0
\$\begingroup\$

I need to create a high pass filter to attenuate frequencies below 50KHz, I intend to use an RC circuit with an operational amplifier.

My question is: How do I set my cutoff frequency? I know that if I set a cutoff frequency of exactly 50K Hz I will get losses in my signal, how do I find the ideal cutoff frequency? Do I Use the Bode Diagram? If so, how does it apply in this situation?

Thanks!!

\$\endgroup\$
  • 1
    \$\begingroup\$ In order for anyone to help you work out a cutoff frequency, or any of a number of other important details about a high pass filter, they will need to know what are you trying to do. Do you imagine that one can just magically set a cutoff frequency knowing nothing more than, "I want a high pass filter?" Do you think there is a universal methodology that always works in every single situation without exception?? And that if there was, that it could be written here (without a huge tome being written?) I've already written more than you. And I'm just commenting, for gosh sake. \$\endgroup\$ – jonk Aug 5 '17 at 20:03
  • \$\begingroup\$ there are always tradeoffs. depending on your requirements, more complex filters could be necessary. But most of the time the 1st order can work just fine. other filter types which trade off the steepness of cutoff have more rippling/distortion in the passband. things to look up: chebyshev (moderate slope, moderate distortion), butterworth (slighter slope, minimal distortion), elliptic (maximal slope, maximal distortion). higher order means more parts but also "better" response generally. \$\endgroup\$ – jbord39 Aug 5 '17 at 20:20
  • \$\begingroup\$ Why do you need a highpass filter in the first place? Let's start with that. \$\endgroup\$ – mkeith Aug 6 '17 at 0:31
1
\$\begingroup\$

Here's a bode plot of a simple RC high pass filter's amplitude response - it contains the formula you need to work out the amplitude at every frequency. Use the formula to decide what cut-off frequency you need and whether the attenuation below 50 kHz is suitable: -

enter image description here

If the filter doesn't attenuate quickly enough below 50 kHz then you need to consider a 2nd order filter or, maybe one that is of a much higher order.

Here's a 2nd order high-pass-filter's response at a cut-off of 48.2 kHz and note that it can produce a peak (resonance) if the Q is too high: -

enter image description here

enter image description here

I've chosen a value of resistance such that the Q is unity - notice the little peak in the response - if R is smaller the peak rises. The benefit of using a 2nd order filter is that instead of a 6 dB per octave fall in gain at low frequencies, the rate is 12 dB per octave (40 dB per decade).

Interactive RLC tool

\$\endgroup\$
1
\$\begingroup\$

To build on top of Andy aka's answer.

$$|V_{out}|=|V_{in}|×\frac{\omega RC}{\sqrt{1+\omega²R^2C^2}}$$

$$H(\omega)=\frac{\omega RC}{\sqrt{1+\omega^2R^2C^2}}$$

\$H(\omega)\$ will give us the absolute value of the filter, since it's already absolute-valued by Andy aka. I'm aware that a more normal way to write it would be \$|H(S)|\$, but that's our \$H(\omega)\$. So we're cool.

There's a couple of free parameters, but I'd start with \$R\$ and set it to \$1kΩ\$ because that's a reasonable value, not too high and not too small. \$\omega\$ is \$2\pi×50×10^3\$. The last parameters we get to choose is the resulting amplitude, how much do you really want?
Do you want \$\frac{\sqrt{2}}{2}≈70\%\$? Or perhaps you want \$95\%\$? I'll assume you want \$95\%\$.

So what we want is the last parameter, \$C\$

After some fiddling, you'll get this equation:

$$C=\frac{\frac{95}{\sqrt{100^2-95^2}}}{\omega R}$$

If we plug in our numbers it will look ugly like this: $$C=\frac{\frac{95}{\sqrt{100^2-95^2}}}{(2\pi×50×10^3) (10^3)}=9.683nF$$

The cutoff frequency (-3dB) will be at: $$\omega_c=\frac{1}{RC}=\frac{1}{(10^3)(9.683×10^{-9})}=103krad/s=16.4kHz$$

You tell me if that's good enough for you. But that's what you get with a first order RC HP filter.

Here's a schematic if you want to mess around or if you don't believe me.

enter image description here

And here's the link.

\$\endgroup\$
  • 1
    \$\begingroup\$ You should start out with C and pick an obtainable value. You can get practically any resistor value you like, but capacitors are much more constrained. \$\endgroup\$ – user207421 Aug 6 '17 at 1:32
  • \$\begingroup\$ You're actually right, I keep messing that up every time I do it. Oh well, that's a really easy fix with the equations above, though I won't edit my answer for that. \$\endgroup\$ – Harry Svensson Aug 6 '17 at 1:50
1
\$\begingroup\$

If you decide to use the circuit below, the cut-off frequency is depend on R3, R4, C1 and C2. When R3 = R4 = 10k ohm, then the value for C1 = C2 = 318pF. However do take note opamps have their limitation of you go to higher frequency, not all opamps can suit your requirements if the frequency can go up to GHz range.

2nd order High Pass filter

\$\endgroup\$
  • 2
    \$\begingroup\$ Not even all passive components like capacitors go up to the Ghz range. \$\endgroup\$ – Harry Svensson Aug 6 '17 at 1:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.