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Everything You Wanted To Know About Microwave
Things, But Not So Much You'd Have To Get Too
Familiar With James Clerk Maxwell
This paper will deal with a variety of microwave information that is not
contained on most data sheets. It involves things like cable attenuation,
power handling capability, etc.
First off, an information tidbit. One of our microwave tools that has not been
completely superseded by the electronic calculator is a cardboard sliderule
called the HP Reflectometer Calculator, (Lit # 5952-0948). It relates SWR to
reflection coefficient to return loss and mismatch loss. It shows you
mismatch losses for power measurements when you know the SWR of the
source and the SWR of the power sensor. And it is a handy reminder for the
numerical vs. the dB ratios of voltage and power.
ATTENUATION AND VOLTAGE AND POWER RATINGS
OF MICROWAVE THINGS
HP sells a variety of accessories that are basically unspecified. These are the
connector adapters, cables, etc. A typical customer question might ask for
the attenuation of the RG-214/U cable used in the lIP 11500A Cable
Assembly. The answer is to look in the Reference Data Book For Radio
Engineers. Cable attenuation is plotted for all sorts of coaxial cables from the
RG-59/U video cable we use with BNC connectors to large 3 1I8-inch rigid
coax, which we don't sell. Incidentally, our lIP 11500A which is 6-feet long
would have attenuation of 2.4 dB at 10 GHz (40 dB per 100 feet).
The ITT tables on the Army-Navy list of preferred Radio-Frequency cables
also shows other useful data. For example, the capacitance of RG-59/U cable
is 21.5 pF per foot. This may be useful if you need to know the shunt loading
capacitance of some BNC video cable which may connect a microwave
detector to the input jack of your oscilloscope, if you want to use it
unterminated, and look at the pulsed RF envelope.
Those same tables show that the minimum operating voltage of the RG-214/U
cable is 5,000 volts rms. In a 50 system, that might imply that you could put
one half megawatt on the cable. But there are two things that limit that
power. The Type-N connector on each end is only rated at 500 volts and if
you look at one, you can see why. It uses air dielectric while the cable itself
uses polyethylene dielectric. That connector rating of 500 volts, of course,
permits about 5000 watts. But if you consider the attenuation of the cable,
that will limit your CW power to much less. Consider the 2.4 dB at X-Band.
That means that about 7 feet of the cable would exhibit 3 dB or a loss of half
the power. I'd guess that dissipating even 100 watts in 7 feet of cable would
be a problem. Think of it as bottling up the heat of a 100 watt light bulb in
that length of cable. If you figured that the 7 foot cable could dissipate 50
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watts on a continuous basis, then at X-band with its 3 dB of loss, the cable
could transmit about 100 watts CWo So you have to use that sort of analysis
to figure power. At lower frequencies, of course, the losses are far less.
Generally, cables can carry up to the connector limit in peak pulsed power;
for example, the 5 kW through the connector, and then use a duty cycle that
limited the average power to an acceptable amount. Of course, this whole
analysis explains why the big transmitting stations use the large rigid coax of
3 1/8 inch or even 6 inch diameter for their CW signals.
In analyzing peak-power-carrying capacity, you need also to consider the
standing wave effects on a line. The real world would seldom have a
perfectly matched load and a flat transmission line. For example, suppose
that a peak pulse in a line hits an antenna with an SWR of 1.5. That means
that a reverse wave of power 14 dB below the incident power heads back
toward the transmitter. I hope you're using your reflectometer sliderule.
(Lit # 5952~0948). Slide in 1.5 on the SWR scale and read the return loss of
-14 dB. At the peak of the standing wave, the peak RF voltage is now 2()OAJ
higher than the forward wave. The 20% comes from the .2 reflection
coefficient also shown on the sliderule. If the phase of the antenna reflection,
or the length of the line is such that the peak of the standing wave pattern
sits right at the position of the connector pair, then of course, it is subject to
120% of the peak voltage of the forward wave. Likewise, a more serious
reflection like a short or open would run up the voltage as high as 20()OAJ for a
full reflection.
The opposite side of that reflection problem is that at the voltage minimums,
the current in the line is greatest. Thus, if the SWR null (current maximum)
happens to occur at the point in the connector where the pins mate, the
contact resistance of the connector pins would get currents as high as twice
the normal forward current of a well-matched line. Remember that dissipated
power is related to the square of the current. The same sort of reasoning has
to be done on rating estimates of things like a slotted line which has
dielectric beads positioning the center conductor in the center of the outer
conductor surface. If the voltage maximums happen to occur at the bead
positions, the excess voltage may cause excess heat dissipation in the poly
beads, to the extent they might melt, even though the line and pin contacts
can handle the load. Dielectric losses often go up as the square of the applied
voltage.
