How do we measure inductance with a DC source?

Question

I just got this question asked by a Staff Engineer for an entry-mid level job at a large company, and I suspect the interviewer might be wrong here. I had this interview on Monday and I got a rejection from him because of lack of experience. This has been bugging me ever since, so I wanted to understand his view and learn from this experience.

                                 

Consider the above circuit. Assume we have a known DC voltage source and a switch we can control (which we're switching on at time \$t_1\$ and switching off again at \$t_2\$). We need to find the value of the inductance by monitoring its behavior. We can probe anywhere we need and measure V or I.

My answer was: to get a reliable measurement, include a known resistor in series with the switch/inductor, probe for voltages, record the decay in transient voltage across the L at times \$t_1\$ and \$t_2\$ (which we'll see as small peaks on the oscilloscope), roughly estimate the time constant, and find L.

My interviewer said: The voltage peaks at times \$t_1\$ and \$t_2\$ won't be a function of the inductance, but based on “the nature of how you're testing it”. His view was to ignore those peaks and focus on the transient current between \$t_1\$ and \$t_2\$.

Our discussion after that revealed that he was assuming the inductance to be in an order of 1-10 \$Hs\$, but I was considering it in an order of \$\mu Hs\$.

The discussion that lead to this conclusion:

I pointed out the voltage across the inductor will only be there if there's a change, else it'll just be steady. He just said “Are you saying that an inductor is a resistor?”. His rationale was that if the current between \$t_1\$ and \$t_2\$ was steady, then the inductor is effectively a resistor. That's when I realised that he was assuming the value of the inductance to be in an order of 1-10 H, but I was assuming this to be in \$\mu H\$s. My thinking was that it's basically a short because of a low value (hence steady), but I didn't say anything since we were out of time.

He drew a curve for “di vs. dt” (instead of I vs t) to demonstrate his method & claimed it's slope would be L (I think this is definitely wrong). He had it shoot up after some time, and asked me if I knew about this phenomenon. To which, I had no idea about why it would shoot up for no reason.

It was a rough diagram, but this is what he drew:

                                   

I think the method he proposed (measuring the slope of the transient current) is unreliable for a couple of reasons:

• Measuring an inductor by only using a DC voltage makes no sense, since we don't know the inductor value. We won't be able to infer the magnitude beforehand, hence relying on the accuracy of the slope is not a good idea
• The amount of current we'd draw is unknown, which would mean we'd be damaging the device if we test it haphazardly. Known resistors would limit the current as well.

I also tried simulating the cases I was concerned about in LTspice, and the simulation follows my intuition:

Current across the inductor L1 (1\$\;\mu H\$) in the circuit below is going up to ~5 kA in less than 10 ms:

Voltage measured across the inductor L2 in my simulation (V(n001)-V(n003) is the voltage across the inductor L2). Notice the peaks I talked about at the edges. Now, this will have a linear drop in voltage as time passes because the current is independent of time, making the I vs T's slope V/L, not L.

So, apart from the miscommunication between us about the order of magnitude, what is the correct way to measure an unknown inductance?

And what was the immediate shoot up in \$di\$ the interviewer was referring to?

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EDIT: Thank you so much for your answers! People have already answered that the shoot-up was due to the saturation of the inductor core (thanks @Andyaka!), but I see that there might be some clarifications about the technical aspects of the original question that need to be made here.

Here are the claims made by the interviewer that are bothering me:

1. The interviewer said to not go with the method of using the time constant at times \$t_1\$ & \$t_2\$ by including a series resistor, and focus on using the transient current across the DUT to find the inductance value.
2. The peaks that I wanted to analyse were only due to experimental inaccuracies, not because of the RL circuitry.

There were other inaccurate claims (like L being the slope of the transient current across an inductor connected to a DC source without any series resistance, when it should clearly be V/L), but I'm not concerned about those inaccuracies.

The simulations I did in the original question to confirm my intuition were to counter these claims. The simulation for L1 shows an excessive current (~ 5kA) flowing through the inductor, refuting Claim (1). The simulation for L2 refuted Claim (2) - we see the peaks I predicted at \$t_1\$ and \$t_2\$ in a simulation free from experimental errors, meaning that those spikes are because of RL time constant.

