More Math for the "SETUP"

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.

...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?

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

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

My trusty HP35s gives 80.61e0 +j1.2813e3

So far, so good.

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.

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.

Your understanding isn't that bad.

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

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)

Yep, I made a poor choice of word.

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?

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.

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.

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

Thanks for your help, Mikek

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.

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

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)

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

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.

Yes.

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.

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.

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

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,

Close up of wire;

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.

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

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)

Just spotted that, my mistake, big Sunday lunch:-)

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

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

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

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

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

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.

Site index.

Thanks 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?

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.

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.

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

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

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

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?

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.

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

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)

That's a feasible value.

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

Let's say, I'm skeptical.

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)