# Inductor impedance-Square wave?

Discussion in 'General Electronics Chat' started by electronice123, Feb 6, 2014.

1. ### electronice123 Thread Starter Senior Member

Oct 10, 2008
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0
Hi everyone,

I am wondering how to calculate the impedance of an inductor when a square wave is applied. I have an idea but I don't know if it's right.

100mH inductor, 10V, 50uS pulse

I would think V=L*di/dt could be rearranged and used to calculate the impedance.

di/dt=V/L

10V/.100H= 100 amp-seconds x .00005 seconds = 5mA
Then could we just use ohms law V/I=Z

So, 10/.005 = 2kΩ

The math works, but is this the correct calculate the impedance of an inductor when a square wave is applied to it?

Last edited: Feb 6, 2014
2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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What you suggest doesn't work as a means to finding an equivalent impedance.

Consider the apparent "impedance" at 25usec for instance - you'll obtain a different value.

Impedance generally has meaning at one particular sinusoidal frequency of interest, which is normally the source frequency - whereas a pulse has an infinite bandwidth. Hence, in the latter case impedance has no specific meaning.

3. ### electronice123 Thread Starter Senior Member

Oct 10, 2008
304
0
Well, I would not call it resistance either, that would make it even more confusing.

I know what you mean though, because if what I said above was correct then that would also mean the impedance (if we can call it that) was also related to the applied voltage (V/L=di/dt).

The formula is correct in determining the rate of current rise for an inductor.
And if that is related to voltage, then why are the calculations not correct in determining "impedance"?

What would be the correct way? Is there a correct way? Or am I just looking at this incorrectly and asking a question with no significance?

4. ### corefinder Member

Oct 6, 2011
55
0
Pulse frequency: 1/50 MHz
Inductance Impedance: X(l)=w*l

w=2*π *f
= 2*3.14*(1/50)*10^6
So, X(l)=w*100*10^-3
X(l)=12.56 kΩ

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
With respect to [linear] impedance there is proportional relationship between current and voltage when one is considering steady-state behavior. So for a given ideal inductance driven by a single frequency AC source this proves to be the case - albeit with the added complexity of phase displacement between terminal voltage and current.

Under transient conditions one applies differential equations or other methods such as Laplace transforms to develop the relationship between stimulus and response. Your context is transient behavior. I guess one can for instance think of a Laplace transform representation of passive energy storage circuit elements as being akin to representing their impedances.

Are you familiar with Laplace Transforms and their use in circuit analysis?

Last edited: Feb 7, 2014
6. ### LDC3 Active Member

Apr 27, 2013
920
160
You made an assumption that may not be correct. Does the voltage across the inductor really change by 10 V in 50 μS? or is it faster?

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Presumably the OP is considering current response in an inductor driven by a DC pulse of magnitude 10V and duration 50usec.

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
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It gets weirder....

Suppose this is an ideal inductor driven by an ideal voltage pulse source. At the completion of 50usec the current would indeed be 5mA. When the source pulse voltage drops to zero at 50usec the current of 5mA flows on forever - assuming this was a single pulse stimulus. What is the apparent "impedance" at t=100usec with a current of 5mA and terminal voltage of zero?

9. ### LDC3 Active Member

Apr 27, 2013
920
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I would expect the driving circuit to be able to produce the desired waveform.
What I was trying to point out was whether the current was still increasing, or had it been stable for some time (say 20 μs).

10. ### WBahn Moderator

Mar 31, 2012
18,087
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That di/dt is the instantaneous current change in the circuit.

10V/100mH = 100 amps PER second, not 100 amp-seconds.

The 50μs time you are grabbing is the width of the pulse, which is NOT the differential time associated with the ramp. You are making the mistake of grabbing anything that happens to have the units you want and just throwing it as an equation. Doesn't work that way.

Your pulse can be written as the sum of a series of sinusoidal signals. Each signal has it's own frequency, amplitude and phase. You can find the response of the circuit to each of those signals, with the impedance of the inductor being calculated for each signal frequency in turn. You can then add the individual responses up to get your total response.

But even then you can't use steady-state techniques to find the response to a single, isolated pulse. You need to use either Laplace transforms or Fourier transforms (or just solve the differential equations directly) to do that.

11. ### WBahn Moderator

Mar 31, 2012
18,087
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Not all that weird at all. This is exactly what you do with a superconducting magnet! You apply a charging voltage to it long enough to get the desired current. You then short the supply leads with a superconducting "persistent-mode" switch and then remove the supply. The current then runs forever (well, okay, it generally drops at the rate of a percent per century or so due to parasitics).

Since it is in DC steady state, the impedance calculation is trivial. X=jωL with ω=0r/s. So the impedance is 0Ω.

12. ### electronice123 Thread Starter Senior Member

Oct 10, 2008
304
0
Ok, Well I have to admit I know nothing about Laplace & Fourier transforms. I only got as far as algebra in school. But I can learn anything.

So, anyone care to go into detail?

I'll start by reading the book section 'Square wave signals'.

Thanks everyone for all the help so far.

Last edited: Feb 7, 2014
13. ### crutschow Expert

Mar 14, 2008
13,496
3,373
The frequency-domain concept of impedance value in ohms is used for sinusoidal excitation and doesn't readily apply in the time-domain for a square-wave or other non-sinusoidal signal unless you use Laplace transforms which can get rather messy. In the frequency-domain the inductive impedance goes from 0 ohms at DC, to infinite ohms for the infinite frequency of the zero rise and fall time of an ideal square-wave. So the easiest way to calculate the resultant current is in the time-domain using V = L*di/dt or di = V/L dt. Thus the application of one 10V, 50μs pulse from an ideal voltage source into a grounded 100mH inductor will cause an increase in the inductor current of 10/100mH * 50μs = 5mA, as previously calculated. If there is no other impedances in the circuit then each additional pulse will increase the current by another 5mA. So you can't really assign a simple impedance value to that type of behavior.

14. ### russ_hensel Distinguished Member

Jan 11, 2009
820
47
Get a cap whose value is know accurately, put in series with a 10 K resistor and the inductor ( resistor on high side )
Scope the voltage across the LC
Excite with a square wave. If you are any where near resonance you will see a ripple that makes it easy to adjust the square wave to the resonate frequency.
Calculate the inductance from the cap and frequency.

I am thinking of doing an article on this technique, but the above should serve.

15. ### davebee Well-Known Member

Oct 22, 2008
539
46
If the wave is square with a high (the pulse) of 50 us, then I'd think it would also have a low of 50 us and would be described as a 10 kHz square wave.

A 10 kHz square wave will have a 10 kHz fundamental sinusoidal component plus odd harmonics at lower strengths, 1/3 the strength of the fundamental at 30 kHz, 1/5 the strength at 50 kHz, and so on.

So your inductor will react most strongly to the strongest 10 kHz component.

Calculating the impedance at 10 kHz gives:

Xl = 2 pi fl = 6.28 * 10000 hz * 0.1 H = 6280 Ohms

That value is probably pretty close to how the part will behave in a circuit. It will present 3 times that impedance to the 30 kHz component, around 19k ohms of impedance, and it will present proportionally higher impedances to the higher frequency components of the square wave, so you'll mostly see a 10 kHz sinewave at the output.

If the inductor is resonated with a capacitor, it will go even farther at filtering the sinusoidal components of the square wave, showing peaks at each sinusoidal component.

16. ### electronice123 Thread Starter Senior Member

Oct 10, 2008
304
0
What if it is a unipolar square wave?

Would this change the Fourier transform?

17. ### WBahn Moderator

Mar 31, 2012
18,087
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As described, it IS a unipolar square wave.

But, yes, if you change the signal you change the Fourier transform of that signal.

Think of the Fourier transform for what, at the end of the day, it is -- nothing more than a different way of writing a function. There is a one-to-one correspondence between the time-domain representation for a function and its representation in the complex frequency-domain (assuming a representation actually exists in both domains). Change one and you change the function. Change the function and you need to change both representations.

18. ### vk6zgo Active Member

Jul 21, 2012
677
85
If it is an ideal Inductor,it doesn't really have an Impedance-----the correct term is Reactance!

19. ### crutschow Expert

Mar 14, 2008
13,496
3,373
But inductors do have impedance. Impedance is a term the includes all the resistance and reactance at the node of interest. So if there is only an ideal inductor then the impedance of the inductor is its inductive reactance. A real inductor will also have some stray capacitance and resistance so its impedance would include those terms also.

20. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
Sure it does. Resistance and reactance are proper subsets of impedance. Just as a pure real number and a pure imaginary number are both members of the set of complex numbers.