All the answers below stick with Claim (1), saying that analyzing the slope of the transient current is much better than analyzing the time constant of the voltage peaks, despite the drawbacks claim (1) has.

So, why is this way of measuring an unknown inductance with a DC source preferred?

I performed another simulation to justify my points:

Below simulation is a clear example of the voltage peaks across the inductor L2 being dependent on the nature of the RL circuit, and not the errors in the experimental setup. Notice how the maximum current across the inductor L1 is being controlled by the inclusion of a known resistor (i.e. L2), which is a responsible way of dealing with an unknown component without damaging it by accident.

Here's an example with a 56 \$\mu H\$ inductor L3 (assumed this to be the SBCP-47HY560B inductor). As my initial simulation showed, an inductor in this order of magnitude will draw a lot of current (~5kA due to it's low parasitic resistance). But simply adding a 10 ohm resistor gave us a reliable measurement we can use to extract L without comprimising the component.

tl;dr Why do "experienced" engineers prefer measuring an unknown inductance with analyzing the transient current when connected to a DC source instead of focusing on the time constant by giving a step response with a known resistor?

Answer 1

My interviewer said: The voltage peaks at times t1 and t2 won't be a function of the inductance, but based on “the nature of how you're testing it”. His view was to ignore those peaks and focus on the transient current between t1 and t2.

The interviewer is correct. V1 is an ideal voltage source, so when the switch is closed at t1, it defines the voltage across the inductor; the voltage across the inductor will be the same as V1 until t2, when the switch is opened.

What happens at t2 when the switch is opened (and after) is far more complex (which we can explore later), but is ultimately not relevant to the aim of the question, which was to establish the value of L. So, yes, the interviewer is correct: ignore the time before t1 and after t2, and focus only on the time between t1 & t2, when the inductor voltage is defined by V1.

I pointed out the voltage across the inductor will only be there if there's a change, else it'll just be steady.

I can see where this is coming from (Lenz's law, perhaps?) but you must be careful when applying these "rules of thumb".

Any time you find yourself looking at a problem, you must identify what are the independent variables, and what are the dependent variables. In this case, the independent variable is the voltage across the inductor - as explained, during the period of interest (t1 < t < t2) this is set by V1 (an ideal voltage source). The dependent variable is now the current through the inductor.

So the classic equation for the relationship between voltage and current for an inductor must be re-arranged to make V appear on the right-hand side:

$$ V = L\dfrac{di}{dt} ⇒ \dfrac{di}{dt} = \dfrac{V}{L}\ ⇒ {i}(t) = \dfrac{1}{L}\int{V(t)}{dt} $$

Since V(t) is constant for (t1 < t < t2), then we can replace it with V1 (a constant), and solve the simple integral to get:

$$ {i}(t) = \dfrac{1}{L}{(V1)}.{t} $$

So the current will be a ramp, starting from 0, and increasing with a constant slope determined by V1 and L.

In practice, there are limits to this simple explanation. The first limit is the series resistance of the circuit. We have assumed that the current is related to the voltage only by the inductance, as per the equations above. However, if there is sufficient resistance present (regardless of where that R exists: the voltage source, the inductor, or the connecting cables), then that assumption is no longer valid, and the relationship between current and time becomes more complicated.

The next limit is the saturation of the inductor. If the opening of the switch is sufficiently delayed, the current will eventually reach a point where the magnetic circuit of the inductor becomes saturated; that is, its permeability drops dramatically. At this point, the value of L dramatically reduces, thus the slope of the current waveform rapidly increases.

The last limit is the output current capability of V1. The current cannot exceed the output current limit of the power supply, V1. In practice, capacitors are placed across V1 and are used to "hold-up" the voltage across the inductor, to allow the current to momentarily reach values much higher than the output capability of the power supply.

Here is a suggested practical inductance tester. The key features are:

1. Current from the power supply is limited by R1.
2. C1 provides most of the energy to build up the current in the DUT (Device Under Test).
3. R2 provides a means to measure the current in the DUT via a scope.
4. D1 provides a path for the current to reset back to zero gracefully after the switch is opened. (Scope is represented by voltmeters VM1, VM2.)
5. UPDATE: The op-amp and NPN BJT provide accurate control over the voltage applied to the DUT, regardless of the voltage across the current measurement resistor R2, and the droop of C1 voltage as it is discharged during the test pulse.

