More Math for the "SETUP"

[amdx]

The “setup” is a board that I'm testing to measure three items:
Input voltage,
Voltage across a 47.5 ohm sense resistor,
Phase difference between these two voltages.

All measurements are on a scope at 3.85MHz. With 10 Vpp signal.

The 47.5 ohm resistor is in series with anything I measure.
It must be subtracted when doing the calculation, this is
very accurate when I measure a 50 BNC termination resistor.
I measured it at 97.9 ohms - 47.5ohms = 50.4 ohms, zero phase. 3.85MHz
When I measured a 3,090 ohm resistor I get 3,119 ohms after subtracting
47.5 ohms for the sense resistor.
That's a 1% error, I'm happy with that, especially sense this
is reading the scope traces.

Now I get to measure reactances.

I have a 55uh inductor, XL = 1329.8 ohms @ 3.85MHz.
Q is 250 so R = 5.3 ohms.
As measured on a 50 year old Boonton 260A.
This is good enough to get me the math I need.

When I read the scope and do the first calculation Z=E/I,
I get 1283.8 ohms this includes the 47.5 ohm sense resistor.
The current and voltage have 62.5 ns phase difference.

The phase difference:
T of 3.85MHz is 259.7 ns and 62.5ns, is 24% of 259.7ns.
Then, 24% of 360* equals a 86.4* phase angle.

I have this:
R = 1283.8 x COS(86.4*) so, R = 80.6 ohms
X 1283.8 x SIN (86.4), making X = 1281.3 ohms.

I want to know the formula that I go through to get the REAL R and Z.
The 47.5 ohm sense resistors effect on R and phase need to be
taken out in the calculation.
I don't know at what point I subtract it out, but I do know it's not
at the end.

I gave you the understanding that I have, please correct me where I made
mistakes.
Pretend, no, assume I'm in 7th grade math. (Algebraically I am).

So where do I start?

Many thanks, Mikek

 

[amdx]

On 2/2/2014 9:18 AM, amdx wrote:

Just wanted to add, I have my two new scope probes, they both read
athe same now. ( I had problem with one of my probes before)
The new probes have smaller tips, I had to build a new board to fit
the probes, but being smaller tightened thing up more, so that was good.

So where do I start?

Many thanks, Mikek

 

[RobertMacy]

On Sun, 02 Feb 2014 08:18:04 -0700, amdx <nojunk@knology.net> wrote:

....snip....just to keep Aioe happy

I have this:
R = 1283.8 x COS(86.4*) so, R = 80.6 ohms
X 1283.8 x SIN (86.4), making X = 1281.3 ohms.

I want to know the formula that I go through to get the REAL R and Z.
The 47.5 ohm sense resistors effect on R and phase need to be
taken out in the calculation.
I don't know at what point I subtract it out, but I do know it's not at
the end.

I gave you the understanding that I have, please correct me where I made
mistakes.
Pretend, no, assume I'm in 7th grade math. (Algebraically I am).

So where do I start?

Many thanks, Mikek

Read a scope trace?! Aren't scopes something like 3% devices!

You should change your nomenclature, it's confusing. Let Z equal the WHOLE
impedance you see with Re being the equivalent series resistance and XL
being the equivalent series reactance of your inductor, thus Z = Re + 1i*XL


the 'formula' you're looking for is

V2=V1*Z/(Z+R) where Z is a complex number

solving for Z: let V2/V1=Vr
Vr*(Z+R)=Z
Vr*R=Z*(1-Vr), note Vr will ALWAYS be less than 1

now,
Z=R*Vr/(1-Vr)

now you're looking for two terms in series, not parallel, to create that
value of impedance, write it out differently using normal imaginary number
type stuff, 1i where i is sqrt(-1), or 1j, your preference:

real(Z)+1i*imag(Z) = Re + 1i*XL = Vr/(1-Vr)

therefore,
Re = real( (Vr/(1-Vr) ) and
XL = imaginary( Vr/(1-Vr) )

you know the real part of Z is the loss, Re, and the imaginary part is the
reactance, XL, is 2pi f * L that you're looking for.

So now just find the real part of that ratio and the imaginary part, and
you're done.

to change Vr/(1-Vr) into being purely an imaginary part and a real part,
multiply the equation by 1, but make that 1 equal to the 'conjugate' of
the denominator

(1+Vr')/(1+Vr'), where conjugate is the real value left intact, but change
the sign of the imaginary value. That way the denominator becomes like the
sum of two squares = square real + square imagineary, but has NO imaginary
terms left in it.

That way you can easily separate out the real and imaginary parts.

Is that what you wanted?

 

[amdx]

On 2/2/2014 9:48 AM, RobertMacy wrote:

On Sun, 02 Feb 2014 08:18:04 -0700, amdx <nojunk@knology.net> wrote:

...snip....just to keep Aioe happy

I have this:
R = 1283.8 x COS(86.4*) so, R = 80.6 ohms
X 1283.8 x SIN (86.4), making X = 1281.3 ohms.

I want to know the formula that I go through to get the REAL R and Z.
The 47.5 ohm sense resistors effect on R and phase need to be
taken out in the calculation.
I don't know at what point I subtract it out, but I do know it's not
at the end.

I gave you the understanding that I have, please correct me where I made
mistakes.
Pretend, no, assume I'm in 7th grade math. (Algebraically I am).

So where do I start?

Many thanks, Mikek

Read a scope trace?! Aren't scopes something like 3% devices!

