# Characteristic impedance of AWG 33 copper magnet wire

1. ### CopyOfA

24
I am working on impedance measurements of close-wound electromagnetic coils. I am using an Agilent LCR meter to measure the impedance of these coils over a frequency range of 20Hz-2MHz. When I perform the measurements, I get impedance magnitude and phase angle in degrees. I would like to construct a Smith chart of these measurements, but I am unsure of how to measure the characteristic impedance of the coils. These are single strand, AWG 33 copper wires - diameter of approximately 180μm including the insulation. Thus, they are not traditional transmission lines, based on my understanding. What is the best method to measure the characteristic impedance of these coils? Is it simply the DC resistance? Thanks for your help.

2. ### Baluncore

3,012
A coil is not really a transmission line, it has an impedance, but not a characteristic impedance. AWG 33 copper magnet wire would only have a characteristic impedance when part of a transmission line such as when in conjunction with another parallel strand or above a ground plane.

The impedance of your coils at a particular frequency will be the inductive reactance in series with the wire resistance. The resistance will be slightly dependent on frequency due to skin effect. There will also be a slight parallel capacitive reactance, (negative), due to lead and terminal capacitance.

Measure the inductance and series resistance with the meter. Compute reactance from inductance.
Normalise the values to the chart reference impedance, then plot them on the Smith Chart.

3. ### CopyOfA

24
Thanks for your help. I'm a bit of a newbie to the whole transmission line theory. I'm hoping to perform some reflectometry measurements on the coil -- do you think this is possible? Or do you have any literature that could point to the use of reflectometry on single wires (i.e. those that are not transmission lines)? I've also read that the reflection coefficient can be derived from the complex impedance. Again, it seems this requires the knowledge of some kind of characteristic impedance...

In performing these impedance measurements on the coils, I've come to some of the same realizations that you pointed out. In the lower frequency range, the coils are dominated by resistance in that the impedance Bode plot is essentially a flat line with no slope. However, as the frequency increases the slope changes to +20 dB/decade, indicating an inductive region. At some point in the response, the coil reaches anti-resonance (maximum value of the impedance), but then begins to decrease at -20 dB/decade, indicating a capacitive region. I think this capacitive region is due to the effects of the turn-to-turn insulation capacitance of the coil.

In order to compute the reactance and resistance could I not simply separate the real and imaginary portions of the complex impedance? If I do this, the resistance is highly frequency dependent.

4. ### Baluncore

3,012
No. Reflectometry requires a transmission line. Unless you do something unusual, such as wide space your coil on a conductive core, you do not actually have a transmission line. You might connect your coil as a termination at the end of a transmission line, then observe the reflection of the line to load mismatch. It won't do you much good.

The resistance is not highly frequency dependent, it will only change slightly over the frequency range. Compute the skin effect to estimate the RF resistance.

You know the inductance from your lower frequency measurements.
You know the parallel capacitance from your higher frequency measurements.
You measured series resistance over the frequency range.

For any frequency you can solve numerically for the complex Z = Parallel( XC, Series(R, XL) )

5. ### CopyOfA

24
Sorry to reply so late; I was out of the country with limited internet access. If you have some time, I have some remaining questions.

I've noticed something entirely different in my experiments. Here are some resistance and reactance measurements from a couple of coils that I've worked with.

I computed these values by splitting real and imaginary impedance measurements taken on an Agilent e4980A LCR meter.

$$Z=\left|Z\right|e^{j\theta} = R + jX$$
where, ##R = \Re\left\{\left|Z\right|e^{j\theta}\right\}## and ##X = \Im\left\{\left|Z\right|e^{j\theta}\right\}##.

Your last response seemed to imply that if I removed the skin effect from the resistance measurements, the resistance would remain essentially constant over the frequency range. Based on my understanding of the skin effect, I would expect the resistance to increase as frequency increases, but that does not happen here. Do you have any thoughts as to the reason for this? Again, thanks for your help.

6. ### Baluncore

3,012
It appears that up to 100 kHz things are well behaved.

Where the reactance changes sign there is a resonance.
There are a couple of resonances, one at about 500 kHz and another at 1 MHz.
The very high Q of the resonances shows the resistance is much lower than the reactance at resonance.

It appears the instrument is being dominated by the reactance at resonance.
A series resonant element in series with the DC resistance will look like resistance.
It is being fooled about the resistive component.
That could be because of the multiple turns and cross coupling.
The instrument can only tell the difference between R and jX by the phase of the returned signal.

I suggest that you read the resistance at 100 kHz, then replace the coil with an equivalent resistor. That will tell you if the resonance is in your instrument / leads.

