DigiKey and others are incorrect about cutoff frequency for an LC filter

[Michael Craft]

(At least I believe this is true.)

If you Google "LC filter" and "cutoff frequency" you will probably see the formula 1/(2π sqrt(LC)). An example would be this online calculator at DigiKey:

https://www.digikey.com/en/resources/conversion-calculators/conversion-calculator-low-pass-and-high-pass-filter



However, I believe this is the resonant frequency, not the -3 dB cutoff frequency:



I made this plot that shows the resonant frequency and -3 dB cutoff frequency for those component values:



So here's my "proof." The voltage transfer function V_OUT/V_IN is:



The resonant frequency f_r is where V_OUT/V_IN gets very large. In other words, when the denominator goes to zero:



The resonant frequency is not the -3 dB cutoff frequency as so many articles & websites claim. The cutoff frequency (f_C) is defined as the frequency where the output power is half the input power:



Please check my math, but I'm pretty confident it is correct. If so, I wonder how many designers have been using the wrong formula for the cutoff frequency for an LC filter?

So to summarize,

[T3sl4co1l]

There is no impedance (or Q factor) so the problem is ill-defined.

The frequency would be better called "unloaded resonant frequency" or something like that.

The frequency and impedance of standard filter prototypes is simply adjusted proportionally to each, if that's any help.

Tim

[Michael Craft]

I agree that the analysis for a practical implementation would also have to take into account the source impedance and (most importantly) the load impedance. None-the-less, you can still come up with a transfer function for just this circuit, along with a resonant frequency and -3 dB frequency. My point is that DigiKey and others are referring to the resonant frequency as the -3 dB frequency, which is obviously incorrect.

[tooki]

I think you’ll find this to be an uphill battle, insofar as every textbook I checked, as well as my own class handouts I got, all give the -3dB point as 1/(2pi x sqrt(LC)). It’s not compensated for impedances*, but bear in mind that the target user of an online calculator is not likely to have a solid understanding of impedance as a concept, never mind know what the actual input and output impedances of their circuit are.

*some sources, like some of my school notes and the German Wikipedia, expressly state to assume an output impedance of 0 ohms on the source and an input impedance of infinity ohms on the load.

[Michael Craft]

tooki, I also found that many online sources use the formula 1/(2pi x sqrt(LC)) when talking about the cutoff frequency for an LC filter. What's amazing is that it's very prevalent but completely wrong:



And yes, I understand this circuit on its own is pretty useless in the real world due to the extreme underdamped response. (A low load impedance will help damp it.) None-the-less, I am still bothered by how many articles get this incorrect.

[RoGeorge]

I remember the cutoff frequency introduced as the intersection point, the knee, between the asymptote to the flat response line, and the asymptote to the cutting slope.  The point where they intersect happens to be at -3dB for a simple RC (or other first order/single pole).

I've always thought about a cutting frequency as the frequency of a pole (or a zero), not in terms of halving (or doubling) the power.

I'm not sure which one is the official definition, I'm tempted to say the one based on the frequency of the poles and zeroes, otherwise the definition based on power ratio would be OK only for first order filters.

The bandwidth, however, I know it defined in terms of power, so relative to the -3dB, and not relative to the frequency of poles or zeroes.

[Michael Craft]

You brought up an interesting point: what is the definition of "cutoff frequency"? If we define it very generally as the maximum (or minimum) usable frequency, then I suppose the resonant frequency could also be referred to as the cutoff frequency (but certainly not the -3 dB frequency). Even then, it would be a poor choice, since it may not be useable close to the resonant frequency. At any rate, I think most people define the cutoff frequency as the half-power (-3 dB) frequency.

[tooki]

tooki, I also found that many online sources use the formula 1/(2pi x sqrt(LC)) when talking about the cutoff frequency for an LC filter. What's amazing is that it's very prevalent but completely wrong:



And yes, I understand this circuit on its own is pretty useless in the real world due to the extreme underdamped response. (A low load impedance will help damp it.) None-the-less, I am still bothered by how many articles get this incorrect.

But is it incorrect or just incomplete? Those aren’t synonyms, after all…

[tooki]

You brought up an interesting point: what is the definition of "cutoff frequency"? If we define it very generally as the maximum (or minimum) usable frequency, then I suppose the resonant frequency could also be referred to as the cutoff frequency (but certainly not the -3 dB frequency). Even then, it would be a poor choice, since it may not be useable close to the resonant frequency. At any rate, I think most people define the cutoff frequency as the half-power (-3 dB) frequency.

Everything I’ve seen defines it as Vin x sqrt(2).

[RoGeorge]

I was referring to Bode plots, which is the frequency response (amplitude & phase), and the asymptotic representation of the amplitude:  https://en.wikipedia.org/wiki/Bode_plot


Source:  Wikipedia, https://en.wikipedia.org/wiki/Bode_plot

It is possible to determine the asymptotes (the black lines) starting from the transfer function alone, by identifying the poles and zeroes in the transfer function.  From there, it is possible to draw the black lines without numerical calculation of each and every point.  It is enough to know the intersections (aka the cutting frequency points - in an RC is a single cutting point, at 100Hz in the example picture).  A little better explained here (first search result, but it looks correct):  https://global.oup.com/us/companion.websites/fdscontent/uscompanion/us/static/companion.websites/9780199339136/Appendices/Appendix_F.pdf

When plotting the true response, and not just the asymptotic representation, we need to compute each point, and it looks like the red curve.  The red curve happens to be at -3dB when intersecting the cutting frequency for an RC.

But for another filter (for example for two RC filters one after another, both with the same cutting frequency), the red curve won't be at -3dB, will be lower.