That's rain on my parade, but I'm going to continue dancing until I
figure out, this dance has no value.

You should change your nomenclature, it's confusing. Let Z equal the
WHOLE impedance you see with Re being the equivalent series resistance
and XL being the equivalent series reactance of your inductor, thus Z =
Re + 1i*XL


the 'formula' you're looking for is

V2=V1*Z/(Z+R) where Z is a complex number

solving for Z: let V2/V1=Vr
Vr*(Z+R)=Z
Vr*R=Z*(1-Vr), note Vr will ALWAYS be less than 1

now,
Z=R*Vr/(1-Vr)

now you're looking for two terms in series, not parallel, to create that
value of impedance, write it out differently using normal imaginary
number type stuff, 1i where i is sqrt(-1), or 1j, your preference:

real(Z)+1i*imag(Z) = Re + 1i*XL = Vr/(1-Vr)

therefore,
Re = real( (Vr/(1-Vr) ) and
XL = imaginary( Vr/(1-Vr) )

you know the real part of Z is the loss, Re, and the imaginary part is
the reactance, XL, is 2pi f * L that you're looking for.

So now just find the real part of that ratio and the imaginary part, and
you're done.

to change Vr/(1-Vr) into being purely an imaginary part and a real part,
multiply the equation by 1, but make that 1 equal to the 'conjugate' of
the denominator

(1+Vr')/(1+Vr'), where conjugate is the real value left intact, but
change the sign of the imaginary value. That way the denominator becomes
like the sum of two squares = square real + square imagineary, but has
NO imaginary terms left in it.

That way you can easily separate out the real and imaginary parts.

Is that what you wanted?

Much of that was over my head, but I don't see where I take out the
47.5 ohms sense resistor from R or it's affect on the phase angle.

Let me try again. I want a formula, so that after reading my scope, I
have voltage, current and phase delay. ( Maybe I don't need to convert
this to Z and phase angle or maybe I do)
I plug the resulting numbers into the formula, somewhere the formula
handles the 47.5 ohm sense resistor problem, then out pops my R and X.

These formulas get me close, but has not handled the 47.5 ohm sense
resistor.
R = Z x COS (angle)
X = Z x SIN (angle)
I could use a 0.01 ohm sense resistor and forget about it, but
then the voltage on the scope is to low to read.

Thank you, Mikek

 

[Fred Abse]

On Sun, 02 Feb 2014 09:18:04 -0600, amdx wrote:

The “setup” is a board that I'm testing to measure three items:
Input voltage,
Voltage across a 47.5 ohm sense resistor,
Phase difference between these two voltages.

All measurements are on a scope at 3.85MHz. With 10 Vpp signal.

The 47.5 ohm resistor is in series with anything I measure.
It must be subtracted when doing the calculation, this is
very accurate when I measure a 50 BNC termination resistor.
I measured it at 97.9 ohms - 47.5ohms = 50.4 ohms, zero phase. 3.85MHz
When I measured a 3,090 ohm resistor I get 3,119 ohms after subtracting
47.5 ohms for the sense resistor.
That's a 1% error, I'm happy with that, especially sense this
is reading the scope traces.

Now I get to measure reactances.

I have a 55uh inductor, XL = 1329.8 ohms @ 3.85MHz.
Q is 250 so R = 5.3 ohms.
As measured on a 50 year old Boonton 260A.
This is good enough to get me the math I need.

When I read the scope and do the first calculation Z=E/I,
I get 1283.8 ohms this includes the 47.5 ohm sense resistor.
The current and voltage have 62.5 ns phase difference.

The phase difference:
T of 3.85MHz is 259.7 ns and 62.5ns, is 24% of 259.7ns.
Then, 24% of 360* equals a 86.4* phase angle.

I have this:
R = 1283.8 x COS(86.4*) so, R = 80.6 ohms
X 1283.8 x SIN (86.4), making X = 1281.3 ohms.

My trusty HP35s gives 80.61e0 +j1.2813e3

So far, so good.

I want to know the formula that I go through to get the REAL R and Z.

Er... let's be a bit pedantic, here. The word "real" has a definite
mathematical meaning. In the above example, 80.61 is the real part, 1281.3
is the imaginary part.

Better to use "true", rather than "real", in the sense that *you* meant it.

The 47.5 ohm sense resistors effect on R and phase need to be
taken out in the calculation.
I don't know at what point I subtract it out, but I do know it's not
at the end.

Continuing, the real part is the resistance (80.61 ohms), and the
imaginary part is the reactance (1281.3 ohms).

From now on, the resistance and reactance are effectively separate.

All you need to do is subtract the value of your sensing resistor from the
*resistive (real)*, component. Always assuming your sensing resistor is,
as near as dammit, a pure resistance.

Hence the resistive component of your DUT is 80.61 - 47.5 = 33.11 ohms.

I gave you the understanding that I have, please correct me where I made
mistakes. You didn't make any mistakes, as far as you got

Your understanding isn't that bad.


> Pretend, no, assume I'm in 7th grade math. (Algebraically I am).

I'll give you a B+ for what you've done so far ;-)

So where do I start?

Way back when, I posted a PDF describing the various ways of expressing
impedance, in response to something you posted in October 2010. Did you
not see it? If you like, I'll repost it.


