Hi all,
how do i design a low pass filter with just one inductor? Can someone point me to the theory of it?Thanks.

Regards

 

Very simple.....one series inductor and one load (shunt) resistor. Your cutoff frequency is the frequency at which the inductive reactance (Xl=2*pi*f*L) is equal to the load resistance.

It's not a GREAT filter, but it fulfills all the basic definitions of a lowpass filter.

Eric

 

You might want to look at Elsie; it's an LC filter design software package. Even in the freeware mode, you can do quite a bit with it.

http://tonnesoftware.com/elsie.html

I threw together a capacitive input 3rd order Chebyshev low-pass 1khz LC filter in a couple of minutes.
Here's the resulting schematic:

Attached is the schematic and transmission & return loss plot.

LC filter design can get mighty deep. I'd rather have the software tools to do it for me than try to do the math by hand.

 

Great pics Wookie!

What does the Return,db mean?

Isnt the bandwidth determined by w3db?
Meaning in this case it should be ~1.3KHz, and not 1KHz.

 

But is it possible to design a filter only with an inductor and nothing else?

 

But is it possible to design a filter only with an inductor and nothing else?

No. Substitute either 0 or ∞ for the load resistance and what you get is not a filter in the sense of having any meaningful properties.

 

Great pics Wookie!

What does the Return,db mean?

That is the reflected energy. It's a measure of the efficiency of the filter; how well it's tuned. In this case, I was only using a single inductor with a capacitor on either side; a classic "pi" configuration. I used a Chebyschev formula for a better cutoff at the cost of more ripple (0.3dB) in the passband.

Isn't the bandwidth determined by w3db?
Meaning in this case it should be ~1.3KHz, and not 1KHz.

Yes. If the Butterworth option is used, it generally hits the -3db point on the nose.

If you allow more ripple in the passband (say, 2db) the Chebyschev option gets much closer to -3db at the specified bandwidth, but return loss suffers quite a bit.

However, try finding other freeware software that has all these features.

Note that input impedance (Rs) specification makes a huge difference in the selection of LC values. If you don't match your impedance values properly, you will wind up with a very poor filter.

 

Thank you!

That is the reflected energy. It's a measure of the efficiency of the filter; how well it's tuned. In this case, I was only using a single inductor with a capacitor on either side; a classic "pi" configuration. I used a Chebyschev formula for a better cutoff at the cost of more ripple (0.3dB) in the passband.

When you say reflected energy, do you mean the total reactive energy exists in the circuit?
Meaning V_source * I_source * sin(total_phase) ?

 

Thank you!

When you say reflected energy, do you mean the total reactive energy exists in the circuit?
Meaning V_source * I_source * sin(total_phase) ?

Please don't make my head hurt.

I don't pretend to understand all of the math behind it. Math isn't one of my stronger points, to put it mildly. That's why I like using the software packages that do the math for me.

It has to do S-parameters; real and imaginary numbers. How's your Calculus? Mine stinks!

The return loss is very useful for determining the efficiency of the filter (how well it's tuned) in the passband.

If you happen to have a network analyzer equipped with an S-parameter test set, the forward transmission is S-12, the forward return loss is S-11, the reverse transmission is S-21, and the reverse return loss is S-22. (If you're lucky, you might pick up a very old, very used network analyzer that's out of calibration with a dim screen and an S-parameter test set for under $1,000. New network analyzers start out around $5,000 and go up to the stratosphere.)

Confusing? Yep. OK, I'll try to 'splain it.

A network analyzer outputs RF across a selectable range of frequencies, and displays the results on the screen.
It has a number of ports, but if an S-parameter test set is included, the ports of primary interest are ports 1 and 2.
S-12 shows the strength of the signal received at port 2 that was transmitted from port 1 (forward transmission).
S-21 shows the strength of the signal received at port 1 that was transmitted from port 2 (reverse transmission).
S-11 shows the strength of the signal received at port 1 that was transmitted from port 1 (forward return loss).
S-22 shows the strength of the signal received at port 2 that was transmitted from port 2 (reverse return loss).

In the plot I posted, only S-12 (forward transmission) and S-11 (forward return loss) were shown. In many cases, this is all that is necessary.

 

Hi all,
assuming i want to design a RL filter ,L being series and R being parellel with Vo.
The gain equation would look something like this Vo/Vi= (1/1+(wL/R)^2) right?
Assuming i select a cutoff frequency of 100Hz, using 2*pi*100=R/L,with L=32e-3 i got R=12.06ohms,but after trying to simulate this with Vin=100 Vpeak 60Hz, how come i don't get 100Vpeak at the output since it should pass the whole 60Hz signal?

