EMRFD Message Archive 10208

Message Date From Subject
10208 2014-08-05 18:41:00 farhanbox@gmail.c... imd of unequal levels

we measure IMD with two signals pf equal level to get the third order intercept. what is the maths of calculating IMD products when the signals are of unequal levels? i know it takes a bit of head acratching with paper and pen, i had seen this worked out in some ham literature. though I am not sure where.

any pointers will ne appreciated. these are needed to co-related the spurs from rf power amps  due to unsupressed LO etc,

- f

10209 2014-08-05 22:12:52 kerrypwr Re: imd of unequal levels
I'm neither expert nor mathematical but I thought the question interesting.

I found this;

http://www.m2global.com/resources/imd-dropin-isolators-circulators/

Kerry VK2TIL.
10210 2014-08-05 22:15:38 kerrypwr Re: imd of unequal levels
And, in a simpler form that I can grasp,see the bottom of this page;

Cascaded Budget 2-Tone, 3rd-Order Compression Point IP3 - RF Cafe

Kerry.

 

10211 2014-08-05 23:13:32 Ashhar Farhan Re: imd of unequal levels
the answer was lying by my pillow. Equation 6.3-9 of Introduction to Radio Frequency Design.

- f


10213 2014-08-06 06:46:19 billw77aaz Re: imd of unequal levels
> And, in a simpler form that I can grasp,see the bottom of this page;
>
> Cascaded Budget 2-Tone, 3rd-Order Compression Point IP3 - RF Cafe
> http://www.rfcafe.com/references/electrical/ip3.htm

GOOD
W7AAZ
10217 2014-08-06 10:29:49 boblarkin02 Re: imd of unequal levels
Let me mention that there is a straightforward method to totally generalize the multisignal IM level.  R G Sea, Proc IEEE Aug 1968 is a note on calculating the level of any product with any number of tones ata any levels!  The assumption is the same as we use for the two tone 3rd IM, and that is that the device is describable by a power series.  This is close to many practical cases.

Sea (and Engle)'s method allows you to find stuff like:
  3rd IM of 2-sig at different levels
  9th order 3 signal IM,  5*f1 - 2*f2 - 2*f3 (or any other)
  7th harmonic of f1
  ...

The tricky part is figuring out what the power series is for the distorting device.  But this method also allows you to deduce the power series by measuring the products (like harmonics) with a spectrum analyzer and adjusting the power series coefficients to make it fit.

When you look at the paper, it may not be obvious at first how to do the calculation.  To help, and also if you want to just get the answers, I posted a Basic program:
http://www.proaxis.com/~boblark/IM_Products.bas

It runs under most basic interpreters, and for sure under SmallBasic:
Zhttp://smallbasic.sourceforge.net/    (remove the leading 'Z')
This is NOT the Small Basic of MS.

Have fun!

73, Bob  W7PUA

 

10218 2014-08-06 12:40:28 ehydra Re: imd of unequal levels
Thanks!

The correct title is:
"An algebraic formula for amplitudes of intermodulation products
involving an arbitrary number of frequencies"
http://www.researchgate.net/publication/2991039_An_algebraic_formula_for_amplitudes_of_intermodulation_products_involving_an_arbitrary_number_of_frequencies

Is there a change somebody can download it?


This paper looks like a extension of the method:
http://e-collection.library.ethz.ch/eserv/eth:34315/eth-34315-02.pdf

- Henry

boblark@proaxis.com [emrfd] schrieb:
> Let me mention that there is a straightforward method to totally generalize the multisignal IM level. R G Sea, Proc IEEE Aug 1968 is a note on calculating the level of any product with any number of tones ata any levels! The assumption is the same as we use for the two tone 3rd IM, and that is that the device is describable by a power series. This is close to many practical cases.
>
10219 2014-08-06 15:43:21 boblarkin02 Re: imd of unequal levels
Henry, yes that is the full title.  As is hinted at in the Basic program, there is an additional short paper related to actually doing the calculation: R., G. Sea and A. G. Vacroux, "On the Computation of Intermodulation Products for a Power Series Nonlinearity," Proceedings of the IEEE, March 1969, pp 337-338.