--
"Design is the reverse of analysis"
(R.D. Middlebrook)

 

[amdx]

On 2/2/2014 12:39 PM, Fred Abse wrote:

On Sun, 02 Feb 2014 09:18:04 -0600, amdx wrote:


The “setup” is a board that I'm testing to measure three items:
Input voltage,
Voltage across a 47.5 ohm sense resistor,
Phase difference between these two voltages.

All measurements are on a scope at 3.85MHz. With 10 Vpp signal.

The 47.5 ohm resistor is in series with anything I measure.
It must be subtracted when doing the calculation, this is
very accurate when I measure a 50 BNC termination resistor.
I measured it at 97.9 ohms - 47.5ohms = 50.4 ohms, zero phase. 3.85MHz
When I measured a 3,090 ohm resistor I get 3,119 ohms after subtracting
47.5 ohms for the sense resistor.
That's a 1% error, I'm happy with that, especially sense this
is reading the scope traces.

Now I get to measure reactances.

I have a 55uh inductor, XL = 1329.8 ohms @ 3.85MHz.
Q is 250 so R = 5.3 ohms.
As measured on a 50 year old Boonton 260A.
This is good enough to get me the math I need.

When I read the scope and do the first calculation Z=E/I,
I get 1283.8 ohms this includes the 47.5 ohm sense resistor.
The current and voltage have 62.5 ns phase difference.

The phase difference:
T of 3.85MHz is 259.7 ns and 62.5ns, is 24% of 259.7ns.
Then, 24% of 360* equals a 86.4* phase angle.

I have this:
R = 1283.8 x COS(86.4*) so, R = 80.6 ohms
X 1283.8 x SIN (86.4), making X = 1281.3 ohms.

My trusty HP35s gives 80.61e0 +j1.2813e3

So far, so good.


I want to know the formula that I go through to get the REAL R and Z.

Er... let's be a bit pedantic, here. The word "real" has a definite
mathematical meaning. In the above example, 80.61 is the real part, 1281.3
is the imaginary part.

Better to use "true", rather than "real", in the sense that *you* meant it.

Yep, I made a poor choice of word.

The 47.5 ohm sense resistors effect on R and phase need to be
taken out in the calculation.
I don't know at what point I subtract it out, but I do know it's not
at the end.

Continuing, the real part is the resistance (80.61 ohms), and the
imaginary part is the reactance (1281.3 ohms).

From now on, the resistance and reactance are effectively separate.
,
All you need to do is subtract the value of your sensing resistor from the
*resistive (real)*, component.

This is not intuitive to me. I was thinking the sense resistor would
need to come out before this. As a thought experiment, let's mentally
draw the circuit. We have an L in series the an R (loss) in series with
another R. (I'm dropping back to Robert Macy's method, not fully
understanding) We measure one voltage across the L and R and a second
voltage across the other sense R. The first RL has a phase shift
compared to the second Sense R.
Just seems to me it needs to be dealt with earlier in the calculations.
Your input?


>Always assuming your sensing resistor is, as near as dammit, a pure resistance.

Yep, metal film, short leads. At one point I even tried an 0802 surface
mount sens resistor. I somewhat understand strays, doing my best to
minimize them. Not sure I'm real good at it.

Hence the resistive component of your DUT is 80.61 - 47.5 = 33.11 ohms.

True R 5.3 = ohms, measured R = 33 ohms.
True XL = 1330 ohms, measured XL = 1,281 ohms.
This is poor to fair. 3.6% error for the L,
Large error for the R.
I just realized, have a Boonton calibration inductor.
It's low frequency, 400kHz or so, but I'll check it when I get time.

I would like you to check the assumption above you made that
I don't find intuitive, and see if I changed your mind enough to
rethink it.
I don't think that will fix my R though.

I gave you the understanding that I have, please correct me where I made
mistakes. You didn't make any mistakes, as far as you got

Your understanding isn't that bad.


Pretend, no, assume I'm in 7th grade math. (Algebraically I am).

I'll give you a B+ for what you've done so far ;-)


So where do I start?

Way back when, I posted a PDF describing the various ways of expressing
impedance, in response to something you posted in October 2010. Did you
not see it? If you like, I'll repost it.

That was last computer/HD, could you repost it.

Thanks for your help, Mikek

 

[Fred Abse]

On Sun, 02 Feb 2014 15:27:44 -0600, amdx wrote:

This is not intuitive to me. I was thinking the sense resistor would
need to come out before this. As a thought experiment, let's mentally
draw the circuit. We have an L in series the an R (loss) in series with
another R. (I'm dropping back to Robert Macy's method, not fully
understanding) We measure one voltage across the L and R and a second
voltage across the other sense R. The first RL has a phase shift
compared to the second Sense R.
Just seems to me it needs to be dealt with earlier in the calculations.
Your input?

You should be simultaneously measuring the voltage across the sense
resistor, and across the DUT plus sense resistor combination. They are in
series,so the current is the same through both.

We will neglect probe capacitance for now. It only matters across the
sense resistor, anyway. 10pF at 3085MHz is about 4k ohm.

The voltage across the sense resistor gives you the amplitude and phase of
the current.

The amplitude across the combination, divided by the amplitude across the
resistor gives you the magnitude of the total impedance, DUT plus sense
resistor.

We now have magnitude and angle of the combination.

We now do a polar to rectangular conversion to give real and imaginary
components of the total impedance, that is, resistance and reactance

Neglecting probe capacitance, the reactance component is due only to the
DUT, the resistance is that of the series combination, so we subtract the
sense resistor from it.

