Topics

Cavity resonator loops impedance vs frequency

Martin Sole
 

Hi,

Note. This is a purely technical RF type question

I have a question and I am sure the vast wealth of expertise here can advise me. I am trying to get a feel for how the impedance looking into a cavity resonator will vary as the frequency of the applied signal changes. I appreciate that I might be over thinking this and that the most apparently obvious effect, at least to me, is indeed what is happening and I can continue with my pondering without much further concern.

Assume a regular VHF cavity bandpass filter, assume also it is terminated on one side with a decent 50 +/-j minimal ohm match. Lets say the filter is tuned to 145Mhz and and a signal is fed into the cavity at 145Mhz.

1. My assumption is that at this tuned frequency the input looks like 50 +/-j0 (near enough) ohms and the length of coax cable connecting the signal to the input coupling loop provides no dramatic variations in this state of affairs.

If  1. above is true what happens when the frequency applied is changed without retuning the filter? Now the filter is no longer resonant as far as the applied frequency is concerned so my next assumption is that the impedance looking into the filter no longer appears as a 50 +/-j0 or thereabouts impedance.

The simplistic result, to me, is that as the frequency decreases, the impedance looking in to the filter becomes capacitive and as it increases it becomes inductive.

If this is correct in the general case can I make an estimate of the amount of change for a given frequency change? I am assuming here Q plays a part. If it's not the case can anyone explain what does take place and how I might wrap my limited knowledge around the goings on.

Many thanks

Martin, HS0ZED

Andy G4JNT
 

It depends very much on the two Qs of the cavity.   The unloaded Q that is a measure of the quality of the cavity and the loaded Q that is a function of the load impedance transformed to effectively sit as a resistor across the cavity resonance.

In direct answer to your question, to a first approximation you can assume unloaded Q is very high, so the frequency response and bandwidth is defined by the loaded Q, QL, which is set by the coupling into the cavity.   The higher the loaded QL, the lower the bandwidth    BW ~    Cfreq / QL
As you go off resonance the impedance will develop a strong reactive component which will rapidly dominate any real term.

You don't state if the coupling into the cavity is loop or probe - I think that will define which way the reactance will go as you move away from resonance.   At low freq below resonance a loop will be inductive, a probe capacitive, and vice versa above resonance

You really can't get a feel for what it will do without actually measuring its frequency response




On Sat, 14 Sep 2019 at 08:30, Martin Sole <hs0zed@...> wrote:
Hi,

Note. This is a purely technical RF type question

I have a question and I am sure the vast wealth of expertise here can advise me. I am trying to get a feel for how the impedance looking into a cavity resonator will vary as the frequency of the applied signal changes. I appreciate that I might be over thinking this and that the most apparently obvious effect, at least to me, is indeed what is happening and I can continue with my pondering without much further concern.

Assume a regular VHF cavity bandpass filter, assume also it is terminated on one side with a decent 50 +/-j minimal ohm match. Lets say the filter is tuned to 145Mhz and and a signal is fed into the cavity at 145Mhz.

1. My assumption is that at this tuned frequency the input looks like 50 +/-j0 (near enough) ohms and the length of coax cable connecting the signal to the input coupling loop provides no dramatic variations in this state of affairs.

If  1. above is true what happens when the frequency applied is changed without retuning the filter? Now the filter is no longer resonant as far as the applied frequency is concerned so my next assumption is that the impedance looking into the filter no longer appears as a 50 +/-j0 or thereabouts impedance.

The simplistic result, to me, is that as the frequency decreases, the impedance looking in to the filter becomes capacitive and as it increases it becomes inductive.

If this is correct in the general case can I make an estimate of the amount of change for a given frequency change? I am assuming here Q plays a part. If it's not the case can anyone explain what does take place and how I might wrap my limited knowledge around the goings on.

Many thanks

Martin, HS0ZED

John Fell
 

Regarding the unloaded Q , the quality of the cavity will be critically influenced by the materials used .
In the case of a Repeater Duplexer application the quarter wave  line resonator and the internal walls of the cavity need to be at least copper plated , Silver better .Notch depth and width will reduce and increase respectively due to poor surface conductivity .
Temperature variation is also an important consideration in cavity design if a high Q application is involved - hence low expansion rates for the bulk of the cavity , with the surface (skin depth) in a low resistance layer .