[Kalvin]

That original LC-circuit is an (unloaded) resonant circuit. However, when you add correct loading resistance to the circuit, you will get an LC low-pass filter with the expected -3dB frequency.

You may want to visit the following page:
http://www.stades.co.uk/RLC%20filters/RLC%20LPF.html
Scroll down to RLC LOW PASS FILTER 2, and enter the following values to the circuit:
R = 0.7071
L = 1 H
C = 1 F
The -3dB frequency will be now the expected 0.1592Hz.

[CaptDon]

If you set up that exact filter which would be defined as low pass and drive it with 8 ohms and load it with 8 ohms it will not show the resonance in your sim. If you drive it from a low impedance source and look at the midpoint as 'signal out' into an unloaded high impedance anything is up for grabs. You have not discovered a new loop hole in filter math and no, digikey and others are not wrong about the cutoff points when the circuit is simulated correctly. There are sims and then there is the real world. In fact that circuit being series resonant would basically show a shorted condition to your drive circuit at resonance. The only limiting factor is the resistance of the components which determine to some extent the overall 'Q' of the circuit.

[dmills]

Lets see, 100uH, 10uF. At resonance (5032Hz)

Xl = j3.1617 Ohms
Xc = -j3.1617 Ohms.

Note the magnitude is the same but they have opposite signs, which identifies them as being at resonance.

There is a trap here in that the natural thing to do is to apply the voltage divider formula Vo/Vi = Xl/(Xl+Xc), but that breaks because there is no real part and at resonance the impedance of the whole thing falls apart because the denominator goes to zero!

However note that the magnitudes of the impedances are equal, so there will be the same voltage dropped across each impedance, but the top arm is at ∠90 being an inductor, so the overall phase shift at the node you are examining will be 45 degrees.   

pretty sure it all drops out to sqrt(2)/2 which is -3dB assuming your driver was infinitely stiff.

[Kalvin]

When you add a correct load R to the circuit's output so that the Q of circuit becomes 0.7071, the -3dB point will be then at frequency 1/(2*PI*sqrt(L*C)).

For example, if we use the original circuit where L=10uH, C=100uF, and we want the -3dB frequency be 5032Hz, the R shall be 0.2236 ohms.

The R can be calculated from RCωn=1/Q=2ζ (see http://www.stades.co.uk/RLC%20filters/RLC%20LPF.html).

Here is LTSpice simulation for the original circuit with proper load R:

[Michael Craft]

Thanks Kalvin!

It's true that, as the load resistance decreases, the -3 dB cutoff frequency approaches the resonant frequency as defined by 1/(2πsqrt(LC)). But it doesn't actually equal it.

I sent an email to DigiKey. They agreed they were in error, and were looking in to fixing it.

[RoGeorge]

It's true that, as the load resistance decreases, the -3 dB cutoff frequency approaches the resonant frequency as defined by 1/(2πsqrt(LC)). But it doesn't actually equal it.

Prove it.   

Hints:
-   not only the -3dB point can equal the resonant frequency, but it can go smaller, when the load resistance is small enough.  If we consider a load resistance together with the LC, the -3dB point can be at a frequency either lower, or higher, than the resonant frequency.
-   there is no such thing as "-3dB cutoff frequency", that's a mixture of terms.  The "cutoff frequency" can have different definitions, depending of the context, and the order of the filter.  And the -3dB is the loosely-speaking name for the term "half-power transfer".  By the way, the value is not exactly -3dB, but very close to 3.  So, if we aim for precisely -3.(0)dB, then all the formulas are wrong, including the formulas for the RC, or for the RL filters.


Look for the definition for:
- corner (knee) frequency
- half-power transfer frequency
- cutoff frequency
- 3dB frequency
- pole frequency

For the 1st order filters, all of the above terms are interchangeable, because they all point to the same frequency.  For higher order filters, not all of those terms are interchangeable.  Wikipedia gives such an example:

Sometimes other ratios are more convenient than the 3 dB point. For instance, in the case of the Chebyshev filter it is usual to define the cutoff frequency as the point after the last peak in the frequency response at which the level has fallen to the design value of the passband ripple. The amount of ripple in this class of filter can be set by the designer to any desired value, hence the ratio used could be any value.

Quote from:  https://en.wikipedia.org/wiki/Cutoff_frequency

RC or RL filters are 1st order type, while an (R)LC filter is second order type, which makes the -3dB point not very important.  In fact, for an RLC, the -3dB can be anywhere between zero and the formula you want to replace the current one with.


To be more specific, I think digikey and others are not incorrect.

The only thing you can complain about the digikey calculator is that they added the "-3dB" to the name of the 3rd textbox.  Should have been named rather "Cutoff Frequency", instead of "-3dB Cutoff Frequency".

The formula you propose is irrelevant for the purpose of that calculator.  The digikey calculator is about finding the transition frequency, so to estimate the frequency at which the behavior change.  It is not about precisely finding the frequency of the -3dB point.

For the 1st order filters the -3dB and the corner frequency happen to be the same.  An LC filter is 2nd order filter, and thus the confusion about the -3dB point.

[Solder_Junkie]

The combination of 100uH and 10uF is bizarre. The missing parts are the Q factor, and terminating load, which makes a big difference to filter loss, although using a 100uH “RF” choke with 10uF is going to fall outside the normal filter configs.

One really useful filter design tool is Elsie, which I have used to design RF filters with great results (actual measured results). It will no doubt also work for audio, but I haven’t used it for that. Don’t forget to change the default Q values to realistic values. The free version is good enough for almost all purposes.

http://tonnesoftware.com/elsie.html

SJ