# Help with Low Pass Filters

Discussion in 'General Electronics Chat' started by Mark3722, Dec 26, 2014.

1. ### Mark3722 Thread Starter New Member

Dec 26, 2014
13
0
Hi All,

Software guy here who has developed an interest in hardware. I started with a RPI then an Arduino. Now I am starting working with basic discreet circuits and going analogue! Mind bending stuff. Not the precise 1's and 0's I am used to working with.

I am trying to get my arms around low and high pass filters with RC components. I have both a signal generator and new Rigol DS1054Z oscilloscope to help me through my learning curve.

To start, I built a circuit targeted at blocking signals above 310Hz using a 510Ohm resistor and a 1uF capacitor (electrolytic) . I built a first order and second order version of the circuit (see below) and then ran tests on the oscilloscope measuring the amplitude in V at different frequencies from 30 to 8K Hz. The results from the tests are below.

I somehow think I messed things up. With both the 1st and 2nd order circuits, the voltage dropped far sooner that the 310Hz target. I guess I was expecting something north of 8.5V at 310Hz the appropriate logarithmic drops for each circuit.

Can anyone give me some guidance here? What am I doing wrong? Be nice Like I said, I am just starting to explore the analogue world. I was also going to try and simulate this in LT Spice but have not got that far...

Thanks,

Mark
-----------------------------------
Here is a representation of the circuit:

9 VAC -------R-------------------R---------------- OUTPUT
| |
C C
| |
GD GD
-----------------------------------

Here is the recorded data:

Exp 1Exp 2
R=513.2,510.8R=513.2
C=1u,1uC=1u
f=310Hf=310H

fV with 2nd order circuitV with 1st order circuit
309.099.24
608.49.1
1205.88.6
2402.97.05
3102.76.42
4801.74.8
9600.5442.8
20000.1821.44
40000.0440.709
80000.0090.306

2. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,066
The forum software swallowed all of your tabs, so your posting is unreadable.

Use the .code tags to escape the normal reformatting. (Come on, software guys should know about this)

3. ### MrChips Moderator

Oct 2, 2009
12,636
3,455
Firstly, your circuit is incorrect. The capacitor is in the wrong place.
The capacitor goes across the output and GND, not the input.

Secondly, your numerical result shown is hard to read. Where does the frequency end and the voltage begin?

4. ### Mark3722 Thread Starter New Member

Dec 26, 2014
13
0
yep. everything looked ok in the editor but did not come out the way I expected. here are some pics.

5. ### Mark3722 Thread Starter New Member

Dec 26, 2014
13
0
here is the table

"

6. ### #12 Expert

Nov 30, 2010
16,704
7,354
I can't quite track the numbers, but I can say: An RC filter is calculated for the -3db point, so it has to start responding early in the frequency sweep to get there by the declared frequency. That's based on the idea that the Xc (reactance of the capacitor) is equal to the resistance of the resistor at, "the frequency". When you get to the frequency of interest, each component is using up part of the voltage.

7. ### crutschow Expert

Mar 14, 2008
13,503
3,376
You can't combine two 1st order filters like that and get what you expect. The second filter significantly loads the first filter thus giving a rolloff at at different frequency then you expected. You can minimize that effect by using a resistor value for the second filter at least 10 times the first resistor and the capacitor at least 10 times smaller.

Also note that the filter corner frequency is the point where the response is -3db down. If you put two identical frequency 1 pole filters in series then the corner frequency will be -6db down.

I suggest you try simulating the circuit in LTspice. It's a lot easier and faster than building and testing the circuit.

8. ### Mark3722 Thread Starter New Member

Dec 26, 2014
13
0
Dear Crutschow,

Thanks for pointing this out on the 2nd filter. I will dive into LTSpice to simulate this.

Just so I am clear, what would the expected voltage be with a 1st order circuit at the corner frequency if the original signal was 9vac? Around 6.5vac? Perhaps you can also show me the math you are using to arrive at this number.

Many thanks,

Mark

9. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,507
512
I did a quick model in MultiSim of the 2nd order RC filter, at 310-311 Hz I get -9 to -10 dB. Like others already said, the normal design would call for -3 dB at the target frequency. So if you want your cutoff frequency to be 310 Hz, then you need to redesign the filter because you are getting -10 dB where you should be getting -3 dB.

Last edited: Dec 26, 2014

Oct 2, 2009
12,636
3,455
11. ### crutschow Expert

Mar 14, 2008
13,503
3,376
The reponse will be down 3dB or .707 so 9Vac will be reduced to about 6.36Vac.
The equation is Vo = Vin * 1/(1+RC2ϖf) where f equals the frequency in Hz.
(Note that your should use Hz for Hertz. H stands for Henry, a unit of inductance.)