The resistance should steadily rise in proportion to the square root of the frequency.
That is from the skin effect equation.
All other changes to resistance estimates are measurement artefacts.

7. ### marcusl

2,138
I wouldn't say that the Q is "very high". The width of the resonance divided by its center freq looks from the plot to be of order 5--well within expected values for a simple coil. BTW, the resonance is caused by capacitance between the turns in parallel with the inductance,

8. ### Baluncore

3,012
We are in the dark here. We do not know the size of these coils, but I would be very surprised if it was a case of adjacent turn resonance. I would expect the capacitance and inductance of two adjacent turns to be very small and so give a broad resonance at hundreds of MHz, not sharp, at 1MHz. I expect the resonance demonstrated here is due to the capacitance of the terminals and connections to the meter, reacting with the bulk inductance of the coil.

The coil is AWG33, close wound. We need to know;
What is the coil length?
What diameter?
How many turns on the coil?
How is the connection made to the meter?

3,973

Debra,

Dave

10. ### The Electrician

856
Try plotting your data with a logarithmic vertical scale.

11. ### CopyOfA

24
Thanks for all the input. I'll try to answer Baluncore's questions.

1) Wire is AWG 33, which implies a wire diameter of ~180μm. The diameter of the copper is ~173μm and there is ~3.5μm of insulation thickness surrounding the copper. The insulation is rated as Class F.

2) Coil length is unknown. However, the resistance of Coil 1 is approximately 11.4Ω and Coil 2 is 11.2Ω. Using the geometry and resistivity of copper (1.678e-8), the approximate length can be determined. I've calculated the lengths as: ~63.9m and 62.8m for Coil 1 and Coil 2, respectively.

3) Inner diameter of the bobbin is 0.568in or 1.44cm

4) How many turns is unknown. There are approximately 2 layers of coil wrapped around the central bobbin. See pictures below.

Coil1:

Coil 2:

5) Connection to the LCR meter -- see pictures below.

This image only shows the connection with Coil 1, but both were connected in the same manner. The blue wires were soldered onto the ends of the magnet wire and the other ends are attached to the LCR meter via the connection box shown. All measurements were made while the coil was placed on the solenoid valve stem as shown. Thus, there is a (soft) magnetic core inside the solenoid.

I've seen this resonance behavior before in much larger coils as well. For example, this coil (shown below) produces a similar impedance response.

The impedance Bode plot is shown below. Sorry, the frequency resolution is not as good with this one, as I was first beginning to use the LCR meter. And the magnitude units should read as dBΩ and not just Ω...

The LCR meter outputs ##\left|Z\right|## measured in Ohms and ##\theta## measured in degrees.

I've made measurements with the coil (only for Coil 1) attached directly to the LCR meter as shown below. The impedance measurements were taken with and without the (softly) magnetic core.

Measurement setup:

Impedance Bode plot with core:

Impedance Bode plot without core:

Thanks again for all the assistance. I really appreciate the input.

Last edited: Jul 23, 2014
12. ### jim hardy

5,321
Hmmm. at 100khz 50 ohms with core, 55 ohms with no core ?

But at 100 hz, maybe 26 with core, 22 without.

interesting.

13. ### Baluncore

3,012
In post #11, the "coil 2" picture shows a metal sleeve inside the bobbin. If that is stainless steel it will reduce the inductance very slightly. These coils have a ferromagnetic conductive core while being tested. Above about 10 kHz, skin effect will eliminate almost all of that ferromagnetic material from the inductance computation.

By length of the coil I meant the solenoid length, not the wire length. I was trying to compute the inductance from the solenoid dimensions.

14. ### jim hardy

5,321
yes, skin effect in the iron as opposed to the copper?

And might it be worth twisting his test leads together to reduce enclosed area?

15. ### Baluncore

3,012
A conductive material in the core prevents the magnetic field from entering that volume of core, which distorts the ideal field, reducing the coupling between turns and that in turn reduces inductance. An iron core will increase inductance significantly at lower frequencies but less as the frequency rises, a brass, aluminium or stainless steel core will actually reduce the inductance below that of an air core.

I expect twisting the leads will probably raise the external lead capacitance and so lower the resonant frequency. The reduction in area will not reduce inductance by much and so not increase the resonant frequency. It would be an interesting experiment to try.

This circuit has a capacitance involved in the resonance that has not yet been positively identified. Where the leads come from the same end of the coil, the capacitance may be “coil end to coil end” capacitance due to the overlaying of the first layer of the coil by the second layer at the common terminal end.

