Subject: glowbugs V1 #236
glowbugs Tuesday, February 3 1998 Volume 01 : Number 236
Date: Mon, 2 Feb 1998 19:27:32 +0000
From: Sandy W5TVW <ebjr@worldnet.att.net>
Subject: "CX" this weekend..
20 meters was slim pickins'! 40 was very active. I managed
to use most of my gear. Was saving the "big gun" forr 80 and also the
'10 Hartley for 80 as well. Unfortunately, we had thunderstorms here
which just about wiped out any hope of any decent 80 meter operations!
My score and fun suffered this time! I also understand there was a
foul-up in some of the "magazine" announcements for CX and some
people worked it Saturday instead of Sunday. Didn't even try any AM
phone due to the noise!
Wait 'till next time!
73,
E. V. Sandy Blaize, W5TVW
"Boat Anchors collected, restored, repaired, traded and used!"
417 Ridgewood Drive
Metairie, LA., 70001
**860 Hartley 'ECO' construction "on hold"****
*** Looking for a TRC-10 transceiver ******
*** Looking for an RAL receiver ***********
Date: Mon, 2 Feb 1998 13:19:32 -0600 (CST)
From: glueck@austin.ibm.com (Glueck)
Subject: AEA AT-300 Switch needed
I got a 300 watt AEA Ant tuner at a swapfest which needs one of two inductor
switches, the one on the TRANSMITTER inductor (appears to be very much like
the other switch on ANTENNA side, but am not for sure). This is a two wafer
switch that is bigger than the standard Centralab switch assemblies that are
reasonably plentiful. The switch is burned up, probably from switching the
taps under power. I had heard that AEA was absorbed by someone else, but do
not know who. Parts may have been available from yet another source, as I
remember.
The unit is model AT-300 with cross-needle meter, three tuning knobs (two
switches and one tuning capacitor), and antenna selector switch plus a couple
of lesser switches to change meter power level and turn on lamp for meter.
If there are folks out there that knows where to get parts, please answer
to either my email address directly (glueck@austin.ibm.com) - the preferred
approach - or to the list.
Thanks,
Howard Glueck / K5ZUA
Date: Mon, 2 Feb 1998 18:49:25 -0400
From: "Brian Carling" <bry@mnsinc.com>
Subject: Attn: JUNO ops!
For all of you e-mail-only guys, here is a helpful hint!
Please pass it along to ALL of the hams you know who use JUNO
or other e-mail-only based web access!!
For callsign-lookup-by-email send a blank message to:
lookup@qrz.com
You will get back the information on how to look up anyone!
Date: Tue, 3 Feb 1998 11:28:05 -0500 (EST)
From: rdkeys@csemail.cropsci.ncsu.edu
Subject: How are CW output power and input power and PEP related?
FOOD FOR THOUGHT AND DISCUSSION.....
(cuz ol' BA Bob has his dunce cap on today and don't haves his books
handy ta latch onto):
Can someone explain the relationship between average CW output power,
and PEP and how we technically measure both? The FCC rates power output
as PEP power. I want to make sure I am measuring things correctly.
I thought I knew how to do that, but now I am not so sure. I want to
make sure I get it exactly right.
The FCC rules stiplulate PEP measurements for output power on all
transmitters.
One of the FCC OET bulletins indicates measuring peak RF voltage across
a calibrated load and equating it to obtain PEP as......
PEP = (Epeak x Epeak)/R.
Is this correct?
ON a CW sine wave (not complex waveform such as SSB modulation) how can
this be done with an average-reading RF current ammeter typically found
in the average glowbug junquebox? I prefer to use the simple RF ammeter
if I can.
If I am interpreting things correctly....
Average Power = PEP/1.414 = 0.707 PEP.
Thus if we are allowed 1500 watts PEP that is actually 0.707 x 1500 = 1060
watts average power (gee they gave us 60 watts). Is this correct?
Thus an old Navy KW CW rig can be loaded up to 1060 watts average power
output on CW and still meet the 1500 watts PEP limit? Is this correct?