73
John
GW0API 

Martin Sole
 

Andy,

Thank you very much for the nicely succinct answer. I rather suspected that Q would be the key player in this, certainly so far as the rate of any change is concerned. QL will be  variable through coupling loop position in this case then.

They are certainly coupling loops though rather than a probe, and as I thought about that aspect I realise that I had the inductive vs capacitive bit bas-ackwards.

I don't have the cavity filter on hand though I  hope to have it sometime this next week. I do have a DG8SAQ VNWA so once I can make some measurements I'm sure I will feel a bit more comfortable about what's going on.


Thanks
Martin, HS0ZED


On 14/09/2019 15:49, Andy TALBOT wrote:
It depends very much on the two Qs of the cavity.   The unloaded Q that is a measure of the quality of the cavity and the loaded Q that is a function of the load impedance transformed to effectively sit as a resistor across the cavity resonance.

In direct answer to your question, to a first approximation you can assume unloaded Q is very high, so the frequency response and bandwidth is defined by the loaded Q, QL, which is set by the coupling into the cavity.   The higher the loaded QL, the lower the bandwidth    BW ~    Cfreq / QL
As you go off resonance the impedance will develop a strong reactive component which will rapidly dominate any real term.

You don't state if the coupling into the cavity is loop or probe - I think that will define which way the reactance will go as you move away from resonance.   At low freq below resonance a loop will be inductive, a probe capacitive, and vice versa above resonance

You really can't get a feel for what it will do without actually measuring its frequency response




On Sat, 14 Sep 2019 at 08:30, Martin Sole <hs0zed@...> wrote:
Hi,

Note. This is a purely technical RF type question

I have a question and I am sure the vast wealth of expertise here can advise me. I am trying to get a feel for how the impedance looking into a cavity resonator will vary as the frequency of the applied signal changes. I appreciate that I might be over thinking this and that the most apparently obvious effect, at least to me, is indeed what is happening and I can continue with my pondering without much further concern.

Assume a regular VHF cavity bandpass filter, assume also it is terminated on one side with a decent 50 +/-j minimal ohm match. Lets say the filter is tuned to 145Mhz and and a signal is fed into the cavity at 145Mhz.

1. My assumption is that at this tuned frequency the input looks like 50 +/-j0 (near enough) ohms and the length of coax cable connecting the signal to the input coupling loop provides no dramatic variations in this state of affairs.

If  1. above is true what happens when the frequency applied is changed without retuning the filter? Now the filter is no longer resonant as far as the applied frequency is concerned so my next assumption is that the impedance looking into the filter no longer appears as a 50 +/-j0 or thereabouts impedance.

The simplistic result, to me, is that as the frequency decreases, the impedance looking in to the filter becomes capacitive and as it increases it becomes inductive.

If this is correct in the general case can I make an estimate of the amount of change for a given frequency change? I am assuming here Q plays a part. If it's not the case can anyone explain what does take place and how I might wrap my limited knowledge around the goings on.

Many thanks

Martin, HS0ZED


geoffrey pike
 

Hi Martin,
As Parky would say "tough question"", but do trawl through the archives at repeater-builder@groups.io or even post the
same question there in.
Its a interesting topic having hand built 2 sets of 50 MHz cavities over the years
cheers
Geoff
GI0GDP

On Saturday, 14 September 2019, 08:30:39 BST, Martin Sole <hs0zed@...> wrote:


Hi,

Note. This is a purely technical RF type question

I have a question and I am sure the vast wealth of expertise here can advise me. I am trying to get a feel for how the impedance looking into a cavity resonator will vary as the frequency of the applied signal changes. I appreciate that I might be over thinking this and that the most apparently obvious effect, at least to me, is indeed what is happening and I can continue with my pondering without much further concern.

Assume a regular VHF cavity bandpass filter, assume also it is terminated on one side with a decent 50 +/-j minimal ohm match. Lets say the filter is tuned to 145Mhz and and a signal is fed into the cavity at 145Mhz.

