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I am trying to derive an equation describing the cutoff frequency of a high pass filter op amp, as seen below: High-Pass Op-Amp

I have tried deriving the cut off frequency from it's transfer function: transfer function

But I can't get the intended result as indicated at the bottom of working? Is anyone able to spot where I am going wrong in my working or method of attack please?

working

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  • \$\begingroup\$ Simple way to analyze ( in your head) is this is DC gain =R2/R1 and HPF breakpoint is Zc(f)=R1 then solve for f but assumes OpAmp an operate below Gnd so split supply or change bias on Vin+ \$\endgroup\$ – Sunnyskyguy EE75 Apr 9 '17 at 11:46
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For this circuit we have

$$H_{(s)} = - \frac{s R_2 C_1}{1 + s R_1 C_1} $$

So we have one Pole at

$$-\frac{1}{R_1 C_1}$$

And one Zero at the origin.

All this means that for low frequency the circuit behaves like an ordinary op-amp based differentiator.

With the gain $$A_V = \omega R_2C_1$$

And the gain reachs \$1 V/V\$ when sinal frequancy is equal to \$Fo=\frac{1}{2 \pi R_2 C_1}\$

As signal frequency increases \$Xc\$ drops and when \$Xc = R_1\$ we have a Pole:

$$\omega = \frac{1}{R_1 C_1}$$

$$Fp=\frac{1}{2 \pi R_1 C_1}$$.

And the the magnitude of a transfer function (voltage gain vs frequency) is equal $$A = \frac{\omega R_2 C_1}{\sqrt{1 + \left ( \omega R_1 C_1 \right )^2}}$$

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  • \$\begingroup\$ @ConfusedCheese Also notice the the cutoff frequency is the frequency when the voltage gain drops -3db below the maximum gain. And this is not equal to \$\frac{1}{\sqrt2}\$. But \$\frac{R2}{R1}*\frac{1}{\sqrt2}\$ \$\endgroup\$ – G36 Mar 12 '17 at 13:37
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That's right, the -3dB point is at R1.C 1/radians, or 1/2.pi.R1.C Hz.

Were you expecting R2 to figure in there?

From inspection, you can notice that the inverting input is at a virtual ground because of R2. This effectively isolates R1 from R2, so only R1 and C affect the frequency response.

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  • \$\begingroup\$ Sorry, i'm confused as how to arrive at that cut off frequency from the transfer function? \$\endgroup\$ – ConfusedCheese Mar 12 '17 at 12:22
  • \$\begingroup\$ So the impedance of the capacitor is low enough above the cut off frequency to allow those voltages through? But below that they are dropped across the impedance of R1 & C? \$\endgroup\$ – ConfusedCheese Mar 12 '17 at 12:37
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Adding to Neil's answer:

The opamp holds its "-" input at a virtual ground.

R1+C1 convert input voltage into a current.

Opamp uses R2 to convert this current back to voltage at the output.

R1, C1 control cutoff. R2 controls gain/attenuation.

Note: Cutoff depends on signal source impedance! Your calculation is only valid for a zero-ohm impedance voltage source. Keep this in mind when using this circuit in the real world...

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