Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

[SOLVED] Passive Low Pass Filter

Status
Not open for further replies.

arunkumar.kunjappan

Newbie level 4
Joined
Feb 21, 2011
Messages
5
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,315
Hi Guys,

I have a RC LPF filter with fc=15.9KHz (R=10K and C=1nF). When I cascaded such two filters to form a second order filter and did an AC sweep, I got the new fc around 6KHz. But if I refer the formula for the fc of a second order filler with
R1=R2 and C1=C2, I should get the same fc (i.e 15.9KHz). Now I intoded a voltage vollower with op amp in between the first stage and second stage. After this I could see the fc changed to around 10KHz. Can anyone explain me the reason behind this behaviour.

Thanks inadvance,
Arunkumar.
 

Hi Guys,

I have a RC LPF filter with fc=15.9KHz (R=10K and C=1nF). When I cascaded such two filters to form a second order filter and did an AC sweep, I got the new fc around 6KHz. But if I refer the formula for the fc of a second order filler with
R1=R2 and C1=C2, I should get the same fc (i.e 15.9KHz). Now I intoded a voltage vollower with op amp in between the first stage and second stage. After this I could see the fc changed to around 10KHz. Can anyone explain me the reason behind this behaviour.
Thanks inadvance,
Arunkumar.

1.) If you connect two identical RC stages (1st order) in series, the second stage loads the first stage and, thus, it influences the voltage/current ratios (that means: the frequency properties) in the 1st stage.
2.) If you use a buffer between both stages they don`t influence each other - nevertheless, the corner frequency changes.
The reason is simple: The gain of each RC stage drops at w1=1/RC by 3 dB. That means: In total you have a gain drop of 6 dB at w=w1. Since the overall corner frequency w2 still is at the -3dB point, of course w1 does NOT equal w2.
3) Question: You refer to the "formula for the fc of a 2nd order filter". Please note that there are several different 2nd order stages with opamps with a fc foermula that is NOT identical for all these stages. What formula do you refer to?
 
You cannot cascade two passive filters like that without one affecting the other. Even if you introduce a buffer between them it is not going to have the same effect as designing a "proper" filter. You will not get a sharp response by just cascading passive RC filters. You need to look at either active filters of LC filters.

Keith.
 

Thanks for explaining me the real cause of my problem......

I refered the formula "1/[2*pi*sqrt(R1R2C1C2)]".
 

Do you have another question? You have simply listed an equation with no circuit or question.

Keith
 

I am extreamly sorry Mr. Keith........

I have quoted the formula in reply to Mr. LvW's question.........
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top