When we get a circuit such as the following:
How do we define the cut-off frequency? Is it still $$f_c = \frac1{2\pi R_1C_1}$$ since \$f_c\$ is defined for \$X_{c1} = R_1\$? Or is it defined so that \$V_{out}\$(the one after OA3) is \$0.707V_2\$?
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\$\begingroup\$ OA1,3 serve no purpose \$\endgroup\$– Tony Stewart EE75May 15, 2018 at 1:18
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3 Answers
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The cutoff frequency is defined as the -3dB point, where 0dB is defined as the amplitude of the signal in the passband. So it's still \$\frac{1}{2πR_1 C_1}\$.
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\$\begingroup\$ You Are forgetting current limit and large signal response \$\endgroup\$ May 15, 2018 at 0:50
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3\$\begingroup\$ @TonyStewartolderthandirt But they didn't ask about that. They just asked about the cutoff frequency of the filter. Which, unqualified, I would take to mean small-signal. \$\endgroup\$– HearthMay 15, 2018 at 1:07
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2\$\begingroup\$ Tony - I think, the definition of the 3dB cut-off frequency is applicable to LINEAR systems only - that means: Current limit and large signal behaviour (slew rate) are not involved! \$\endgroup\$– LvWMay 15, 2018 at 7:42
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It's defined to be the half-power point. Since power is proportional to \$V^2\$ (and \$I^2\$ for that matter), one half power is when \$V_\text{OUT}=\frac{V_\text{IN}}{\sqrt{2}}\approx 0.7071\cdot V_\text{IN}\$.
There are other definitions. Different filter types may set the bar elsewhere (Chebyshev, for example.)
My own way of looking at it is that the critical point is when the \$2^\text{nd}\$ derivative of phase with respect to frequency goes through zero. But that's my arbitrary choice and it incorporates the effects of nearby poles and zeros. So just ignore me on that point.
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\$\begingroup\$ Wait, so in my case, the cutoff frequency is affected by OA3? \$\endgroup\$– BeeMay 15, 2018 at 6:56
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\$\begingroup\$ @Bee I hadn't looked, but I see that you are heavily loading the output (actually, everything.) Chances are, the middle opamp won't handle the load. You should plan no more than about \$5\:\text{mA}\$ on the output of your middle opamp. Are you suggesting lighter loading, perhaps? \$\endgroup\$– jonkMay 15, 2018 at 7:17
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How do we define the cutoff frequency in an active Op-Amp filter?
For a simple RC filter the so-called cut-off frequency is when the impedance of the capacitor equals the resistance of the resistor i.e.: -
\$\dfrac{1}{2\pi f C} = R\$
Re-arranging we get \$f = \dfrac{1}{2\pi C R}\$
In your circuit you do have op-amps but, they are only providing "gain" and this does not alter the relationship between cut-off frequency, C and R except in the case when the cut-off frequency is so high that the op-amps can no-longer provide that gain.
