The source is 50 ohms, I am feeding through a 1000 ohm resistor to
my tuned circuit at 30mhz, and the load is over 5000 ohms (FET input).

I want to use a parallel resonant circuit to ground to form a pass
filter. If I want a Q of around 5, would I need 5000 ohm XL/XC ??

This would seem nearly correct. Have I calculated correctly?

In article ,
(gudmundur) writes:

The source is 50 ohms, I am feeding through a 1000 ohm resistor to
my tuned circuit at 30mhz, and the load is over 5000 ohms (FET input).

I want to use a parallel resonant circuit to ground to form a pass
filter. If I want a Q of around 5, would I need 5000 ohm XL/XC ??

This would seem nearly correct. Have I calculated correctly?

Not quite. Loaded Q is dependent on ALL losses in parallel
with a parallel-resonant circuit. If you want a Q of around 5
then the R:Xc or R:Xl ratio is 5.

A problem seems to be the loading of the 50 Ohm source plus
a series (?) resistance of 1000 Ohms. That alone puts 1050
Ohms in parallel with the resonant circuit. If the total parallel
R is 1050 Ohms, the total from FET input and coil's own Q is
5000 Ohms, then the total parallel R will be 868 Ohms. The
inductive reactance would have to be 1/5th of that or equivalent
to 0.921 uHy. The Xl of 0.921 uHy is 173.6 Ohms at 30 MHz.

The problem is compounded by using a series resistor to feed
the parallel resistance. If the resonant circuit has a total parallel
resistance of 5000 Ohms, the magnitude of the impedance is
that value, 5000 Ohms. But, with a series resistance from the
50 Ohm source there's a power loss from the equivalent voltage
divider action equal to 5:6 or about 1.6 db at 30 MHz resonance.

It would be better to use another way to couple the 50 Ohm
source, such as a tapped inductor. Tapping down at 1/10th of
the turns has an approximate 1:100 impedance change which
would transform the source to be about 5000 Ohms across the
parallel L-C. That would change the needed inductance to get
the loaded Q but there would be no extra losses involved. The
new inductance value at resonance would be about 2.65 uHy for
an Xl of 500 Ohms (1/5th of 5000 in parallel with 5000 Ohms).

Note: At resonance (and only at that frequency), the impedance
across the L-C is entirely resistive. At any other frequency it is
a complex number (both R and jX).



retired (from regular hours) electronic engineer person

In article ,
says...

In article ,
(gudmundur) writes:

The source is 50 ohms, I am feeding through a 1000 ohm resistor to
my tuned circuit at 30mhz, and the load is over 5000 ohms (FET input).

I want to use a parallel resonant circuit to ground to form a pass
filter. If I want a Q of around 5, would I need 5000 ohm XL/XC ??

This would seem nearly correct. Have I calculated correctly?

Not quite. Loaded Q is dependent on ALL losses in parallel
with a parallel-resonant circuit. If you want a Q of around 5
then the R:Xc or R:Xl ratio is 5.

A problem seems to be the loading of the 50 Ohm source plus
a series (?) resistance of 1000 Ohms. That alone puts 1050
Ohms in parallel with the resonant circuit. If the total parallel
R is 1050 Ohms, the total from FET input and coil's own Q is
5000 Ohms, then the total parallel R will be 868 Ohms. The
inductive reactance would have to be 1/5th of that or equivalent
to 0.921 uHy. The Xl of 0.921 uHy is 173.6 Ohms at 30 MHz.

The problem is compounded by using a series resistor to feed
the parallel resistance. If the resonant circuit has a total parallel
resistance of 5000 Ohms, the magnitude of the impedance is
that value, 5000 Ohms. But, with a series resistance from the
50 Ohm source there's a power loss from the equivalent voltage
divider action equal to 5:6 or about 1.6 db at 30 MHz resonance.

It would be better to use another way to couple the 50 Ohm
source, such as a tapped inductor. Tapping down at 1/10th of
the turns has an approximate 1:100 impedance change which
would transform the source to be about 5000 Ohms across the
parallel L-C. That would change the needed inductance to get
the loaded Q but there would be no extra losses involved. The
new inductance value at resonance would be about 2.65 uHy for
an Xl of 500 Ohms (1/5th of 5000 in parallel with 5000 Ohms).

