Reduce the L to reduce the resistive loss - the essence of L
is the energy stored in its current carrying, and it is the current that
causes I^2 R losses. The energy stored in the C is static. (Yes, there
are some losses in polarising the dielectrics but these are small enough
to be ignored)

"Paul Burridge" wrote in message
...
ISTR that one can improve Q in resonant tanks by having a low L-C
ratio. Or was it high L-C ratio. I can't remember but need to know.

Reduce the L to reduce the resistive loss - the essence of L
is the energy stored in its current carrying, and it is the current that
causes I^2 R losses. The energy stored in the C is static. (Yes, there
are some losses in polarising the dielectrics but these are small enough
to be ignored)

"Paul Burridge" wrote in message
...
ISTR that one can improve Q in resonant tanks by having a low L-C
ratio. Or was it high L-C ratio. I can't remember but need to know.

Hi guys,

ISTR that one can improve Q in resonant tanks by having a low L-C
ratio. Or was it high L-C ratio. I can't remember but need to know.
Can any kind soul help me out? Thanks.

p.
--

The BBC: Licensed at public expense to spread lies.

Airy,
What you said would be relevant only if you were trying to determine circuit
losses due to "unloaded" Q of the components. I believe Paul is trying to
determine the 'loaded" Q in order to obtain best selectivity (narrowest
bandwidth). Is this true Paul?

In order to obtain maximum selectivity, the loaded Q needs to be as high as
possible. In the case of a resonant 'tank', the tank reactances are loaded
by the external environment. The circuit Q (or 'loaded' Q) in this case is
Q=R/X. In order to maximize loaded Q, the X term (reactance) needs to be
minimized. This means low L and high C.

In any case, the actual circuit losses will be a function of the ratio of
unloaded Q (Q of the components) to loaded Q. The higher the unloaded Q of
the components, the lower the losses in the circuit.

Joe
W3JDR


"Airy R. Bean" wrote in message
...
Reduce the L to reduce the resistive loss - the essence of L
is the energy stored in its current carrying, and it is the current that
causes I^2 R losses. The energy stored in the C is static. (Yes, there
are some losses in polarising the dielectrics but these are small enough
to be ignored)

"Paul Burridge" wrote in message
...
ISTR that one can improve Q in resonant tanks by having a low L-C
ratio. Or was it high L-C ratio. I can't remember but need to know.

Airy,
What you said would be relevant only if you were trying to determine circuit
losses due to "unloaded" Q of the components. I believe Paul is trying to
determine the 'loaded" Q in order to obtain best selectivity (narrowest
bandwidth). Is this true Paul?

In order to obtain maximum selectivity, the loaded Q needs to be as high as
possible. In the case of a resonant 'tank', the tank reactances are loaded
by the external environment. The circuit Q (or 'loaded' Q) in this case is
Q=R/X. In order to maximize loaded Q, the X term (reactance) needs to be
minimized. This means low L and high C.

In any case, the actual circuit losses will be a function of the ratio of
unloaded Q (Q of the components) to loaded Q. The higher the unloaded Q of
the components, the lower the losses in the circuit.

Joe
W3JDR


"Airy R. Bean" wrote in message
...
Reduce the L to reduce the resistive loss - the essence of L
is the energy stored in its current carrying, and it is the current that
causes I^2 R losses. The energy stored in the C is static. (Yes, there
are some losses in polarising the dielectrics but these are small enough
to be ignored)

"Paul Burridge" wrote in message
...
ISTR that one can improve Q in resonant tanks by having a low L-C
ratio. Or was it high L-C ratio. I can't remember but need to know.

On Sun, 21 Mar 2004 16:02:18 GMT, "W3JDR" wrote:

Airy,
What you said would be relevant only if you were trying to determine circuit
losses due to "unloaded" Q of the components. I believe Paul is trying to
determine the 'loaded" Q in order to obtain best selectivity (narrowest
bandwidth). Is this true Paul?

Some clarification is necessary!
The application is the tank in a frequency multiplier.
I am seeking to select for the 5th harmonic. Therefore, the tank needs
to have as little loss as possible given the fact that the 5th will be
way down dB-wise on the fundamental. I can't afford to attenuate it
too much as it's already weak to begin with. Ergo, I need the lowest
loss components and the best selectivity for the desired 5th harmonic.
Thanks,

p.
--

The BBC: Licensed at public expense to spread lies.

On Sun, 21 Mar 2004 16:02:18 GMT, "W3JDR" wrote:

Airy,
What you said would be relevant only if you were trying to determine circuit
losses due to "unloaded" Q of the components. I believe Paul is trying to
determine the 'loaded" Q in order to obtain best selectivity (narrowest
bandwidth). Is this true Paul?

Some clarification is necessary!
The application is the tank in a frequency multiplier.
I am seeking to select for the 5th harmonic. Therefore, the tank needs
to have as little loss as possible given the fact that the 5th will be
way down dB-wise on the fundamental. I can't afford to attenuate it
too much as it's already weak to begin with. Ergo, I need the lowest
loss components and the best selectivity for the desired 5th harmonic.
Thanks,

p.
--

The BBC: Licensed at public expense to spread lies.

High L/C ratio increases Q.

High L/C ratio increases Q.

Oops - L/R ISTR!

"K9SQG" wrote in message
...
High L/C ratio increases Q.