2
\$\begingroup\$

Real inductors can be represented as perfect inductor plus a series resistance \$R_s\$.

The impedance of this component is \$Z_{L+R_s}=R_s+j \omega L\$ therefore $$\frac{1}{Z_{L+R_s}}=\frac{1}{R_p}-\frac{j}{Z_{Lp}}$$

With \$R_p=\frac{R_s^2+(\omega L)^2}{R_s}\$ and \$Z_{L_p}=\frac{R_s^2+(\omega L)^2}{\omega L}\$

Which means that the same system can be seen as the parallel of a perfect inductor \$Z_{L_p}\$ and a resistance \$R_{p}\$.


I never tried actually, but if I use a RLC bridge or simply a ohmeter to measure the resistance of a real inductor (the same way as if it was a resistor), what do I measure? \$R_s\$ or \$R_p\$? Or I do not measure anything?

If I measure \$R_s\$ will the measurement depend on the frequency of the RLC bridge?

\$\endgroup\$
2
  • \$\begingroup\$ With an ohmmeter $\omega$ is 0, so $Z_{L_p}$ is undefined. So the system cannot be seen as the parallel of a perfect inductor and a resistance. (Also can anyone tell me what I'm doing wrong with Mathjax?) \$\endgroup\$
    – user253751
    Feb 14 '17 at 0:27
  • \$\begingroup\$ @immibis as $\omega$ gets very small, $Z_{L_p}}$ gets very large, and $R_p$ gets very close to $R_s$. So the $Z_{L_p}}$ has negligible effect on the parallel impedances and the limit of the impedance when $\omega = 0$ is $R_s$, as it should be. But all this is irrelevant to measuring $R_s$ - just use an ohmmeter! \$\endgroup\$
    – alephzero
    Feb 14 '17 at 3:07
3
\$\begingroup\$

If I measure \$R_s\$ will the measurement depend on the frequency of the RLC bridge?

Generally, your bridge measurement will be larger than an ohmmeter measurement. How much larger depends on the quality of the inductor.

  • Skin resistance of the inductor's wire will increase at higher frequencies.
  • Losses in an iron core usually increase with frequency.
  • B-H nonlinearity of core material affect both inductance and resistance.

Some bridges allow measurements of either series RL, or parallel RL models. The more common setup is series RL.
Note that you can detect core saturation by varying the bridges' AC amplitude. You may find that small amplitude yields high-Q (small series resistance), but increasing amplitude kills inductor Q, and series resistance increases.

\$\endgroup\$
3
\$\begingroup\$

A common Ohmmeter measures with a constant voltage or current. So after a short time - at usual values less than a second - you get Rs.

\$\endgroup\$
0
2
\$\begingroup\$

It depends on how you make the measurement.

Lets stop and look at your calculation there a little. You've shown that a perfect inductor in series with a perfect resistor has the same impedance as a perfect inductor in parallel with a (different) perfect resistor.

Any real measurement system measures impedance. A single impedance measurement doesn't tell you whether you're measuring a parallel or series arrangement. If you're making a measurement with an LCR meter, you can normally choose to have the measurement displayed in one of several ways: a complex impedance; or as a resistance and capacitance/inductance. If you select the second option, then the meter works out the resistance using a series or parallel model, and you, the operator, have to tell it which model to use.

What about an handheld mulitmeter? Well, they have only one frequency they can measure at, 0Hz, also known as DC. If you look back at your maths, then when \$\omega\$ is zero, \$Rs=Rp\$, and that is the number which will appear on the meter.

So how can you tell the difference between a series and parallel system? You have to take more than one measurement, at different frequencies. If you take several impedance measurements, then calculate the inductance and resistance using one model, and get the same result each time then you have chosen the right model. If you get different results, then try the other model. If neither model fits, then the system you are measuring is more complicated than just two components.

For a real indictor, the model of a perfect inductor in series with a perfect resistor is pretty good. For a real capacitor, the model of a perfect capacitor in paralell with a perfect resistor is good. If you measure up to very high frequencies, then you'll need a more complicated model to accurately represent the real component.

\$\endgroup\$
0
\$\begingroup\$

Measuring the resistance of a real inductor?

with a regular ohm meter?

simply a ohmeter to measure the resistance of a real inductor (the same way as if it was a resistor), what do I measure?

the resistance of that real inductor.

Under some conditions, using the voltage range, you can measure the impedance of an inductor.

\$\endgroup\$
0
\$\begingroup\$

Real inductors can be represented as perfect inductor plus a series resistance \$R_s\$.

At low frequencies where skin effect and inter-turn capacitance are negligible that is a reasonable approximation of a real inductor. Always remember though that it is only an approximation.

Which means that the same system can be seen as the parallel of a perfect inductor \$Z_{L_p}\$ and a resistance \$R_{p}\$.

Not really.

I mean sure you can calculate those numbers but they vary wildly with frequency. They aren't a useful representation of the component. For whatever reason we can (at room temperature at least) make much better insulators than we can conductors and the characteristics of our inductors (which fundamentally are coils of insulated wire) reflect that.

I never tried actually, but if I use a RLC bridge or simply a ohmeter to measure the resistance of a real inductor (the same way as if it was a resistor), what do I measure? \$R_s\$ or \$R_p\$? Or I do not measure anything?

Your multimeter works at DC so it will measure the DC resistance of the inductor.

Your impedance bridge will measure the overall impedance of the component at a given frequency. How it presents this to you is up to the designer of the measurement system but most likely it will express it as a series combination as that is what typically makes the most sense.

\$\endgroup\$
1
  • \$\begingroup\$ Impedance bridge will give a+jb; both configurations have the same impedance. \$\endgroup\$
    – Chu
    Feb 14 '17 at 18:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.