I can measure the CW average power by using a simple RF ammeter into a
calibrated dummy load of say 50 ohms, and relate the measured average
RF current to power as:
Average Power = (Iav x Iav) x R.
PEP = 1.414 Average Power.
For example, using a simple RF ammeter, I measure 1.5 amps into a 50 ohm
dummy load. Then:
Average Power = (1.5 x 1.5) x 50 = 112.5 watts output average
PEP = 1.414 x 112.5 = 159.1 watts peak envelope power output
Is this correct?
What needs to be put together for a proper RF voltmeter to measure
PEP by the FCC method? (Hint... look in the 69 handbook in the test
instruments section, perhaps.) A pair of rectifier diodes as a full
wave into a small 100ua meter with proper capacitive coupling should
do the trick, but is not as accurate as a VTVM or fet voltmeter, and
it would give the average RF voltage, not the instantaneous RF voltage.
But, taking that average reading into account, and the small error
likely to be there, it will probably do for the average ham measurement.
It might be better to measure the waveform on an oscilloscope perhaps?
But, I don't always have an oscilloscope handy but should always at least
have an averaging RF ammeter or RF voltmeter handy with the dummy load.
If using the averaging RF voltmeter, then:
Peak RF voltage = 1.414 RF average voltage.
Thus for proper PEP measurement and calculating using our averaging
RF voltmeter (the FW diode RF voltmeter from the 69 handbook):
PEP = ((1.414 x Eav) x (1.414 x Eav))/R.
Is this correct?
Our glowbugs probably don't usually qualify as big-gun equipment, but,
one or two might, and it is probably good we all review how to properly
measure PEP on our glowbugs.
Am I all screwed up today, or am I missing the forest for the trees?
Thanks.....
Bob/NA4G
Date: Tue, 3 Feb 1998 11:34:57 -0600
From: mack@mails.imed.com (Ray Mack)
Subject: Re: How are CW output power and input power and PEP related?
BA Bob:
The answer to the first question is that RMS Power (what you call average power)
is Erms * Erms/R.
PEP is Epeak * Epeak/R.
Now for the fun part: Epeak = 1.414 * Erms
Substituting in the PEP equation above: PEP = Erms * 1.414 * Erms * 1.414/R
multiply 1.414 by 1.414 and the reduced equation is: PEP = 2 * Erms * Erms/R
Simplifying again: PEP = 2 * Prms
The answer to question 2:
You can, indeed measure the peak of the waveform and do the calculation as
described in the OET bulletin.
The answer to question 3:
See answer 1 above. Take your RMS (if it uses a thermocouple) ammeter reading
for CW and square the current, multiply by 2, and then divide by R.
For a 50 Ohm system PEP = I * I * 2/50
The answer to question 4:
No it is not correct! You need to multi[ply by 1.414 one more time.
The answer to question 5:
Bob, your description of an RF voltmeter is a *peak* reading meter!!!
There are 2 sets of circuits in the figure (My book is 1977, but should be
the same as your 1969). The top ones without capacitors give a good
approximation to average voltage. Notice from the drawings that this is AVERAGE
and *NOT* RMS. The 2 are different!! The circuits at the bottom of the figure
use capacitors and give *PEAK* readings.
RMS is *MUCH* harder to do than peak unless you are sure you have a pure sine
wave.
Remeber that you are making a circuit that is exactly the same as a capacitor
input power supply when doing the peak circuits, except you are getting the AC
from your transmitter instead of a power transformer on the AC mains. I suspect
that the ease of making a peak reading RF voltmeter over a true RMS is the
reason for the method of measuring. The main source of error is the slight
droop in the voltage due to the loading of the microammeter. Using a DVM or
VTVM as the readout will give more accurate peak readings. It will also give a
longer time constant so you can actually read the meter before the volage falls
off.
I did some quick calculations to see how important the diode drops are in the
calculation. At 10 W PEP the peak voltage in a 50 Ohm system is 22 Volts. At
1500 W PEP the peak voltage is 193 volts!!! Better use some hefty diodes and
resistors in that meter. Or better yet, use a 20 dB T pad attenuator ahead of
the meter.
Answer to question 6:
These calculations suffer from the same errors as in the question 4 section.
