How to measure the reactive part of output impedance?

In summary, an oscilloscope can be used to measure the open circuit voltage and the reactive component of the output impedance.
  • #1
Abdullah Almosalami
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Given a sinusoidal voltage source attached to a load, like the following,

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I want to get both the resistive ## R_o ## and the reactive ## X_o ## components of the output impedance ## R + jX_o ##. Measuring the open circuit voltage will give ##V##. If I connect a test resistance ##R_1## at the output, and measure the magnitude of the resulting voltage across it ##V_o1## with an oscilloscope, and then do the same again with another test resistance ##R_2## and its corresponding voltage magnitude ##V_o2##, then using the voltage division equations, I can obtain values for ##Ro## and ##Xo^2##, and to walk through the Math (note the ##\Rightarrow [X]## at the end of some of the lines are just to label the equations for easy reference):

Testing Resistance ##R_1##:
## V_{o1} = V * \frac {R_1} {R_1 + R_o + jX_o} \Rightarrow [1]##
## \Rightarrow R_1 + R_o + jX_o = R_1 * \frac {V} {V_o1} ##
(taking magnitude of both sides)
## \Rightarrow \sqrt {(R_1 + R_o)^2 + X_o^2} = R_1 * \frac {|V|} {|V_o1|}##
(squaring both sides)
## \Rightarrow (R_1 + R_o)^2 + X_o^2 = (R_1 * \frac {|V|} {|V_o1|})^2 \Rightarrow [2]##

Likewise with ##R_2##:
## (R_2 + R_o)^2 + X_o^2 = (R_2 * \frac {|V|} {|V_o1|})^2 \Rightarrow [3]##

Now subtracting both equations and getting ride of the ##X_o^2## term,
## (R_1 + R_o)^2 - (R_2 + R_o)^2 = (R_1 * \frac {|V|} {|V_o1|})^2 - (R_2 * \frac {|V|} {|V_o1|})^2 \Rightarrow [4]##
(expanding the two expressions on the LHS)
## R_1^2 + 2R_1R_o + R_o^2 - R_2^2 - 2R_2R_o - R_o^2 = (R_1 * \frac {|V|} {|V_o1|})^2 - (R_2 * \frac {|V|} {|V_o1|})^2 ##
(rewriting the ##R_1^2## and ##-R_2^2## terms as a difference of squares and factoring ##2R_o## from the other terms)
## (R_1 - R_2)(R_1 + R_2) + 2R_o(R_1 - R_2) = (R_1 * \frac {|V|} {|V_o1|})^2 - (R_2 * \frac {|V|} {|V_o1|})^2 ##
(dividing out ##(R_1 - R_2)## and ##2##)
## \frac {R_1 + R_2} {2} + R_o = \frac {1} {2(R_1 - R_2)} * (R_1 * \frac {|V|} {|V_o1|})^2 - (R_2 * \frac {|V|} {|V_o1|})^2 ##
(finally solving for ##R_o##)
## R_o = \frac {1} {2(R_1 - R_2)} * (R_1 * \frac {|V|} {|V_o1|})^2 - (R_2 * \frac {|V|} {|V_o1|})^2 - \frac {R_1 + R_2} {2} \Rightarrow[5]##

And then from either of the two original voltage division equations ##[2]## or ##[3]##, you can solve for ##X_o^2## to get the following:
## X_o^2 = (R_{1 or 2} * \frac {|V|} {|V_o1|})^2 - (R_{1 or 2} + R_o)^2 \Rightarrow [6]##

Now for ##X_o##, we can get the magnitude by taking the square root of ##[6]## but the sign could go either way. My thinking was, alright, calculate the current by using ## \frac {V_{1or2}} {R_{1or2}}## and then the phase of the current will indicate whether the output impedance is capacitive or inductive, but wait, we don't know the phase relative to the source, and there is no way of directly measuring both the ##Vo## and ##V## at the same time on a scope and seeing the phase difference with just the output terminals. And likewise to measure short circuit current and then divide the open circuit voltage by that doesn't work either, since they have to be measured at the same time, which is impossible. So, how can I determine the nature of the reactance of the output impedance?
 

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  • #2
What is your goal with this question? To calibrate an instrument? To understand your options in calibrating a measurement? To just understand some math?

Depending on your goal, there is probably an optimum method to find the Zout components with the minimum number of measurements...
 
