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:

                                    enter image description here

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:

Current through L1 Positive edge for I(L1) zoomed in

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.

enter image description here enter image description here

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?

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.

enter image description here

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.

enter image description here

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?

  • 1
    \$\begingroup\$ RE: the graph image on paper of "di vs dt": I suspect the axes on the graph should be i (current) on the vertical, and t (time) on the horizontal; not "di" and "dt". And in real-life, it would not have a flat plateau after the ramp; the plateau comes after the "spike up", and is the current limit of the power supply. Cheers. \$\endgroup\$ Commented Jun 22 at 21:53
  • \$\begingroup\$ Rohinb97, I suspect they wanted to know if you were ever driven into doing anything like this. When I got my first scope (Tek 2104) perhaps one of the first things I wanted to learn to do was to measure capacitance and inductance. (So many years ago, such instruments were way over my expense account limit.) If you have ever done much DIY -- if you truly love electronics as a hobby and not just as a paycheck -- then you would have been there, done that, and found more than one way to do this kind of measurement. They asked. You told them it's not a burning passion. They got the message. \$\endgroup\$ Commented Jun 23 at 8:40
  • \$\begingroup\$ @periblepsis It is a passion (I have a lab at home), and this is how I do it for both inductances as well as capacitances. I like to think ahead instead of intentionally damaging components. If you're using an oscilloscope with an 8-bit ADC then your LSB becomes ~20 mV on a 5V reading, which isn't a good enough margin to leave if you end up with a seemingly constant voltage level or a ramp you can't accurately calculate the slope for because of the error in measurement \$\endgroup\$
    – Rohinb97
    Commented Jun 23 at 15:57
  • \$\begingroup\$ @Rohinb97 1. Try marking your sims so we can tell what the waveforms are, for example, what is [V(n001) - V(n003)]. 2. Try reducing the duration of the PULSE on time from 2 seconds to something that is about 3 to 5 times the L/R time constant. 3. Careful with LTspice, the model for L has a default value of series R (something like 1mΩ from memory - read the help). 4. Using a PULSE is not exactly the same using a switch (the the first circuit you drew). \$\endgroup\$ Commented Jun 23 at 17:01
  • 1
    \$\begingroup\$ Moderators have had several flags on this from more than one point of view. This is a nobody-wins sort of situation where all may feel that their point has not been taken. The ideal is that if a question is modified that it is not substantially modified in intent (although some modification is usually unavoidable) and that the changes made by modification are well identified AND that all parties (those who may be affected by the modifications, those who make them, the moderators, ...? ) seek to minimise adverse consequences, feelings and reactions. || In a perfect world ... :-) . \$\endgroup\$
    – Russell McMahon
    Commented Jun 30 at 23:28

5 Answers 5


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


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:






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.


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:




  • \$\begingroup\$ The first comment you quoted me for was in context or the RL circuit (assuming a lossy inductor with a parasitic R in series). The interviewer said that the peaks I showed at the ends (L2 in my sim) were because of the measurement setup, which I'm claiming is clearly wrong \$\endgroup\$
    – Rohinb97
    Commented Jun 23 at 16:07
  • \$\begingroup\$ As for why I said the voltage would be "steady" was because in my mind I assumed the value of L to be low (I do HW design & IC design, so this is where my mind went), which basically meant that you're shorting an inductor. This is the entire reason I asked this question since plugging an unknown value & damaging the L/DC source doesn't make sense to me. Frankly, I think I deserved credit because I realized these issues before he could reveal his answer & suggested a method that has no drawbacks like this. +1 for the practical tester \$\endgroup\$
    – Rohinb97
    Commented Jun 23 at 16:13
  • \$\begingroup\$ But I do see your point about this being the spirit of the question: I have an MS + 3 YoE, and the questions he asked before were pretty basic as well... \$\endgroup\$
    – Rohinb97
    Commented Jun 23 at 16:18
  • \$\begingroup\$ A couple of points: 1. RE: your sims: I cannot interpret the results; Waveforms labelled as, for example, (V(n001) - V(n003), are not marked on the schematic - we cannot tell what is node (n001) and (n003). 2. What are these "peaks" you are referring to? Voltage, or current? And for which component? \$\endgroup\$ Commented Jun 23 at 16:31
  • 1
    \$\begingroup\$ 3. Your comment: "This is the entire reason I asked this question since plugging an unknown value & damaging the L/DC source doesn't make sense to me. " Most lab DC supplies are current limited, and cannot be damaged by a heavy load or even a short-circuit. Also, inductors are pretty robust devices, yes, they can saturate but that does not of itself cause damage. My "suggested practical inductance tester" shows various means by which an inductor can be tested without causing damage to itself or the test equipment. \$\endgroup\$ Commented Jun 23 at 16:35

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.

  • 1
    \$\begingroup\$ I think "shoot up" is a reference to the saturation effect. \$\endgroup\$
    – Dave Tweed
    Commented Jun 22 at 19:54
  • \$\begingroup\$ @DaveTweed that's how I read it. \$\endgroup\$
    – Andy aka
    Commented Jun 22 at 20:03
  • \$\begingroup\$ @Andyaka please flag before taking matters into your own hands \$\endgroup\$
    – Voltage Spike
    Commented Jun 30 at 2:14
  • \$\begingroup\$ @Rohinb97 It looks like you changed the question after answers were posted, this can be a problem because it can make the answers invalid and invalidate with that people spent on the answers. It's best to start a new question and reference the old one. In one of the edits there was also conversation (please reserve that for the comments section) \$\endgroup\$
    – Voltage Spike
    Commented Jun 30 at 2:17
  • \$\begingroup\$ @VoltageSpike No one's answers are being "invalidated" here. There were only two answers before I made the edits & I was getting repeated comments about the timescale & the original question. I had to make these edits & do new simulations to actually clarify what I meant in the actual question. The new question doesn't change in any way, I still kept everything the original question had & clearly mentioned the Q is edited. AndyAka's answer was never relevant - even the original question was about measuring transient slope vs RL TC and he still didn't acknowledge why the RL approach was wrong \$\endgroup\$
    – Rohinb97
    Commented Jun 30 at 8:05

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

enter image description here

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.

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.

  • \$\begingroup\$ +1 for mentioning my simulation isn't representative of the results of the switch setup \$\endgroup\$
    – Rohinb97
    Commented Jun 23 at 23:15

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.


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}

enter image description here


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