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Lowpass Filter (LC) resonance problem

Hello everyone!

I am looking to design a low pass filter that has a cutoff frequency of 20 kHz. The filter needs to be able to handle about 20A of current so the dir ection I have taken is an LC filter. (Inductor with a high current rating).

I simulated the low pass filter tonight (LTspice) using the values of 7.9 u H and 15 uF suggested from some of the online lowpass filter calculators.

From the simulation I see 0 dB in most of the passband but then I see a hug e resonance point of +32 dB at 14.49 kHz!

Here is a link to the simulation result:

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F_zps5421958c.jpg

1.) Is there a way to eliminate this resonance?

2.) Is there a different filter that would have less peaking but still able to handle the high current for this application (20A)?

3.) There is an H-brige that will be modulated at 200 kHz (pwm) and this wi ll going through the 20kHz lowpass filter to given a DC voltage without all the pulses. If the resonance point is kept as it is what type of issues w ould this cause? Is this even a problem?

4.) From the simulation I see a cutoff frequency of 22.7 kHz instead of 20 kHz. What is the best way to tune this? Maybe to keep the capacitor fixed (

15 uF) and then sweep the inductance value in simulation?

Thank you you so much for your thoughts.

-Mike

Reply to
mikemillernow
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** You need to specify a load impedance for a simple LC filter to work sensibly.

The values you came up with suggest a load of about 0.5ohms.

** Depends if the PWM modulation contains frequency components near 15kHz.

Cos that LC network presents a dead short at resonance.

** Try an LC filter that matches your load at resonance.

.... Phil

Reply to
Phil Allison

You have simulated a tank circuit (LC resonator) which has no load. I don't see any load (or other) resistance anywhere in your simulated circuit.

It's not at all surprising that the node located between the L and the C is bouncing around to extremely high voltages. If the simulator is treating the L and C as "ideal" components, with no resistance in them, I think you'd have an infinitely-high Q in your circuit.

What you should do, for a meaningful simulation, is figure out what your load resistance is going to be. Add a resistor which simulates this load, in parallel with C1.

Then, see how Vout looks. I think you'll be happier with it.

Reply to
David Platt

Probably add some series resistance to the inductor as well.

Reply to
miso

I don't think you're understanding how an LC filter works. The operation of the thing is pretty meaningless without a load, and (as Phil pointed out) the load impedance that you've designed for sounds pretty low (but maybe it makes sense in the context of your 20A??).

We'll help you, but you need to tell us more. Specifically, we're going to need to know enough to figure out what the design load impedance should be, and for that we need to know what the load is.

--
Tim Wescott 
Control system and signal processing consulting 
www.wescottdesign.com
Reply to
Tim Wescott 

--
Here's something to play with: 

Version 4 
SHEET 1 880 680 
WIRE 144 112 80 112 
WIRE 272 112 224 112 
WIRE 352 112 272 112 
WIRE 80 160 80 112 
WIRE 272 160 272 112 
WIRE 352 160 352 112 
WIRE 80 272 80 240 
WIRE 272 272 272 224 
WIRE 272 272 80 272 
WIRE 352 272 352 240 
WIRE 352 272 272 272 
WIRE 80 368 80 272 
FLAG 80 368 0 
SYMBOL ind 240 128 M270 
WINDOW 0 32 56 VTop 2 
WINDOW 3 5 56 VBottom 2 
SYMATTR InstName L1 
SYMATTR Value 12¦ 
SYMBOL res 336 144 R0 
SYMATTR InstName R1 
SYMATTR Value 1 
SYMBOL voltage 80 144 R0 
WINDOW 3 19 129 Invisible 2 
WINDOW 123 19 101 Left 2 
WINDOW 39 0 0 Left 2 
SYMATTR InstName V1 
SYMATTR Value SINE(0 20 20k) 
SYMATTR Value2 AC 1 
SYMBOL cap 256 160 R0 
SYMATTR InstName C1 
SYMATTR Value 10¦ 
TEXT 88 320 Left 2 !.ac oct 1024 10 100k 
TEXT 88 296 Left 2 !;tran 1m
Reply to
John Fields

Is this for class D amplifier? This is well known problem. There could be huge resonant peak in response depending on particular load. Speakers are very much inductive at high frequency. You could handle this with Zobel (R-C) network at the output; however power dissipation could be an issue. Diodes from filter output to power rails to divert overvoltage into rails is other option. I would use a combination of diodes and "light" R-C network.