Answer to question 7:
Not all screwed up, but probably overwhelmed by the algebra involved! It's good
to do a sanity check from time to time. I know I frequently miss one of the
small details then remember "oh, yeah. I knew that" :<)
Why life is the way it is:
I did the calculations when I was looking at just exactly how does AM vs SSB
compare. In the old days we were limited to 1000 W DC input for the carrier.
If we got 60% efficiency, we had 600 W RMS (1200 W PEP) out just for the
carrier! When you add another 1200 W PEP for the modulator power, we were
getting 2400 W PEP for AM when going full out and staying legal.
Today with the 1500 W PEP limit, we are limited to 375 W RMS of carrier.
Obviously, those of us who like Admirable Modulation got stuck with a sizeable
power drop with the new regulations.
Ray Mack
WD5IFS
mack@mails.imed.com
Friendswood (Houston), TX
______________________________ Reply Separator _________________________________
Subject: How are CW output power and input power and PEP related?
Author: rdkeys@csemail.cropsci.ncsu.edu at mails
Date: 2/3/98 11:28 AM
FOOD FOR THOUGHT AND DISCUSSION.....
(cuz ol' BA Bob has his dunce cap on today and don't haves his books
handy ta latch onto):
//question 1
Can someone explain the relationship between average CW output power, and
PEP and how we technically measure both?
<snip>
//question 2
One of the FCC OET bulletins indicates measuring peak RF voltage across
a calibrated load and equating it to obtain PEP as......
PEP = (Epeak x Epeak)/R.
Is this correct?
//question 3
ON a CW sine wave (not complex waveform such as SSB modulation) how can
this be done with an average-reading RF current ammeter typically found
in the average glowbug junquebox? I prefer to use the simple RF ammeter
if I can.
<snip>
//question 4
<snip>
//question 5
What needs to be put together for a proper RF voltmeter to measure PEP by
the FCC method?
<snip>
//question 6
Is this correct?
//question 7
Our glowbugs probably don't usually qualify as big-gun equipment, but,
one or two might, and it is probably good we all review how to properly
measure PEP on our glowbugs.
Am I all screwed up today, or am I missing the forest for the trees?
Thanks.....
Bob/NA4G
Date: Tue, 3 Feb 1998 10:18:22 -0800 (PST)
From: Ken Gordon <keng@uidaho.edu>
Subject: 304TL socket...
I need ONE of these. I thought I had two, but can only find one.
I really can't understand why these things should be so hard to find.
Several models of Nuclear Magnetic Resonance instruments as used in
Chemistry Departments, Physics Departments, and Hospitals used up to 12
ea. 304TLs as shunt regulators, and most of those machines are now no
longer being used.
Wonder where they all went?
Anyone got a socket they would part with for a reasonable sum?
Ken W7EKB
Date: Tue, 3 Feb 1998 18:48:14 +0000
From: Sandy W5TVW <ebjr@worldnet.att.net>
Subject: WTB: "White" boxes
Does anyone know of a source of miniature and other size
"white boxes" for tubes other than Antique Electronic Supplies?
73,
E. V. Sandy Blaize, W5TVW
"Boat Anchors collected, restored, repaired, traded and used!"
417 Ridgewood Drive
Metairie, LA., 70001
**860 Hartley 'ECO' construction "on hold"****
*** Looking for a TRC-10 transceiver ******
*** Looking for an RAL receiver ***********
Date: Tue, 03 Feb 1998 20:01:50 +0100
From: Jan Axing <janax@algonet.se>
Subject: Re: How are CW output power and input power and PEP related?
rdkeys@csemail.cropsci.ncsu.edu wrote:
[shortened to save some BW]
>
> FOOD FOR THOUGHT AND DISCUSSION.....
> (cuz ol' BA Bob has his dunce cap on today and don't haves his books
> handy ta latch onto):
>
> Can someone explain the relationship between average CW output power,
> and PEP and how we technically measure both? The FCC rates power output
> as PEP power. I want to make sure I am measuring things correctly.
> I thought I knew how to do that, but now I am not so sure. I want to
> make sure I get it exactly right.
