What would be the proper value for Choke (Inductor) and Capacitor to reject 60Hz frequency? I could not find on google. I tried with LT-Spice but I got the capacitor value equals to 0.1F (not 0.1uF). That is not proper to me and my inductor value was 0.2H.

Any suggestions will be helpful to me. I am trying to find more and more.
Thank you!!!

 

Hi,
How did you calculate 2.375 VAC at 38ohm resistor?

If the reactance of the inductor is 3.8 KΩ, it forms an AC voltage divider with the 38Ω resistor.

240 VAC * [ 38 / (3.8K + 38)] = 240 * (38 / 3838) ≈ 2.376 VAC

 

What would be the proper value for Choke (Inductor) and Capacitor to reject 60Hz frequency? I could not find on google. I tried with LT-Spice but I got the capacitor value equals to 0.1F (not 0.1uF). That is not proper to me and my inductor value was 0.2H.

Any suggestions will be helpful to me. I am trying to find more and more.
Thank you!!!

To create a 60 Hz notch you create a parallel combination of L and C that is resonant at 60 Hz. The condition at resonance is that the reactances are equal.

ωL = 1 / ωC implies
(ω^2)LC = 1 implies
ω = 1 / √LC implies
2π60 = 1 / √LC implies
√LC = 1 / 377 implies
LC = 7.035 e-6

Two things multiplied together equaling a constant is described by a conic section called a hyperbola. Any pair of values for L and C that lie on the hyperbola will work so there are an infinite number of solutions.

Let's take your numbers:

.1F implies 1 / (2π60*.1) = 0.0265Ω
.2H implies 2π60*.2 = 75.4Ω

The parallel combination of those two is not resonant and won't reject 60 Hz.

How about 2500 uF and 2.8 mH
Now 1 / (377*2500e-6) = 1.06Ω
and 377 * 2.8 e-3 = 1.05Ω

 

To create a 60 Hz notch you create a parallel combination of L and C that is resonant at 60 Hz. The condition at resonance is that the reactances are equal.

ωL = 1 / ωC implies
(ω^2)LC = 1 implies
ω = 1 / √LC implies
2π60 = 1 / √LC implies
√LC = 1 / 377 implies
LC = 7.035 e-6

Two things multiplied together equaling a constant is described by a conic section called a hyperbola. Any pair of values for L and C that lie on the hyperbola will work so there are an infinite number of solutions.

Let's take your numbers:

.1F implies 1 / (2π60*.1) = 0.0265Ω
.2H implies 2π60*.2 = 75.4Ω

The parallel combination of those two is not resonant and won't reject 60 Hz.

How about 2500 uF and 2.8 mH
Now 1 / (377*2500e-6) = 1.06Ω
and 377 * 2.8 e-3 = 1.05Ω

Thanks for your effort.
Actually my L and C values were for the cut-off frequency at 1Hz (not 60Hz).
So that the values what I choose for 1Hz and so it will not allow to pass any frequency beyond 1Hz. Is it correct?

Your values are for f(cut-off) equals to 60Hz and that means that filter allows 60Hz frequency.

I don't want filter to pass any frequency greater than 1Hz. By doing this, I can get almost none frequency (with harmonics) at the output of the filter.

Is my thinking correct?

Thanks again for your time.

 

No

The cutoff frequency of a filter defines the beginning of the transition from the passband to the stopband. The LTspice file in the next post shows how the parallel combination of an L and a C produces a notch.

If you want a low pass filter with a cutoff frequency of 1 Hz then you can consult a table of coefficients for a Butterworth filter. So a 3-pole low pass would consist of 2 capacitors and an inductor in the arrangement in the LTSpice file below.

You can see that the insertion loss is 6 dB and at 1 Hz the response is 3 dB down, and at 10 Hz you are 65 dB down

 

No

The cutoff frequency of a filter defines the beginning of the transition from the passband to the stopband. The LTspice file in the next post shows how the parallel combination of an L and a C produces a notch.

If you want a low pass filter with a cutoff frequency of 1 Hz then you can consult a table of coefficients for a Butterworth filter. So a 3-pole low pass would consist of 2 capacitors and an inductor in the arrangement in the LTSpice file below.

You can see that the insertion loss is 6 dB and at 1 Hz the response is 3 dB down, and at 10 Hz you are 65 dB down

Thanks a lot!!
I would like to learn in detail so next time I can try myself. If you don't mind can you explain where I was wrong.
In my filter, I have choose L=0.2H and C=0.1F. By formula cut-off frequency is 1.12Hz. Now as I know cut-off means from that frequency (here 1.12Hz), on-wards it will start to attenuate. When I simulate in LTSpice, it gives me linear straight line.
If you have something that can explain in detail then suggest me. (Any paper/book/article.)
Thank you!!!

 

What is the reason behind putting R1 and R2 in the circuit?

 

Without the filter the source has an impedance of 1 Ω and the load has an impedance of 1 Ω. This situation results in a maximum power transfer from source to load. The prototype filter is matched to this condition. you can of course scale your final design to match any source and load impedance. Now when you insert the filter you can see the "insertion loss" in the response.

The response of the filter is down 3 dB at the "corner" or "cutoff" frequency. Adding that 3 dB to the 6 dB insertion loss means the output is down a total of 9 dB with respect to the source at the "cutoff" or "corner" frequency. You can of course adjust the filter so that at 1 Hz. it will only be down say 0.5 dB which pushes the actual cutoff frequency up.

 

how to define transfer function for the circuit? Can you please give me any kind of material so that I can study.

 

The transfer function for the circuit is a Butterworth Polynomial.

The Butterworth filter has the property that its passband response is maximally flat. What that means, is that the passband response is monotonically decreasing.

Start with the following Wiki

http://en.wikipedia.org/wiki/Butterworth_filter