Vladimir Vassilevsky DSP and Mixed Signal Designs

formatting link

Reply to
Vladimir Vassilevsky

"Vladimir Vassilevsky"

** With a 0.5 ohms load ?

** Avoided by setting the filter resonance higher than 20kHz.
** That is not a correct generalisation.

Until the frequency exceeds 50kHz.

.... Phil

Reply to
Phil Allison

Could be a car audio, for example.

Until someone checks frequency response with signal generator.

Good point.

VLV

Reply to
Vladimir Vassilevsky

** Bose once made 0.5ohm drivers for car audio, but that was unusual.

The norm is to use step-up inverters as part of the amplifier and regular 8 ohm speakers.

** Any smart designer includes a sharp low pass filter at the input to the power amp - with a turnover frequency about an octave lower than the output filter resonance.
** FYI:

Speaker voice coils are poor inductors due to severe eddy current losses in the iron pole piece - some even refer to them as "semi inductors" since the impedance only doubles for each TWO octaves rise in frequency.

Phase angle of the current is about 45 to 50 degrees rather than 90 .

... Phil

Reply to
Phil Allison

The 0.5 Ohm woofers are not uncommon as step-up guarantees problems with EMC unacceptable for OEMs. But even retail stepped-up car amplifiers 8 Ohm is too much. I'd say 2 Ohm is usual.

Reply to
Vladimir Vassilevsky

"Vladimir Vassilevsky"

** Bullshit - they are very uncommon.

** Total bullshit.

Practically all 12V car amplifiers use step-up supplies.

** I mentioned 8 ohm speakers - not 8 ohm rated amplifiers.
** 2 ohm rated amplifiers and 4 ohms rated in bridge mode with *hundreds* of watts of output.

That does not happen without a step up inverter.

You stupid, over-snipping f*****ad.

.... Phil

Reply to
Phil Allison

Is that phase angle caused by the eddy current losses? Is there any contribuion at all from launching audio energy into the 'room'?

Reply to
RobertMacy

"RobertMacy"

** Yep

** Nope.

... Phil

Reply to
Phil Allison

You're right. Ignore what Phyllis has to say. He's off his meds, again.

BTW, dual voice coils are quite common, as well. 2x2ohm voice coils gets close to 100W without a boost. That's good enough for 90% of the market. 2x1ohm buys more power but causes other problems. It's done but not often.

Reply to
krw

Have you noticed how these people show up with a question, and then they never follow up?

Reply to
miso

New names on the group don't always do that. But this does seem to be one of those times.

--
Tim Wescott 
Control system and signal processing consulting 
www.wescottdesign.com
Reply to
Tim Wescott

The 20A coil will need to be a large air core to avoid saturation effects. THe simulation needs to be shown with a 8 Ohm load or whatever. The peak gain can be estimated easily as the ratio of Reactance to Resistance. Sinc e there is no load shown impedance of L or C at resonance to the source res istance must be almost 100 to get almost 40 dB gain. Once loaded with 8 Oh ms this also reduces the Q or peak gain.

Reply to
Anthony Stewart

"Anthony Stewart"

** Why is that - Tony ??

You read it in a post on a web forum somewhere ?

You plucked it out you arse ?

** Yeah - whatever ..........

PISS OFF you PITA Google Groups cretin.

.... Phil

Reply to
Phil Allison

The reason ferrite and iron core chokes are NOT used in high current audio filters such as speaker crossovers is prevent the THD and IMD products from their non-linearity.

Inductors drop in value with rising DC current or Bass freq current. The rated current is where the Inductance is 50% L at 0 Amp current.

This is why ALL crossovers for Bass will use air core coils often with 4" diameter.

{Phil u should know better than to question my judgement.}

Reply to
Anthony Stewart

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