Well, I have seen many definitions of what PEP is. Here in SM back before
1994 when new rules came in effect, power was defined as the RMS power
of the carrier at the peak of the modulation envelope. The power should
be measured with a current meter with a maximum specified time constant.
Power back then was max 500W input.
Now after 1994, maximum power is 1000W (150W on 30 meter), no definition,
no PEP no nothing. It's up to us to interpret it... I guess I can run
an AM TX with 1000W carrier over here legally.
Back to the subject, I think that pre-94 definition was PEP even if it
was not mentioned as such. With the same definition, CW would be allowed
1500W output since we measure the RMS at keydown. The OOK modulation
envelope is an uneven square wave and the peak of the envelope will be
the carrier itself.
For AM, peak RMS power at 100% is 4 times the carrier RMS power, thus the
carrier would be limited to 375W RMS to stay within the 1500W limit.
The other definition most common I've seen is PEP=the peak power at the
peak of the modulation envelope (not the RMS power). This does not make
sense to me since then, by definition, CW would be limited to 1060W RMS.
That is also the power dissipated in the dummy load if you put out 1500W
PEP using this definition. Additionally, AM carrier would be limited to
265W RMS. RMS is what we usually see on our mechanical meters, DVM's
show average which for a pure sine wave happens to be RMS.
The formula PEP = Epeak x Epeak / R is correct. We just have to define
what Epeak is. Is it the RMS voltage at the peak of the envelope or
is it the peak voltage at the peak of the envelope?
IMO, the first definition above (RMS at the peak of the envelope) is the
correct one. The second definition above just don't make sense to me.
The legal definition is another matter. It depends on the technical skill
of the people who wrote the rule. If the correct definition is the peak
in the peak, I find it rather silly. Or am I screwed up today?
Jan, SM5GNN
Date: Tue, 3 Feb 1998 11:03:19 -0800 (PST)
From: Ken Gordon <keng@uidaho.edu>
Subject: 3579 QRG, last night.
Worked Jack, W7QQQ, and his beautiful sounding 813 crystal oscillator last
night (Monday). Band conditions not too hot. He had LOTS of static, and
my 50 watts were difficult copy for him. HE was 579x here. Also heard
W7ZFB later in the evening, but he didn't hear me, and N4QY, same story.
Several other un-identified signals tuning up at various times, which
would have been Q-5 copy here.
It was good to hear the activity.
Ken W7EKB
Date: Tue, 3 Feb 1998 15:39:10 -0500 (EST)
From: rdkeys@csemail.cropsci.ncsu.edu
Subject: Re: How are CW output power and input power and PEP related?
This is where I was trying to go.....
> Take your RMS (if it uses a thermocouple) ammeter reading
> for CW and square the current, multiply by 2, and then divide by R.
>
> For a 50 Ohm system PEP = I * I * 2/50
Using a simple junk box WWII surplus RF ammeter (thermocouple type
with the separate or built-in thermocouple) and measuring the RF
into the classic 50 ohm dummy.
FCC's PEP == I x I x 2/50.
I knew I was not losing it entirely, but forgot the twice the
square root of two thingie and how to get there correctly......(:+\\.....
Geesh, I should know better.....
Thanks
Bob/NA4G
>
> The answer to question 4:
>
> No it is not correct! You need to multi[ply by 1.414 one more time.
>
> The answer to question 5:
>
> Bob, your description of an RF voltmeter is a *peak* reading meter!!!
> There are 2 sets of circuits in the figure (My book is 1977, but should be
> the same as your 1969). The top ones without capacitors give a good
> approximation to average voltage. Notice from the drawings that this is AVERAGE
> and *NOT* RMS. The 2 are different!! The circuits at the bottom of the figure
> use capacitors and give *PEAK* readings.
>
> RMS is *MUCH* harder to do than peak unless you are sure you have a pure sine
> wave.
