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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?
$begingroup$
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?
impedance impedance-matching
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.
$endgroup$
add a comment |
$begingroup$
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?
impedance impedance-matching
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.
$endgroup$
add a comment |
$begingroup$
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?
impedance impedance-matching
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
impedance impedance-matching
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.
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.
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New contributor
Dominik Heidler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
"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
add a comment |
$begingroup$
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.


$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 )
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
"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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
"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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
"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
add a comment |
$begingroup$
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.


$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.


$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.


$endgroup$
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.


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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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 )
$endgroup$
add a comment |
$begingroup$
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 )
$endgroup$
add a comment |
$begingroup$
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 )
$endgroup$
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 )
answered 9 hours ago
Scott Earle♦Scott Earle
2,7731922
2,7731922
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add a comment |
Dominik Heidler is a new contributor. Be nice, and check out our Code of Conduct.
Dominik Heidler is a new contributor. Be nice, and check out our Code of Conduct.
Dominik Heidler is a new contributor. Be nice, and check out our Code of Conduct.
Dominik Heidler is a new contributor. Be nice, and check out our Code of Conduct.
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