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I know the basics of self-inductance, and how an inductor creates a voltage/current that opposes the increase or decrease in current.

I can't find anything on how it reacts to/blocks certain frequencies.

Can anyone tell me how inductors select or block certain frequencies?

share|improve this question
    
Induced emf (or 'back emf'), which opposes the applied voltage, is proportional to rate of change of current. As the frequency of a sinusoidal current increases the rate of change at the steepest part of the sine wave increases, hence the magnitude of the back emf increases with frequency, and thus the opposition to the applied voltage increases with frequency. – Chu Mar 1 at 0:51
6  
Take a small cart and push it back and forth a short distance, once a second. Now load the cart full of bricks and try it again. Can you manage the same frequency or does it feel like a ton of bricks? How about at a lower frequency? The bricks creating a force that opposes a change in the speed of the cart. :) – hobbs Mar 1 at 7:08
up vote 7 down vote accepted

Ideal inductor's impedance is given by:

$$Z_L = j\omega L = j2\pi f L$$

This means that its impedance (which can be interpreted as "current opposition") is proportional to signal frequency. Therefore it acts as short circuit to DC current (\$f = 0\$, and \$Z = 0\$) and its impedance increase linearly with frequency. The larger the inductance \$L\$, the "faster" this increase goes.

If you're interested in the physical background of this, such behavior model can be derived from Maxwell's equations, particularly from the third equation (Faraday's law of induction).

An example application of using inductors in filters is given in the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

This is a low-pass filter, with cut-off frequency given by

$$f_c = \frac{1}{2\pi\sqrt{LC}}$$

Other filter types can be obtained exploring the same properties (high-pass from switching L and C, bandpass with other topologies, etcetera).

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If I simulate this circuit, \$f_c\$ is the resonance frequency. The cutoff-frequency is higher. Am I missing something? – sweber Mar 2 at 16:32

If you put a voltage across an inductor, it is going to take that energy and start to 'wind up' the magnetic field around it. The magnetic field can't wind up forever. It 'winds up' with the voltage being proportional to the derivative of the current. $$V_L = L\frac{dI}{dt}$$ This also means that higher frequencies will not pass through the inductor because a sudden change will be impeded by the magnetic field.

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For my opinion, THIS is the best answer that deserves more points. The reason: The given formula LdI/dt is the KEY for deriving the inductive impedance Z=jwL. – LvW Mar 1 at 15:16
    
Add to that that an answer in the time domain is easier to understand. High frequencies mean fast changing currents, therefore high voltage drops across the inductor. The link between time and frequency domain is not obvious to everyone, and requires at least a basic understanding of Fourier and/or Laplace transforms. 1 upvote for you. – Bart Mar 1 at 15:42
    
I'm not keen on it, as the last paragraph requires more explanation than the given equation. – Scott Seidman Mar 2 at 12:09

An inductor does not "select or block certain frequencies". The voltage across it is the negated time-derivative of the current applied to it multiplied by a constant (the inductivity).

The negated time-derivative of sin(2 pi f t) is - 2 pi f cos(2 pi f t), so you get a phase shift of 90 degrees, and a frequency-dependent factor of - 2 pi f.

Taken together, one calls this a complex impedance of 2 pi j f L (j being the engineers' name for sqrt(-1) as i is already taken for currents).

This just favors admitting higher frequencies but does not single out any frequency. To actually single out frequencies, you need to combine the inductor with other components. Resistors are always involved in real circuits (any non-superconducting elements have finite resistance). But to get actual selectivity with passive components, you need to work with capacitors as well.

If you put an inductor and a capacitor in series, there will be one frequency where the voltages across both capacitor and inductor, given the same current, will just cancel out. This will effectively form a shortcircuit for this frequency, admitting arbitrary currents for a small applied voltage, basically only limited by parasitic resistance. The individual voltages across inductor and capacitor will still be large, requiring them to be specified accordingly.

Similarly, if you put an inductor and a capacitor in parallel, there will be one frequency where the currents across both capacitor and inductor, given the same voltage, will just cancel out. This will effectively block currents for this frequency. Again, inductor and capacitor will still have to withstand the current through the individual components even though the net current will be almost zero.

Filters specialized on a single frequency (either admitting or blocking it) can be made very selective with few components as long as those components are close to ideal capacitors/inductors without large parasitic resistances. However, crossover networks (like used in loudspeakers) have a whole passband and stopband rather than single frequencies to be let through or blocked. In this situation, you'll need more components for sharper transitions, regardless of the components' quality.

For HF filtering purposes, the inductors are the most conspicuous components and those that are used for tuning the filters (by screwing the ferrit core further in or out). But they won't do the trick without accompanying capacitors either.

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The OP may have meant low pass and high pass filters as well, it might be worth adding a bit on that. – Mister Mystère Mar 1 at 11:16

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