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how can I calculate the cut off frequency of the non inverting op amp circuit? I calculated the one for the inverting op amp so far.

I calculated the unity gain bandwidth A0 = G1 + G2 / G2.

should I assume that s is wp?

thanks

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  • \$\begingroup\$ Without an external capacitor, you'll be relying on the gain-bandwidth product (and possibly parasitic capacitance) of the op-amp, which varies significantly from chip to chip. \$\endgroup\$ Commented Sep 5, 2019 at 17:33
  • \$\begingroup\$ Also, note that the op-amp itself does not have a single transfer function since it is a nonlinear device. \$\endgroup\$ Commented Sep 5, 2019 at 20:36

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Without any reactive components (ie. capacitors or inductors) in your feedback circuitry, the cutoff frequency will be set almost entirely by the characteristics of the op-amp itself. As Caleb commented, the gain bandwidth product from the datasheet will give you an estimate.

If you have a specific op-amp in mind I'd directly measure your cutoff frequency. Change the frequency of the input signal until you find the frequency where the gain drops -3dB and you've found your cutoff frequency.

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  • \$\begingroup\$ this is only an exam exercise. no other details were provided. is the calculation correct for wp = w1 * G2 / (G1 + G2) at least? \$\endgroup\$
    – tairit
    Commented Sep 6, 2019 at 8:50
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    \$\begingroup\$ (G1+G2)/G2 is the correct gain (if G2 is the feedback resistor's conductance - ie. not the resistor from inverting-in to ground). However then the pole would occur when G1 = 0 or R2 = 0 (this makes the denominator zero). The cutoff frequency of single pole circuit is equal to the frequency of the pole (which can be shown with some algebra), so if R2 isn't actually a resistor and can have an impedance of zero at some frequency then that frequency would be your cutoff frequency. Assuming R2 is a resistor and the op-amp is ideal I don't believe there is a cutoff frequency, gain should be linear. \$\endgroup\$
    – Giesbrecht
    Commented Sep 7, 2019 at 17:53

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