Also, the titles use the term  "Intermodulation," but the method includes all products such  as 1 signal product at 0 frequency (e.g., the DC output of a detector) and the single frequency output at the input frequency (gain compression).

Lastly, here is a simple example of a run of the program for the low-side third-order IM product for uneven inputs. The  larger input is assumed to be at the lower frequency:
-----------------------------------------------
  Number of Signal inputs, M? 2
  Signal level 1? 1
  Signal level 2? 0.5
  Highest degree of power series, Nps? 3
  Coefficient 1? 15
  Coefficient 2? 0
  Coefficient 3? -2
  There are M of the following multipliers, but some can be zero
  The sum of their absolute value us the order of theproduct of interest
  Enter all zeros to find dc term
  Multiplier 1? 2
  Multiplier 2? -1
  V = -0.75
----------------------------------------------
All levels and the output V refer to the coefficient A as in "A cos(2 pi f t)," i.e., the peak voltage. The multipliers as shown are for the IM product 2*f1 - f2 where f1 and f2 are the frequencies in the order the levels are entered. Since we assume that f1 < f2, the IM product being calculated will be on the low side of the two inputs.  Of course, we never enter the actual frequencies, as they are not relevant.

If you rerun the program with the multipliers reversed, the level of the IM product, now on the high frequency side, will drop to half of that shown, or -0.375 Volts.  That's what happens with uneven input levels.

There are lots of things one can play with here.  For example, try a power series with degree 5 (or 7, etc.) and look at the resulting compression and 3rd order IM.  It shows why the 3rd IP need not be 10.6 dB above the 1 dB compression point.  And on and on...

73, Bob  W7PUA
10220 2014-08-06 17:26:00 Ashhar Farhan Re: imd of unequal levels
thanks all,

this is much simpler that I thought. I will, time permitting, rewrite the program as a javascript calculator.

- f


10223 2014-08-07 11:29:48 ehydra Re: imd of unequal levels
> There are lots of things one can play with here. For example, try a power series with degree 5 (or 7, etc.) and look at the resulting compression and 3rd order IM. It shows why the 3rd IP need not be 10.6 dB above the 1 dB compression point. And on and on...
>
> 73, Bob W7PUA

And that is what is done by harmonic simulation in RF packages like ADS
or some SPICE variants?

- H.
10436 2014-11-12 12:41:48 billw77aaz shielding a toroid
I've built a PIN diode 1.8-30 MHz T/R switch with slightly over 100 dB of
receiver isolation from the antenna/transmitter. I want more isolation.
There are inductors feeding current/voltage to the diodes. It appears,
despite complete shielding and RF filtering of wires, that I have
INDUCTIVE coupling between a 2" toroidal coil in the transmit diode area
and a 1/2" toroidal coil in the receive area. They are separated by a
distance of 1.25 inches and a 1/16" thick aluminum partition with large,
tightly bolted joints.

Would putting the planes of the two coils at right angles produce a
substantial improvement at radio frequencies? I'm contemplating wrapping a
copper strap around the OD of the toroids, making an external "shorted
turn", but wondering if that is a poor choice.

Bill W7AAZ
10437 2014-11-12 13:09:23 Tayloe, Dan (NSN ... Re: shielding a toroid

That is something that I have noted with DC receivers.  My first attempts used a passive L/C low pass filter for the audio like the KK7B receivers.  What I found was that these AF coils, despite being ferrite self “shielded”, picked up the 60 Hz “hum” of the power supply transformer, especially with 60 dB of gain after the filter. Aluminum is not a magnetic shield, and neither is copper.  Since it was a separate PS, I found I could orient the receiver to minimize the hum.  Hum is one reasons why I decided to go with the more complex R/C active filter route for my DC receiver audio chain.

 

Since these chokes are in the same unit as the toroid core, it does seem like their orientation could be made to minimize the magnetic coupling.

