Inductive low-pass filter - where does the filtered energy go?

Kurious

Joined Sep 15, 2017
10
This is a self study question, so there's no deadline.

In a filter design such as the first one in the AC unit chapter 8 of the Kuphaldt book, where does the filtered current go? It's not being grounded out or shorted to anywhere. Is it dissipated as heat, radiated off into space, and/or simply stored in the inductor's magnetic field?

Thank you very much.

crutschow

Joined Mar 14, 2008
28,536
where does the filtered current go?
There is no "filtered current".
The inductive reactance of the inductor simply increases as the frequency increases, which reduces the current to, and the voltage across the load, as shown in the LTspice simulation below:
The output voltage falls to the -3dB point (0.7V) when the inductive reactance equals the load resistance at 53Hz. Last edited:

Kurious

Joined Sep 15, 2017
10
Okay, thank you. I believe I get it. The electromotive force simply isn't allowed to push as much current through the inductor at a higher frequency because it meets more impedance at that frequency. The lower frequency is what's desired, so we let the higher frequency be attenuated.

Comparing to the inductive high-pass filter then, shorting the lower frequencies away from the load is really shifting the ratio of where the voltage is allowed to push current so more of the desired higher frequencies go where we want them. That perhaps makes more sense in the form of equations than the words I'm coming up with right now. However, is this a proper understanding?

MrAl

Joined Jun 17, 2014
8,612
Hi,

If you really want to understand this then it's not a good idea to take the point of view where we have the inductor acting like just a frequency sensitive resistor because it's not that. It's an energy storage element and that is different like the difference between a computer memory and a computer display monitor/keyboard.

You can understand that the inductor current is out of phase with the voltage source, and that means that sometimes the inductor is supplying current to the source, not always getting current from the source. It stores a little energy that it can put back into the source. As the frequency gets higher, the phase shfit gets bigger so it puts more and more energy back into the source so the output gets less and less energy into the resistor. If there was no resistor the inductor would store energy and then put it back into the source as the cycle progresses. That's with an ideal inductor of course. So over one cycle the inductor takes in energy and then releases some of it into the resistor and some back into the source, and the amount into the resistor gets larger as the frequency goes down as the phase shift decreases.

Do some plots showing the energy in all three components.

The capacitor LPF will put a little energy back into the source also but some is lost in the series resistor too which accounts for some of the filtering action.

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• pmd34 and nsaspook

Picbuster

Joined Dec 2, 2013
1,027
MrAi is correct,
you should replace coil with a frequency depended impedance.
changing the frequency will change the impedance (resistance) and will also change the current.

Iresistor = V1 / (impedance+resistor)
Vresistor = Iresistor x resistor.
energy in resistor = (iresistor) ^2 x resistor

Total energy ( from V1) = V1* Iresistor ( sum of both impedance and resistor)
now you are able to calculate power in impedance.( this will allow you to calculate the wire size and core related to the used frequency)

Picbuster

Kurious

Joined Sep 15, 2017
10
...sometimes the inductor is supplying current to the source, not always getting current from the source. It stores a little energy that it can put back into the source. As the frequency gets higher, the phase shfit gets bigger so it puts more and more energy back into the source so the output gets less and less energy into the resistor
This is helpful. The inductor is, more or less, pushing back while a resistor is just getting in the current's way.

So the higher frequencies are being sent back to where they came from. If the inductor is parallel to the load, the path back to source is also a path to load. And that's the design of an inductive high-pass filter. I hope I understand correctly, because that makes a lot of sense.

• pmd34

Picbuster

Joined Dec 2, 2013
1,027
correct!
next step is to create an amplification( creating a resonator circuit by adding parallel a capacitor to the coil).
What happened? when the resonance frequency is reached the impedance goes high and the voltage over the coil goes up and over the resistor down.
This will produce a dip in the bandwidth measured over the resistor for that frequency.

Picbuster

nsaspook

Joined Aug 27, 2009
9,076
correct!
next step is to create an amplification( creating a resonator circuit by adding parallel a capacitor to the coil).

Picbuster
Amplification is not the correct word because there is no such thing as amplification by passive resonance. The energy of oscillation is exactly the energy (minus losses) of the external forces supplied because of energy conservation. It is possible to have Entrainment where coupled (electrical) systems feedback energy to reinforce the oscillation.

crutschow

Joined Mar 14, 2008
28,536
Amplification is not the correct word because there is no such thing as amplification by passive resonance.
But the voltage amplitude can be increased by resonance.

nsaspook

Joined Aug 27, 2009
9,076
But the voltage amplitude can be increased by resonance.
Sure, you can do that with a simple non-resonant transformer too. We don't consider that amplification either. I understood the point he was making, resonance can be a energy storage system that can modify the effects and characteristics of energy flow in a circuit as it transforms from magnetic to electric and back at resonance.

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