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  • #3
There is a lot of information on the web about this type of measurement. Much of it is in relation to measuring the input and/or output impedance of power supplies.
The measurement can be done directly with a frequency response analyzer and a voltage and current probes attached to the output. There are some important details to get the analysis of the data right, but these are well described. All of this work follows from the work of R.D Middlebrook et. al.
All of these methods rely on some sort of perturbation of the port being measure with a signal source, because you may not have control of the perturbation from the D.U.T. or, as you pointed out, you may not be able to measure it.
For example, look at these sites (there are many others too):
http://ecee.colorado.edu/~ecen5797/course_material/Lecture26.pdf
http://www.deltartp.com/dpel/dpelconferencepapers/33.3_10275.pdf
https://venable.biz/tech-pubs/Source-Load Interactions in Multi-Unit Power Systems.pdf
 
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  • #4
@berkeman My goal is to complete this as a lab assignment. All we have available to measure is an oscilloscope.

@DaveE Before I go through the papers you linked, is this possible with just an oscilloscope? I also am not familiar with a frequency response analyzer but I just pulled up a page to read through and see what it's about. Cool!
 
  • #5
If you have a dual trace scope you can use two resistors in series. Use one channel to measure the output voltage of the generator and the other channel to measure the voltage drop across (current thru) one of the resistors. Then compare the phases.

Cheers,
Tom

EDIT: this could also be done using two single trace scopes as long as they trigger on the same signal.
 
  • #6
Tom.G said:
If you have a dual trace scope you can use two resistors in series. Use one channel to measure the output voltage of the generator and the other channel to measure the voltage drop across (current thru) one of the resistors. Then compare the phases.

Cheers,
Tom

EDIT: this could also be done using two single trace scopes as long as they trigger on the same signal.
The assumption is you don't have access to the voltage from the generator. You need to simply use the output terminals of the load. As if you had access to just two wires come out of some mysterious black box that was generating some voltage for itself and you need to find out its output impedance.
 
  • #7
Didn't you indicate you are supplying the testing resistances R1 and R2? I saw no restriction that the testing resistance could not be made up of two resistors.

Did I miss something?
 
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  • #8
If you have an oscilloscope with an input for both vertical and horizontal axes, connect the horizontal input to the source and the vertical input across the R. The resulting figure will tell you a lot.
 
  • #9
Svein said:
If you have an oscilloscope with an input for both vertical and horizontal axes, connect the horizontal input to the source and the vertical input across the R. The resulting figure will tell you a lot.
I think the nearest OP can get to the "source" under load is going to be the voltage across R.
 
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  • #10
Tom.G said:
Didn't you indicate you are supplying the testing resistances R1 and R2? I saw no restriction that the testing resistance could not be made up of two resistors.

Did I miss something?
What I got from your suggestion was to compare the voltage source's waveform with the voltage across one of the resistors' waveform, and from that, I can get the phase difference, but the problem is I can't use the voltage source's waveform or measure it in any way. Unless I misunderstood what you said.
 
  • #11
Does your test load have to be a resistor? What if you also use a reactive load and compare the resulting amplitudes.
Or, can you adjust the frequency of the source (without changing it's amplitude) and repeat your measurement?
 
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  • #12
I took the block diagram in your first post meaning that "R" is your test resistance and everything inside the box is unknown and not accessible.
If this is the case, manipulate "R" in any way needed to get your data. You may even use some "L" or "C" in combination with "R" to get usable results.

Does that work?

I see DaveE and I had the same thought at the same time, but he typed faster.
 
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  • #13
If you can trigger the oscilloscope to show what happens during the transient when the load is added or removed you should be able to see the phase shift due to the output impedance.
 
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  • #14
Abdullah Almosalami said:
Now for XoXoX_o, we can get the magnitude by taking the square root of [6][6][6] but the sign could go either way. My thinking was, alright, calculate the current by using V1or2R1or2V1or2R1or2 \frac {V_{1or2}} {R_{1or2}} and then the phase of the current will indicate whether the output impedance is capacitive or inductive, but wait, we don't know the phase relative to the source

i think you'll have to connect a load that has some reactance.

Apply in turn equal ohms of capacitive then inductive reactance.
One should lower and the other raise Vout as it adds to or subtracts from the voltage drop across internal X.
 
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  • #15
jim hardy said:
i think you'll have to connect a load that has some reactance.

