Lock-in demodulation of inductive component

B

bitrex

Guest
With respect to this quadrature demodulation inductance measurement
circuit we've discussed before, could you please remind me of the
phasing relationships? I'm having difficulty visualizing how the
current/voltage relationships look.

So you force a sinusoidal current thru the L and ESR, the voltage leads
the current, and then multiply with a sine wave that is...the text says:

<https://www.qsl.net/va3iul/L_meter/L_meter.htm>

"The amplitude of the quadrature component of the voltage across Lx may
be recovered by synchronous demodulation, with the demodulator's
reference signal provided by a +90Âş phase-shifted version of the current
source signal." That is a little unclear to me do they mean shifted with
respect to the voltage driving the Howland current source, or the
current thru the L and R itself?

A diagram of what the phase relationships between inductor voltage,
inductor current, +90 degree quadrature demodulation signal, and
in-phase signal should look like with respect to each other would be
helpful if anyone has it
 
On Saturday, September 14, 2019 at 1:31:41 PM UTC-4, bitrex wrote:
With respect to this quadrature demodulation inductance measurement
circuit we've discussed before, could you please remind me of the
phasing relationships? I'm having difficulty visualizing how the
current/voltage relationships look.

So you force a sinusoidal current thru the L and ESR, the voltage leads
the current, and then multiply with a sine wave that is...the text says:

https://www.qsl.net/va3iul/L_meter/L_meter.htm

"The amplitude of the quadrature component of the voltage across Lx may
be recovered by synchronous demodulation, with the demodulator's
reference signal provided by a +90Âş phase-shifted version of the current
source signal." That is a little unclear to me do they mean shifted with
respect to the voltage driving the Howland current source, or the
current thru the L and R itself?

A diagram of what the phase relationships between inductor voltage,
inductor current, +90 degree quadrature demodulation signal, and
in-phase signal should look like with respect to each other would be
helpful if anyone has it
Click to expand...

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 :)
 
On Sat, 14 Sep 2019 13:31:36 -0400, bitrex <user@example.net> wrote:

With respect to this quadrature demodulation inductance measurement
circuit we've discussed before, could you please remind me of the
phasing relationships? I'm having difficulty visualizing how the
current/voltage relationships look.

So you force a sinusoidal current thru the L and ESR, the voltage leads
the current, and then multiply with a sine wave that is...the text says:

https://www.qsl.net/va3iul/L_meter/L_meter.htm

"The amplitude of the quadrature component of the voltage across Lx may
be recovered by synchronous demodulation, with the demodulator's
reference signal provided by a +90ş phase-shifted version of the current
source signal." That is a little unclear to me do they mean shifted with
respect to the voltage driving the Howland current source, or the
current thru the L and R itself?

A diagram of what the phase relationships between inductor voltage,
inductor current, +90 degree quadrature demodulation signal, and
in-phase signal should look like with respect to each other would be
helpful if anyone has it
Click to expand...

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.
 
On 9/14/19 4:21 PM, jlarkin@highlandsniptechnology.com wrote:
On Sat, 14 Sep 2019 13:31:36 -0400, bitrex <user@example.net> wrote:

With respect to this quadrature demodulation inductance measurement
circuit we've discussed before, could you please remind me of the
phasing relationships? I'm having difficulty visualizing how the
current/voltage relationships look.

So you force a sinusoidal current thru the L and ESR, the voltage leads
the current, and then multiply with a sine wave that is...the text says:

https://www.qsl.net/va3iul/L_meter/L_meter.htm

"The amplitude of the quadrature component of the voltage across Lx may
be recovered by synchronous demodulation, with the demodulator's
reference signal provided by a +90Âş phase-shifted version of the current
source signal." That is a little unclear to me do they mean shifted with
respect to the voltage driving the Howland current source, or the
current thru the L and R itself?

A diagram of what the phase relationships between inductor voltage,
inductor current, +90 degree quadrature demodulation signal, and
in-phase signal should look like with respect to each other would be
helpful if anyone has it

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.
Click to expand...

