Impedance ratio vs. SWRHow can I safely transmit without an antenna tuner or SWR meter?If two antennas of 50 Ω and 377 Ω have VSWR=1:1, then which one is more efficient?Understanding coax impedanceResistive impedance with frequencyWhy does “high SWR” damage transmitters, instead of “impedance mismatch”?What kind of losses do you get from an LC network matching the antenna's impedance?What is the output impedance of a typical solid state ham transmitter?How do I set up my radio for an antenna tuner?Impedance effect of using 75 ohm coax between transmitter and an EFHW antennaExactly why do some SWR meters give a changing reading depending on the length of coax used to connect to an antenna?

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Impedance ratio vs. SWR

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Impedance ratio vs. SWR


How can I safely transmit without an antenna tuner or SWR meter?If two antennas of 50 Ω and 377 Ω have VSWR=1:1, then which one is more efficient?Understanding coax impedanceResistive impedance with frequencyWhy does “high SWR” damage transmitters, instead of “impedance mismatch”?What kind of losses do you get from an LC network matching the antenna's impedance?What is the output impedance of a typical solid state ham transmitter?How do I set up my radio for an antenna tuner?Impedance effect of using 75 ohm coax between transmitter and an EFHW antennaExactly why do some SWR meters give a changing reading depending on the length of coax used to connect to an antenna?













4












$begingroup$


The specs of the Elecraft T1 tuner say, that it can match a 10:1 SWR.



  • What does that mean in terms of impedance ratio?

  • What antenna impedance can it match to 50 Ohm then?

  • Or is it not that easy?









share|improve this question









New contributor



Dominik Heidler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$
















    4












    $begingroup$


    The specs of the Elecraft T1 tuner say, that it can match a 10:1 SWR.



    • What does that mean in terms of impedance ratio?

    • What antenna impedance can it match to 50 Ohm then?

    • Or is it not that easy?









    share|improve this question









    New contributor



    Dominik Heidler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$














      4












      4








      4





      $begingroup$


      The specs of the Elecraft T1 tuner say, that it can match a 10:1 SWR.



      • What does that mean in terms of impedance ratio?

      • What antenna impedance can it match to 50 Ohm then?

      • Or is it not that easy?









      share|improve this question









      New contributor



      Dominik Heidler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$




      The specs of the Elecraft T1 tuner say, that it can match a 10:1 SWR.



      • What does that mean in terms of impedance ratio?

      • What antenna impedance can it match to 50 Ohm then?

      • Or is it not that easy?






      impedance impedance-matching






      share|improve this question









      New contributor



      Dominik Heidler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.










      share|improve this question









      New contributor



      Dominik Heidler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.








      share|improve this question




      share|improve this question








      edited 9 hours ago









      Marcus Müller

      7,9421031




      7,9421031






      New contributor



      Dominik Heidler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.








      asked 9 hours ago









      Dominik HeidlerDominik Heidler

      233




      233




      New contributor



      Dominik Heidler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.




      New contributor




      Dominik Heidler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          Short detour:



          There's a so-called reflection coefficient $Gamma$ that says "OK, for this mismatch, so and so much of the power is reflected back where it came from".



          We can calculate it as, based on load impedance $Z_L$ and conduction line impedance $Z_0$:



          $$ Gamma =fracZ_L - Z_0Z_L + Z_0$$



          Also, the VSWR is a result of things getting reflected back:



          beginalign
          DeclareMathOperatorvswrVSWR
          vswr &= frac1+lvertGammarvert1-lvertGammarvert\
          &implies\
          lvertGammarvert &= fracvswr-1vswr +1\
          &iff\
          leftlvertfracZ_L - Z_0Z_L + Z_0 rightrvert &=fracvswr-1vswr +1
          endalign



          Assuming $Z_L>Z_0$:



          beginalign
          fracZ_L - Z_0Z_L + Z_0 &=fracvswr-1vswr +1&& lvertvswr=10, Z_0 = 50,mathrmOmega\
          &=frac911\
          &implies\
          fracZ_L - Z_0Z_L + Z_0 - frac911 &= 0 && lvertcdot (Z_L+Z_0)\
          0 &=Z_L - Z_0 - frac911(Z_L + Z_0) \
          &=frac211Z_L -frac2011Z_0 &&lvertcdot11:2\
          &=Z_L -10 Z_0\
          Z_L&= 10 Z_0 \
          &= 500,Ω
          endalign