WAVEGUIDE AND COAXIAL POWER RATINGS
Generally speaking, our equipment is imbedded and built for
instrumentation-type power levels of a few hundred milliwatts or perhaps in
the case of some power sensors, 25 watts. On the other hand, we understand
that customers often want to put one of our directional couplers into a
transmission line carrying transmitter power and sample off 20 dB or 1/100th
of the power to measure.
Let's talk a bit about power handling in waveguide. For the rest of this note,
well be using the back-page reference data section of the Microwave Test
Accessories Catalog, (Lit. #5091-4269E), August, 1992 which I will,
henceforth, refer to as the MTA Catalog. If you look in the waveguide
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standard data chart you'll see theoretical CW power rating column. That was
calculated by computing the maximum E-vector in the fundamental
transmission mode of the guide (TE 10) and using a 15,000 volt per
centimeter dry air breakdown at sea level, in addition to allowing a 2 to 1
safety factor. Note that the power rating varies from one end of the frequency
band to the other even in the same guide.
Take X-band for example. At 8.2 GHz, standard guide will rate out at 206 kW
peak. That implies a clean guide with no little protrusions or solder bumps
inside that could concentrate the E-field. Every new FE gets to ask the
question, how much power can our lIP 281-Series Coax-Waveguide Adapters
handle? Well, obviously, they won't go 206 kW because not only do they have
a full short on the backwall, but about one-quarter wavelength out from the
backwall, protrudes a small1-cm antenna from the Type-N connector on the
flat side. Around the antenna is a polystyrene cylinder which helps shape the
field from the cross-guide field into a field that has the little antenna stub as
the center conductor of a coax line (the Type-N) connector.
So how much power? Well, we earlier decided that the Type N connector
handles 5 kW peak if everything is clean. But we're pretty sure that the
comers and sharp edges of the little poly cylinders might cause some field
bunching. So we might recommend a safety factor of 2 for 2.5 kW, peak. And
to allow for the fact that the transmission line or the load may not be
well-matched, I'd allow another 2 to 1 derating for that for a suggested rating
of 1 kWpeak.
Now, for another matter on CW rating. We know that such adapters have
about a 0.1 to 0.2 dB loss around X-band. If you look at your handy sliderule,
you can see on the voltage/power ratio scales on the bottom, that a 0.2 dB
reads out at about 1.045 power ratio. (l didn't actually read it to 3 decimal
places, but in microwave work, you learn that for power ratios below 1 dB,
Broad wall low-power
you can approximately double the 0.2 and multiply that by 10 and get the
4.5% power loss.
Now let's try to put a CW signal with 1 kW through the adapter with a 4.5%
power loss. That works out to 45 watts, and you can imagine it heating up too
hot to touch. Let's estimate that the small plastic part can dissipate about 5
watts, unless it's well heat-sinked (consider a small 7.5 watt light bulb),
which isn't bad considering that it might have a good heat sink on the run of
other waveguide in the system. That would then translate upward to about
100 watts of RF rated CW signal.
Narrow wall higher power
Naturally, the rating depends entirely on the loss in the adapter and certainly
at S-band, there is less loss in the 2.6 to 3.95 GHz band. There may only be
0.05 dB loss in that band. Do you see how it's done?
How about power ratings of something like a multi-hole directional coupler
like the lIP 752-Series? There are two problems with that one. First, these
"broadwall" couplers have rows of perhaps 30 holes drilled in the top wall to
couple off power to the secondary line for a precise 3, 10, or 20 dB factor.
Those holes won't allow much high power to flow down the main line
without breakdown. Cross-guide "Riblet"
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The only coupler to recommend for high power in waveguide are called
narrow-wall couplers. They use holes drilled in the narroW wall where there
is no E-vector, but they lose some precision. Typically they have 40 dB
coupling (1/10,000 power into the secondary line) because they are intended
for lines carrying 100 kW and would thus have 10 watts in the secondary.
Several companies including NardaILoral and Systron carry these
narrow-wall couplers. The advantage of the multi-hole coupler remains the
fact that coupling is flat to ◦ Jabse Service Manual Search 2024 ◦ Jabse Pravopis ◦ onTap.bg ◦ Other service manual resources online : Fixya ◦ eServiceinfo