Apply in turn equal ohms of capacitive then inductive reactance.
One should lower and the other raise Vout as it adds to or subtracts from the voltage drop across internal X.
I think if you place an LC circuit in series with the load you can find everything.
First find Voc with no load. Then insert LC and RL.
At resonance the voltage across the load will be Voc (Rg)/(Rg+RL). This gives you generator resistance.
Then measure voltage across either L or C at resonance. This will give you the volt drop across the generator reactance when the current is determined by resistances only. I = Voc / (Rg + RL), and Vxg = I Xg.
I also think it can be done by placing a parallel resonant circuit across the load but the maths are a bit harder.


.
 
  • #16
tech99 said:
I think if you place an LC circuit in series with the load you can find everything.

It'd be difficult for me as a home experimenter to find the exact values needed for resonance
as i just have a junkbox full of old motor run capacitors to choose from. A lab is doubtless better equipped.

and a caution-- beware of voltage gain in series resonant circuits

but that approach has a certain elegance about it
and might even have a lesson about ferroresonance embedded in it !

old jim
 
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  • #17
I noticed a slight correction to what I said. At resonance, the reactance of the LC combination is numerically equal and opposite sign to the unknown reactance. So it is possible to measure the voltage across the external LC combination at resonance to find the unknown reactance. The sign of the reactance can be found by the direction in which LC has to be altered to obtain resonace when alone. Alternatively, by comparing the voltages across the external C and L.
I hope you will be told the approximate values, whether pF or Farads for instance!
 
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  • #18
jim hardy said:
i think you'll have to connect a load that has some reactance.

Apply in turn equal ohms of capacitive then inductive reactance.
One should lower and the other raise Vout as it adds to or subtracts from the voltage drop across internal X.
Honestly that seems so obvious and simple to me now, I should've just tried it right away! I'll see today what that gives me.
 
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  • #19
@DaveE @jim hardy @tech99 your suggestions worked out, at least to say whether the output impedance was net capacitive or net inductive. I knew that the output impedance of the setup should be net capacitive (because that's what I built) and when I put an inductor on the output, it had a higher voltage amplitude than if I put a capacitor, so the net output impedance was capacitive and Xo would be the negative square root of my equation for ##X_o^2##.

For some reason, though, I get wildly different values for the magnitudes of ##R_o## and ##X_o## depending on what test resistors I used, and sometimes I get a negative value for ##X_o^2##, which would mean that there would be no reactive component (??). It might come down to inaccuracies in measurements and tolerances of the components, or I made errors in the procedure, but yeah. Definitely wildly different readings, and the net result I got was that the output resistance of my circuit was somewhere in the range of 300 to 1.5k and the output reactance somewhere in the range of -j400 to -j2k. The actual values should have been Ro = 1.066k and Xo = -436.=, so close enough?
 
  • #20
Abdullah Almosalami said:
The actual values should have been Ro = 1.066k and Xo = -436.=, so close enough?
i'd be guessing.

So you built something from real parts?

Where'd you buy the ideal voltage source ?

upload_2019-2-19_20-0-20.png


I need one...
 

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  • #21
Abdullah Almosalami said:
@DaveE @jim hardy @tech99
For some reason, though, I get wildly different values for the magnitudes of ##R_o## and ##X_o## depending on what test resistors I used, and sometimes I get a negative value for ##X_o^2##, which would mean that there would be no reactive component (??).
Did you use a wirewound resistor? They need to be non inductive.
 
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  • #22
just a thought
Do you think you could be approaching resonance ?
Might voltage gain of a resonant circuit foil assumptions about voltage division ?

Increase series resistance somewhat to lower the Q and see if things change ?

old jim
 
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  • #23
jim hardy said:
i'd be guessing.

So you built something from real parts?

Where'd you buy the ideal voltage source ?

View attachment 239021

I need one...

Well, all this stuff is from parts in a university lab. The power supply was a function generator and I verified its values with the oscilloscope I used. Maybe the other components where a little off though. I did not verify their nominal values with a multimeter or anything.

Also, same. I would love one in my room. Get me a scope and a generator and I'll never leave my room haha.
 
  • #24
tech99 said:
Did you use a wirewound resistor? They need to be non inductive.
No, I believe the resistors I used are the basic typical "carbon film resistor."
2.png

However, I do think that perhaps the nominal values of the components were a little off, and perhaps the resistors did have higher than would be inductance and capacitance?
1.png

Anyways, at least the general idea did work, so exact values would be expected to be off.
 