          Assuming $Z_L<Z_0$:



          beginalign
          fracZ_0 - Z_LZ_L + Z_0 &=fracvswr-1vswr +1&& lvertvswr=10, Z_0 = 50,mathrmOmega\
          &=frac911\
          &implies\
          fracZ_0 - Z_LZ_L + Z_0 - frac911 &= 0 && lvertcdot (Z_L+Z_0)\
          0 &=Z_0 - Z_L - frac911(Z_L + Z_0) \
          &=frac211Z_0 -frac2011Z_L &&lvertcdot11:2\
          &=Z_0 -10 Z_L\
          frac110Z_0&= Z_L \
          &= 5,Ω
          endalign



          So, 5 Ω to 500 Ω are "specification-wise" matchable.



          This is assuming a real-valued antenna impedance. That's often not given. For the complete region of applicable values, see (and upvote!) Cecil's answer.



          However, impedance matching doesn't happen on a "reflection coefficient level"; it happens by having an adjustable matching network (typically takes the shape of an adjustable LC filter). Things actually get rather interesting there, because the matches you get from that typically aren't wideband and typically aren't all real and typically aren't all within a "nice" shape in the real world. So, assuming the best, the SWR range is but a thing that the manufacturer has tested to work, and special complex impedances outside the circle from Cecil's answer can be matched too.






          share|improve this answer











          $endgroup$












          • $begingroup$
            "So, 5 Ω to 50 Ω are specification-wise matchable." — I'm pretty sure you mean "5 Ω to 500 Ω" right?
            $endgroup$
            – natevw - AF7TB
            5 hours ago











          • $begingroup$
            yeah, @natevw-AF7TB
            $endgroup$
            – Marcus Müller
            4 hours ago


















          8












          $begingroup$

          As you can see from the following equation, it is definitely not that easy. What I would do is draw a 10:1 SWR circle on a Smith Chart and assume that your tuner can match all of the infinite number of impedances inside that 10:1 SWR circle. If you don't know how to read impedances from a Smith Chart, it would be worth your while to learn how. The green area below on the Smith Chart normalized to 50 ohms is the advertised matching range for your tuner.



          enter image description here



          enter image description here






          share|improve this answer











          $endgroup$












          • $begingroup$
            Yes, this is the 10:1 circle. But given a basic Pi network, limited component values and a wide range of frequencies, it's probably more like an amoeba that changes shape over the bands, sometimes touching 10:1 !
            $endgroup$
            – tomnexus
            2 hours ago


















          2












          $begingroup$

          When a tuner in a radio says it can match 3:1 and below, they sometimes say that it can match 16.7-150 ohms.



          I would take this to mean that a tuner that can match 10:1 and below, can match 5-500 ohms.



          There is an example of a tuner that can match some bands to 10:1 on this page. they specify:



          Frequency               Typical Matching Range and Power Limit



          3 — 30 MHz            600W into 5 to 500 Ohms (10:1 SWR)



                                           1000W into 16 to 150 Ohms (3:1 SWR )






          share|improve this answer









          $endgroup$













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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            Short detour:



            There's a so-called reflection coefficient $Gamma$ that says "OK, for this mismatch, so and so much of the power is reflected back where it came from".



            We can calculate it as, based on load impedance $Z_L$ and conduction line impedance $Z_0$:



            $$ Gamma =fracZ_L - Z_0Z_L + Z_0$$



            Also, the VSWR is a result of things getting reflected back:



            beginalign
            DeclareMathOperatorvswrVSWR
            vswr &= frac1+lvertGammarvert1-lvertGammarvert\
            &implies\
            lvertGammarvert &= fracvswr-1vswr +1\
            &iff\
            leftlvertfracZ_L - Z_0Z_L + Z_0 rightrvert &=fracvswr-1vswr +1
            endalign



            Assuming $Z_L>Z_0$:



            beginalign
            fracZ_L - Z_0Z_L + Z_0 &=fracvswr-1vswr +1&& lvertvswr=10, Z_0 = 50,mathrmOmega\
            &=frac911\
            &implies\
            fracZ_L - Z_0Z_L + Z_0 - frac911 &= 0 && lvertcdot (Z_L+Z_0)\
            0 &=Z_L - Z_0 - frac911(Z_L + Z_0) \
            &=frac211Z_L -frac2011Z_0 &&lvertcdot11:2\
            &=Z_L -10 Z_0\
            Z_L&= 10 Z_0 \
            &= 500,Ω
            endalign



