# High-Pass Filter

1. Feb 4, 2008

### DefaultName

Hi, I need to construct a filter to pass 5Hz and above. I'm also using this circuit to prevent any DC signals from passing (I have a DC offset that I need to get rid of). I'm using the equation f = 1 / 2*pi*RC (in which I found after derivation).

As of right now, I have two designs:

1) Series RC - works fine in SPICE.
2) Differentiator (OpAmp config) - can't get working on SPICE. I have no idea why it's not working. I've tried connecting pins 1 and 5 of the opamp to GND, as well.

This is what I have below:

http://img138.imageshack.us/img138/3764/14914103um3.png [Broken]

Code (Text):
* Schematics Netlist *

V_V2         $N_0001 0 5v V_V3 0$N_0002 5v
R_R1         $N_0004$N_0003  79.5k
C_C1         $N_0005$N_0004  .1u
R_R2         $N_0003$N_0006  79.5k
X_U2         0 $N_0003$N_0001 $N_0002$N_0006 uA741
C_C2_BYPASS         $N_0001$N_0002  .01u
V_V_INPUT         $N_0005 0 DC 0 AC 1 +SIN 2.4 50m 100 0 0 0 R_R3 0$N_0006  10k

I am using PSpice Schematics, Evaluation Version 9.1 - Web Update 1 for my simulations.

Last edited by a moderator: May 3, 2017
2. Feb 4, 2008

3. Feb 4, 2008

### user101

Oh... so a regular series RC will do then. As far as picking a practical value of R and C... how do I do that if I want to just block DC values? This means I would have to pick something obviously above a few Hz for it to properly filter... Frequency_cutoff > 0. If a real engineer were to design this, how would he/she go about this?

So one of the advantages of using an opamp for filtering is for a higher range of frequencies?

4. Feb 5, 2008

### rbj

DC will always be blocked. but you want your RC product to be large enough that none of the low frequencies that you are interested in are attenuated more than you want them to be. that's what that formula:

$$f_0 = \frac{1}{2 \pi R C}$$

$$H(s) = \frac{RCs}{1 + RCs} = \frac{1}{1 + \frac{1}{RCs}}$$

if you plug in $s = j 2 \pi f$, evaluate the transfer function for that s, and compute the magnitude, you get for a frequency response

$$H(j 2 \pi f) = \frac{1}{1 + \frac{1}{j 2 \pi RC f}} = \frac{1}{1 - j \frac{f_0}{f}}$$

or

$$|H(j 2 \pi f)|^2 = \frac{1}{1 + \left( \frac{f_0}{f} \right)^2 }$$

or, in dB,

$$20 \log_{10}|H(j 2 \pi f)| = -10 log_{10} \left(1 + \left( \frac{f_0}{f} \right)^2 \right)$$

when f=f0, then you're down to -3.01 dB. at what frequency are you willing to be attenuated by 3 dB? when you decide that, you will know what the RC product (sometimes called a "time constant") is.

5. Feb 5, 2008

6. Feb 9, 2008

### user101

Thank you.

If I were to send a small signal through the circuit, something along the lines of a sine wave with a 100 mV peak to peak signal, would this work at such a low amplitude? This wave will be traveling at f = 200 Hz. I'm planning on using a metal-film resistor and a electrolytic capacitor of some sort. I won't have access to lab until two weeks later, so I was just curious :/ It works on spice... but that isn't as realistic.

7. Feb 9, 2008

### mheslep

Sure it will work if you are measuring the output of your filter w/ a high impedance tool like and oscilloscope. However if you actually want to do anything with the filter - like feed it into some small load, say an 8 ohm speaker, then the cut off frequency you originally calculated above will change drastically. To avoid this use the op amp.

8. Feb 12, 2008

### edmondng

When choosing RC for filter, are there any difference between values R and C other than tolerance?

For example :
10uF and 8.2k produces cutoff of 1.94Hz
1uF and 82k produces cutoff of 1.94Hz.

The reason i ask is because i was trying to remove a DC offset for my square wave and it didn't do it until i increase the capacitance to 20uF and reduce the resistance to 4.7k. Even so, my square wave goes from -1.45V to 1.85V rather than -1.65V to 1.65V
How can i make it symmetric? Input going in looks good, 0-3.3V.

I have no idea why..?? If i change the resistance to 47k or 82k and keep 20uF it remove just a little like -0.8V to 2.5V.

Thanks

9. Feb 12, 2008

### mheslep

Im assuming you are building a simple RC circuit, no op amp. No it doesnt make any difference, theoretically. However, the output and input impedances of the filter will differ so that, practically speaking, when you connect the filter to some load the total (R+Rload)C will be different and you'll have a different cut off.

Something is wrong in your setup, probably some kind of ground loop. Any value of series of capacitance should block the DC voltage.

10. Feb 12, 2008

### edmondng

capacitance block DC but if you just put a cap in series with the input and no resistor there is no path for the DC to flow to ground.

The setup is just simple. Square wave coming out from an op amp, goes through the RC to remove the DC offset then into another amplifier to increase the amplitude. I have a ground plane so with the via it goes straight to ground. Its puzzling for me too....

11. Feb 13, 2008

### mheslep

Im assuming the RC circuit shown in post https://www.physicsforums.com/showpost.php?p=1597214&postcount=2". If you connect your biased square wave generator to the input the DC part will not enter the circuit. In reference to the ground plane loop I mentioned, imagine an additional 'hidden resistance' somewhere in the ground return path. The effect of this is that as the ground current flows through this resistance it creates a voltage drop across the hidden resistance which appears as bias on top of your idea output.

Last edited by a moderator: Apr 23, 2017