>
> Remeber that you are making a circuit that is exactly the same as a capacitor
> input power supply when doing the peak circuits, except you are getting the AC
> from your transmitter instead of a power transformer on the AC mains. I suspect
> that the ease of making a peak reading RF voltmeter over a true RMS is the
> reason for the method of measuring. The main source of error is the slight
> droop in the voltage due to the loading of the microammeter. Using a DVM or
> VTVM as the readout will give more accurate peak readings. It will also give a
> longer time constant so you can actually read the meter before the volage falls
> off.
>
> I did some quick calculations to see how important the diode drops are in the
> calculation. At 10 W PEP the peak voltage in a 50 Ohm system is 22 Volts. At
> 1500 W PEP the peak voltage is 193 volts!!! Better use some hefty diodes and
> resistors in that meter. Or better yet, use a 20 dB T pad attenuator ahead of
> the meter.
>
> Answer to question 6:
>
> These calculations suffer from the same errors as in the question 4 section.
>
> Answer to question 7:
>
> Not all screwed up, but probably overwhelmed by the algebra involved! It's good
> to do a sanity check from time to time. I know I frequently miss one of the
> small details then remember "oh, yeah. I knew that" :<)
>
> Why life is the way it is:
>
> I did the calculations when I was looking at just exactly how does AM vs SSB
> compare. In the old days we were limited to 1000 W DC input for the carrier.
> If we got 60% efficiency, we had 600 W RMS (1200 W PEP) out just for the
> carrier! When you add another 1200 W PEP for the modulator power, we were
> getting 2400 W PEP for AM when going full out and staying legal.
>
> Today with the 1500 W PEP limit, we are limited to 375 W RMS of carrier.
> Obviously, those of us who like Admirable Modulation got stuck with a sizeable
> power drop with the new regulations.
>
> Ray Mack
> WD5IFS
> mack@mails.imed.com
> Friendswood (Houston), TX
>
> ______________________________ Reply Separator _________________________________
> Subject: How are CW output power and input power and PEP related?
> Author: rdkeys@csemail.cropsci.ncsu.edu at mails
> Date: 2/3/98 11:28 AM
>
>
> FOOD FOR THOUGHT AND DISCUSSION.....
> (cuz ol' BA Bob has his dunce cap on today and don't haves his books
> handy ta latch onto):
>
> //question 1
> Can someone explain the relationship between average CW output power, and
> PEP and how we technically measure both?
> <snip>
> //question 2
> One of the FCC OET bulletins indicates measuring peak RF voltage across
> a calibrated load and equating it to obtain PEP as......
>
> PEP = (Epeak x Epeak)/R.
>
> Is this correct?
> //question 3
> ON a CW sine wave (not complex waveform such as SSB modulation) how can
> this be done with an average-reading RF current ammeter typically found
> in the average glowbug junquebox? I prefer to use the simple RF ammeter
> if I can.
> <snip>
> //question 4
> <snip>
> //question 5
> What needs to be put together for a proper RF voltmeter to measure PEP by
> the FCC method?
>
> <snip>
> //question 6
> Is this correct?
>
> //question 7
> Our glowbugs probably don't usually qualify as big-gun equipment, but,
> one or two might, and it is probably good we all review how to properly
> measure PEP on our glowbugs.
>
> Am I all screwed up today, or am I missing the forest for the trees?
>
> Thanks.....
>
> Bob/NA4G
>
>
Date: Tue, 03 Feb 1998 14:01:01 -0600
From: Conard Murray <ws4s@InfoAve.Net>
Subject: Re: How are CW output power and input power and PEP related?
>This is where I was trying to go.....
>
>> Take your RMS (if it uses a thermocouple) ammeter reading
>> for CW and square the current, multiply by 2, and then divide by R.
>>
>> For a 50 Ohm system PEP = I * I * 2/50
>
When I have used the hot wore ammeters for power measurement and then
compared results with a wattmeter, I find I get the proper reading assuming
the meter is reading peak current rather than rms current ......I have been
using I^2*R. This works with the meter in the TCS anyway. I guess it is all
in how you mark the meter face. I don't remember seeing anything stating
whether the calibration was in rms or peak ... just RF amps.
Anyone know if these RF ammeters are calibrated in RMS or peak?
73 de Conard, WS4S
End of glowbugs V1 #236
***********************