16. ### The Electrician

856
The capacitance involved in the resonance(s) is not in the leads or the instrument; it's the distributed capacitance of the coil itself. There is capacitance from each turn in the coil to ALL the other turns. See here for info:

http://www.g3ynh.info/zdocs/refs/Medhurst/Med35-43.pdf

I wound an inductor similar to the OP's, but with 30 gauge wire rather than 33 gauge. There are a total of 210 turns, roughly in 2 layers:

Then I generated a plot of the impedance and phase angle of the impedance over a frequency range of 20 Hz to 5 MHz. The vertical scale of the phase goes from -100° to 100° and is linear. The vertical scale for the impedance is logarithmic as shown. Phase is yellow and impedance is green:

There are several self resonances as the sweep approaches 5 MHz; this is typical for a coil of this type.

Here's another sweep, but showing AC resistance in green:

A question for the OP. What is your purpose (goals?) in all this?

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17. ### Baluncore

3,012
Your coil shown has two layers, so it is wound back on itself. The significant capacitance is not distributed amongst all turns, but between the coil ends where they lie against each other.

That paper is considering only single layer space wound coils on pre-threaded formers. Those coils are then grounded at one end, so it is not surprising that each turn has a small capacitance to one end of the coil, and all nearby grounds.

If the coils were close wound as a single layer, then the resonance would be at a higher frequency. An orderly winding would have normal adjacent turn capacitance, but with a significant degree of capacitive isolation between turns at opposite ends of the coil. That is because the turns at the ends of the coil cannot directly “see” each other electrostatically. Only adjacent turns would effect each other as electrostatic capacitors. The resonance would be that of the capacitance of two adjacent turns with the inductance of those two turns alone, but with many of those resonant elements in series.

We now see that the coils are scramble wound with no attempt made to keep the ends or layers of the coil apart. The ends of the coil can capacitively influence each other very strongly which will have a completely different self capacitance and resonance behaviour to an orderly close or space wind. It is the two layer wind that has brought the resonant frequency significantly down, by bringing the terminal ends together.

If the coil is for an iron-cored electromagnet, self resonance will matter little, especially when the magnetic circuit is closed. The forgotten art of winding coils with low self-capacitance need not be revived.

18. ### CopyOfA

24
The sleeve inside the bobbin is non-conductive. It is used to direct the magnetic field into the stem -- these coils are used to operate a solenoid valve and so directing the field into the stem (where the actuator lies) provides more efficiency.

Sorry about that... The length of the coils are: Coil 1 - 1.023 in., Coil 2 - 0.853 in.

I too believe that the capacitance is due to the turn-to-turn interaction of the voltage differences (between the turns) with the magnet wire insulation. See http://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=602562. If the link doesn't work the paper is: "Self-Capacitance of Inductors" by Massarini and Kazimierczuk.

Also see post #11, picture 4. This coil was produced as a component of a Parker solenoid valve -- not hand wound. The impedance response shows a similar behavior though the two ends of the coil do not see each other electrostatically. Below is the impedance response for this coil.

Forgive me if this sounds completely ignorant, but what is AC resistance and how is it different from impedance?

19. ### Baluncore

3,012
What is it made from, iron powder or ferrite? Does it pass all the way through the coil or is it only at the ends?

Complex impedance, Z = R + j X, where R is resistance, X is the reactance.
DC resistance is R at DC = zero frequency. AC resistance is R at non-zero frequency.

What might DC reactance be, is it always zero, – infinity or maybe undefined?

20. ### The Electrician

856
Here's a sweep of a 1 meter long piece of 30 gauge wire. This image shows the reactance (imaginary part of the impedance) in green and the resistance (real part of the impedance) in yellow. The reactance increases as the frequency increases because it is due to the inductance of the wire. The resistance is another phenomenon due to losses in the copper to the flow of electricity. Those losses occur whether the current is DC or AC. Even though the reactance increases with frequency, one might think the resistance should be constant with frequency, but due to skin effect, it does increase when the frequency gets high enough. We see a slight increase in resistance as the frequency approaches 1 MHz:

This image shows the magnitude of the impedance and AC resistance. Since impedance is denoted Z = R + jX, with R being the resistance and X the reactance, the magnitude of the impedance is given by SQRT(R^2+X^2); its value can't be less than either component (R or X). Thus, the impedance curve is coincident with the resistance curve at low frequencies, whereas the reactance goes below the resistance curve:

The next two images show the same two type sweeps but of a 1 meter piece of 10 gauge wire. The skin effect comes into play at a much lower frequency, and the AC resistance of the wire at 1 MHz is about 10 times what it is at 100 Hz:

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