True R 5.3 = ohms, measured R = 33 ohms.
True XL = 1330 ohms, measured XL = 1,281 ohms.
This is poor to fair. 3.6% error for the L,

How did you determine your "true" L? If it was measured at a different
frequency, its self-capacitance can introduce an error.


> Large error for the R.

The AC resistance of an inductor is a function of frequency, and can be
much higher than that measured at DC, or a lower frequency. At what
frequency did you measure your "true" R?

X=1281, and R=33 is a Q of 427, that's a pretty outstanding inductor, at
3.85MHz, even air-cored. What was it?

Welcome to the world of measurement ;-)

That was last computer/HD, could you repost it.

Will do.

--
"Design is the reverse of analysis"
(R.D. Middlebrook)

 

[amdx]

On 2/2/2014 4:57 PM, Fred Abse wrote:

On Sun, 02 Feb 2014 15:27:44 -0600, amdx wrote:

This is not intuitive to me. I was thinking the sense resistor would
need to come out before this. As a thought experiment, let's mentally
draw the circuit. We have an L in series the an R (loss) in series with
another R. (I'm dropping back to Robert Macy's method, not fully
understanding) We measure one voltage across the L and R and a second
voltage across the other sense R. The first RL has a phase shift
compared to the second Sense R.
Just seems to me it needs to be dealt with earlier in the calculations.
Your input?

You should be simultaneously measuring the voltage across the sense
resistor, and across the DUT plus sense resistor combination. They are in
series,so the current is the same through both.

That is correct, measuring input voltage and voltage across sense
resistor.

We will neglect probe capacitance for now. It only matters across the
sense resistor, anyway. 10pF at 3085MHz is about 4k ohm.

It's worse than that, the probes are 15pf, frequency is 3.85MHz so 2,757
ohms.

Yes, in the final test, the sense resistor should be as small as
possible and still be able to make a reasonable amplitude on the scope.
This would minimize error. Or put it all in a spreadsheet and calculate
everything including the effect of measuring equipment.

The voltage across the sense resistor gives you the amplitude and phase of
the current.

Yes.

The amplitude across the combination, divided by the amplitude across the
resistor gives you the magnitude of the total impedance, DUT plus sense
resistor.

I think you meant was "The amplitude across the combination, divided by
the amplitude across" (Rsense / the ohms of R sense) "gives you the
magnitude of the total impedance, DUT plus sense
resistor."

Yes, but won't they be a vector?
Or maybe I'll get a better understanding with this,
Does Vtotal = Vdut + Vsense ?

Ok, I've been googling and found in a series R L circuit,
Vtotal = Vdut + Vsense.

I had this confused with LC circuits.

This was a big block, it will help having that straighten out.

We now have magnitude and angle of the combination.

We now do a polar to rectangular conversion to give real and imaginary
components of the total impedance, that is, resistance and reactance

Neglecting probe capacitance, the reactance component is due only to the
DUT, the resistance is that of the series combination, so we subtract the
sense resistor from it.

True R 5.3 = ohms, measured R = 33 ohms.
True XL = 1330 ohms, measured XL = 1,281 ohms.
This is poor to fair. 3.6% error for the L,

How did you determine your "true" L? If it was measured at a different
frequency, its self-capacitance can introduce an error.

Good point, my Boonton was at the bottom of the internal capacitor, so
I measured at a lower frequency. I remeasured as high as I could go in
frequency 3.80Mhz and got 57.9uh and a Q of 196.

So XL = 1400 ohms and R = 7.1 ohms using 3.85Mhz in the calculation.

Large error for the R.

The AC resistance of an inductor is a function of frequency, and can be
much higher than that measured at DC, or a lower frequency. At what
frequency did you measure your "true" R?

Previously, at 3 Mhz
Just now, in the numbers above, 3.85MHz as noted.
I'm using Q to calculate Rloss. 1400/196 = 7.1 ohms

X=1281, and R=33 is a Q of 427, that's a pretty outstanding inductor, at
3.85MHz, even air-cored. What was it?

You should try that calculation again 1281/33 = 38.8, but we know the
33 ohms is bogus.

The inductor is one I put together, 7/16" polypropylene tube wrapped
with 90 turns of 660/46 litz wire, I have an 8" ferrite rod that I slide
in or out to adjust inductance. It is variable from 8uH to 247uH.
The Q at 472kHz is 550!
Looks like this,

http://i395.photobucket.com/albums/pp37/Qmavam/P1010008.jpg
Close up of wire;
http://i395.photobucket.com/albums/pp37/Qmavam/P1010009.jpg

> Welcome to the world of measurement ;-)

The point of the exercise is to learn the math required to perform the
calculations needed to find XL, XC, and R of complex circuits while
using the "setup" and to refine the "setup" so I can rely on it's accuracy.
So far, I think I have the RL figured out, but the "setup" is
questionable. I don't know why, yet. It does great on resistors only,
even at 10 MHz.
I'm going to test my Boonton Standard inductor, the calibrated data is
only good 450kHz, but it's the only calibrated inductor I have.
Going to the Orlando Hamfest next weekend maybe I can find some other
standard inductors. And capacitor measurement are next, but I'm hoping
if it works for inductors it will work for caps.

That was last computer/HD, could you repost it.

Will do.

Tomorrow I'll try to write up what I think I know about doing
these calculations. and think about why reactances don't measure
with the accuracy I want. Yet.