 

-          Dan, N7VE

 

10438 2014-11-12 13:30:01 Ed Manuel Re: shielding a toroid
You might experiment with a tin-plated steel shield.  Just a thought, not something I've noticed before.

Ed, N5EM


10439 2014-11-12 13:59:07 kb1gmx Re: shielding a toroid
Passive filters have problems with shielding depending on the inductors used.
Some of the high value ferrite loaded (more than 1000uH) are not well shielded 
despite their claims.  Orientation is is important for those.  Even then wiring 
running around on the board or off it with 80+db of gain following will also pick up
hum is the hum source is close/strong.

Despite that I find that passive filters often have far greater dynamic range
than active filters.  So where they are in the gain plan makes a huge difference 
and the opamps used and power supply voltages selected are a factor.


Mu-metal (tinplated iron is worth a try) is the king for shielding.  Used a lot of 
it over they years for things like O'scope crts and 7360/6AR8 (sheet beam tubes)
in the battle of power supply magnetically induced hum.  Copper or aluminum is 
good at RF and for electrostatic fields but for magnetic fields you need iron 
or similar.

Note how a power transformer is constructed makes a huge difference in it 
radiated magnetic field with cheap transformers in plastic or aluminum cases
being a major offender.

Allison
10440 2014-11-12 15:16:34 billw77aaz shielding a toroid: HELP WES
Thanks for the input guys.

After removing the RX blocking diode, I disconnected the DC feed to the
inductor that feeds DC to that (now missing) diode. No effect. So
remaining leakage isn't primarily signal leakage on that DC wiring. When I
disconnected the cold (bypassed) end of that inductor (no change in
capacitive pickup) the signal leakage dropped approximately 15 dB. So I
concluded a major source of leakage is direct magnetic pickup on this
inductor from the TX coil on the other side of the bulkhead.

Both inductors are toroids. What orientation of two toroidal inductors
results in minimum coupling between them? Putting them on two mutually
perpendicular planes? Or doesn't the "perpendicular axes" of two solenoids
have a corresponding position of minimum coupling with toroids? It was a
bitch, but I rotated the RX coil approximately 90 degrees. Little/no
effect. But they aren't in the same plane and that might explain that.
Here's a place where I'd like to understand the theoretical coupling of
two toroids.

I'm sure that without the aluminum bulkhead in place the isolation would
not be the 106 dB I'm getting at 30 MHz. At 2-30 MHz aluminum cabinets or
even plastic with a conductive spray coating is effective for shielding.
Even though I don't think ferromagnetic material is useful at this
frequency, sheets of .010" tinned steel have zero effect (playing around
with shielding material in a box with 750 volts from a 10 uF filter cap
can lead to shocking results). The box is built, so I can't simply add
another layer of shielding around the whole transmit (or receive)
sections of the TR switch.

I'm down to the "short strokes". I won't tear it down to try something
without a clear theoretical reason to justify the attempt. I bet Wes
knows...but he may not read every post, especially ones like this. I'll
change the title, heh heh.

W7AAZ




> You might experiment with a tin-plated steel shield. Just a thought,
> not something I've noticed before.
>
> Ed, N5EM
>
>
>
10441 2014-11-12 16:32:24 w7zoi Re: shielding a toroid: HELP WES
Hi Bill, and gang,

Yes, I do read them all.  I try to avoid answering, even when there is a comment asking me for an answer, for it always seems that when I do post something, further posts on the subject seem to go away.  Well, I should say that they are attenuated.   The last thing I want to do is to shut down the discussions.

When you already are working at the 100 dB level, there are all kinds of things that can get in the way.  The other folks have done a great job of summarizing them.   

Let's start with a solenoid.   If the length is much less than the diameter, the inductance is (nearly)  proportional to the square of the number of turns.   This indicates that almost all of the magnetic flux from one turn goes through all turns.   But as you increase the length, this simple "law" goes away.   By the time length is two or three diameters, L is almost linear with the number of turns.    This is like placing a bunch of magnetically isolated inductors in series with each other where the flux in one L does not couple to the next.  The long solenoid must have a lot of the flux from one region leaking out between or through the wires, never linking to the rest of the turns.