1. My assumption is that at this tuned frequency the input looks like 50 +/-j0 (near enough) ohms and the length of coax cable connecting the signal to the input coupling loop provides no dramatic variations in this state of affairs.

If  1. above is true what happens when the frequency applied is changed without retuning the filter? Now the filter is no longer resonant as far as the applied frequency is concerned so my next assumption is that the impedance looking into the filter no longer appears as a 50 +/-j0 or thereabouts impedance.

The simplistic result, to me, is that as the frequency decreases, the impedance looking in to the filter becomes capacitive and as it increases it becomes inductive.

If this is correct in the general case can I make an estimate of the amount of change for a given frequency change? I am assuming here Q plays a part. If it's not the case can anyone explain what does take place and how I might wrap my limited knowledge around the goings on.

Many thanks

Martin, HS0ZED

David
 

Martin
Martin
When you couple 50 Ohms to a coaxial cavity, this is translated into a
high resistance in parallel with the top of the inner. It is like a
parallel LC circuit, where this high resistance is in parallel to L
and C. If a small loop is closely couple to the cavity, it resembles
another LC circuit having the same Q and bandwidth, but with 50 Ohms
across it, a very high value of C and very small value of L.
If the coupling is loose, some of the magnetic field lines do not link
the inner and the loop. In this case we see a reactance in series with
the loop, but it can be tuned out by re-tuning slightly. It is
equivalent to a smaller loop.
73
Dave
G3PVH

On 9/14/19, geoffrey pike via Groups.Io <gi0gdp=yahoo.co.uk@groups.io> wrote:
Hi Martin,As Parky would say "tough question"", but do trawl through the
archives at repeater-builder@groups.io or even post thesame question there
in.Its a interesting topic having hand built 2 sets of 50 MHz cavities over
the yearscheersGeoffGI0GDP
On Saturday, 14 September 2019, 08:30:39 BST, Martin Sole
<hs0zed@...> wrote:

Hi,

Note. This is a purely technical RF type question

I have a question and I am sure the vast wealth of expertise here can
advise me. I am trying to get a feel for how the impedance looking into a
cavity resonator will vary as the frequency of the applied signal changes. I
appreciate that I might be over thinking this and that the most apparently
obvious effect, at least to me, is indeed what is happening and I can
continue with my pondering without much further concern.

Assume a regular VHF cavity bandpass filter, assume also it is terminated
on one side with a decent 50 +/-j minimal ohm match. Lets say the filter is
tuned to 145Mhz and and a signal is fed into the cavity at 145Mhz.

1. My assumption is that at this tuned frequency the input looks like 50
+/-j0 (near enough) ohms and the length of coax cable connecting the signal
to the input coupling loop provides no dramatic variations in this state of
affairs.

If  1. above is true what happens when the frequency applied is changed
without retuning the filter? Now the filter is no longer resonant as far as
the applied frequency is concerned so my next assumption is that the
impedance looking into the filter no longer appears as a 50 +/-j0 or
thereabouts impedance.

The simplistic result, to me, is that as the frequency decreases, the
impedance looking in to the filter becomes capacitive and as it increases it
becomes inductive.

If this is correct in the general case can I make an estimate of the amount
of change for a given frequency change? I am assuming here Q plays a part.
If it's not the case can anyone explain what does take place and how I might
wrap my limited knowledge around the goings on.

Many thanks

Martin, HS0ZED





Alan
 

Martin

If you wanted to “have a play” you could emulate the cavity with a parallel tuned circuit with coupling loops each side  ( or each end) of the coil or capacitors to the hot end if you like c coupling. Then you can sweep it with  your VNA.

 

The Q will be around a tenth of a cavity but the coil and capacitor costs pennies and is much more accessible for playing.

 

Not many people Know.....  Louis Essen used cavity resonators to estimate the speed of light with very carefully made cavity resonators at the NPL, Q’s were >10,000 and probe depth was minute.