Note: At resonance (and only at that frequency), the impedance
across the L-C is entirely resistive. At any other frequency it is
a complex number (both R and jX).



retired (from regular hours) electronic engineer person


I see my error, the XL/XC is one fifth, not five times the R.
I see this now in tank circuits where Q=10 R=2000 XL/XC=200.

O.K., now I can calculate my parts. This is for a bandpass
preselector, and loss is no problem, there is plenty of drive
available. I just need some relative measure of if any power
in the total signal level happens to be at 30mhz plus or minus
about 250kc at perhaps a 3db slope.

TNX KD3SH (WN3SOH-WB3BIO)

The loading on the resonator is 1050 ohms in parallel with 5000 ohms, which
is 868 ohms. A coil or capacitor reactance value that is 868 / 5 ohms
(173.6 ohms) produces a Q of 5.

The initial value of L is

L = 173.6 / (2 * pi * 30MHz) = 0.92 uH

An additional loading is due to the resonator itself, which is the product
of the coil Q times coil reactance.Let's assume a coil Q of 200. The
resistance value of the coil itself is

Q * XL = 200*173.6 = 38720 ohms

The resistance loading by the capacitor can usually be neglected if it is
very small.

The loading on the output side by the combination of the 5000 ohm load and
the 38720 is 4428 ohms.

Go back to the beginning and get a slightly revised value of the resistive
loading = 849 ohms that includes the loading by the coil itself.
Recalculate L and C for this slightly different value of loading.

This iterative process can be stopped here.

Connecting the 50 ohm source directly to the coil through a 1000 ohm
resistor produces an additional power loss which is equal to

20 * log[(1000+4428) / 4428)] = 1.8 dB.

This additional loss may not be acceptable in some applications. A low-loss
coupling transformer (e.g. tapped coil) or low-loss tuned network is a
better approach.

Bill W0IYH

"gudmundur" wrote in message
...
The source is 50 ohms, I am feeding through a 1000 ohm resistor to
my tuned circuit at 30mhz, and the load is over 5000 ohms (FET input).

I want to use a parallel resonant circuit to ground to form a pass
filter. If I want a Q of around 5, would I need 5000 ohm XL/XC ??

This would seem nearly correct. Have I calculated correctly?

Well. the 1050 ohm source in parrallel with the 5000 ohm load puts the
resistance seen by the tank at around 867 ohms.

The circuit reactances would need to be RL/Q = 174 ohms.

I think you multiplied instead of divided. The calculations above assume
that the Q of the elements that you use is large enough not to impact the
calculations. You can calculate the equivilent parrallel resistance of the
inductor you select, and add it in parrallel with the 867 ohms to assess the
impact of a finite-Q inductor.

Don

On Sat, 05 Feb 2005 15:42:57 GMT, "William E. Sabin"
wrote:

The loading on the resonator is 1050 ohms in parallel with 5000 ohms, which
is 868 ohms. A coil or capacitor reactance value that is 868 / 5 ohms
(173.6 ohms) produces a Q of 5.

The initial value of L is

L = 173.6 / (2 * pi * 30MHz) = 0.92 uH

An additional loading is due to the resonator itself, which is the product
of the coil Q times coil reactance.Let's assume a coil Q of 200. The
resistance value of the coil itself is

Q * XL = 200*173.6 = 38720 ohms

The resistance loading by the capacitor can usually be neglected if it is
very small.

The loading on the output side by the combination of the 5000 ohm load and
the 38720 is 4428 ohms.

Go back to the beginning and get a slightly revised value of the resistive
loading = 849 ohms that includes the loading by the coil itself.
Recalculate L and C for this slightly different value of loading.

This iterative process can be stopped here.

Connecting the 50 ohm source directly to the coil through a 1000 ohm
resistor produces an additional power loss which is equal to

20 * log[(1000+4428) / 4428)] = 1.8 dB.

This additional loss may not be acceptable in some applications. A low-loss
coupling transformer (e.g. tapped coil) or low-loss tuned network is a
better approach.

