I want to know the cut off frequency of this circuit, given that the GBW is constant (1.9MHz) and R1=8.2kOhm and R2=114.6KOhm. I did 19e6/2*\$\pi\$*R2 (since it doesn't have a capacitor) but my calculations aren't right.
What I am doing wrong?
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\$\begingroup\$ What you are looking for would be given by \$\frac {GBW}{Gain}\$ but as I understand it that's not a good way to make a filter. It is less a filter and more an upper limit to the usefulness of the circuit. \$\endgroup\$– JRECommented May 29, 2023 at 17:45
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\$\begingroup\$ and the Gain in this case is -R2/R1, right? \$\endgroup\$– Patricia VieiraCommented May 29, 2023 at 17:50
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2 Answers
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Get some lin-log graph paper and do this: -
The 3 dB point of your amplifier will be about 130 kHz. If you need a more exact answer, turn the graph into a formula.
answered May 29, 2023 at 17:58
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\$\begingroup\$ it was 135.95 Khz, Thank you very much Andy! \$\endgroup\$ Commented May 29, 2023 at 18:05
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2\$\begingroup\$ Shouldn't the gain on the Y-axis be ~15? I.e. (1+R2/R1) for bandwidth purposes? Not a big difference in the result though. eng.libretexts.org/Bookshelves/Electrical_Engineering/… \$\endgroup\$– John DCommented May 29, 2023 at 18:15
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\$\begingroup\$ @JohnD it's a gain of minus 14 and, when the sign is ignored and 14 is converted to decibels you get 22.9 dB. Also, the formula you used was for a non-inverting amplifier. \$\endgroup\$– Andy akaCommented May 29, 2023 at 18:22
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\$\begingroup\$ @PatriciaVieira If we are done here, please take note of this: What should I do when someone answers my question. If you are still confused about something then leave a comment to request further clarification. \$\endgroup\$– Andy akaCommented May 29, 2023 at 18:23
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\$\begingroup\$ @Andyaka Right, the formula I gave is for a non-inverting amplifier, but it's the "beta" or feedback factor that counts for calculating bandwidth, not the inverting gain. (See link in my previous comment or answer below.) \$\endgroup\$– John DCommented May 30, 2023 at 21:56
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You have accepted an incorrect answer. John D's comment is correct.
Whether we are considering a non-inverting amp or an inverting amp, the bandwidth is given by GBW/Noise Gain = GBW/(1+R2/R1) = GBW*beta where beta = 1/Noise Gain.
So, assuming that the phase of the openloop gain is -90 degrees (falling at -20 dB/decade) where the horizontal Noise Gain plot (1/beta plot) crosses the openloop gain plot then, at this frequency, the closed loop gain of a non-inverting amplifier and the closed loop gain of an inverting amplifier will both be down 3 dB from there low frequency values. This assumes that the two amplifiers have the same Noise Gains (1/betas). Of course this leaves you with the characteristic of the inverting amplifier that it has less closed loop signal gain than the non-inverting amplifier for the same bandwidth. For example if an inverting amplifier is configured for unity closed loop signal gain with a Noise Gain of 2 then a non-inverting amplifier with the same Noise Gain will have a closed loop signal gain of 2. Both amplifiers have the same Noise gain and closed loop bandwidth but the non-inverting amplifier has double the closed loop signal gain compares to that of the inverting amplifier.
Strictly speaking this assumes that the open loop gain is falling at -20 dB/decade = -6 dB/octave which is approximately true for most of the amplifier's openloop bandwidth.
And so in your case the closed loop bandwidth = 1.9 MHz/14.98 = 126.9 kHz.
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1\$\begingroup\$ There is only one answer (to be accepted), you can not accept a comment as an answer. If your answer proves to be more valid than the other answer, then the accepted answer can be changed. Also, there is no need to address your answer with an
@as the OP will receive a notification and see your answer regardless. You may want to edit and amend your answer by removing the first couple of lines (which would have served better as a comment under the other answer). \$\endgroup\$ Commented May 29, 2023 at 21:27 -
1\$\begingroup\$ A short explanation to the given expression (GBW/Noise gain): This expression results from the general definition for the closed-loop bandwidth of a system with feedback: The 3dB bandwidth is defined at a frequency where the magnitude of the loop gain is unity, \$\endgroup\$– LvWCommented May 30, 2023 at 7:58
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1\$\begingroup\$ That's right LvW. At the closed loop gain's -3 dB frequency, the amplification through the forward part of the loop is cancelled out by the attenuation through the feedback network giving Loop Gain = Aol*beta = 1. Although strictly speaking the loop gain would be equal to -Aol * beta, the minus sign being included as a result of the inversion at the input's summing junction. \$\endgroup\$– user340992Commented May 30, 2023 at 10:46
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1\$\begingroup\$ Further clarification: My comment applies to a first-order system only (thats what we can assume for a unity-gain stable opamp) \$\endgroup\$– LvWCommented May 30, 2023 at 18:33