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As I predicted in post #11.
Second order passive low pass filter design
- Thread starter pinkyponky
- Start date
Hi,
I calculate the -3db point to be at:
f=79581.45 Hz
The response is not very 2nd order-ish though. To see some real 2nd order action lower the value of R1, but then recheck the frequency may have to adjust L and/or C. You also have to be careful you dont run into a more bandpass like response if that's not what you want and that means keeping the overshoot to an acceptable level although probably not zero to get a sharper response.
I am afraid that the thread opener does not know how to set up a specification for a filter (he wants an attenuation "as much as possible").
Hello Everyone,
Here I have shown my circuit and also attached the simulation file below. Please could you help me that I want to design the RLC filter circuit with the cut off frequency of 100Khz. I think everyone is happy now.
Here I have shown my circuit and also attached the simulation file below. Please could you help me that I want to design the RLC filter circuit with the cut off frequency of 100Khz. I think everyone is happy now.
Hello Everyone,
Here I have shown my circuit and also attached the simulation file below. Please could you help me that I want to design the RLC filter circuit with the cut off frequency of 100Khz. I think everyone is happy now.
View attachment 277384
Well you still have to answer one question, why is R1 so high in value? That makes L1 ineffective which reduces the sharpness of the filter.
If you look at the attachment, you can see that the -3db point is nearly constant for any value of L1 and that is because L1 doesnt do much at all.
You may do well to look at some of the canonical filter types on Wikipedia to figure out how sharp you want to the filter to be.
I think we all assume that if you place an inductor in a circuit you want it to do something for you. In other words, if you lower the value of R1 you will get a better filter out of it.
I also see that you are dividing the response by approximately 3 and then multiplying the response by 3, is that to protect against over voltage at the input of the op amp while maintaining an overall gain of 1 (to act as buffer) ?
Even with R1=0, L1 will need to be a few mH, because of the 3.3k load resistance.
Hello there,
Well im not sure how much the load resistance will affect it, you mean to get the cutoff frequency lower?
Did you mean lower the 3.3k or increase it's value?
I also figure we should start with R1 but if you see another problem then we have to address that too.
Taking a Butterworth filter as a example.
Coefficients Cl and CC are both √2
Then C=Cc/(2πfR)
L=Cl.R/(2πf)
for a 3.3k load, zero source impedance and 100kHz cutoff, L=7.4mH C=680pF
for other types of filter, and different source impedance, I’ll have to get out my copy of Zverev.
How your saying that the load resistance is 3.3Kohms?. Where did you see this value?.
The resistance between the output of the filter and ground is R4+R5
I have selected R1 as a high resistance value because I have an input impedance to the RLC filter and also I have an impedance on the Inductor, so these will affect the signal if I chosen low resistance as a R1, thats's why I have selected the high value of R1.
The higher the resistances in the filter, the larger the inductor must be, so the less effective it will be at high frequencies due to self-capacitance.
I'm not clearly understand.
Do you mean that, do I need to chose larger inductance If I chosen the high resistance as I did.
If so, then what value can I choose?. and why high inductance value is required?.
The formulae for inductance and capacitance values are in post #22, repeated in post #130
How to know those coefficient values as a CC and CI?.
You look them up in a table of filter coefficients, as explained in post #22
Hello,Hello Everyone,
Here I have shown my circuit and also attached the simulation file below. Please could you help me that I want to design the RLC filter circuit with the cut off frequency of 100Khz. I think everyone is happy now.
View attachment 277384
I'm trying to modify the design by following the formulas in post #22. While I'm doing this I have some questions are:
According to my above design, I have load resistance of RLC filter is 3.3k ohms, which is RL. That's fine.
According to my above design, which value can I consider as the source resistance (Rs) either R1=1k ohm or Rser=15 ohms or R1+Rser= 1.015k ohms ?.
In post #22, the image says that the resistor (in the RLC) is 1/Rs. In this case, I have already chosen this value as a 1k ohm. So, as per the image (in the post #22) it would be like 3.3k/1K ohm, if so, then the value would be 3.3ohm. Is that the resistance value to be chose as a R1 in my circuit?. or what?.
Last edited:
1k or 1.015k isn't going to make much difference, it's 1.5% and you'll not get an inductor that accurate.
Going back to the table in post #22, if you want to use a 1k source impedance and a 3.3k load impedance, then you should use 8th row (load/source = 3.33) which gives 0.2447 for Cl and 5.31 for Cc.
That gives an inductance of 1.28mH
I found this one
https://4donline.ihs.com/images/Vip...1-1.pdf?hkey=6D3A4C79FDBF58556ACFDE234799DDF0
and it has a self-resonant frequency of 1.3MHz, which means that above that frequency it behaves like a capacitor.
That means that if you want to remove frequencies above 1.3MHz it won't work, and a single-order RC filter would work better (and an 2nd order RCRC filter would work better still).
