB
situation, too) that talking about peak power has only theoretical value
and almost no practical value and is confusing. You can only measure it on
a scope and S-meter readings will be subject to the damping factor in the
mechanical needle, electronic fudging by so-called "peak reading" meters
including the bar graph things, and any asymmetry in the voice waveform,
and distortions and non-linear characteristics in the rest of the
electronics. I have heard guys on the air using the same amplifier in a
linear class mode who switch from AM mode to SSB mode and my S meters (on
lots of receivers) show the same peak value on SSB as the steady value on
AM, plus or minus maybe one or two db, at most.

Guys would be best off talking about final amp measured DC input watts
(continous) and/or final amp measured carrier (continuous) RF output watts
and not say too much about their modulator power unless they have a scope
on the instantaneous modulator output voltage and current and can actually
make a real, valid, representative measurement of both peak and average
power and be able to say "Oh, my actual, real, measured-on-a-scope
instantaneous peak whatever is X plus or minus Z accuracy."

Now, how about peak power input to your 100 watt incandescent light bulb
at home? Remember your house AC line voltage (117 VAC) is measured and
speced as RMS (root mean square), so peak voltage is 1.41 X 117, peak
current is 1.41 x 1 amp, and peak power is thus 1.41 X 1.41 X 100 watts?
About 200 watts? Is that meaningful? No, because it isn't. Think DC and
why RMS specs are always used for AC circuits and VOM voltmeter scales.

Thanks for your attention, sorry to be long-winded.

Art, W4PON

P= E squared / R

141.4 x 141.4 = 19994
19994 / 40 = 400 watts

73
Gary K4FMX

Let's review some definitions to start:

AVERAGE POWER

Average power is found by squaring RMS voltage and dividing by
resistance. Or RMS voltage times RMS current.


PEP

It helps to fully understand exactly what PEP is in an SSB
transmitter. Then it is easier to see in an AM transmitter.

Peak envelope power is important because that is how the FCC defines
how much power we can run.

Let's look at the FCC definition of Peak Envelope Power:

"Peak envelope power output of a transmitter is the AVERAGE power at
the crest of the modulation envelope over at least one rf cycle."

NOT TO BE CONFUSED WITH AVERAGE POWER READ ON A METER as it swings
around!

If you think about what that is saying it will make sense. If you
transmit just a carrier with an SSB transmitter of say 100 watts. That
is 100 watts average power output. It is also 100 watts PEP output.
(in this case the envelope is infinitly long) If you were to key it on
and off in the CW mode the power relationship would be the same. 100
watts PEP on each CW dash or dot sent.

If you now switch to the SSB mode and modulate the transmitter so that
the peaks on the scope looking at that signal reach the same height,
the transmitter will be putting out 100 watts PEP.

If you were to modulate that same SSB transmitter with 2 equal
amplitude tones you would get a scope pattern that looks similar to an
AM signal modulated with a single tone. The crest (or peak) of the
waveform represents 100 watts PEP same as with voice but with the
tones it is easier to see as the waveform will be stable.

If you were to increase the speed of the time base on the scope and
spread the waveform out you would see that each crest of the audio
wave form has within it many cycles of the RF frequency. These many
cycles of RF are the AVERAGE power contained in the signal.

You will note that the maximum AVERAGE power is only reached for
several RF cycles at the crest of each audio cycle.

This is what is known as PEAK ENVELOPE POWER. (see definition above
again)


PEP WATTMETERS

A true PEP reading watt meter will show the peak envelope power of the
above signals as described.
There are a lot of so called PEP watt meters on the market. Not all
are able to properly read. Even the Bird meters have problems with
some types of wave forms.


S METER READINGS

S meter readings will vary according the particular receiver being
used but most all S meters are peak responding circuits. Most will
read pretty close to the peak values, depending on the decay time of
the circuit some may not hang up there like others do. If you think
about it if you have ever been plagued with pulse noise like ignition
noise it only takes a very narrow pulse occurring at a rather slow
rate to hold the S meter up high. Increasing the rate will not
increase the meter reading.

Likewise with an SSB signal, once the station is transmitting his peak
power on a regular basis, increasing mike gain or increasing
compression will not raise the S meter reading maximum.

The AVC circuit in the receiver must respond to the peaks or the
receiver would overload the detector if the gain was not cut back when
a peak was received. The S meter reads AVC voltage.


AM TRANSMITTER

It is best to try and understand the output signal of the AM
transmitter before trying to coralate it with what goes on at the
input side. Swapping back and forth can be confusing.