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:

<https://www.researchgate.net/figure/Mirrored-Enhanced-Howland-Current-Source-with-differential-amplifiers-MEHCS-DIF_fig6_329867525>

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.
 
On 9/14/19 5:32 PM, bitrex wrote:
On 9/14/19 4:21 PM, jlarkin@highlandsniptechnology.com wrote:
On Sat, 14 Sep 2019 13:31:36 -0400, bitrex <user@example.net> wrote:

With respect to this quadrature demodulation inductance measurement
circuit we've discussed before, could you please remind me of the
phasing relationships? I'm having difficulty visualizing how the
current/voltage relationships look.

So you force a sinusoidal current thru the L and ESR, the voltage leads
the current, and then multiply with a sine wave that is...the text says:

https://www.qsl.net/va3iul/L_meter/L_meter.htm

"The amplitude of the quadrature component of the voltage across Lx may
be recovered by synchronous demodulation, with the demodulator's
reference signal provided by a +90Âş phase-shifted version of the current
source signal." That is a little unclear to me do they mean shifted with
respect to the voltage driving the Howland current source, or the
current thru the L and R itself?

A diagram of what the phase relationships between inductor voltage,
inductor current, +90 degree quadrature demodulation signal, and
in-phase signal should look like with respect to each other would be
helpful if anyone has it

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.
Click to expand...

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
100kHz at those voltages.

<http://www.ti.com/product/OPA322>
 
On 9/14/19 3:54 PM, Whoey Louie wrote:
On Saturday, September 14, 2019 at 1:31:41 PM UTC-4, bitrex wrote:
With respect to this quadrature demodulation inductance measurement
circuit we've discussed before, could you please remind me of the
phasing relationships? I'm having difficulty visualizing how the
current/voltage relationships look.

So you force a sinusoidal current thru the L and ESR, the voltage leads
the current, and then multiply with a sine wave that is...the text says:

https://www.qsl.net/va3iul/L_meter/L_meter.htm

"The amplitude of the quadrature component of the voltage across Lx may
be recovered by synchronous demodulation, with the demodulator's
reference signal provided by a +90Âş phase-shifted version of the current
source signal." That is a little unclear to me do they mean shifted with
respect to the voltage driving the Howland current source, or the
current thru the L and R itself?

A diagram of what the phase relationships between inductor voltage,
inductor current, +90 degree quadrature demodulation signal, and
in-phase signal should look like with respect to each other would be
helpful if anyone has it

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 :)
Click to expand...

Yep, that's making sense.
 
On Sat, 14 Sep 2019 17:34:59 -0400, bitrex <user@example.net> wrote:

On 9/14/19 5:32 PM, bitrex wrote:
On 9/14/19 4:21 PM, jlarkin@highlandsniptechnology.com wrote:
On Sat, 14 Sep 2019 13:31:36 -0400, bitrex <user@example.net> wrote:

With respect to this quadrature demodulation inductance measurement
circuit we've discussed before, could you please remind me of the
phasing relationships? I'm having difficulty visualizing how the
current/voltage relationships look.

So you force a sinusoidal current thru the L and ESR, the voltage leads
the current, and then multiply with a sine wave that is...the text says:

https://www.qsl.net/va3iul/L_meter/L_meter.htm

"The amplitude of the quadrature component of the voltage across Lx may
be recovered by synchronous demodulation, with the demodulator's
reference signal provided by a +90ş phase-shifted version of the current
source signal." That is a little unclear to me do they mean shifted with
respect to the voltage driving the Howland current source, or the
current thru the L and R itself?

A diagram of what the phase relationships between inductor voltage,
inductor current, +90 degree quadrature demodulation signal, and
in-phase signal should look like with respect to each other would be
helpful if anyone has it

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.

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
100kHz at those voltages.

http://www.ti.com/product/OPA322
Click to expand...

There are simpler current sources.
 