A toroid is what happens when we bend a solenoid around in a loop and close the thing upon itself.   Most of the magnetic flux is contained within the interior of the structure.    But not all of it.   Just as we saw with the raw solenoid, some of the magnetic flux escapes.    We can model the toroid (again, an approximation) as two magnetic structures.   One has all flux within the core and completely isolated from the rest of the world.   The other is a single turn  with a diameter of about that of the core.   That one is free to link to about anything else in the vicinity.     Rotation of the core moves this single turn around, controlling it's linkage to other things.   Remember that the wires into and out of the core are major contributors to this single turn.

I don't really have any wisdom to contribute here, as most of you already realize.   I'm shooting from the hip, as they say.   This is especially the case when you are talking about 100 dB isolation levels.    You are clearly doing the right experiments, for you are making progress.    Allison and others have already commented
10442 2014-11-12 16:50:30 Todd F. Carney / ... Re: shielding a toroid: HELP WES
Just a naive question here, but if the OP already has 100dB of isolation, isn't that more than sufficient? I mean, that's a lot of isolation. 

73,

Todd
------------------------------------------------------------------------------------
K7TFC / Medford, Oregon, USA / CN82ni / UTC-
7
 (P
​D
T)
------------------------------------------------------------------------------------


10443 2014-11-12 18:00:53 Thomas S. Knutsen Re: shielding a toroid: HELP WES
I want to way in on the radiating toroids as this have been a continuing issue here and I have done some experiments to try to determine how it behaves. 

The physics say  that all toroidal cores have their magnetic field contained inside the core, and for a large coil this is mostly true. 
When the number of turns on the core and the permability of the core its wound  gets low then the single turn effect gets more prominent. With ferite cores, the single turn effect is not as noticeable as it is with the iron powder cores.  This is mentioned briefly in F.W. Grover, Inductance Calculations (ISBN 0-486-47440-2 ) from 1946.

An fairly simple experiment I did to determine this for a number of different cores that where avaible to me is to wind a transformer on the core, with an number of bifilar wound turn, then determine the coupling ratio. This can be done with a simple inductance measurment of oposing and aiding connections of the transformer. 
The data from this shows that the low permability cores like -0 phenolic cores (solid air) radiates quite well, with an average coupling (k)  factor of 0.56, on a scale where 1 = no radiation, and 0 = all radiated. Most of the iron powder cores we use radiates quite well, I measured the T50-2 to have a k factor around 0.7, and the T37-6 at around 0.65. 

The ferrite cores in the same number of turns shows clearly less radiation. for the FT37-43 I measured a k factor in the 0.98-0.99 range. Binocular cores, like the BN-2402-43 I had shows a somewhat better k factor than the toroidal cores. I had expected these cores to be worse, due to them having a non closed form. 

The inductance measurments needs to be done carefully. I used an VNA, and S11 RF I/V measurments to do the inductance measurments at the wanted operating frequency. The jig that is used to do the measurments needs to be carefully designed, as to avoid both resonance and radiation from the measuring jig. 
If anyone wants to see my data, I can make a scanned copy of the lab notebook avaible. 

I did a couple of experiments to determine how the radiation appears around the core, but all my experiments so far have been inconclusive.
Also note that if any of the conductors on the cores cary DC, the inductance will be decreased due to the DC saturation effect.  

Obtaining 100dB isolation in the PIN switch is admirable, the next 10dB is probably the same amount of work as the first 100. At this level the layout is the most important thing to obtain that isolation, most common types of coax have an isolation somwhere around this level, and anyone who have tried obtaining 100dB dynamic range from an network or spectrum analyzer project can tell you how important the choise of good coax and connectors and screening is at this level. Its after all an ratio of 1/10000000000!

I hope that my ramblings in the middle of the nigth makes sence, if not then let me know and I'l try to explain better. I had a plan to write up some of these experiments for an article, but life just seems to get in the way. 

73 de Thomas LA3PNA.


