Sensing small inductances

I doubt you go back to your monitor when you build, you use your paper schematic. If you are going to post schematics and we want you to, why not make them readable by all those people that don't know how to adjust their monitors. It's really not hard. Is it just that you have had so many people tell you about the quality of your schematics that you're not going to let them tell you what to do? What, your 13 yrs old.

You're deflecting, your schematics are hard to read

Now help me on my inductor ferrite choice! :-) With love, Mikek

Sounds like you could build an oscillator and measure frequencies.

Only slightly related, here's my latest oscillator design:

This (and some FPGA logic) replaces an obsolete Maxim tapped delay line. It only needs to oscillate for five cycles.

No, more like my sarcasm. It's not a thing I would say, but it is a thing oppressed right-wing minorities like former Google employee James Damore would write a whole paper on

So why did you say it?

You meant it.

the client was hoping to measure small changes in inductance directly from frequency counting but the raw differences at the lower frequencies aren't large enough for his hardware to detect well. the tank needs to be high Q to oscillate at all, reliably, but a high Q tank oscillation frequency is insensitive to small changes in the inductance, unless the inductance is very small with respect to the capacitance. but too large a cap with respect to the inductance wrecks the Q, also. it's an irritating prison of constraints that are annoying to try to work around, to force it that way.

I think changing the whole resonant frequency by doing capacitor swap and a "differential" measurement doing the math in software as Jan suggested will likely work.

I'll gladly tell my girl friend a right wing engineering professional on the internet said I was a misogynist, I expect she'll probably roll her eyes and laugh in the way she usually does when from time to time I show her some of the right-wing rants that pop up here lol. Hold up I'm gonna show her the whole thread right now. brb.

A high-Q LC oscillator follows the resonance equation even closer than a low-Q one. w = 1/root(LC)

w = 1/root(LC)

With infinite Q.

For ordinary inductors, the ratio of the actual resonant frequency to the thoretical resonant frequency is root(1 - 1/(4Q^2)). In practise, a Q of 10 gives an actual frequency of 99.87% of theoretical.

See Radiotron_Designers_Handbook_1954.pdf 90.6MB

See EQ 3 on page 449. This explains the variation in resonant frequency with Q. I have not been able to find this information anywhere else.

Right what I mean is that you can "nudge" the frequency of a low-Q oscillating tank made with an inductor that has intrinsically low Q in that frequency area, around easier to wider deviation by say physically stretching or compressing an air-core coil, assuming there's enough overall amplifier gain in the bandwidth of interest to keep it spinning, because the loop gain skirt is relatively broad. this also means that its long term frequency stability and short-term phase noise sucks.

Or you can use negative resistance or something to boost the lossy inductor's intrinsic Q up, and thusly the Q of the tank. however since Q is a function of frequency a fixed negative R only really works good in the area of a single resonant frequency. A high-Q tank's gain skirt plunges very quickly off-resonance, the ESR-lossy rapidly dominates again and it drops out of oscillation.

a more elegant solution on paper is to use a negative capacitance instead of a negative resistance to boost the Q of a lossy tank. but all real-world negative impedance circuits love to mis-behave, all of these negative impedance circuits are academic exercises and useless for real work as-drawn:

It's all a matter of perspective. If you work out the impedance of an oscillator circuit from the viewpoint of the resonator, you'll get a negative resistance.

I don't see the point of negative capacitance. Increasing the Q implies reducing or compensating losses. A reactive component doesn't do that.

Jeroen Belleman

A problem with measuring inductance by making the DUT part of an oscillator tank and measuring frequency is the square root relationship works against you by compressing sensitivity.

What you could try is adapting the Boonton 72 capacitance meter topology to measure inductance. A fixed frequency low current source feeds the DUT into a calibrated resonant LC in the test instrument, amplify and measure with quadrature synchronous detector. RF techniques rather than time domain!

The Boonton 72 can resolve tiny capacitance changes so I expect a C to L transformed version might also be capable of resolving tiny inductance changes?

piglet

only if it's already oscillating - if the coil is so low Q that it won't even start then there's no negative-nothing, nowhere!

a negative capacitance has to be powered to operate; the charge goes down but the voltage (and thus 1/2CV^2 energy) goes up, that requires it to get some energy from somewhere the system didn't have before. No such thing as a passive negative capacitance that behaves just like a positive passive capacitance with its sign flipped and still conserves the total energy of the system that I know of. so long as it's not all returned to the source in a purely reactive system that excess is then available to do work.

in an electronic negative capacitance circuit the physical capacitor in the feedback loop is of course always acting like a regular capacitance so the idea of "charge" is kind of a metaphor I think but the whole system is behaving like a negative one.

For small-signal analysis I guess it's OK to treat them that way, small signal analysis assumes an infinitesimal signal and so the energy required is an infinitesimal too.

But, any LC oscillator only has 0.5% frequency change for a 1% inductance change. It's a weaker dependence than, say, an L-R oscillator.

An oscillator, LR circuit, and thermal measure of the resistor temperature rise would be pretty much foolproof, and if the inductive conductance is high, the back-emf will have square-law effect on the resistor heating.

Sigh.

A negative capacitance indeed has to be a powered active circuit, however, to provide nett energy, it is necessary to have a negative real component in the impedance. A purely reactive impedance, negative or not, does not provide or absorb nett work. That's the

OK.

Jeroen Belleman

The lousy contrast wouldn't be so bad if the illumination was a bit more uniform, but the combination of uneven lighting, crumpled paper and dark top right corner defeats most histogram equalisation. The JPEG artifacts and low light colour noise do it no favours at all.

It would also be around 200k as a greyscale PNG if correctly exposed.

Anyone who wants to read the circuit diagram should try convert to greyscale, histogram equalisation to make the (presumed) white paper white followed by unsharp masking to bring out the faint pencil lines.

Expect to lose some detail in the dark areas...

On Aug 25, 2019, Steve Wilson wrote (in article):

I have seen it elsewhere, in old textbooks. It may be in Terman.

Joe Gwinn

There are some naturally-occurring structures e.g. in ferroelectric crystals that also exhibit negative capacitance they require energy from somewhere, too

who cares what the academic definition of a "purely reactive" negative capacitance is, they don't exist, the electronic ones can compensate tank circuit losses just fine just like a negative resistance but they don't have DC gain which can be a nice feature to have