^{simulate this circuit – Schematic created using CircuitLab}

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UPDATE #1:

Responding to OP comment below, this section:

Why should we measure an unknown L by measuring the slope of transient current with a DC voltage when we can use time constants?

1. Because it is easier to measure constant voltages, and slopes of lines (rise over run), and then do some simple calculations, rather than try to estimate time-constants of exponential curves. Further, this process can be automated: the current waveform (a rising line with a relatively constant slope) can be put into a simple differentiator to extract the slope.

2. One does not first have to guess a suitable R-value to put in series with the DUT, one can simply adjust the output V of the DC supply to get a slope compatible with the measurement system (amplitude, time-base).

3. The saturation current is easily observed and measured.

4. Non-linear magnetisation (non-constant L value) is more easily observed and quantified. This point is quite important; many magnetic materials, particularly distributed air-gap materials, show distinct non-linear magnetisation; this means the inductance value changes with current (usually L reduces as current increases). This feature is sometimes exploited in switch-mode converters to reduce the ripple current at light load thus extending the region of continuous current operation. Swinging inductors are also exploited in filters, where they provide reduced size and weight compared to non-swinging magnetics for a given degree of relative filtering. Refer links below:

https://www.mag-inc.com/Media/Magnetics/File-Library/Product%20Literature/Ferrite%20Literature/fc-s4.pdf

https://patentimages.storage.googleapis.com/49/f2/91/807a9ccdb6c5c5/US3605003.pdf

https://patentimages.storage.googleapis.com/0f/9b/f1/08c4a8188ef4be/US20150016162A1.pdf

https://library.e.abb.com/public/56a1ca38920d46ddc125746b0027f201/SwingingChoke-US-10.pdf

Continuing:

Obviously we'd be picking the known resistors (starting from bigger) & continue going towards smaller Rs to get to a workable reading if we're using the time constant method.

For this method, it would seem you would need to:
(a) have suitable resistors readily available, and
(b) make several attempts with different resistor values.

The method your instructor was probing you on (and what I proposed) does not require you to install different resistors; you merely adjust the output voltage of the (current-limited) DC supply.

I chose that curve as a clear example to disprove the claim the interviewer made & demonstrate the "peaks" aren't because of experimental errors but due to the inherent nature of the RL circuit.

Question: Are you referring to the example you have posted, where a very narrow "peak" of less than 1mV is the response you see from a circuit that is stimulated with a signal that is over 5,000 times higher than that (5V)?

If "yes", then:
I would not characterise such a waveform as being a "peak" - unless I was looking at a low-noise and very sensitive detector or similar. This is not that type of circuit; this circuit is intended to test an inductor at current levels that will cause it to saturate, which requires substantial energy. If I had seen these waveforms, then at first inspection, I would have assessed this "peak" as being most probably "an artifact of the simulation", or the experiment, and buried in the noise. I would suggest this is not an effective measurement system. It would be far too easy to miss this response (too low compared to the stimulus), or measure it inaccurately, leading to an inaccurate estimate for the inductance value.

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UPDATE #2:

My suggested tester can be improved by adding a series-pass transistor and a feedback controller to regulate the voltage across the DUT to be constant. This means the voltage across L is kept constant, regardless the voltage drop across the current measurement resistor, and the "sag" of the capacitor as it drains. I have updated my schematic to show this.

You may enjoy these links regarding power inductor testers:

https://sound-au.com/project250.htm

https://hackaday.io/project/194478-inductor-saturation-current-tester

https://ludens.cl/Electron/lmeter/lmeter.html

Answer 2

If you apply a DC constant voltage across an inductor that was previously carrying no current, the current initially ramps up at a rate of \$\frac{V}{L}\$. This comes from the standard formula of an inductor: -

$$V = L\dfrac{di}{dt}$$

In other words, \$\dfrac{di}{dt} = \dfrac{V}{L}\$.

Eventually (usually after a very short period of time) the current reaches a level where the inductor's core starts to magnetically saturate and current rises much more rapidly. This is because in saturation, the inductance starts to drop dramatically.