Thanks for all the help, Mikek

 

[Fred Abse]

On Sun, 02 Feb 2014 20:19:32 -0600, amdx wrote:

I think you meant was "The amplitude across the combination, divided by
the amplitude across" (Rsense / the ohms of R sense) "gives you the
magnitude of the total impedance, DUT plus sense
resistor."

My bad.

The amplitude across the combination, divided by the amplitude across the
sense resistor, divided by the value of the sense resistor.

Sorry. More follows.

I'm going to bed;-)

--
"Design is the reverse of analysis"
(R.D. Middlebrook)

 

[Fred Abse]

On Sun, 02 Feb 2014 20:19:32 -0600, amdx wrote:

You should try that calculation again 1281/33 = 38.8, but we know the
33 ohms is bogus.

Just spotted that, my mistake, big Sunday lunch

33.8 is in the ballpark for the inductor you describe, at your test
frequency.

The inductor is one I put together, 7/16" polypropylene tube wrapped
with 90 turns of 660/46 litz wire, I have an 8" ferrite rod that I slide
in or out to adjust inductance. It is variable from 8uH to 247uH. The Q
at 472kHz is 550!

A ferrite core should bring the Q *down*, significantly.

--
"Design is the reverse of analysis"
(R.D. Middlebrook)

 

[amdx]

On 2/3/2014 3:13 AM, Fred Abse wrote:

On Sun, 02 Feb 2014 20:19:32 -0600, amdx wrote:

You should try that calculation again 1281/33 = 38.8, but we know the
33 ohms is bogus.

Just spotted that, my mistake, big Sunday lunch

33.8 is in the ballpark for the inductor you describe, at your test
frequency.

We can disagree on that. As measured on my Q meter the Q at 3.85MHz
is 196.

The inductor is one I put together, 7/16" polypropylene tube wrapped
with 90 turns of 660/46 litz wire, I have an 8" ferrite rod that I slide
in or out to adjust inductance. It is variable from 8uH to 247uH. The Q
at 472kHz is 550!

Sounds like you don't believe 550 at 475kHz.
It is in that ball park. To the error of my Q meter.

> A ferrite core should bring the Q *down*, significantly.

I agree, ferrites add loss, but there is a trade off between the extra
wire, the additional interwinding capacitance caused by more wire and
the additional losses caused by proximity effect because of more turns
on an air core.
Ben Tongue has a ferrite rod inductor that has a Q over 875 over the
entire AMBCB.

See table 2.
http://www.bentongue.com/xtalset/29MxQFL/29MxQFL.html

Site index.
> http://www.bentongue.com/xtalset/xtalset.html

Thanks Mikek

 

[RobertMacy]

On Sun, 02 Feb 2014 09:23:46 -0700, amdx <nojunk@knology.net> wrote:

...snip...

Much of that was over my head, but I don't see where I take out the
47.5 ohms sense resistor from R or it's affect on the phase angle.

Let me try again. I want a formula, so that after reading my scope, I
have voltage, current and phase delay. ( Maybe I don't need to convert
this to Z and phase angle or maybe I do)
I plug the resulting numbers into the formula, somewhere the formula
handles the 47.5 ohm sense resistor problem, then out pops my R and X.

These formulas get me close, but has not handled the 47.5 ohm sense
resistor.
R = Z x COS (angle)
X = Z x SIN (angle)
I could use a 0.01 ohm sense resistor and forget about it, but
then the voltage on the scope is to low to read.

Thank you, Mikek

Aioe access locked me out yesterday, and could not reply!

If that happens again, you're in the middle of a project, and you want an
answer NOW, feel free to email directly.

How do you measure current?

Are you working with V(1) as drive through Rsense of 47.5 ohms, to V(2)
with inductor to GND?

 

[RobertMacy]

On Sun, 02 Feb 2014 09:23:46 -0700, amdx <nojunk@knology.net> wrote:

ARRRGG!! I just reread your first posting which CLEARLY said what you're
doing!

Define V1 as the input voltage that you adjust to around 10 Vpp at
3.85MHz. Then you have Rs=47.5 ohms that you trust connected to the
inductor to measure, with a probe there, V2 in reference to V1 obviously
has some drop and some phase shift to it.

Note: I usually use the scope as an indicator, not a measuring device, the
accuracy is so poor. For example, take a known cap, place in parallel with
inductor, note resonant point, measure freq [freq meters are ACCURATE!]
You can tell where resonance is by looking for that little 'spike' your
function generator puts in there at 'turn around' point. That squiggle is
a very accurate indication of the phase of your resonance. You can
measure, kind of the cap by using a resistor in series and note when you
get EXACTLY sqrt(2)/2, 0.0707 of input value, [use a VTVM, even though its
accuracy is questionable, it may track well enough for a comparison
measurement] you can get about 1% to 0.1% accuracy here. Now knowing the
resonant frequency you get the L of the inductor, by noting the Q of the
resonance, you get the equivalent resistance of the inductor AND the cap,
but remember you used a good cap, so its resistance can be ignored. Thus,
without using the scope as a meter you have measured, fairly accurately,
the L and the R of the inductor.

