# Inductance Measurement: Error Due To Series Resistance

#### BlackMelon

Joined Mar 19, 2015
168
Hi!

Please refer to the word file. I am thinking of measuring inductance of an inductor by applying a DC voltage onto it. Then, calculate L = V/(di/dt).
I wonder if the series resistance (R) of the voltage source plus of the inductor under test would cause much error.
So, I derive the transient responses of the cases with and without R.
I calculated the error between the two cases and got the indeterminate form.
I would like to know how to properly analyze this problem?

Thanks!
BlackMelon

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#### Ian0

Joined Aug 7, 2020
9,533
Hi!

Please refer to the word file. I am thinking of measuring inductance of an inductor by applying a DC voltage onto it. Then, calculate L = V/(di/dt).
I wonder if the series resistance (R) of the voltage source plus of the inductor under test would cause much error.
So, I derive the transient responses of the cases with and without R.
I calculated the error between the two cases and got the indeterminate form.
I would like to know how to properly analyze this problem?

Thanks!
BlackMelon
The further you get into the test the larger the error!
At t=∞ I=V/R, the inductance is no longer in the equation.
Use the segment starting from t=0. The smaller the time interval the less the contribution from the series resistance.
Just to add to your problems, the inductance will vary with the DC current (which is nothing to do with the series resistance)

#### crutschow

Joined Mar 14, 2008
34,075
I would like to know how to properly analyze this problem?
The easiest way is to simulate it in a Spice type simulator and let the computer do the calculations.

Example below for a 1H inductor and a series resistance of 10mΩ, 100mΩ, and 1Ω;

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#### MisterBill2

Joined Jan 23, 2018
17,885
The short answer as to will the resistance cause any error in the inductance measurement is YES BUT. Using the method described in post #1 there will be some error due to the resistance being in series with the inductance, as the inductor model shows.
There is a work-around that is probably simpler to implement. That would be to measure both the DC resistance and the AC impedance, and then vector subtract the resistance from the impedance, which will give you the inductive reactance at the test frequency. From the reactance and the frequency you can calculate the inductance.