T
Tim Williams
- Jan 1, 1970
- 0
Ok, and sorry you ran with my mistake, I calculated Q with 100Mhz
instead of the correct 50Mhz.
So 33/ 80 = 0.4125. Then 0.4125 x 1.4 = 0.5775 ohms.
Sure would be nice to have equipment to measure such things.
Mikek
I think you are already in trouble..
#16 at 100 Mhz?
I need to dig up the skin formula.
Jamie
Ya, I've done the poor man's trick many times, but at lowerThe equipment at Semco was a bridge type of meter that could do C or L.
You first zero everything, including the peak pot..
Put the DUT on the jig, and then turn the main dial back the res
again, and the turn the main POT to bring the peak back to the null
point.
The main pot was just that, a potentiometer, that would give you the
Q of the coil, because a correction had be made from the original zero
point. As for the frequency offset, that would thus report L or C, what
ever you were measuring.
I've done the poor mans trick. You need the signal generator and a high
Z RF millvolt meter. In your case, a 24Pf cap with the 105nh coil. Sweep
the generator and do the math at the skirts of the bw window.
I think you are already in trouble..
#16 at 100 Mhz?
I need to dig up the skin formula.
Jamie
So 33/ 80 = 0.4125. Then 0.4125 x 1.4 = 0.5775 ohms.
Sure would be nice to have equipment to measure such things.
Mikek
Mikek
Q often will start to decrease in a 240uh coil near 1.5Mhz, and it is not near SRF.
Mirek,
FYI - SI Units based on a person's name are always capitalized.
So, MHz (not Mhz), uH (not uh), etc...
I think you are already in trouble..
#16 at 100 Mhz?
I need to dig up the skin formula.
Jamie
Easy to remember just remember you need the number 2, and three terms,
permeability, conductivity, and frequency
In the MKS system skin depth is the square root of the whole quantity of 2
over the quantity of permeability times conductivity times frequency
sd = sqrt(2/(p*c*w), easier to visualize this equation then remember it.
p, permeability is in absolute value, like free space is 4 pi 10-7
c, conductivity is in S/m, copper is 58 MS/m, or 58e6 [58 x 10-6]
w, is 'frequency' but in radians per second, or 2*pi*f [f is frequency in
Hz]
sd, skindepth in meters
amdx said:Not trouble, maybe wasted copper, the Q is better the bigger the wire,
because the larger circumference provides more copper even if most of
the current is in the outer 0.001" in depth.
My opinion, until I get better info.
If the aspect ratio and everything is good. A coil too tight (pitch near
wire diameter) has lower Q because the field can't circulate around each
wire, or equivalently, the wires induce eddy currents in each other,
causing more loss than the tight winding increases inductance. Typically
pitch is double the wire diameter for best results. And solenoid diameter
is some times the wire diameter, for similar reasons.
Tim
Q often will start to decrease in a 240uh coil near 1.5Mhz, and it is not near SRF.
Mirek,
FYI - SI Units based on a person's name are always capitalized.
So, MHz (not Mhz), uH (not uh), etc...
Don't mean to derail your thread.
At least you're consistent.![]()
Yes, but due to all the conversion terms, it's very difficult to remember!Yes, that is the one but something tells me mine was in feet/inches
Jamie
I think you are already in trouble..
The 'silly' chart on my wall says skin depth of copper at 1 MHz is a bit less than 100 um. 100 MHz will be a bit less than 10um.#16 at 100 Mhz?
I need to dig up the skin formula
The 'silly' chart on my wall says skin depth of copper at 1 MHz is a bit
less than 100 um. 100 MHz will be a bit less than 10um.
There must be a table on the web
George H.
Interesting chart, mine says 66 um, and 6.6um for excellent copper [58MS/m
conductivity]
and 80um at 1MHz for PCB copper-like stuff. [40MS/m conductivity]
Doesn't matter a lot, because skin depth is based upon a planar wave
hitting infintiely large planar material perpendicularly to its surface.
Don't get a lot of that around here.![]()
That seems about right... for me a factor of two can be "a bit".
(I like to let 2*pi = 10)
100um looks to be closer to 300 kHz, (it's not a very good chart.)
so ~sqrt(3) off...
Ok I reread the thread, here are 4 answers I got.
1) If L and R would remain constant, the reactance and Q would double.
They probably won't stay exactly the same.
2) air core copper inductors (solenoids, etc.) go as Q ~
sqrt(F).
3) Double the frequency and you double the inductive reactance, while
keeping the resistance constant. So, the Q @ 100 MHz should be 160.
4) usually sqrt(f/f0) times the losses.
So instead of a non answer, how about being helpful.
If you knew the answer you could have copied and pasted it for me.
I think the real answer is, it is not an easy calculation, it gets into
the changes in skin effect, proximity effect, interwinding capacitance
and eddy currents when going from 50Mhz to 100Mhz.
Mikek
btw, What is your answer?
Grin,10?! now THAT is rounding off.
Oh thanks, but that's not necessary. I don't use skin depth very often."...not a very good chart."! is accurate, more like 440kHz to yield 100um
If you have octave [free MatLab clone] I can send you the function given
frequency, [conductivity, and [permeability]] brackets mean optional
entries, in *.m text script form. Or, the function that returns frequency
given skin depth, etc.
It's set up to accept an array of frequencies too.