But back to your original goal:

you have Rs, V1, and V2, FIND L and Re of the inductor:

[you know magnitude of V2 and phase angle, theta, of V2]

current through the resistor is
I = ( V1-V2 )/Rs, preserve V2's phase shift

now armed with current you can calculate Z:

Z = V2/I again, preserve ALL phase shifts

now you can separate out L and Re

the math is simply complex math. far easier to get a copy of LTspice and
stick in the values and find what you want, or if you want to follow the
math step by step, get a copy of octave and perform the complex
calculations [a great way to keep track to make sure you have NOT made an
error] but in case you want the laborious path:

I = ( V1-V2 )/Rs
think of I in terms of
I = re(I) + 1i*im(I)
where
re(I)=(V1-V2*cos(theta))/Rs
im(I)=(V1-V2*sin(theta))/Rs

of course your unknown impedance is
Z = V2/I
but again think in terms of
V2 = re(V2) + 1i*im(V2) = V2*cos(theta) + 1i* V2(sin(theta)

so you now have Z
Z = V2/I = ( V2*cos(theta) + 1i* V2(sin(theta) )/(
(V1-V2*cos(theta))/Rs+1i*(V1-V2*sin(theta))/Rs )
looks a bit daunting, eh?
to cleverly get rid of the complex numbers in the denominator, multiply by
1, but a special 1
1 = ( (V1-V2*cos(theta))/Rs-1i*(V1-V2*sin(theta))/Rs )/(
(V1-V2*cos(theta))/Rs-1i*(V1-V2*sin(theta))/Rs )

note that these values are the complex conjugate of your denominator,
where you simply change the sign of the complex value.

now, after you multiply and collect like terms you will have one value
that's real and one value that's imaginary, you're done.

Since Aioe won't let me post too many lines, I'll have to reply to this
reply with the reduced formula.

 

[amdx]

On 2/3/2014 8:06 AM, RobertMacy wrote:

On Sun, 02 Feb 2014 09:23:46 -0700, amdx <nojunk@knology.net> wrote:

...snip...

Much of that was over my head, but I don't see where I take out the
47.5 ohms sense resistor from R or it's affect on the phase angle.

I have just learned again, that all series Rs just add together
and cause the reduction from a 90* phase shift caused by a reactance.

Let me try again. I want a formula, so that after reading my scope,
I have voltage, current and phase delay. ( Maybe I don't need to
convert this to Z and phase angle or maybe I do)
I plug the resulting numbers into the formula, somewhere the formula
handles the 47.5 ohm sense resistor problem, then out pops my R and X.

To reiterate, I no longer think Rsense needs special handling, it is
just part of the total R in the circuit.

These formulas get me close, but has not handled the 47.5 ohm sense
resistor.
R = Z x COS (angle)
X = Z x SIN (angle)
I could use a 0.01 ohm sense resistor and forget about it, but
then the voltage on the scope is to low to read.

Thank you, Mikek


Aioe access locked me out yesterday, and could not reply!

If that happens again, you're in the middle of a project, and you want
an answer NOW, feel free to email directly.

How do you measure current?

Are you working with V(1) as drive through Rsense of 47.5 ohms, to V(2)
with inductor to GND?


See last line, you already figured this out.

No,
V1 is directly across the input. Call it a coaxial input, it is a panel
mount BNC. From the center conductor, I connect one end of my DUT,
the other end of DUT connects to Rsense (47.5 ohm resistor) and the
other end of Rsense is connected to the shield of the BNC connector.
V2 is measure across Rsense.
All grounds are common, (sig gen and both scope probes.)

Well I just read your followup before sending this, so you already know.

Thanks, Mikek

 

[amdx]

Ok guys, I think you have told me what I need to know to solve
what I was after.
I thank you for all the help.
I wish I had known where my interests would lie 42 years ago when I
avoided learning algebra in High School.

I still don't know where the errors in my "setup" come from.
It seems very accurate on resistance, but when I measure and inductor
the XL is close but the R has high error.

I haven't looked at capacitors yet.

If you have any ideas where the errors are creeping in,
other than scope magnitude calibration, scope timebase calibration,
errors in reading the scope, Rsense tolerance and parasitics, and not
knowing the real characteristics of my DUT coils.
If you know of any other source of error, "you know"something that could
matter.
I'd like to hear.
Mike

 

[RobertMacy]

On Mon, 03 Feb 2014 09:40:23 -0700, amdx <nojunk@knology.net> wrote:

...snip...
V1 is directly across the input. Call it a coaxial input, it is a panel
mount BNC. From the center conductor, I connect one end of my DUT,
the other end of DUT connects to Rsense (47.5 ohm resistor) and the
other end of Rsense is connected to the shield of the BNC connector.
V2 is measure across Rsense.
All grounds are common, (sig gen and both scope probes.)

Well I just read your followup before sending this, so you already
know.

Thanks, Mikek

V2 the input driving voltage? I just number from left to right.

But I'll change V2 as drive and reference input into Rsense. V1 is voltage
across the inductor being measured.

you can't just 'add' Rsense to inductor's resistance, doesn't work quite
like that.

Really, seriously grab a copy of LTspice, it's educational beyond belief
for learning this stuff. And for the math get a copy of octave it'll act
as an 'automatic' check as you write out equations.

Just curious, I did a calculation of the value of reactance of the scope
probe's capacitance at that frequency as you measured the 3k resistor. You
did notice that parallel capacitive reactance is bogging down your reading?

 

[RobertMacy]

On Mon, 03 Feb 2014 08:25:52 -0700, RobertMacy <robert.a.macy@gmail.com>
wrote:

...snip....
Since Aioe won't let me post too many lines, I'll have to reply to this
reply with the reduced formula.

new tac, use the fact that the sense resistor is in series with the
equivalent resistance of the inductor, so the total current is based upon
the total, but the voltage you measure is only across the inductance and
its resistance.