            Assuming $Z_L<Z_0$:



            beginalign
            fracZ_0 - Z_LZ_L + Z_0 &=fracvswr-1vswr +1&& lvertvswr=10, Z_0 = 50,mathrmOmega\
            &=frac911\
            &implies\
            fracZ_0 - Z_LZ_L + Z_0 - frac911 &= 0 && lvertcdot (Z_L+Z_0)\
            0 &=Z_0 - Z_L - frac911(Z_L + Z_0) \
            &=frac211Z_0 -frac2011Z_L &&lvertcdot11:2\
            &=Z_0 -10 Z_L\
            frac110Z_0&= Z_L \
            &= 5,Ω
            endalign



            So, 5 Ω to 500 Ω are "specification-wise" matchable.



            This is assuming a real-valued antenna impedance. That's often not given. For the complete region of applicable values, see (and upvote!) Cecil's answer.



            However, impedance matching doesn't happen on a "reflection coefficient level"; it happens by having an adjustable matching network (typically takes the shape of an adjustable LC filter). Things actually get rather interesting there, because the matches you get from that typically aren't wideband and typically aren't all real and typically aren't all within a "nice" shape in the real world. So, assuming the best, the SWR range is but a thing that the manufacturer has tested to work, and special complex impedances outside the circle from Cecil's answer can be matched too.






            share|improve this answer











            $endgroup$












            • $begingroup$
              "So, 5 Ω to 50 Ω are specification-wise matchable." — I'm pretty sure you mean "5 Ω to 500 Ω" right?
              $endgroup$
              – natevw - AF7TB
              5 hours ago











            • $begingroup$
              yeah, @natevw-AF7TB
              $endgroup$
              – Marcus Müller
              4 hours ago















            3












            $begingroup$

            Short detour:



            There's a so-called reflection coefficient $Gamma$ that says "OK, for this mismatch, so and so much of the power is reflected back where it came from".



            We can calculate it as, based on load impedance $Z_L$ and conduction line impedance $Z_0$:



            $$ Gamma =fracZ_L - Z_0Z_L + Z_0$$



            Also, the VSWR is a result of things getting reflected back:



            beginalign
            DeclareMathOperatorvswrVSWR
            vswr &= frac1+lvertGammarvert1-lvertGammarvert\
            &implies\
            lvertGammarvert &= fracvswr-1vswr +1\
            &iff\
            leftlvertfracZ_L - Z_0Z_L + Z_0 rightrvert &=fracvswr-1vswr +1
            endalign



            Assuming $Z_L>Z_0$:



            beginalign
            fracZ_L - Z_0Z_L + Z_0 &=fracvswr-1vswr +1&& lvertvswr=10, Z_0 = 50,mathrmOmega\
            &=frac911\
            &implies\
            fracZ_L - Z_0Z_L + Z_0 - frac911 &= 0 && lvertcdot (Z_L+Z_0)\
            0 &=Z_L - Z_0 - frac911(Z_L + Z_0) \
            &=frac211Z_L -frac2011Z_0 &&lvertcdot11:2\
            &=Z_L -10 Z_0\
            Z_L&= 10 Z_0 \
            &= 500,Ω
            endalign



            Assuming $Z_L<Z_0$:



            beginalign
            fracZ_0 - Z_LZ_L + Z_0 &=fracvswr-1vswr +1&& lvertvswr=10, Z_0 = 50,mathrmOmega\
            &=frac911\
            &implies\
            fracZ_0 - Z_LZ_L + Z_0 - frac911 &= 0 && lvertcdot (Z_L+Z_0)\
            0 &=Z_0 - Z_L - frac911(Z_L + Z_0) \
            &=frac211Z_0 -frac2011Z_L &&lvertcdot11:2\
            &=Z_0 -10 Z_L\
            frac110Z_0&= Z_L \
            &= 5,Ω
            endalign



            So, 5 Ω to 500 Ω are "specification-wise" matchable.



            This is assuming a real-valued antenna impedance. That's often not given. For the complete region of applicable values, see (and upvote!) Cecil's answer.