A bit later, the power supply can't deliver any more current and you hit the end-stops.

what is the correct way to measure an unknown inductance?

Given what you said was available I would look at the rate of rise of current and calculate inductance from that (knowing V).

And, what was the immediate shoot up in di the interviewer was referring to?

I'm not a mind-reader but it does sound you got the wrong end of the stick on this.

Answer 3

The amount of current we'd draw is unknown, which would mean we'd be damaging the device if we test it haphazardly. Known resistors would limit the current as well

True, but not in the spirit of the question. I think you're being asked to demonstrate that you know \$\frac{di}{dt}=\frac{V}{L}\$, and how to apply it.

I had no idea about why it would shoot up for no reason

There is a reason. Eventually, at some maximum current \$i\$, the core saturates, meaning that it no longer contributes to a further increase of magnetic field with current. At this point the inductor behaves as if it had no core. Inductance \$L\$ falls rapidly, and slope \$\frac{di}{dt}=\frac{V}{L}\$ increases commensurately.

My thinking was that it's basically a short because of a low value (hence steady)

I understand your concern that eventually current will be huge, but again in the spirit of the question, remember that \$\frac{di}{dt}=\frac{V}{L}\$ is never infinite. There's nothing constraining \$V\$. You can apply any voltage you like across an inductor, instantaneously, and the inductor will not oppose that, since it will be "far" into the future that the inductor will eventually be drawing significant current. We can assume that the experiment is over, once the slope \$\frac{di}{dt}\$ has been measured, and we can disconnect the source long before the voltage source is starved of current, or the core saturates.

Whether or not there is resistance in the loop, the relationship \$\frac{di}{dt}=\frac{V}{L}\$ holds true. The effect of resistance will be to slowly reduce the voltage \$V\$ across the inductor (as the resistor develops an increasing voltage across itself), which will of course reduce the slope \$\frac{di}{dt}\$. However, with or without resistance, initially current \$i\$ will rise at the same rate, since initially the voltage across the inductor is the same:

^{simulate this circuit – Schematic created using CircuitLab}

V1 applies 10V at time \$t=1\mu s\$, and you can see that immediately following that, both inductor currents \$i_1\$ and \$i_2\$ rise at a rate of:

$$ \frac{di}{dt} = \frac{10V}{1mH} = 10mA/\mu s $$

Knowing \$V=10V\$, it is sufficient to measure \$\frac{di}{dt}\$ immediately following time \$t_1\$ to determine \$L\$, regardless of loop resistance.

What happens after that measurement is taken, either due to current limits being reached, components melting, or core saturating, is moot. You just need to ensure that \$t_2\$ (source disconnection) occurs before any of these issues arise.

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Just a reminder, for your next interview, you should mention that at time \$t=t_2\$, when the switch opens, there's a huge voltage spike across the inductor, as loop impedance rises to infinity. It's a great way to destroy the measurement equipment.

Answer 4

I know my answer will sound simplistic (compared to the other ones in here), but I am making an inductance meter with Arduino and came up with a (not at all new) idea of having an inductor be the main component in a 555 oscillator, then measure said oscillator frequency with an Arduino.

The inductance would then be calibrated from the frequency measured. I will use this 555 oscillator circuit: https://www.edn.com/rld-based-astable-555-timer-circuit/

Edit: I do not answer your question fully, but do answer how would one measure an inductance. My approach is frequency based.

Answer 5

You seem to have been led astray by not considering an appropriate time interval for your plots of current vs time. When 5 V is applied across a 1 uH inductor, the rate of change of current will be 5 MA/s. If you want to look at a change in current of just a few amperes, you need to look at a much shorter interval than 5 s, or even 1 s. Here's a plot taken from the same LTspice schematic as you used, but with voltages of \$+5\$ V, 0 V, and \$-5\$ V, and looking over a 3.3 us span. You'll see at a glance that the magnitude of the inductance is the slope of the diagonal current traces.

\begin{align} L &= \frac{V}{dI/dt} \\ &= \frac{\text{5 V}}{(\text{2.5 A})/(\text{500 ns})} \\ &= \text{1 uH}. \end{align}