A simple RC low pass filter

Go To Last Post
8 posts / 0 new
Author
Message
#1
  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Hi, I'm not an expert in constructing a simple passive low pass filter and I need help from someone who has an experience with this.
I need to make a simple low pass filter with R and C, like on the picutre, for filtering AC power voltage, with fc=500 Hz.
In theory fc = 1/(2*pi*R*C), and I can get the approximately right fc with using R=33ohm and C=10uF or R=68ohm and C=4.7uF or R=680ohm and C=470nF or R=3k3 and C=100nF ..., but which configuration will I use in project?
Signal Vin is a max 5V AC (reduced from 230V AC power network) and signal Vout is then connected to the a/d of atmega48.

Attachment(s): 

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Choose any combination for which R is large compared to the source impedance/resistance.

Jim

Jim Wagner Oregon Research Electronics, Consulting Div. Tangent, OR, USA http://www.orelectronics.net

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Ok, my input circuit is shown on the picture below. Is it better to put the filter at pins Uin or Uout, where Uin is from a transformer (I don't have data about impedance/resistance). If I connect a filter at Uout pins does it mean that R in low pass filter need to be greater than R2 and how many?

Attachment(s): 

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Note that the effective resistance of the divider as seen by the output (Thevenin equivalent) is the parallel combination of R1 and R2, i.e. R1*R2/(R1+R2). So you need to subtract this value from the R used in the RC network. If you select the values for the divider and C you can eliminate R altogether.

If you think education is expensive, try ignorance.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Ok so Uin is a direct connectionto 230V mains?

I would think twice about proceeding with if I had to ask questions about RC filters..

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Yes, it's connected directly to 230V mains and it works ok :D , but since I'm using algorithm with Fourier series I must have a simple anti-aliasing filter.

@emuler
Excellent, if I understand correct, then I have a simple solution, I just need to compute adequate capacitor with respect to the voltage divider, that is

R= (R1*R2)/(R1+R2)

C = 1/(2*pi*R*fc)

thanks for your answers!

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Beware:

If the input signal is AC, then the output signal will go negative. A Mega48 (or any AtMega processor) ADC CANNOT handle negative inputs!

Jim

Jim Wagner Oregon Research Electronics, Consulting Div. Tangent, OR, USA http://www.orelectronics.net

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

I would not suggest direct mains connection either. Use a isolation transformer at least, or better yet, any transformer that already outputs something lower than mains voltage. [edit: you had transformer already, good. I thought someone just said direct connection there.]

Also put a half supply voltage DC bias to transformer so you get AC over VCC/2 DC component and you can measure the full wave.