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  • #25
jim hardy said:
just a thought
Do you think you could be approaching resonance ?
Might voltage gain of a resonant circuit foil assumptions about voltage division ?

Increase series resistance somewhat to lower the Q and see if things change ?

old jim
I am a little shaky on resonance to be honest, so I would have to come back to that. But I can say that I did try out various R combinations, and found that when I did try out a combination based on what I knew the actual output resistance and reactance should be, I did find that using a slightly higher resistance value made for more power indeed.
 
  • #26
Abdullah Almosalami said:
I get wildly different values for the magnitudes
Yes, that is why in practice people use more sophisticated instruments to make these measurements. The way to get good data is to inject a known signal into the circuit and measure the resulting perturbation at the frequency of the source using tracking narrow band voltmeters (essentially like lock-in amplifiers) to filter out unrelated noise. This also allows you to make small perturbations to the system, so it remains linear and operating in the same mode. This can also be done with network analyzers, which are much more expensive, but more common. The network analyzers don't work as well as frequency response analyzers because the receivers in them are too sensitive to large out of band noise, which is common in real world systems (power supplies, control systems etc.), they also tend not to work at low frequencies.
One thing you could try with your oscilloscope is to get a stable trigger on the source frequency and then set it to an averaging mode, this will make a low performance version of a lock-in amplifier and will give you a cleaner signal. Signals that aren't synchronous with the trigger source will tend to average out. You could also build a filter that is centered on the frequency of interest and put it between the measuring instrument and the circuit. It will, of course, also be part of the load impedance in the circuit.
 
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  • #27
DaveE said:
The way to get good data is to inject a known signal into the circuit and measure the resulting perturbation at the frequency of the source using tracking narrow band voltmeters (essentially like lock-in amplifiers) to filter out unrelated noise.
This also allows you to change the measurement frequency to get more data about how the impedance varies with frequency. Even with noisy data you can fit it to an impedance model over a range of frequencies.
 
  • #28
Abdullah Almosalami said:
I am a little shaky on resonance to be honest,
if you find a voltage greater than your open circuit reading you're probably seeing effect of resonance.
 
  • #29
It may be a bit out of date but your math has a mistake (I don't know if you solved it yet) but you use the same voltage Vo1 when you change from resistors R1 and R2 or whatever impedances you are considering - if they are not equal, which I think it is required you must consider different voltages when solving the equations.

cheers,
carlos
 
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  • #30
Carlos Leitao said:
It may be a bit out of date but your math has a mistake (I don't know if you solved it yet) but you use the same voltage Vo1 when you change from resistors R1 and R2 or whatever impedances you are considering - if they are not equal, which I think it is required you must consider different voltages when solving the equations.

I was going to post that too. Also, if the source is hidden you'll have to do the measurement at multiple frequencies because otherwise the phase will have no reference time and you can't determine the imaginary part of the impedance. If you do it at multiple frequencies and multiple loads then you should be able to get enough equations to find all the coefficients in the output impedance.

Of course, if you only care about exactly one frequency, and there is no global timing reference to be concerned with, then I guess it doesn't matter too much.
 
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1. What is the reactive part of output impedance?

The reactive part of output impedance refers to the portion of impedance that is caused by the reactive components in a circuit, such as capacitors and inductors. It is measured in units of ohms and represents the resistance to the flow of alternating current.

2. Why is it important to measure the reactive part of output impedance?

Measuring the reactive part of output impedance is important because it allows us to understand the behavior of a circuit when exposed to alternating current. It also helps us to design and optimize circuits for specific applications.

3. How is the reactive part of output impedance measured?

The reactive part of output impedance can be measured using a variety of techniques, such as using an impedance analyzer or a vector network analyzer. These instruments apply a known alternating current signal to the circuit and measure the resulting voltage and current, allowing for the calculation of impedance.

4. What factors can affect the measurement of the reactive part of output impedance?

There are several factors that can affect the measurement of the reactive part of output impedance, including the frequency of the alternating current signal, the accuracy of the measurement instrument, and the presence of other components in the circuit that may affect the impedance.

5. How can the accuracy of the measurement of the reactive part of output impedance be improved?

To improve the accuracy of the measurement of the reactive part of output impedance, it is important to use high-quality measurement instruments and to carefully control the frequency and other parameters of the alternating current signal. It may also be helpful to use calibration techniques to account for any potential errors in the measurement setup.

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