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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!!
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).
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.
And here's the link.
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.