            However, impedance matching doesn't happen on a "reflection coefficient level"; it happens by having an adjustable matching network (typically takes the shape of an adjustable LC filter). Things actually get rather interesting there, because the matches you get from that typically aren't wideband and typically aren't all real and typically aren't all within a "nice" shape in the real world. So, assuming the best, the SWR range is but a thing that the manufacturer has tested to work, and special complex impedances outside the circle from Cecil's answer can be matched too.






            share|improve this answer











            $endgroup$












            • $begingroup$
              "So, 5 Ω to 50 Ω are specification-wise matchable." — I'm pretty sure you mean "5 Ω to 500 Ω" right?
              $endgroup$
              – natevw - AF7TB
              5 hours ago











            • $begingroup$
              yeah, @natevw-AF7TB
              $endgroup$
              – Marcus Müller
              4 hours ago













            3












            3








            3





            $begingroup$

            Short detour:



            There's a so-called reflection coefficient $Gamma$ that says "OK, for this mismatch, so and so much of the power is reflected back where it came from".



            We can calculate it as, based on load impedance $Z_L$ and conduction line impedance $Z_0$:



            $$ Gamma =fracZ_L - Z_0Z_L + Z_0$$



            Also, the VSWR is a result of things getting reflected back:



            beginalign
            DeclareMathOperatorvswrVSWR
            vswr &= frac1+lvertGammarvert1-lvertGammarvert\
            &implies\
            lvertGammarvert &= fracvswr-1vswr +1\
            &iff\
            leftlvertfracZ_L - Z_0Z_L + Z_0 rightrvert &=fracvswr-1vswr +1
            endalign



            Assuming $Z_L>Z_0$:



            beginalign
            fracZ_L - Z_0Z_L + Z_0 &=fracvswr-1vswr +1&& lvertvswr=10, Z_0 = 50,mathrmOmega\
            &=frac911\
            &implies\
            fracZ_L - Z_0Z_L + Z_0 - frac911 &= 0 && lvertcdot (Z_L+Z_0)\
            0 &=Z_L - Z_0 - frac911(Z_L + Z_0) \
            &=frac211Z_L -frac2011Z_0 &&lvertcdot11:2\
            &=Z_L -10 Z_0\
            Z_L&= 10 Z_0 \
            &= 500,Ω
            endalign



            Assuming $Z_L<Z_0$:



            beginalign
            fracZ_0 - Z_LZ_L + Z_0 &=fracvswr-1vswr +1&& lvertvswr=10, Z_0 = 50,mathrmOmega\
            &=frac911\
            &implies\
            fracZ_0 - Z_LZ_L + Z_0 - frac911 &= 0 && lvertcdot (Z_L+Z_0)\
            0 &=Z_0 - Z_L - frac911(Z_L + Z_0) \
            &=frac211Z_0 -frac2011Z_L &&lvertcdot11:2\
            &=Z_0 -10 Z_L\
            frac110Z_0&= Z_L \
            &= 5,Ω
            endalign



            So, 5 Ω to 500 Ω are "specification-wise" matchable.



            This is assuming a real-valued antenna impedance. That's often not given. For the complete region of applicable values, see (and upvote!) Cecil's answer.



            However, impedance matching doesn't happen on a "reflection coefficient level"; it happens by having an adjustable matching network (typically takes the shape of an adjustable LC filter). Things actually get rather interesting there, because the matches you get from that typically aren't wideband and typically aren't all real and typically aren't all within a "nice" shape in the real world. So, assuming the best, the SWR range is but a thing that the manufacturer has tested to work, and special complex impedances outside the circle from Cecil's answer can be matched too.






            share|improve this answer











            $endgroup$



            Short detour:



            There's a so-called reflection coefficient $Gamma$ that says "OK, for this mismatch, so and so much of the power is reflected back where it came from".



            We can calculate it as, based on load impedance $Z_L$ and conduction line impedance $Z_0$:



            $$ Gamma =fracZ_L - Z_0Z_L + Z_0$$



            Also, the VSWR is a result of things getting reflected back:



            beginalign
            DeclareMathOperatorvswrVSWR
            vswr &= frac1+lvertGammarvert1-lvertGammarvert\
            &implies\
            lvertGammarvert &= fracvswr-1vswr +1\
            &iff\
            leftlvertfracZ_L - Z_0Z_L + Z_0 rightrvert &=fracvswr-1vswr +1
            endalign



            Assuming $Z_L>Z_0$:



            beginalign
            fracZ_L - Z_0Z_L + Z_0 &=fracvswr-1vswr +1&& lvertvswr=10, Z_0 = 50,mathrmOmega\
            &=frac911\
            &implies\
            fracZ_L - Z_0Z_L + Z_0 - frac911 &= 0 && lvertcdot (Z_L+Z_0)\
            0 &=Z_L - Z_0 - frac911(Z_L + Z_0) \
            &=frac211Z_L -frac2011Z_0 &&lvertcdot11:2\
            &=Z_L -10 Z_0\
            Z_L&= 10 Z_0 \
            &= 500,Ω
            endalign