Take our 100 watt carrier output transmitter again. Measuring the
output voltage of the RF we find that it is 70.7 volts RMS across our
50 ohm load. P = I squared / R so 70.7 x 70.7 = 5000. 5000/50 = 100
watts.

Let's modulate 100% with a single audio tone. We get out of it a 100
watt carrier and two 25 watt side bands. 3 distinct signals. As you
stated before the carrier always remains constant.

If we look at the output signal on our scope we will see that it looks
similar to the SSB signal that was modulated with 2 tones. We see the
modulation envelope. We can again expand the scope's time base and
look at the RF cycles within each modulation peak. Same as with the
SSB signal, at the crest of the modulation envelope is the peak
envelope power of the composite signal.

Now let's get back to measuring that PEP. We know that the carrier
alone had a power of 100 watts which produced 70.7 volts across a 50
ohm load. If we look at the scope with and without modulation we see
that the voltage output doubles with modulation so it will be 141.4
volts RMS at the crest of the modulation wave form.
Again P = E squared /R. 141.4 x 141.4 = 20000. 20000 / 50 = 400 watts
PEP.

If we were to measure this with a good PEP wattmeter we would see the
meter also indicate 400 watts PEP.
Again some so called PEP meters do not do well on this type of wave
form. The carrier tends to confuse the meter as it causes an offset
in the voltage being read by the meter and the meter tries to average
it so the net result is some gets subtracted from the reading. It is
due to the way in which the particular meter circuit operates.


CONVERTING RMS TO PEAK

It would seem at first glance that you could find the peak power of
the 25 watt side bands and add things together but that doesn't work.
You can not convert power. THERE IS NO RMS IN POWER!

There is a wide misconception that there is something called RMS
power. There is no such thing! There is only AVERAGE power and PEAK
power. (note the FCC definition of PEP)

You find average power by using RMS voltage. But once you multiply or
divide, RMS term, the RMS goes away.
So once you have power you can not multiply it by 1.414 to find peak
power.

To find peak power you must first add the voltages together or find
the peak voltage of an rms voltage by multiplying by 1.414 then
finding power.


AM LINEAR

Operating an SSB transmitter and amplifier in the AM mode, if properly
set up, will produce exactly the same looking output signal as a plate
modulated AM transmitter.

If we have an SSB amplifier that will put out 1000 watts PEP on SSB it
will also put out 1000 watts PEP on AM.
But in order to do so the carrier output must be limited to 250 watts
output. It must be tuned up in the CW mode for 1000 watts output. Then
when switching to AM the carrier is reduced to 250 watts output
without touching any tuning controls. The amplifier must still be
tuned to be able to produce the 1000 watt peak envelope output.

When we modulate the 250 watt carrier with AM the peak envelope power
at 100% modulation will reach 1000 watts pep (or 4 times the carrier)
just as it did with the plate modulated AM transmitter. Looking at the
output with a scope we will see the voltage double with modulation
verses just the carrier.


POWER IN SIDE BANDS

As a note is seems that having 2 side bands with the same information
in an AM signal is useless but it is not.
In the detector of the receiver the energy in both side bands combine
and add together. So rather than using only 1 side band of 25 watts
you are really using 50 watts of side band energy. So a 3 db addition.
There is also another 3 db gained in the detector because of the
voltage doubling with the side bands being coherent in the detector.
So the carrier is really the only thing wasted.


PLATE CURRENT AND VOLTAGE DOUBLING

It is easiest to see with a triode tube that is plate modulated.
Doubling the plate voltage will cause the plate current to also
double. That is if the tube is capable of providing enough emission.
This must be a linear function in order to avoid distortion when
modulating.

Tubes that are weak may not be able to provide this. That is one
reason that PEP may not fully reach 4 times the carrier power with
100% modulation.

Screen grid tubes are not linear in this respect. Plate current is
somewhat independent of plate voltage. That is why you must also
partly modulate the screen along with the plate when using a screen
grid tube in the final. You want to have a linear plate voltage to
plate current relationship.

This is also why a lot of broadcast transmitters use triodes in the
final. Easier to maintain linear modulation.


HANDBOOK

All this can be found in the AM section in some of the older
handbooks. The newer ones do not cover AM very well.

73
Gary K4FMX

Gary (and anyone else who cares), since my last post, which responded to
several other posts on the topic of PEP in an AM transmitter, I looked up
some things and cleared up a major misunderstanding in my own mind. I will
add that as comments to the part of your post, below, which is relevant to
the issue. As far as all of your definitions below, PEP wattmeters,
S-metes, SSB signals are concerned, I think you made a lot more mistakes
than you realize. However, I'm going to delete all these irrelevant parts
(most of what you said) and concentrate on the source of the confusion.
I may make comments in a separate post on the parts I deleted fro this
one.