2014-11-13 1:32 GMT+01:00 wesw7zoi@gmail.com [emrfd] <emrfd@yahoogroups.com>:
 

Hi Bill, and gang,


Yes, I do read them all.  I try to avoid answering, even when there is a comment asking me for an answer, for it always seems that when I do post something, further posts on the subject seem to go away.  Well, I should say that they are attenuated.   The last thing I want to do is to shut down the discussions.

When you already are working at the 100 dB level, there are all kinds of things that can get in the way.  The other folks have done a great job of summarizing them.   

Let's start with a solenoid.   If the length is much less than the diameter, the inductance is (nearly)  proportional to the square of the number of turns.   This indicates that almost all of the magnetic flux from one turn goes through all turns.   But as you increase the length, this simple "law" goes away.   By the time length is two or three diameters, L is almost linear with the number of turns.    This is like placing a bunch of magnetically isolated inductors in series with each other where the flux in one L does not couple to the next.  The long solenoid must have a lot of the flux from one region leaking out between or through the wires, never linking to the rest of the turns.

A toroid is what happens when we bend a solenoid around in a loop and close the thing upon itself.   Most of the magnetic flux is contained within the interior of the structure.    But not all of it.   Just as we saw with the raw solenoid, some of the magnetic flux escapes.    We can model the toroid (again, an approximation) as two magnetic structures.   One has all flux within the core and completely isolated from the rest of the world.   The other is a single turn  with a diameter of about that of the core.   That one is free to link to about anything else in the vicinity.     Rotation of the core moves this single turn around, controlling it's linkage to other things.   Remember that the wires into and out of the core are major contributors to this single turn.

I don't really have any wisdom to contribute here, as most of you already realize.   I'm shooting from the hip, as they say.   This is especially the case when you are talking about 100 dB isolation levels.    You are clearly doing the right experiments, for you are making progress.    Allison and others have already commented
10444 2014-11-12 21:28:16 billw77aaz Re: shielding a toroid: HELP WES
Thanks Wes, Thomas and everyone else.

I am familiar with measuring aiding and opposing inductance to compute the
coupling coefficient between two isolated inductors. I have an extremely
high resolution Boonton 63H bridge that I've used for that in the past.
Actually, assuming the coupled signal of -106 dBc was all due to inductive
coupling, I could compute what the coupling coefficient must be. While
that would be interesting, I don't really care what it is: I want it to be
ZERO!

I was looking for a miracle: some tried-and-true methods someone knows to
orient the two toroids to minimize the coupling. Alternatively, shielding
techniques at HF.

The TX coil, being u=10 powdered iron wound with #16 wire with about 4-5
wire diameter spacing between turns, undoubtedly has significant flux
leakage. There are three cores in the RX compartment, only one candidate
in the TX compartment, so I'd about arrived at the conclusion that the
first thing to do is wrap a thick strap of copper around that T200-2
toroid in the TX compartment. It will be open at the ends, but at 2.5"
wide the strap will completely surround the toroid. That seems to
correlate with what you suggested, Wes.

A decade or two ago, the 5.2 MHz Motorola xtal filters I disassembled to
build filters for my STAR transceiver had matching networks whose toroid
was slipped inside a tinned copper container. I discarded them, but
recognized that copper can was to control leakage to/from that toroid. If
the copper strap increases the attenuation, but not enough, I think I can
fold the 2.5" wide .031" thick copper strap I have into boxes for the
toroids in the RX compartment. They're 0.5" diameter and there's more room
for that in the TX compartment. If I'm lucky maybe that won't be needed.

OK, no miracles, but thanks again for the input.

W7AAZ
10445 2014-11-12 22:07:59 billw77aaz Re: shielding a toroid: HELP WES
> Just a naive question here, but if the OP already has 100dB of isolation,
> isn't that more than sufficient? I mean, that's a lot of isolation.
>
> 73,
>
> Todd

That's a perfectly rational question.