 

 

73

Alan

G8LCO

 

Sent from Mail for Windows 10

 

 

David
 

By the way, why do we call a coaxial line resonator a cavity resonator
when it isn't?
Dave
G3PVH

On 9/14/19, Alan <g8lco1@...> wrote:
Martin
If you wanted to “have a play” you could emulate the cavity with a parallel
tuned circuit with coupling loops each side ( or each end) of the coil or
capacitors to the hot end if you like c coupling. Then you can sweep it with
your VNA.

The Q will be around a tenth of a cavity but the coil and capacitor costs
pennies and is much more accessible for playing.

Not many people Know..... Louis Essen used cavity resonators to estimate
the speed of light with very carefully made cavity resonators at the NPL,
Q’s were >10,000 and probe depth was minute.


73
Alan
G8LCO

Sent from Mail for Windows 10






Andy G4JNT
 

We don't.
What specific case are you thinking of ?



On Sat, 14 Sep 2019 at 12:49, David <davidjohnsumner@...> wrote:
By the way, why do we call a coaxial line resonator a cavity resonator
when it isn't?
Dave
G3PVH

On 9/14/19, Alan <g8lco1@...> wrote:
> Martin
> If you wanted to “have a play” you could emulate the cavity with a parallel
> tuned circuit with coupling loops each side  ( or each end) of the coil or
> capacitors to the hot end if you like c coupling. Then you can sweep it with
>  your VNA.
>
> The Q will be around a tenth of a cavity but the coil and capacitor costs
> pennies and is much more accessible for playing.
>
> Not many people Know.....  Louis Essen used cavity resonators to estimate
> the speed of light with very carefully made cavity resonators at the NPL,
> Q’s were >10,000 and probe depth was minute.
>
>
> 73
> Alan
> G8LCO
>
> Sent from Mail for Windows 10
>
>
>
>
>
>
>



Martin Sole
 

I've only ever known them as cavity filters.

I think having a play with some loops and a parallel tuned circuit sounds fun and I have just the bits to try that on hand at least.

Martin, HS0ZED


On 14/09/2019 18:49, David wrote:
By the way, why do we call a coaxial line resonator a cavity resonator
when it isn't?
Dave
G3PVH

On 9/14/19, Alan <g8lco1@...> wrote:
Martin
If you wanted to “have a play” you could emulate the cavity with a parallel
tuned circuit with coupling loops each side  ( or each end) of the coil or
capacitors to the hot end if you like c coupling. Then you can sweep it with
 your VNA.

The Q will be around a tenth of a cavity but the coil and capacitor costs
pennies and is much more accessible for playing.

Not many people Know.....  Louis Essen used cavity resonators to estimate
the speed of light with very carefully made cavity resonators at the NPL,
Q’s were >10,000 and probe depth was minute.


73
Alan
G8LCO

Sent from Mail for Windows 10










Alan
 

Common usage. Every repeater group I know calls their duplexing filters “cavities” not coaxial line resonators. But often the filters are used as High pass or low pass notch filters not simple bandpass filters.

 

73

Alan

G8LCO

 

 

 

Sent from Mail for Windows 10

 

From: David
Sent: 14 September 2019 12:49 PM
To: RSGB-Workshop@groups.io
Subject: Re: [RSGB-Workshop] Cavity resonator loops impedance vs frequency

 

By the way, why do we call a coaxial line resonator a cavity resonator

when it isn't?

Dave

G3PVH

 

 

 

Peter Chadwick
 





------ Original Message ------
From: "Alan" <g8lco1@...>
To: "RSGB-Workshop@groups.io" <RSGB-Workshop@groups.io>
Sent: Saturday, 14 Sep, 19 At 20:52
Subject: Re: [RSGB-Workshop] Cavity resonator loops impedance vs frequency

Common usage. Every repeater group I know calls their duplexing filters “cavities” not coaxial line resonators. But often the filters are used as High pass or low pass notch filters not simple bandpass filters.

73

Alan

G8LCO

Sent from Mail for Windows 10

From: David
Sent: 14 September 2019 12:49 PM
To: RSGB-Workshop@groups.io
Subject: Re: [RSGB-Workshop] Cavity resonator loops impedance vs frequency

By the way, why do we call a coaxial line resonator a cavity resonator

when it isn't?

Dave

G3PVH