Bill W0IYH


interesting to see that you came to the same conclusion alltogether,
but I believe it is another problem to make a coil with Q=200, usually
the Q will be lower for readily available components, so the better
way is to make the coil first, measure it and then start the
calculation, hi

73
Jan-Martin, LA8AK
---
J. M. Noeding, LA8AK, N-4623 Kristiansand
http://home.online.no/~la8ak/c.htm

I have had no trouble with Q=200 coils. I have a Boonton Q meter that
verifies this when the correct core material (especially in a toroid) is
used. At 30 MHz we do have to be somewhat persistent, though.

Best regards,

Bill W0IYH

"J M Noeding" wrote in message
...
On Sat, 05 Feb 2005 15:42:57 GMT, "William E. Sabin"
wrote:

The loading on the resonator is 1050 ohms in parallel with 5000 ohms,
which
is 868 ohms. A coil or capacitor reactance value that is 868 / 5 ohms
(173.6 ohms) produces a Q of 5.

The initial value of L is

L = 173.6 / (2 * pi * 30MHz) = 0.92 uH

An additional loading is due to the resonator itself, which is the product
of the coil Q times coil reactance.Let's assume a coil Q of 200. The
resistance value of the coil itself is

Q * XL = 200*173.6 = 38720 ohms

The resistance loading by the capacitor can usually be neglected if it is
very small.

The loading on the output side by the combination of the 5000 ohm load and
the 38720 is 4428 ohms.

Go back to the beginning and get a slightly revised value of the
resistive
loading = 849 ohms that includes the loading by the coil itself.
Recalculate L and C for this slightly different value of loading.

This iterative process can be stopped here.

Connecting the 50 ohm source directly to the coil through a 1000 ohm
resistor produces an additional power loss which is equal to

20 * log[(1000+4428) / 4428)] = 1.8 dB.

This additional loss may not be acceptable in some applications. A
low-loss
coupling transformer (e.g. tapped coil) or low-loss tuned network is a
better approach.

Bill W0IYH


interesting to see that you came to the same conclusion alltogether,
but I believe it is another problem to make a coil with Q=200, usually
the Q will be lower for readily available components, so the better
way is to make the coil first, measure it and then start the
calculation, hi

73
Jan-Martin, LA8AK
---
J. M. Noeding, LA8AK, N-4623 Kristiansand
http://home.online.no/~la8ak/c.htm

On Sun, 06 Feb 2005 12:46:05 GMT, "William E. Sabin"
wrote:

I have had no trouble with Q=200 coils. I have a Boonton Q meter that
verifies this when the correct core material (especially in a toroid) is
used. At 30 MHz we do have to be somewhat persistent, though.

Best regards,

Bill W0IYH


Bill,
what I meant is that "when you want to make a coil with Q=200 - not
180 or 190" you may have to try some times if you haven't any
experience from which materials gives at least Q=200, a higher Q may
always be reduced to 200 with a resistor
Yes, I have a Radiometer Q-meter - quite old, almost as old as yours,
with thermionic tubes from the 30's and 50's, hi

73,
jan-martin
LA8AK
---
J. M. Noeding, LA8AK, N-4623 Kristiansand
http://home.online.no/~la8ak/c.htm

Yes, I agree. Thanks.

Bill W0IYH

"J M Noeding" wrote in message
...
On Sun, 06 Feb 2005 12:46:05 GMT, "William E. Sabin"
wrote:

I have had no trouble with Q=200 coils. I have a Boonton Q meter that
verifies this when the correct core material (especially in a toroid) is
used. At 30 MHz we do have to be somewhat persistent, though.

Best regards,

Bill W0IYH


Bill,
what I meant is that "when you want to make a coil with Q=200 - not
180 or 190" you may have to try some times if you haven't any
experience from which materials gives at least Q=200, a higher Q may
always be reduced to 200 with a resistor
Yes, I have a Radiometer Q-meter - quite old, almost as old as yours,
with thermionic tubes from the 30's and 50's, hi

73,
jan-martin
LA8AK
---
J. M. Noeding, LA8AK, N-4623 Kristiansand
http://home.online.no/~la8ak/c.htm