 

another question would be which kind of low pass filter would be better,RL or LC?Thanks

 

You would be far better off with an LC filter.

You will need to know what the output impedance of your filter's input source is.

If an LC filter seems too complex, you could build an active filter using opamps and RC values.

With really low lowpass filters, you will probably be better off with (an) active filter(s). Otherwise, you might need VERY large values for L1.

Texas Instruments has free FilterPro software that you can download from their site.
Google "TI Filter Pro"

 

No. Substitute either 0 or ∞ for the load resistance and what you get is not a filter in the sense of having any meaningful properties.

But the article here suggested that it is possible.As f increases,the impedence of L increases therefore L blocks high frequency signals.Can you please clarify?thanks

 

If you substitute 0 for the load resistance, you wind up with an analog delay line.

 

Hi all,
assuming i want to design a RL filter ,L being series and R being parellel with Vo.
The gain equation would look something like this Vo/Vi= (1/1+(wL/R)^2) right?
Assuming i select a cutoff frequency of 100Hz, using 2*pi*100=R/L,with L=32e-3 i got R=12.06ohms,but after trying to simulate this with Vin=100 Vpeak 60Hz, how come i don't get 100Vpeak at the output since it should pass the whole 60Hz signal?

Someone pls answer this question.Thanks.

 

Please don't make my head hurt.

I don't pretend to understand all of the math behind it. Math isn't one of my stronger points, to put it mildly. That's why I like using the software packages that do the math for me.

It has to do S-parameters; real and imaginary numbers. How's your Calculus? Mine stinks!

The return loss is very useful for determining the efficiency of the filter (how well it's tuned) in the passband.

If you happen to have a network analyzer equipped with an S-parameter test set, the forward transmission is S-12, the forward return loss is S-11, the reverse transmission is S-21, and the reverse return loss is S-22. (If you're lucky, you might pick up a very old, very used network analyzer that's out of calibration with a dim screen and an S-parameter test set for under $1,000. New network analyzers start out around $5,000 and go up to the stratosphere.)

Confusing? Yep. OK, I'll try to 'splain it.

A network analyzer outputs RF across a selectable range of frequencies, and displays the results on the screen.
It has a number of ports, but if an S-parameter test set is included, the ports of primary interest are ports 1 and 2.
S-12 shows the strength of the signal received at port 2 that was transmitted from port 1 (forward transmission).
S-21 shows the strength of the signal received at port 1 that was transmitted from port 2 (reverse transmission).
S-11 shows the strength of the signal received at port 1 that was transmitted from port 1 (forward return loss).
S-22 shows the strength of the signal received at port 2 that was transmitted from port 2 (reverse return loss).

In the plot I posted, only S-12 (forward transmission) and S-11 (forward return loss) were shown. In many cases, this is all that is necessary.

Hey,
Thank you

I didnt mean to ask whats the mathematical meaning of the returned energy, but rather what is its physical meaning.

Capacitors and inductors store energy but they dont dissipate it, but rather return this energy to the source line, isnt it?

Meaning, if in every cycle:
* the source transmits 1KJoul.
* the resistive loads dissipate 0.3KJoul
=> in every cycle, 0.7KJoul are returned to the source.

Is it correct?

 

But the article here suggested that it is possible.As f increases,the impedence of L increases therefore L blocks high frequency signals.Can you please clarify?thanks

For a real inductor -- two things:

1. Inductive reactance does not increase linearly without limit
2. There is a self resonant frequency where the reactance disappears and then becomes capacitative.

As was mentioned, and inductor all by itself may be interesting to you, but it is not a filter.

 

Basically.

The energy that passes through the filter is dissipated in the termination load.

The energy that is not within the range of the filter gets dissipated at the source load.

If the filter is not tuned well, some to most of the energy that is desired to be transmitted will be dissipated in the source load.

 

Someone pls answer this question.Thanks.

Please post a schematic!! your description is not clear..... Is it a first order RL circuit? Please advice

 

Hi all,
Assuming that i have an inverter supplying an input voltage V1 containing harmonics to the power system V2 through the LC filter just like this:
http://img15.imageshack.us/img15/681/81397851.jpg
What i want is to only pass the fundamental voltage of 60Hz across to the power system. If i define the transfer function and obtained the values of LC to make cutoff frequency to be around 100Hz(considering the first harmonic to be at 180hz) ,is that it to design a low pass filter? I read on some articles that i need to look into the S-plane as well,why is that so? Is it possible to make V1 and V2 to be in phase even if we use the LC filter? How do i go about designing it?THanks