            Assuming $Z_L<Z_0$:



            beginalign
            fracZ_0 - Z_LZ_L + Z_0 &=fracvswr-1vswr +1&& lvertvswr=10, Z_0 = 50,mathrmOmega\
            &=frac911\
            &implies\
            fracZ_0 - Z_LZ_L + Z_0 - frac911 &= 0 && lvertcdot (Z_L+Z_0)\
            0 &=Z_0 - Z_L - frac911(Z_L + Z_0) \
            &=frac211Z_0 -frac2011Z_L &&lvertcdot11:2\
            &=Z_0 -10 Z_L\
            frac110Z_0&= Z_L \
            &= 5,Ω
            endalign



            So, 5 Ω to 500 Ω are "specification-wise" matchable.



            This is assuming a real-valued antenna impedance. That's often not given. For the complete region of applicable values, see (and upvote!) Cecil's answer.



            However, impedance matching doesn't happen on a "reflection coefficient level"; it happens by having an adjustable matching network (typically takes the shape of an adjustable LC filter). Things actually get rather interesting there, because the matches you get from that typically aren't wideband and typically aren't all real and typically aren't all within a "nice" shape in the real world. So, assuming the best, the SWR range is but a thing that the manufacturer has tested to work, and special complex impedances outside the circle from Cecil's answer can be matched too.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 4 hours ago

























            answered 8 hours ago









            Marcus MüllerMarcus Müller

            7,9421031




            7,9421031











            • $begingroup$
              "So, 5 Ω to 50 Ω are specification-wise matchable." — I'm pretty sure you mean "5 Ω to 500 Ω" right?
              $endgroup$
              – natevw - AF7TB
              5 hours ago











            • $begingroup$
              yeah, @natevw-AF7TB
              $endgroup$
              – Marcus Müller
              4 hours ago
















            • $begingroup$
              "So, 5 Ω to 50 Ω are specification-wise matchable." — I'm pretty sure you mean "5 Ω to 500 Ω" right?
              $endgroup$
              – natevw - AF7TB
              5 hours ago











            • $begingroup$
              yeah, @natevw-AF7TB
              $endgroup$
              – Marcus Müller
              4 hours ago















            $begingroup$
            "So, 5 Ω to 50 Ω are specification-wise matchable." — I'm pretty sure you mean "5 Ω to 500 Ω" right?
            $endgroup$
            – natevw - AF7TB
            5 hours ago





            $begingroup$
            "So, 5 Ω to 50 Ω are specification-wise matchable." — I'm pretty sure you mean "5 Ω to 500 Ω" right?
            $endgroup$
            – natevw - AF7TB
            5 hours ago













            $begingroup$
            yeah, @natevw-AF7TB
            $endgroup$
            – Marcus Müller
            4 hours ago




            $begingroup$
            yeah, @natevw-AF7TB
            $endgroup$
            – Marcus Müller
            4 hours ago











            8












            $begingroup$

            As you can see from the following equation, it is definitely not that easy. What I would do is draw a 10:1 SWR circle on a Smith Chart and assume that your tuner can match all of the infinite number of impedances inside that 10:1 SWR circle. If you don't know how to read impedances from a Smith Chart, it would be worth your while to learn how. The green area below on the Smith Chart normalized to 50 ohms is the advertised matching range for your tuner.



            enter image description here



            enter image description here






            share|improve this answer











            $endgroup$












            • $begingroup$
              Yes, this is the 10:1 circle. But given a basic Pi network, limited component values and a wide range of frequencies, it's probably more like an amoeba that changes shape over the bands, sometimes touching 10:1 !
              $endgroup$
              – tomnexus
              2 hours ago















            8












            $begingroup$

            As you can see from the following equation, it is definitely not that easy. What I would do is draw a 10:1 SWR circle on a Smith Chart and assume that your tuner can match all of the infinite number of impedances inside that 10:1 SWR circle. If you don't know how to read impedances from a Smith Chart, it would be worth your while to learn how. The green area below on the Smith Chart normalized to 50 ohms is the advertised matching range for your tuner.



            enter image description here



            enter image description here






            share|improve this answer











            $endgroup$












            • $begingroup$
              Yes, this is the 10:1 circle. But given a basic Pi network, limited component values and a wide range of frequencies, it's probably more like an amoeba that changes shape over the bands, sometimes touching 10:1 !
              $endgroup$
              – tomnexus
              2 hours ago













