Without load resistor I want to design a RLC filter. As I told you my load resistor is 30G ohm.

And, as you worked out, the inductor required would then be 660000 Henries.

 

told you my load resistor is 30G ohm.

No it isn’t. This is why you cannot design the details of a circuit without knowing the big picture. There are engineers here with hundreds of years of combined experience telling you that your approach is backwards. Why won’t you listen?

Bob

 

There are engineers here with hundreds of years of combined experience

that’s how old this thread is making me feel.. . . .

 

hi pp,
This is what LTS shows for your filter.

E
BTW: When you say G, thats short for Giga, please confirm.?

 

hi pp,
This is what LTS shows for your filter.

E
BTW: When you say G, thats short for Giga, please confirm.?
View attachment 247511

 

Please answer the question.
BTW: When you say G, thats short for Giga, please confirm.?

 

I want to have graph as shown in red, I'm trying to achieve this.

So you want 42dB/octave roll-off? That’s not a 2nd order filter. That’s a 7th order filter. You will need four inductors and three capacitors. (And for a 7th order filter, being matched to the load impedance is critically important)
Would you like me to post the next page from Zverev, with 5th, 6th and 7th order filter tables?

 

So you want 42dB/octave roll-off? That’s

He has no idea what he wants, or even why he wants it. It is a fool’s errand to try to help him.

Bob

 

So you want 42dB/octave roll-off? That’s not a 2nd order filter. That’s a 7th order filter. You will need four inductors and three capacitors. (And for a 7th order filter, being matched to the load impedance is critically important)
Would you like me to post the next page from Zverev, with 5th, 6th and 7th order filter tables?

No, I don't want to go higher order than the 2nd order. I want to design the filter with 2nd order only. Not exactly -40db, but I want to attenuate the signal as fast as possible.

Please, let me know one thing here:
From above graphs, As far I design the circuit, Fc=5kHz at -3db. So, as I understand from this is that the frequency will cut off higher than the 5kHz at -3db and rest (>5kHz) of the frequency will be attenuate. Am I right?. Or else, the above 5kHz frequency will also be pass, as shown in blue in below image.

 

<<5kHz the line will be flat.
>>5kHz the line will slope down at 6dB/octave or 20dB/decade multiplied by the number of the order.
i.e. 2nd order = 12dB/octave
4th order = 24dB/octave

Around 5kHz, what happens depends on the type of filter.
for Butterworth, it’s the closest you can get to two straight lines intersecting at the cutoff frequency (flat line below cut off, sloping line above cut off).
for Chebyshev, there will be a peak (or several peaks) in the response at around the cutoff frequency, the roll off will start steeper then straighten to 12dB/octave (For 2nd order)
for critically damped and Bessel, the curve will be gentler, but will also straighten to 12dB/octave
so, about three octaves above cutoff, all types of 2nd order filter will be the same, at 12dB/octave.
if you need a steeper slope, you need a higher order.

 

It was not my intention to join this thread or post a Like. That was in error while reading threads on a mobile device.
Now that I am in, here are my comments.

1) Your request is a purely hypothetical one with no clear objective or specification in mind. Hence there is no viable solution.

2) You need to learn about the characteristics of passive filters beginning with a 1st order filter. Read about this first before requesting a 2nd order filter. There is a concept known as the pass band and the stop band. In the stop band, frequencies are never totally eliminated which is what you are seeking, "Ideal Brick Wall Response". It will not happen. The best a filter can do is to attenuate frequencies following a known attenuation pattern, 6dB/octave for a 1st order filter, 12dB/octave for 2nd order filter, as examples. There are undesirable consequences of any filter and it gets worse as the order (as in nth-order) gets higher.



References:
https://www.electronics-tutorials.ws/filter/filter_2.html
https://www.electronics-tutorials.ws/filter/filter_8.html

 

I would like to design the filter that will cut of the frequency above 5kHz. So, I have designed RLC circuit. Please can you have a look and tell me that the circuit design is correct or not?.
View attachment 247498

Hello again,

It looks right to me except for a tiny difference in the cutoff. I calculate a theoretical 4870.6 Hz vs your 5000Hz which is close enough i would think. That's with R2 (the load resistor not shown in your schematic) =30GOhms.
As R2 goes up to infinity it doesnt change much.
As R2 goes down however, the limiting value to get exactly 5000Hz is about 40 Ohms and then the cutoff firequency starts to rise up to a peak around 7800Hz at R2=about 0.4 Ohms.
So for R2 anything over 40 Ohms should be close enough. You also have to check the amplitude to see if it is good enough too. At R2=30 Gigaohms it is about 0.7071 of the max and at R2=40 Ohms it is about 0.67 of the max, both should be acceptable.

I dont know why there were some replies that said it would not work but maybe they had some practical reason in mind like some parasitic effect. It would be good if they elaborated a little.

 

High damping factor = low Q
Low damping factor = high Q
The higher the Q the more it oscillates.

In order to achieve the critical damping (where equilibrium position occurs), which value is to be chosen ζ = 0.7071 or ζ = 1?. Which one is to achieve the critical damping?. Because the websites describes in different ways, some of them considered 0.7071 and other are 1.

 

In order to achieve the critical damping (where equilibrium position occurs), which value is to be chosen ζ = 0.7071 or ζ = 1?. Which one is to achieve the critical damping?. Because the websites describes in different ways, some of them considered 0.7071 and other are 1.

ζ = 1 or Q=0.5 is critical damping.

 

In order to achieve the critical damping (where equilibrium position occurs), which value is to be chosen ζ = 0.7071 or ζ = 1?. Which one is to achieve the critical damping?. Because the websites describes in different ways, some of them considered 0.7071 and other are 1.

Just to expand on what Ian said...

If you look at the general 2nd order transfer function you would see:
wo^2/(s^2+2*d*s+wo^2), where 'd' here is the damping factor.

If you take just the denominator and solve for 's' using the solution to a quadratic equation you would get what is called the 'discriminant'.
When the discriminant is real the response is overdamped.
When the discriminant is complex the response is underdamped (has sinusoidal components).
When the discriminant is exactly 0 (would be rare to be exactly zero but could be close enough) then the response is critically damped. It just so happens that the discriminant for that is sqrt(d^2-1) and so when d=1 it becomes exactly zero,
Note that because the discriminant can be just slightly above zero and real (0.001) or slightly complex (1+0.001*j) that means some values of the damping factor 'd' may result in a transfer function that is very close to being critically damped. This is more like we would see in real life (outside the lab) because component values make up this constant and component values are never exact.

 

It is interesting to note that the highest Q that can be achieved in a passive low-pass or high-pass filter without using inductors is 0.5 (critical damping).

 

ζ = 1 or Q=0.5 is critical damping.

ζ = 0.707 0r Q = 0.7 is not a critical damping?.

 

ζ = 0.707 0r Q = 0.7 is not a critical damping?.

No. That’s a Butterworth.
it is a maximally flat response in the frequency domain, but the time domain response to a pulse or edge has overshoot and ringing.

 

Hello Everyone,

I have design the first order RLC low pass filter with the targeting the cut-off frequency is around 100Khz. I need confirm with you guys that the one which I have designed is correct or not?. See the below image.

 

Hello Everyone,

I have design the second order RLC low pass filter with the targeting the cut-off frequency is around 100Khz. I need confirm with you guys that the one which I have designed is correct or not?. See the below image.

View attachment 277309

No. That’s a first order filter. You can tell because the slope is 6dB/octave and the phase shift is 90°.
From your component values (100nH and 2nF) it doesn’t become second order until above 11MHz.