# Help with passive filters

Discussion in 'General Electronics Chat' started by mike., Feb 24, 2013.

1. ### mike. Thread Starter New Member

Feb 24, 2013
3
0
Hi,

New here and a little stuck on a problem on passive filters, I'm hoping someone may be able to help.

I need to work out the cut off frequency, and understand the basic formula is 1/2(pi)RC

However, for example in the low pass filter, I have 2 additional resistors, each individually in parallel with the capacitor.

How do these additional resistances effect the COF?

Mike

2. ### blah2222 Distinguished Member

May 3, 2010
581
38
So remember, in the equation of $f_{cutoff} = \frac{1}{2\pi RC}$, the resistance R is the equivalent resistance path that the capacitor C sees.

3. ### #12 Expert

Nov 30, 2010
18,078
9,617
You can calculate a single stage filter much like a resistive voltage divider.
Figure the impedance of the capacitive part compared to the series resistor.

4. ### mike. Thread Starter New Member

Feb 24, 2013
3
0

So, the additional resistances would not have an effect on the COF then presumably?

5. ### #12 Expert

Nov 30, 2010
18,078
9,617
They don't change the frequency. They change the Q. The response slope gets more horizontal.

6. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
727
Not much if they are in parallel with the capacitor.

7. ### crutschow Expert

Mar 14, 2008
19,830
5,547
The resistances do affect the COF. You have to calculate the equivalent value of all the resistances and then use that to calculate the COF as blah2222 noted. They will also attenuate the signal in the flat portion of the filter response. They have no affect on the filter Q.

8. ### mike. Thread Starter New Member

Feb 24, 2013
3
0
This is what I thought when I first posted, but I am not sure how to do this.

Any help appreciated

Mike

9. ### tubeguy Well-Known Member

Nov 3, 2012
1,157
198
The capacitor acts like a frequency variable resistor. Its AC resistance continues to drop as frequency increases and it eventually 'swamps out' the resistors in parallel with it.

10. ### Jaguarjoe Active Member

Apr 7, 2010
770
91
You have 2 R's in parallel. Calculate their equivalent R.
Now you only have 1 R across the C.
Calculate the reactance of the C at your frequency of interest.
Calc the equivalent resistance of the C&R in parallel.
Now add the 2nd (or is it 3rd?) R into the mix as the bottom leg of a voltage divider and slide into home.

Divide and conquer.

11. ### #12 Expert

Nov 30, 2010
18,078
9,617
I guess I'm going to have to go back and study filters again.
I was under the impression that a resistance in series with an inductor lowered the Q, a resistance in parallel with a capacitor lowers the Q, and a flatter line on graph of the frequency response indicates a lower Q.

12. ### Jaguarjoe Active Member

Apr 7, 2010
770
91
Take it to the extreme-

If the series R was very big, there wouldn't be much, if any effective capacitance which would then make the Q very small.

13. ### crutschow Expert

Mar 14, 2008
19,830
5,547
Q can only vary in a passive circuit with both and L and C. If you have only R and C, or R and L then the Q is fixed.

#12 likes this.
14. ### crutschow Expert

Mar 14, 2008
19,830
5,547
You can take it to a bizillion ohms if you like but that still won't affect the Q or effective capacitance. A very large R will just make the RC time-constant very large and the consequent COF go to a very low frequency.

15. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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727
Maybe I am picturing this wrong. The calculation is for a series R, parallel C, and it seems the OP is referring to parallel R and parallel C, with no series R.

Did I miss something?

16. ### crutschow Expert

Mar 14, 2008
19,830
5,547
In his original post he stated
That, to me, implies there is also a resistor in series for the low pass function, but I'm open to other interpretations.

17. ### blah2222 Distinguished Member

May 3, 2010
581
38
That is what I thought as well.