V1/V2=Re+1i*XL/(Rs+Re+1i*XL)

=(Re+1i*XL)*(Rs+Re-1i*XL)/( (Rs+Re)^2+XL^2 )

( Re*Rs+Re^2+XL^2 + 1i*(Rs*XL) )/( (Rs+Re)^2+XL^2 )

V1/V2*cos(theta)=Re

now find Re and XL !! good luck with that one!

go back to the original approach:
V2/V1 = ( Rs+Re+1i*XL)/(Re+1i*XL)=Rs/(Re+1i*XL)+1
V2/V1-1=(V2-V1)/V1=Rs/(Re+1i*XL)
V1/(V2-V1)=Re/Rs+1i*XL/Rs
Re = Rs*real( V1/(V2-V1) )
XL = Rs*imag( V1/(V2-V1) )

by definition V2 is real, being the reference
here is a potential source of confusion, the voltage measured, V1, is a
complex number
V1=V1*( cos(theta)+1i*sin(theta) )

V2-V1 is actually V2-V1*( cos(theta)+1i*sin(theta) )
conjugate of V2-V1, is really,
V2-V1*( cos(theta)-1i*sin(theta) )
multiplying leave only the real part of denominator:
(V2-V1*cos(theta))^2-V1^2*sin(theta)^2
remember that cos^2+sin^2 = 1
V2^2+V1^2-2*V1*V2*cos(theta)
denominator imaginary part cancels and disappears

the denominator done longhand is:
=( V2-V1*cos(theta)-1i*V1*sin(theta) )*( V2-V1*cos(theta)+1i*V1*sin(theta)
)
=( V2^2 + V1^2 - 2*V1*V2*cos(theta) ), and the imaginary part cancels

numerator is
( V1*cos(theta)+1i*V1*sin(theta) )*( V2-V1*cos(theta)+1i*V1*sin(theta) )
real part of numerator is
V2*V1*cos(theta)-V1^2*cos(theta)^2-V1^2*sin(theta)^2=V1^2+V2*V1*cos(theta)
or another way
V1*( V1+V2*cos(theta) )
therefore, Re which is the real part is:
Re = Rs*(V1*(V1+V2*cos(theta))/( V2^2+V1^2-2*V1*V2*cos(theta) )
and to find the inductive reactance, the imaginary part is:
=1i*(V1^2*cos(theta)*sin(theta)-V1^2*cos(theta)*sin(theta)+V2*V1*sin(theta)
therefore the inductor's reactance is:
XL=Rs*( V2*V1*sin(theta) )/( V2^2+V1^2-2*V1*V2*cos(theta) )

thus knowing the magnitude of V2, magnitude and phase of V1, and sense
resistor
you can find Re and XL of the inductor

if your measured impedance gets high in comparison to the capacitive
impedance of the probe, you can have a problem with the impact of that
probe's impedance upon your measurements.

I posted the derivation in case I made a mistake [which is likely] and you
can see where everything came from so you will trust the 'final' formula a
bit.

 

[amdx]

On 2/4/2014 6:52 AM, RobertMacy wrote:

On Mon, 03 Feb 2014 08:25:52 -0700, RobertMacy <robert.a.macy@gmail.com
wrote:

...snip....
Since Aioe won't let me post too many lines, I'll have to reply to
this reply with the reduced formula.

new tac, use the fact that the sense resistor is in series with the
equivalent resistance of the inductor, so the total current is based
upon the total, but the voltage you measure is only across the
inductance and its resistance.

That is almost correct.
I'm not measuring the voltage across the Inductance and it's
resistance. I'm measuring the voltage a cross the sense resistor.
And V2 the voltage across the whole system, (the input voltage).
You can solve for the voltage across the Inductance and it's
resistance. V2 - V1 @ 61*= V across inductor and it's resistance
@ angle.

Thanks, Mikek

V1/V2=Re+1i*XL/(Rs+Re+1i*XL)

=(Re+1i*XL)*(Rs+Re-1i*XL)/( (Rs+Re)^2+XL^2 )

( Re*Rs+Re^2+XL^2 + 1i*(Rs*XL) )/( (Rs+Re)^2+XL^2 )

V1/V2*cos(theta)=Re

now find Re and XL !! good luck with that one!

go back to the original approach:
V2/V1 = ( Rs+Re+1i*XL)/(Re+1i*XL)=Rs/(Re+1i*XL)+1
V2/V1-1=(V2-V1)/V1=Rs/(Re+1i*XL)
V1/(V2-V1)=Re/Rs+1i*XL/Rs
Re = Rs*real( V1/(V2-V1) )
XL = Rs*imag( V1/(V2-V1) )

by definition V2 is real, being the reference
here is a potential source of confusion, the voltage measured, V1, is a
complex number
V1=V1*( cos(theta)+1i*sin(theta) )

