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Re: Wich one the get ? nanoVNA F or H4
On Mon, Mar 16, 2020 at 02:51 PM, Lee wrote:
... but when I use polarizedOn a NanoVNA-H, a linear polarizer, like most polarized sunglasses, has relatively little affect on intensity, regardless of the orientation. But in conjunction with pair of "3D" glasses, crossed with the sunglasses, one can get complete extinction as the display is rotated, and increased transmittance relative to the sunglasses, with optimum orientation. Awkward position for the polarizers, though, so leave it to a mechanical wizard to devise a contraption. Complete the fashion ensemble with an aluminum hat! -- I_B_Nbridgema |
Re: Analysing Input Impedance Matching Circuit for the NE602
Kerr,
Here are some tests I did with 1 full turn on primary with 5 full turns on secondary. Primary wire comes out one side and secondary out the other. Ratio of 5 turns gives impedance transformation of 5 square = 25 so 50:1250 First graph is impedance with transformer terminated in 1200 ohms. Second is VSWR with 1200 ohm termination. Tests were done on another type of analyzer. |
Re: Analysing Input Impedance Matching Circuit for the NE602
I have removed the two transformers from my test circuit and attached the NanoVNA to the primary of the first one, the two transformers secondaries (8 turns) are connected together and a 50 ohm resistor is attached to the other end of the second transformer.
The results of this test are attached. |
Re: Analysing Input Impedance Matching Circuit for the NE602
Physically, yes. Circuit-wise and considering magnetics, its is a full
turn. Dave - W?LEV On Tue, Mar 17, 2020 at 6:16 PM Jim Allyn - N7JA <jim@...> wrote: The BN cores aren't actually toroids, they are "balun" cores or-- *Dave - W?LEV* *Just Let Darwin Work* *Just Think* |
Re: Analysing Input Impedance Matching Circuit for the NE602
Hi Roger,
When I do one full turn on the primary (through the first hole and back down the second) I get the results as shown in the attached images. This is why I removed part of a turn to see what would happen. When I measure the inductance of one turn I get 7.38uH which is as close as I can get to my calculated value of 4.5uH (using 4 x 50 ohms for the impedance). If I have a full turn I get about 12.5uH inductance which is about three times too large when compared to the calculated value. |
Re: Analysing Input Impedance Matching Circuit for the NE602
Kerr,
Something is off with your results. I don't know how you are winding the binocular cores. You need full turns and one turn is a complete loop through one hole and back again. Try measuring VSWR and insertion loss by making two transformers and connecting them back to back with a 50 ohm load on the second core. |
Re: Analysing Input Impedance Matching Circuit for the NE602
Ok, understood, but a picture would help to clarify it.
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El 17/03/2020 a las 19:16, Jim Allyn - N7JA escribi¨®:
The BN cores aren't actually toroids, they are "balun" cores or "binocular" cores that have two holes through them. So, if you wind through only one hole, you have a half turn. --
El software de antivirus Avast ha analizado este correo electr¨®nico en busca de virus. |
Re: Dumb question: What are REAL, IMAG, and PHASE good for?
#general_vna
REAL and IMAG apply to reflection coefficient ¦£, in its complex form (a+j.b). That's why values are always in the [-1,1] interval, without any associated unit. When REAL=-1 and IMAG=0, it is the Short circuit situation. When REAL=1 and IMAG=0, it is the Open circuit situation. When REAL=0 and IMAG=0, it is the normal Loaded (50 ohms) situation.Thanks, this is what I was trying to figure out. And thanks for those files! -- KV0A - Robert |
Re: Analysing Input Impedance Matching Circuit for the NE602
hi Kerr,
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I really don't understand the 0.5 to 8 ratio and the "full turn" comment. In toroidal cores, one pass through the hole counts as a full turn, there are not things as half turns, so probably your "full turn" is in fact 2 turns. Regards, Ignacio EB4APL El 17/03/2020 a las 16:55, Kerr Smith escribi¨®:
I have received my new toroids (BN-73-202) and have been having a look at the difference between them and my FT37-43 ones. In this reply I have attached three images. --
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Re: Analysing Input Impedance Matching Circuit for the NE602
To this reply I have attached the images from my NanoVNA, there are two VSWR plots and two smith chart plots.
When using the FT37-43 toroids on the input and output I get about a 1.58 to 1 VSWR at 7.15MHz When using the BN-73-202 toroids I get 1.093 to 1 VSWR (When using a full turn on the primary the VSWR is 2.6 to 1 at 7.15MHz and increases rapidly, at 15MHz it is about 6 to 1) From these two tests I can see the match is much better when using the BN-73-202 toroids and these work well up to 30MHz (VSWR 1.45:1). Looking at my results so far I can see I get a good match when using 2 x BN-73-202 toroids but the gain out of the NE602 is reduced quite a bit. ![]()
FT37-43 on input and output VSWR.png
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BN-73-202 on input and output VSWR.png
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FT37-43 on input and output smith chart.png
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BN-73-202 on input and output smith chart.png
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Re: Analysing Input Impedance Matching Circuit for the NE602
I have received my new toroids (BN-73-202) and have been having a look at the difference between them and my FT37-43 ones. In this reply I have attached three images.
The first is the NE602 with an FT37-43 on the input and output - with -30dB in at 10MHz the output is -15.79dB The second image has a BN-73-202 on the input and an FT37-43 on the output - here the output is -23.69dB The third image has a BN-73-202 on both input and output - here the output is -33.98dB From these tests I can see that lots of the harmonics disappear when using two BN-73-202 toroids but the gain drops quite a bit. I did try adding and removing turns on the BN-73-202 but the best ratio seems to be 0.5 to 8 (1 full turn on the primary reduces the gain and matching). I will post the NanoVNA results in the next reply. ![]()
FT37-43 on input and output.png
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BN-73-202 on input and FT37-43 on output.png
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BN-73-202 on input and output.png
|
Re: ON7DQ_NanoVNA_Presentation_English
Well, I think that the right number is? 1000000 or 1E6
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Regards, Ignacio EB4APL El 17/03/2020 a las 9:52, F6EGK - Jean-Roger escribi¨®:
Dear Luc, --
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Re: ON7DQ_NanoVNA_Presentation_English
Dear Luc,
Before, congratulations for this nice presentation. On page 52 (TDR measurement), there is a small typo error with the example in the formula. With 1 MHz frequency step, divider must be 1.10EXP06 and not 9.10EXP06. Just a last remark : if final result is right (198 m), the round trip of the signal on the cable allows at VNA level a maximum length measurement of (198/2) = 99 m 73 - Jean-Roger - F6EGK |
Re: can't make a Thru calibration?
#calibration
On Mon, Mar 16, 2020 at 06:37 PM, Hern¨¢n Freschi wrote:
Hmm that seemed to work! I wonder why? ============================================= ±á±ð°ù²Ô¨¢²Ô, That shows that your cable is o.k., CH0 is putting out a signal, and CH1 is receiving it. Either something is amiss with your calibration procedure or you need to perform a "clearconfig 1234" from a terminal program to get rid of a corrupt configuration. - Herb |
Re: can't make a Thru calibration?
#calibration
One more thing to try. You can do a thru calibration without doing an OSL calibration if all you are interested in is the S21 loss/gain. I often do that if I am checking an attentuator or filter over a different range than I calibrated. Try doing a "THRU" and "DONE" and see if you get a zero reference across the screen.
-Herb |
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