On Sat, 14 Sep 2019 18:55:40 -0400, bitrex <user@example.net> wrote:

On 9/14/19 6:41 PM, jlarkin@highlandsniptechnology.com wrote:
On Sat, 14 Sep 2019 17:34:59 -0400, bitrex <user@example.net> wrote:

On 9/14/19 5:32 PM, bitrex wrote:
On 9/14/19 4:21 PM, jlarkin@highlandsniptechnology.com wrote:
On Sat, 14 Sep 2019 13:31:36 -0400, bitrex <user@example.net> wrote:

With respect to this quadrature demodulation inductance measurement
circuit we've discussed before, could you please remind me of the
phasing relationships? I'm having difficulty visualizing how the
current/voltage relationships look.

So you force a sinusoidal current thru the L and ESR, the voltage leads
the current, and then multiply with a sine wave that is...the text says:

https://www.qsl.net/va3iul/L_meter/L_meter.htm

"The amplitude of the quadrature component of the voltage across Lx may
be recovered by synchronous demodulation, with the demodulator's
reference signal provided by a +90ş phase-shifted version of the current
source signal." That is a little unclear to me do they mean shifted with
respect to the voltage driving the Howland current source, or the
current thru the L and R itself?

A diagram of what the phase relationships between inductor voltage,
inductor current, +90 degree quadrature demodulation signal, and
in-phase signal should look like with respect to each other would be
helpful if anyone has it

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.

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
100kHz at those voltages.

http://www.ti.com/product/OPA322

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,
Click to expand...

In that L-meter schematic, the voltage at the bottom of Lx is zero.
 
On Saturday, September 14, 2019 at 7:13:14 PM UTC-4, jla...@highlandsniptechnology.com wrote:
On Sat, 14 Sep 2019 18:55:40 -0400, bitrex <user@example.net> wrote:

On 9/14/19 6:41 PM, jlarkin@highlandsniptechnology.com wrote:
On Sat, 14 Sep 2019 17:34:59 -0400, bitrex <user@example.net> wrote:

On 9/14/19 5:32 PM, bitrex wrote:
On 9/14/19 4:21 PM, jlarkin@highlandsniptechnology.com wrote:
On Sat, 14 Sep 2019 13:31:36 -0400, bitrex <user@example.net> wrote:

With respect to this quadrature demodulation inductance measurement
circuit we've discussed before, could you please remind me of the
phasing relationships? I'm having difficulty visualizing how the
current/voltage relationships look.

So you force a sinusoidal current thru the L and ESR, the voltage leads
the current, and then multiply with a sine wave that is...the text says:

https://www.qsl.net/va3iul/L_meter/L_meter.htm

"The amplitude of the quadrature component of the voltage across Lx may
be recovered by synchronous demodulation, with the demodulator's
reference signal provided by a +90Âş phase-shifted version of the current
source signal." That is a little unclear to me do they mean shifted with
respect to the voltage driving the Howland current source, or the
current thru the L and R itself?

A diagram of what the phase relationships between inductor voltage,
inductor current, +90 degree quadrature demodulation signal, and
in-phase signal should look like with respect to each other would be
helpful if anyone has it

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.

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
100kHz at those voltages.

http://www.ti.com/product/OPA322

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,

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

I think he means the voltage at the bottom where the reference signal is,
where it says "zero" in the schematic.
 
On 9/14/19 6:41 PM, jlarkin@highlandsniptechnology.com wrote:
On Sat, 14 Sep 2019 17:34:59 -0400, bitrex <user@example.net> wrote:

On 9/14/19 5:32 PM, bitrex wrote:
On 9/14/19 4:21 PM, jlarkin@highlandsniptechnology.com wrote:
On Sat, 14 Sep 2019 13:31:36 -0400, bitrex <user@example.net> wrote:

With respect to this quadrature demodulation inductance measurement
circuit we've discussed before, could you please remind me of the
phasing relationships? I'm having difficulty visualizing how the
current/voltage relationships look.