V2-V1 is actually V2-V1*( cos(theta)+1i*sin(theta) )
conjugate of V2-V1, is really,
V2-V1*( cos(theta)-1i*sin(theta) )
multiplying leave only the real part of denominator:
(V2-V1*cos(theta))^2-V1^2*sin(theta)^2
remember that cos^2+sin^2 = 1
V2^2+V1^2-2*V1*V2*cos(theta)
denominator imaginary part cancels and disappears

the denominator done longhand is:
=( V2-V1*cos(theta)-1i*V1*sin(theta) )*(
V2-V1*cos(theta)+1i*V1*sin(theta) )
=( V2^2 + V1^2 - 2*V1*V2*cos(theta) ), and the imaginary part cancels

numerator is
( V1*cos(theta)+1i*V1*sin(theta) )*( V2-V1*cos(theta)+1i*V1*sin(theta) )
real part of numerator is
V2*V1*cos(theta)-V1^2*cos(theta)^2-V1^2*sin(theta)^2=V1^2+V2*V1*cos(theta)
or another way
V1*( V1+V2*cos(theta) )
therefore, Re which is the real part is:
Re = Rs*(V1*(V1+V2*cos(theta))/( V2^2+V1^2-2*V1*V2*cos(theta) )
and to find the inductive reactance, the imaginary part is:
=1i*(V1^2*cos(theta)*sin(theta)-V1^2*cos(theta)*sin(theta)+V2*V1*sin(theta)
therefore the inductor's reactance is:
XL=Rs*( V2*V1*sin(theta) )/( V2^2+V1^2-2*V1*V2*cos(theta) )

thus knowing the magnitude of V2, magnitude and phase of V1, and sense
resistor
you can find Re and XL of the inductor

if your measured impedance gets high in comparison to the capacitive
impedance of the probe, you can have a problem with the impact of that
probe's impedance upon your measurements.

I posted the derivation in case I made a mistake [which is likely] and
you can see where everything came from so you will trust the 'final'
formula a bit.

 

[Fred Abse]

On Mon, 03 Feb 2014 10:53:50 -0600, amdx wrote:

If you have any ideas where the errors are creeping in,
other than scope magnitude calibration, scope timebase calibration,
errors in reading the scope, Rsense tolerance and parasitics, and not
knowing the real characteristics of my DUT coils.

Let us look at it from a different angle. What *should* your inductor look
like on your test setup, assuming that your Boonton measurements are
accurate.

Firstly, we'll allow for the probe capacitance. You say that is 14pF.

14pF in parallel with 47.5 ohms , at 3.85MHz looks like 47.49ohms, in
series with 54.1 nF. (Series to parallel impedance conversion). It's
always a good idea to start with everything series, it simplifies
calculation.

That's an resistance of 47.49 ohms, in series with a capacitive
reactance of 0.7641 ohms.

That's an impedance of 47.49 - j0.7641.

We'll call that Zs

Your 5 uH inductor has an inductive reactance of 1330.4645ohms, in series
with a resistance of 1330.4645/250 = 5.2319 ohms.

That's an impedance of 5.2319 + j1330.4645.

We'll call that Zl

The whole setup is a potential divider, whose ratio is Zs/(Zl+Zs)

With 10V input, the voltage at the probe is 10Zs/(Zl+Zs).

That is 10 * (47.49 -j0.7641)/(5.2319 +j1330.4645 +47.49 -j0.7641 ) volts

Or 10 * (47.49 -j0.7641)/(52.8119 +j1329.7004) volts

Which comes to 10 * (0.00084 +j0.03568) volts

Or .0084 +j0.3568 volts, which equals 0.3569volts, angle 88.6473 degrees.

That's what you *should* measure at the sensing resistor.

E.&O.E.

I did a spice simulation as well (You can have it if you like). It agrees
with the above hand calculation to within 3 decimal places. Using both
parallel probe capacitance, and the series conversion.

So, what's the problem? I see three possibilities:

Scope calibration, it's possible that the two channels you are using,
don't match. Swapping them should show that. Have you checked both
channels against a known voltage? DC will do. Don't trust the scope's own
calibrator.

Reading error, How did you determine that the delay, or phase you're
measuring was accurate, IOW, how did you establish zero crossings?


Boonton Q meter inaccurate. That's what I'd go for, first. Have you
verified its calibration? Don't trust 50-year-old standard inductors, the
proper way to do it involves only measuring voltages. See the manual.



--
"Design is the reverse of analysis"
(R.D. Middlebrook)

 

[Fred Abse]

On Mon, 03 Feb 2014 07:40:28 -0600, amdx wrote:

On 2/3/2014 3:13 AM, Fred Abse wrote:
On Sun, 02 Feb 2014 20:19:32 -0600, amdx wrote:

You should try that calculation again 1281/33 = 38.8, but we know the
33 ohms is bogus.

Just spotted that, my mistake, big Sunday lunch

33.8 is in the ballpark for the inductor you describe, at your test
frequency.

We can disagree on that. As measured on my Q meter the Q at 3.85MHz
is 196.

That's a feasible value.

The inductor is one I put together, 7/16" polypropylene tube wrapped
with 90 turns of 660/46 litz wire, I have an 8" ferrite rod that I slide
in or out to adjust inductance. It is variable from 8uH to 247uH. The Q
at 472kHz is 550!

What spacing and length. I'd like to try to FEMM it.

Sounds like you don't believe 550 at 475kHz.
It is in that ball park. To the error of my Q meter.

Let's say, I'm skeptical.

A ferrite core should bring the Q *down*, significantly.

I agree, ferrites add loss, but there is a trade off between the extra
wire, the additional interwinding capacitance caused by more wire and
the additional losses caused by proximity effect because of more turns
on an air core.
Ben Tongue has a ferrite rod inductor that has a Q over 875 over the
entire AMBCB.

See table 2.
http://www.bentongue.com/xtalset/29MxQFL/29MxQFL.html

Never heard of the guy before. He states that FEMM doesn't accommodate
litz wire. My copy certainly does...

--
"Design is the reverse of analysis"
(R.D. Middlebrook)