On Wed, 25 Jan 2006, Gary Schafer wrote:


Let's review some definitions to start:

AVERAGE POWER

Average power is found by squaring RMS voltage and dividing by
resistance. Or RMS voltage times RMS current.


PEP

deleted

PEP WATTMETERS

deleted


S METER READINGS

deleted


AM TRANSMITTER

deleted


CONVERTING RMS TO PEAK

deleted


AM LINEAR

deleted


POWER IN SIDE BANDS

deleted


PLATE CURRENT AND VOLTAGE DOUBLING

Here is the crux of the problem. Earlier today I looked in my old RCA
receiving tube manual and transmitting tube manuals at the transfer
characteristics of many dozens of tubes and I looked at them with this
question of PEP for an AM signal. I will incorporate some of what I
learned as comments on your comments. The basic fact that I was not aware
of is that there is an apparent conflict between the relationship
between plate current and plate voltage if you look at the curves that
show plate current as independent of plate voltage and then ask how do you
get, on modulation, a peak input power four times the unmodulated input
power so you can get a peak, on modulation, output power that is four
times the unmodulated output power.

It is easiest to see with a triode tube that is plate modulated.

Nah, "easiest" has nothing to do with it. Triode has nothing to do with
it.

The issue is that all of the triode transfer characteristics curves I saw
showed plate current to be _proportional_ to plate current (but with
offsets and some non-linearities, which are mostly unimportant).
When I looked at all the tetrodes and pentode curves, then, yes, they all
showed plate current independent of plate voltage. However, at any given
plate voltage, plate current was also _proportional_ to screen voltage
(also with and offset and some non-linearities). Now, it makes sense that
if screen voltage is made proportional -- in some fashion (usually a
screen voltage dropping resistor connected to the modulated plate supply)--
to plate voltage, then plate current will increase, or decrease, in
parallel with plate voltage as modulator voltage adds, and subtracts, from
the B+ plate voltage (all as the modulator output signal varies with audio
input waveform)

Doubling the plate voltage will cause the plate current to also
double.

From the curves, the relationship between plate current and plate current
might not always be exactly a 1:1 relationship, but to an approximation
this doubling is an acceptable understanding. And, that is how, on peak
input from modulation one gets four times unmodulated input, and output
will be proportional to input which can be looked at as average or peak,
but the peak output on modulation will also be four times unmodulated
output.

That is if the tube is capable of providing enough emission.

That is a separate issue and anyone designing a circuit and sellecting a
tube for use needs to understand the specifications in the manuals.

This must be a linear function in order to avoid distortion when
modulating.

Almost nothing is perfectly linear. All audio circuits will have
measureable distortion (IM, harmonic, and others). The only criterion is
whether the distortion is acceptable.

Tubes that are weak may not be able to provide this. That is one
reason that PEP may not fully reach 4 times the carrier power with
100% modulation.

I think for this issue one needs at least an oscilloscope to even start
measuring and investigating what is going on (and they need to be
wideband or sampling scopes, too). "Meters" are just indicators.

Screen grid tubes are not linear in this respect. Plate current is
somewhat independent of plate voltage. That is why you must also
partly modulate the screen along with the plate when using a screen
grid tube in the final.

There is an equally important reason why you must, and preferably, fully
modulate the screen voltage as well as the plate voltage (and this is
almost never discussed). If you ever have screen voltage above plate
voltage, then screen current will go up dramatically and so will screen
heat dissipation. You could melt the screen grid with just one word into
the microphone. You can blow the screen grid almost instantly just by
accidentally having screen voltage present without plate voltage.

You want to have a linear plate voltage to
plate current relationship.

This is also why a lot of broadcast transmitters use triodes in the
final. Easier to maintain linear modulation.

I think, if you looked at as many transfer characteristics, as I did
earlier today, for transmitting tubes, you might appreciate that there
is more heterogeneity between triodes than tetrodes or pentodes in terms
of plate I/V relationships. Broadcast AM transmitters never gave us any
kind of high fidelity so linearity was never that much of an issue. In
broadcasst FM transmitters, power and voltage linearity anywhere in the
RF chain was irrelevant.


HANDBOOK

All this can be found in the AM section in some of the older
handbooks.

I was never very satisfied with much in the handbooks, whether early or
late.

The newer ones do not cover AM very well.

They are covering tubes and analog subjects less well, too.
Everything is going digital, solid stae, chips, and software.

Art, W4PON

73
Gary K4FMX