Lock-in demodulation of inductive component

rent

I'll take a stab at it. The voltage of a pure inductor leads the current by 90 deg. To figure out the inductance of an unknown inductor, the idea here is to put a current of a particular frequency, which will generate a voltage across the inductor that is proportional to the inductance, so by measuring that voltage, we can compute the inductance. So, all we would have to do is measure the voltage on the inductor, so why the complexity of the demodulation part?

The problem is no inductor is pure, so there will be some resistance and the phase shift across the inductor will not be exactly 90 deg. How much resistance component there is will vary depending on the inductor. You have a complex impedance. By running the voltage through that demodulator, the output then tracks only the 90 deg in phase component, ie any resistance dependent part of the voltage is gone. So they run 90 deg shifted reference signal and the voltage from the inductor being measured into the demodulator.

The reference signal leads the inductor current by 90 deg. The measured voltage across the inductor leads the inductor current by close to 90 deg, it would be exactly 90 if it there was no resistance and it was a pure inductor. Let's say the actual phase is 88 deg. The demodulator produces a voltage output that is only dependent on the portion of the voltage that is in phase with the 90 deg reference signal, ie the true inductance portion.

That's how I see it. But then I'd buy a meter :)

Do you run LT Spice? It's great for plotting stuff like that.

Given a sine wave current that peaks at some time, the voltage across a resistor will peak simultaneously with the current. The voltage across an inductor will be a sine that peaks 90 degrees sooner. For current I, the voltage across the inductor is 2*pi*F*L.

If the Howland is suitably wideband, it should make a current that is in phase with its input signal.

Yep, that's what it does in the sim. I'm experimenting with running the Howland with two op-amps their outputs moving "push pull" to get more current swing with a low supply voltage a la:

And keep any DC off the transducer L.

If you replace the Zload with a say 10 ohm R and 10u L the voltage at Vl- at the top output leads the current thru the load. The voltage at the bottom approximately follows along.

To extract the L "information" and drop the R I know I need to multiply with something, but not sure what phase relation sine it should be in that kind of arrangement.

The analog multiplier is too rich for my blood I was hoping I could square up the leading and quadrature signals and XOR as a substitute.

The op amps I'd like to use are the OPA322 1.8 volt CMOS type, LTSpice doesn't have a model for those of course but seem wideband enough for

Yep, that's making sense.

There are simpler current sources.

Looks to me like with that one the voltage signal at the bottom of the LR-series load is always 90 degrees leading the current, while the signal at the top leads a little less proportional to the ESR, so with a diff amp/precision rectifier you give you could dump the DC offset (single supply), add gain, and sample a voltage proportional to just the inductance

An inductance meter with just one quad op amp seems pretty parsimonious to me

In that L-meter schematic, the voltage at the bottom of Lx is zero.

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I think he means the voltage at the bottom where the reference signal is, where it says "zero" in the schematic.

Sorry, I think I forgot to post the schematic of the type of current source I'm referring to, the load floats:

Correction, with this current source above I'd have to put current sense resistors on both ends of the L and sense differentially across it to sense the inductor voltage amplitude, with it leading the current 90 degrees.

IDK why you'd need them on both ends. Whatever current goes in one end of the inductor comes out the other. You have the voltage across it and the current through it.

I have two big constraints on the project I'm working on, low supply voltage (couple volts) and the DUT will be located some distance away (couple feet) from the detector electronics so stray lead inductance and resistance becomes a problem.

So I was thinking I could kill two birds with one stone here and drive the inductor current "push-pull" to get more swing, and then do a quasi-Kelvin sensing arrangement where the L voltage is sensed deferentially right at its terminals and the voltage signal run back, to exclude the parasitics, a la:

green trace is output of op amp U2, the AD8515 isn't really fast enough for square waves at 100kHz but there are faster low voltage RRIOs available. The inputs should probably be capacitively coupled. Then that and the quadrature signal goes into an XOR phase detector.

With appropriate choice of resistors the dual-Howland's current output into a floating load is independent of the load impedance.

If I'm not wrong what you want to have when it's all done, with square wave drive and an XOR detector is a pulse train whose time average amplitude is proportional to only the inductive portion of the load like: