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Why does the impedance of inductor change with frequency?

  1. Oct 14, 2011 #1
    The power stored in an inductor is E=0.5LI2 which is independent of frequency
    However, the impedance is ωL.
    So, what makes the impedance increase with frequency?
  2. jcsd
  3. Oct 14, 2011 #2
    It helps to look at an inductor's response to a step input in the time domain to get a physical "feel" for it behaves. An inductor initially looks like a high resistance (open circuit) that drops to a low resistance (short circuit) over time. So, for a low frequency signal, which takes a lot of time, it behaves more like a short circuit. For a high frequency input, it looks more like a high resistance component.

    A capacitor is the exact opposite, a low resistance (short circuit) that becomes a high resistance (open circuit) over time.

    There's more to it of course but once you get a physical feel for these components, you can use the equations freely without wondering why they're valid.
  4. Oct 14, 2011 #3
    I understand the behavior of inductor but i just puzzle what is opposing to the change of current in high frequency.

    Originally, I guess it is because there are more energy stored in form of magnetic field in high frequency range. But the formual just says the opposite. The energy remains the same regardless of the frequency. So, what matter plays the tricks?
  5. Oct 14, 2011 #4


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    An inductor works off of the principal of Lenz's Law. A changing current induces a changing magnetic field that works to oppose the change in the current. The stronger the change in current, the stronger the changing magnetic field (and vice-versa). Obviously if we were to increase the frequency of the current through an inductor we would see an associated increase in the strength of the induced magnetic fields that work to oppose the change in the currents. This translates to an increase in the apparent impedance of the inductor with frequency.
  6. Oct 14, 2011 #5
    While the equation is correct, the impression it gives isn't. You can see this by taking two transmission lines: a low Z feed line connected to a high Z line terminated in a short circuit. The shorted high Z line is the inductor. Send a step down the feed line. When it reaches the connection to the inductor's line there's a strong positive reflection and some transmission into the inductor. The transmitted part travels down to the terminating short and reflects 100% negatively. It then hits the connection to the low Z feed line. A part of that negative reflection transmits into the feed line (where it slightly reduces the total voltage on the feed line) while the rest reflects negatively back into the inductor. This constant back and forth in the inductor builds up exponentially over time until the part transmitted into the feed line is the inverse of the input, resulting in zero total voltage on the input -- i.e. it looks like a short.

    You can analyze this mathematically and experimentally and it gives the correct results, along with a much more reasonable physical picture. Personally, I think it removes a bunch of the "magic" in electronics (the change in xxxx mysteriously opposes a change in yyyy, etc.), allowing one to use the higher level equations with confidence in their validity.
  7. Oct 16, 2011 #6


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    A transmission line is not an inductor though. It's a distributed circuit model of a waveguide and incorporates both capacitive, resistive and inductive behavior. As long as you are talking about a transmission line you have to assume that it will exhibit both inductive and capacitive behavior regardless of how you try to set your impedance. Second, the reason for the reflection of power is a different set of physics. Since the transmission line is actually a simplified representation of a waveguide, the reflections are due to physical changes in the waveguide. Either a change in the structure, composition, or termination result in these reflections which are the reflections of the electromagnetic waves off of the boundary. Third, the input impedance seen here is independent of time as you seem to be stating. For example, you can easily set the length of the transmission line connected to the short so that it looks like an open circuit to the low impedance transmission line independent of the actual impedance of the high impedance transmission line. Finally, I fail to see how this explains the frequency dependence of the inductor that the OP is having trouble with. This is a circuits problem, no reason to further convolute it by using transmission lines and waveguide theory.

    There isn't any magic in Lenz's Law here. Lenz's Law simply states that as a consequence of Faraday's law of induction that, if I may borrow the expression from Wikipedia, given a magnetic flux through N coils of wire
    [tex] \mathcal{E} = - N \frac{\Delta \Phi}{\Delta t} = -N \frac{ \partial \Phi}{\partial t}[/tex]
    A very basic inductor, which is actually how most inductors are made, is simply a coil of wire (usually with a ferrite core to magnify the effect). We know that a solenoid gives rise to a magnetic field through the center of the coils and thus we see that an inductor creates its own magnetic field. For an ideal solenoid, the magnetic field produced is
    [tex] B = \frac{\mu NI}{\ell}[/tex]
    where [itex]\ell[/itex] is the length of the solenoid. And we note that the EMF is the negative voltage and that the flux [itex]\Phi[/itex] is the area of the coil times the magnetic field B (assuming constant field through the coil's area) we thus see
    [tex] V = N \frac{\mu N A}{\ell} \frac{\partial I}{\partial t} = L \frac{\partial I}{\partial t} [/tex]
    So we can easily recover the circuit equations for an inductor by looking at the self-inductance of a solenoid. So as the change in current increases in time (ie: increasing frequency) we expect that the back voltage due to the inductance increases in strength. That's why we find that the transformed complex impedance scales with frequency.
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