Owen Duffy wrote in
:

....
Some thoughts about inductor loss and self capacitance:

Consider and ideal coil (ie lossless, no distributed capacitance) in
series with a small ideal resistor to represent its loss, the
combination having high Q. Connect it to a constant voltage source at
some frequency and observe that the current lags the voltage by almost
90 deg.

Now shunt that combination coil+resistor with a small lossless
capacitor, and note that the current in the capacitor will be small in
magnitude, and leading the applied voltage by 90 degrees.

The effect of the capacitor is to reduce the total current, and not
change its phase slightly. So the combination of coil & series
resistance, & shunt capacitance draws less current and at slightly
lower (lagging) phase, so it appears to be a smaller but lossier
inductor.

A workup at 10MHz of some numbers for a 10uH inductance in series with 10
ohms loss resistance gives Z=10+j628, Q is 62.8.

When this is shunted by a 2pf ideal capacitor, the impedance is now 11.8
+j682, Q is 58, apparent inductance is 10.9uH in series with 11.8 ohms of
resistance.

The small shunt capacitor has increased the apparent inductance, and
decreased the Q.

Where has this newfound loss come from? The current in the coil's loss
resistance is higher than the current from the source, so whilst the two
terminal equivalent has a higher impedance, the higher internal current
is generating larger loss from the smaller resistance. This is the
"circulating current" people are talking about.

Owen

Note

In message , Owen Duffy
writes
Owen Duffy wrote in
:

...
Some thoughts about inductor loss and self capacitance:

Consider and ideal coil (ie lossless, no distributed capacitance) in
series with a small ideal resistor to represent its loss, the
combination having high Q. Connect it to a constant voltage source at
some frequency and observe that the current lags the voltage by almost
90 deg.

Now shunt that combination coil+resistor with a small lossless
capacitor, and note that the current in the capacitor will be small in
magnitude, and leading the applied voltage by 90 degrees.

The effect of the capacitor is to reduce the total current, and not
change its phase slightly. So the combination of coil & series
resistance, & shunt capacitance draws less current and at slightly
lower (lagging) phase, so it appears to be a smaller but lossier
inductor.

A workup at 10MHz of some numbers for a 10uH inductance in series with 10
ohms loss resistance gives Z=10+j628, Q is 62.8.

When this is shunted by a 2pf ideal capacitor, the impedance is now 11.8
+j682, Q is 58, apparent inductance is 10.9uH in series with 11.8 ohms of
resistance.

The small shunt capacitor has increased the apparent inductance, and
decreased the Q.

Where has this newfound loss come from? The current in the coil's loss
resistance is higher than the current from the source, so whilst the two
terminal equivalent has a higher impedance, the higher internal current
is generating larger loss from the smaller resistance. This is the
"circulating current" people are talking about.

Owen

Note

Just out of interest, if you increased the inductance to 10.9uH by
increasing the number of turns, what effect would it have on the Q?
Ian.
--

Ian Jackson wrote in
:

....
Just out of interest, if you increased the inductance to 10.9uH by
increasing the number of turns, what effect would it have on the Q?
Ian.

Ian, that depends on the type of coil.

A very simple view (eg if a toroidal core was used) would be that it would
take a (10.9/10)^0.5 increase in turns (4.4%), inductive reactance would
increase by 9% and R would increase by 4.4%, Q would increase by 4.4%.

I don't really understand the relevance of the questions.

Owen