# Design high pass filter

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!!

• 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.
– jonk
Aug 5, 2017 at 20:03
• 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. Aug 5, 2017 at 20:20
• Why do you need a highpass filter in the first place? Let's start with that. Aug 6, 2017 at 0:31

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: - 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: -  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

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.  