So you force a sinusoidal current thru the L and ESR, the voltage leads
the current, and then multiply with a sine wave that is...the text says:

https://www.qsl.net/va3iul/L_meter/L_meter.htm

"The amplitude of the quadrature component of the voltage across Lx may
be recovered by synchronous demodulation, with the demodulator's
reference signal provided by a +90Âş phase-shifted version of the current
source signal." That is a little unclear to me do they mean shifted with
respect to the voltage driving the Howland current source, or the
current thru the L and R itself?

A diagram of what the phase relationships between inductor voltage,
inductor current, +90 degree quadrature demodulation signal, and
in-phase signal should look like with respect to each other would be
helpful if anyone has it

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.

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
100kHz at those voltages.

http://www.ti.com/product/OPA322

There are simpler current sources.
Click to expand...

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
 
On 9/14/19 7:32 PM, Whoey Louie wrote:
On Saturday, September 14, 2019 at 7:13:14 PM UTC-4, jla...@highlandsniptechnology.com wrote:
On Sat, 14 Sep 2019 18:55:40 -0400, bitrex <user@example.net> wrote:

On 9/14/19 6:41 PM, jlarkin@highlandsniptechnology.com wrote:
On Sat, 14 Sep 2019 17:34:59 -0400, bitrex <user@example.net> wrote:

On 9/14/19 5:32 PM, bitrex wrote:
On 9/14/19 4:21 PM, jlarkin@highlandsniptechnology.com wrote:
On Sat, 14 Sep 2019 13:31:36 -0400, bitrex <user@example.net> wrote:

With respect to this quadrature demodulation inductance measurement
circuit we've discussed before, could you please remind me of the
phasing relationships? I'm having difficulty visualizing how the
current/voltage relationships look.

So you force a sinusoidal current thru the L and ESR, the voltage leads
the current, and then multiply with a sine wave that is...the text says:

https://www.qsl.net/va3iul/L_meter/L_meter.htm

"The amplitude of the quadrature component of the voltage across Lx may
be recovered by synchronous demodulation, with the demodulator's
reference signal provided by a +90Âş phase-shifted version of the current
source signal." That is a little unclear to me do they mean shifted with
respect to the voltage driving the Howland current source, or the
current thru the L and R itself?

A diagram of what the phase relationships between inductor voltage,
inductor current, +90 degree quadrature demodulation signal, and
in-phase signal should look like with respect to each other would be
helpful if anyone has it

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.

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
100kHz at those voltages.

http://www.ti.com/product/OPA322

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,

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

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

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

<https://content.sciendo.com/view/journals/joeb/9/1/graphic/j_joeb-2018-0011_fig_005.jpg>
 
On 9/14/19 10:06 PM, bitrex wrote:

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,

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

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:

https://content.sciendo.com/view/journals/joeb/9/1/graphic/j_joeb-2018-0011_fig_005.jpg
Click to expand...

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.
 
On Sunday, September 15, 2019 at 2:41:46 AM UTC-4, bitrex wrote:
On 9/14/19 10:06 PM, bitrex wrote:

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,

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

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:

https://content.sciendo.com/view/journals/joeb/9/1/graphic/j_joeb-2018-0011_fig_005.jpg


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.
Click to expand...

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.
 
On 9/15/19 10:18 AM, Whoey Louie wrote:
On Sunday, September 15, 2019 at 2:41:46 AM UTC-4, bitrex wrote:
On 9/14/19 10:06 PM, bitrex wrote:

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,

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

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:

https://content.sciendo.com/view/journals/joeb/9/1/graphic/j_joeb-2018-0011_fig_005.jpg


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.
Click to expand...

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:

<https://imgur.com/a/uv6GhYR>
 
On 9/15/19 10:18 AM, Whoey Louie wrote:
On Sunday, September 15, 2019 at 2:41:46 AM UTC-4, bitrex wrote:
On 9/14/19 10:06 PM, bitrex wrote:

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,

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

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:

https://content.sciendo.com/view/journals/joeb/9/1/graphic/j_joeb-2018-0011_fig_005.jpg


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.
Click to expand...

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:

<https://imgur.com/a/mPulo3z>

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.
 
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