I only have 100 watts, but have run the legal limit in the past and I'm
using 1500W as my TX power. So if key down power is 1500 watts (+61.7 dBm)
with 100 dB of isolation on 10m, the signal out of the T/R switch will be
-38.3 dBm. I could have a Z10043A in line, which has 11 dB of gain so in
the worst case the receiver will see -27.3 dBm or a little over half a
volt.

That is 5.8 milliwatts, 80 dB over S9. It won't burn out anything but it
will saturate the Panadaptor and make a wide blob on the frequency
display. I anticipate that will take time to dissipate so I can see nearby
peanut whistles again. Conceptually I'd like to get it down to S9 from the
T/R switch but I doubt that is possible on 10m. I want to make it better
now, if possible, so I can button this project up and move on.

W7AAZ
10446 2014-11-12 23:02:13 John Marshall Re: shielding a toroid: HELP WES
Your math lost me there, Bill. I'm with you on +61.7dBm -100dB +11dB = -27.3dBm but that's not a half a volt at 50 ohms. -27.3 dBm is 1.9 microwatts or about 10 millivolts at 50 ohms or 45.7 dB over S9. That's still a biggish signal on the panadaptor but way short of 80 over S9. It's late here and I should be in bed but did I miss 34 dB somewhere?

John, KU4AF
Pittsboro, NC

10447 2014-11-12 23:48:51 mvs_sarma@ymail.c... Re: imd of unequal levels
if you observe the toroid ' induction signals after any active elements like amplifier etc,
 we could assume that the DC voltage line itself is perhaps having signal as noise due to poor placement of filter caps and dc voltage pick up pints.
 
 it would be better to follow star type  DC distribution if you use  so called "ugly" or " manhaattan styles of construction.
the toroids are believed to be free from direct induction.
 I suppose, at some stage you might also be having normal RFC type inductors-a few of them. if so try to shield them with a u clamp lie fitting.
Finally whteher the phasing of the concerned toroid- say clock or anti-clock wise , change would help, i wonder.  one has to experiment.
 
 
10448 2014-11-13 05:01:32 Tayloe, Dan (NSN ... Re: shielding a toroid: HELP WES

Remember that the shield of coax has about 30 dB of attenuation.  It may be that when you get this on the air, that direct radiation from the antenna through the coax itself on the other side of the TR switch may swamp the loss through the switch itself.

 

-          Dan, N7VE

 

10449 2014-11-13 08:48:53 kb1gmx Re: shielding a toroid: HELP WES
-27dBm is much less than a half volt. At 50 ohms that's about 10millivolts.

Thats far enough below the damage threshold for most LNAs for safety.
Most can stand 15dBm without damage and if the bias system used is 
reasonable the recovery time is insignificant (for most nominal uses).
As to the panadaptor, I'd think the recovery time would be fairly fast
as t has to be able to react to the presence of a signal in a small fraction
of the scan time (BW/total sweep divided by the sweep time).  However 
the panadaptor will likely display noise or other artifacts for the portion 
of time the TX is active. Its very hard to RX during TX without a lot of 
selectivity and two separate antennas positioned for minimal interaction.

I'd think by one to two sweep times the panadaptor will have recovered
and a good design should be within a short time (in milliseconds or less)
after TX has disappeared.

Then again the time from teh senders QRZ to the replies trying to get them
is around -2 seconds...  Yes they start often before the poor DX has released 
the TX button.  ;)


Allison


10450 2014-11-13 08:48:59 kb1gmx Re: shielding a toroid: HELP WES
I'd argue with that 30dB number.  

I did testing a few years back with a parallel pair of RG316 (teflon rg174) coax
and while there was coupling it was more than 75db down with properly 
terminated and would fall off depending on frequency and severity of the 
bad termination(s).  That was good into the low uhf range.  Switching to 
RG316 dual braid it was at the limit of the PNA at over 100DB.  Same for 
copper UT141.

FYI the same test with parallel pair twin line was done and with good matches 
at both end s and a distance of more than 2x the spacing the coupling was
well over 40db down.  At 3X the pair spacing it was around -50db over HF to VHF.
Again mismatch degraded this.

My call is if he is seeing 100DB then attention to terminations, layout, and shielding 
will be needed to get better than that.  By that I mean multilayer (controlled impedance) 
with buried microstrip transmisstion lines.  That and shielded relays with terminations 
to the null port.   That usually incurs problems of loss and power handling as 
considerations never minding cost.  Its usually at the switching points relay or diode
where the degradation happens.


Allison
10451 2014-11-13 23:04:45 billw77aaz Re: shielding a toroid: HELP WES
> Remember that the shield of coax has about 30 dB of attenuation. It may
> be that when you get this on the air, that direct radiation from the
> antenna through the coax itself on the other side of the TR switch may
> swamp the loss through the switch itself.
> - Dan, N7VE

That's not the case here, Dan. My patch cables are RG214 on the TX
N-connector and LMR240 from the TX connector to the hp 8560E. The T/R
floor is -130 dB on 160m with all the cables connected. I've then pulled
the coax from the RX connector and touched its ground to the TX connector,
T/R chassis and see nothing on the 8560E on any band with -20 dBm full
scale and the generator at +20 dBm.

My closest antenna is 250 feet away, 71 feet up, fed with buried LDF7
Heliax. The 80/160m vertical is another 150 feet from there, fed with
buried LDF4 Heliax. Unselected receivers aren't living in a high RF
environment on any band and shielding of my coax won't be a limitation.

The only leakage I've detected while operating is from the Transco coax
switch that currently selects both transceivers and receivers. When I'm
transmitting the QS1R can pick up the K3 connected to another position on
the same switch. When I put the PIN T/R in service that Transco switch
will only have exciters connected to it. The QS1R, RA6790 and the K3 and
its auxiliary receiver will all be fed from the RX port on the T/R switch
using 18-GHz-rated coaxial relays.

I wrapped the nearest TX T200-2 toroid with a 2.5" wide strap of .030"
copper with a single point ground. Then I wrapped the nearest RX toroid in
a 1.5" wide .030" thick copper strap with a single point ground. I
reconnected the receiver blocking diode (two 1N4007 in series) and measure
98 db RX isolation at 28 MHz. It improves at 11.5 dB/octave on 20m and 40m
to a floor on 160m of 130 dB below the signal on the TX connector.

There is another series diode in the network between the TX and TX ports
so I expected the 12 dB/octave slope. If that's NOT the diodes it is
something that produces the same 12 dB/octave slope. And it dominates the
attenuation at the high end of the HF bands.

Decoupling networks on the diode feeds are the poorest on 160m (lowest
series XL, highest shunt XC) but the floor on 160m is fine.

I'm not satisfied, but I've run out of ideas and do have other things to
do besides stare at the T/R switch. I can live with the attenuation its
producing in the sense I won't burn anything out, but I'm leaving a
handful of dB on the table....somewhere.

Bill W7AAZ
10452 2014-11-13 23:16:04 billw77aaz Re: shielding a toroid: HELP WES
You're right John: my arithemtic was correct, but I blundered when I
punched the keys on the calculator I computed the voltage for +27.3 dBm!
Furthermore my internal sanity checks weren't running.

When I use averaging to lower the noise floor and a really big signal
comes on I can see those who have very hard keying and that persists for
more than "a millisecond or two" because of the averaging. I assume if I'm
watching my own frequency I'll see TX noise sidebands, opposite sideband
on SSB, etc. But you're right: I won't blow anything up.

Bill



> Your math lost me there, Bill. I'm with you on +61.7dBm -100dB +11dB =
> -27.3dBm but that's not a half a volt at 50 ohms. -27.3 dBm is 1.9
> microwatts or about 10 millivolts at 50 ohms or 45.7 dB over S9. That's
> still a biggish signal on the panadaptor but way short of 80 over S9. It's
> late here and I should be in bed but did I miss 34 dB somewhere?
>
> John, KU4AF
> Pittsboro, NC
>
>
10463 2014-11-14 10:59:22 Brooke Clarke Re: shielding a toroid: HELP WES
Hi:

The effectiveness of shielding depends