question about the Quality factor of an air coil

[samy555]

Hi
If I wound a 100nH air coil to place in a 100MHz oscillator, how can I measure its Q?
Q= XL/R, where R is the ohmic resistance of the coil

thanks

 

[MrAl]

Hi,

Maybe you can use it in a filter circuit and then measure the response, then calculate the Q.
100MHz might be a little difficult though.
The Q wont exactly be XL/R because of other factors at that frequency.

 

[unclejed613]

at 100Mhz the value of R is higher than R at DC. this is due to skin effect.
if you know the wire material (usually copper) you can calculate the skin depth at 100Mhz. if you know the skin depth and the wire diameter, you can figure out the cross-section of the conducting layer and figure out the resistance per length. you wound the coil so you should have a clue what the length is.
a description of skin effect can be found here: https://www.rfcafe.com/references/electrical/skin-depth.htm and they have a calculator for skin depth here: https://www.rfcafe.com/references/calculators/skin-depth-calculator.htm

simply put, skin effect is caused by the electromagnetic field around a wire interacting with the current in the wire, and causes the current to flow mostly on the surface of the wire. the higher the frequency, the thinner the layer.

 

[Diver300]

There is also the proximity effect (https://en.wikipedia.org/wiki/Proximity_effect_(electromagnetism)) where the current flow in coils is mostly on the surfaces of the coil. That increases the resistance more than the skin effect alone.

 

[MrAl]

Hi,

And he said he wants to *measure* it.

 

[Diver300]

The Q is a measure of the damping in the tuned circuit. That almost all comes from the coil, and it is easier to measure.

There are various ways to measure it. You can measure with an oscilloscope how fast the waveform decays. Q about twice the inverse of the decay per cycle, so if the waveform decays to 90% in one cycle, it decays by 10%, and the Q is about 2*1/0.1 = 20

You can also work out the impedance at resonance. That will be about Q * ωL and for your circuit ωL = 2*π*100e6*100e-9 = 63.
So if you have a resonant circuit, and you find that a 1kΩ resistor in parallel causes the amplitude to halve, the impedance of the resonant circuit must be around 1kΩ, so the Q is about 1000/63 = 16

In my experience, you won't get a Q more than 20 at 100 MHz without a lot of effort.

 

[unclejed613]

there are a couple of ways used to improve the Q of coils.using stranded wire helps up to about 100khz. the use of Litz wire is effective up to about 3Mhz. silver plating the wire reduces the resistance where it really matters, on the surface of the wire, and this method is used all the way into the microwave region.

 

[JimB]

Samy555 wrote:

If I wound a 100nH air coil to place in a 100MHz oscillator, how can I measure its Q?

The quick and facitious answer is "it depends on what test equipment you have available"

To make measurements at 100Mhz is not a trivial task.
Probably the best way would be to use a resonance method using a "Q meter", or some improvised method with a signal generator and high impedance voltmeter.

Have a look here for details of a home brewed Q Meter:
http://users.tpg.com.au/users/ldbutler/QMeter.htm

JimB