            8












            8








            8





            $begingroup$

            As you can see from the following equation, it is definitely not that easy. What I would do is draw a 10:1 SWR circle on a Smith Chart and assume that your tuner can match all of the infinite number of impedances inside that 10:1 SWR circle. If you don't know how to read impedances from a Smith Chart, it would be worth your while to learn how. The green area below on the Smith Chart normalized to 50 ohms is the advertised matching range for your tuner.



            enter image description here



            enter image description here






            share|improve this answer











            $endgroup$



            As you can see from the following equation, it is definitely not that easy. What I would do is draw a 10:1 SWR circle on a Smith Chart and assume that your tuner can match all of the infinite number of impedances inside that 10:1 SWR circle. If you don't know how to read impedances from a Smith Chart, it would be worth your while to learn how. The green area below on the Smith Chart normalized to 50 ohms is the advertised matching range for your tuner.



            enter image description here



            enter image description here







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 8 hours ago

























            answered 8 hours ago









            Cecil - W5DXPCecil - W5DXP

            1,37517




            1,37517











            • $begingroup$
              Yes, this is the 10:1 circle. But given a basic Pi network, limited component values and a wide range of frequencies, it's probably more like an amoeba that changes shape over the bands, sometimes touching 10:1 !
              $endgroup$
              – tomnexus
              2 hours ago
















            • $begingroup$
              Yes, this is the 10:1 circle. But given a basic Pi network, limited component values and a wide range of frequencies, it's probably more like an amoeba that changes shape over the bands, sometimes touching 10:1 !
              $endgroup$
              – tomnexus
              2 hours ago















            $begingroup$
            Yes, this is the 10:1 circle. But given a basic Pi network, limited component values and a wide range of frequencies, it's probably more like an amoeba that changes shape over the bands, sometimes touching 10:1 !
            $endgroup$
            – tomnexus
            2 hours ago




            $begingroup$
            Yes, this is the 10:1 circle. But given a basic Pi network, limited component values and a wide range of frequencies, it's probably more like an amoeba that changes shape over the bands, sometimes touching 10:1 !
            $endgroup$
            – tomnexus
            2 hours ago











            2












            $begingroup$

            When a tuner in a radio says it can match 3:1 and below, they sometimes say that it can match 16.7-150 ohms.



            I would take this to mean that a tuner that can match 10:1 and below, can match 5-500 ohms.



            There is an example of a tuner that can match some bands to 10:1 on this page. they specify:



            Frequency               Typical Matching Range and Power Limit



            3 — 30 MHz            600W into 5 to 500 Ohms (10:1 SWR)



                                             1000W into 16 to 150 Ohms (3:1 SWR )






            share|improve this answer









            $endgroup$

















              2












              $begingroup$

              When a tuner in a radio says it can match 3:1 and below, they sometimes say that it can match 16.7-150 ohms.



              I would take this to mean that a tuner that can match 10:1 and below, can match 5-500 ohms.



              There is an example of a tuner that can match some bands to 10:1 on this page. they specify:



              Frequency               Typical Matching Range and Power Limit



              3 — 30 MHz            600W into 5 to 500 Ohms (10:1 SWR)



                                               1000W into 16 to 150 Ohms (3:1 SWR )






              share|improve this answer









              $endgroup$















                2












                2








                2





                $begingroup$

                When a tuner in a radio says it can match 3:1 and below, they sometimes say that it can match 16.7-150 ohms.



                I would take this to mean that a tuner that can match 10:1 and below, can match 5-500 ohms.



                There is an example of a tuner that can match some bands to 10:1 on this page. they specify:



                Frequency               Typical Matching Range and Power Limit



                3 — 30 MHz            600W into 5 to 500 Ohms (10:1 SWR)



                                                 1000W into 16 to 150 Ohms (3:1 SWR )






                share|improve this answer









                $endgroup$



                When a tuner in a radio says it can match 3:1 and below, they sometimes say that it can match 16.7-150 ohms.



                I would take this to mean that a tuner that can match 10:1 and below, can match 5-500 ohms.



                There is an example of a tuner that can match some bands to 10:1 on this page. they specify:



                Frequency               Typical Matching Range and Power Limit



                3 — 30 MHz            600W into 5 to 500 Ohms (10:1 SWR)



                                                 1000W into 16 to 150 Ohms (3:1 SWR )







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 9 hours ago









                Scott EarleScott Earle

                2,7731922




                2,7731922




















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