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I was doing a question related to the dependence of op amps on frequency.This question is from 2nd year electrical engineering micro electronics by sedra smith.

Here is the question: enter image description here

So there is a non inverting amplifier and it has a gain of 96. it 3db frequency is 8kHz. I can find the unity gain frequency from here from the given quantities. But for what?. Then they mention that the system is required to have a unity gain frequency/bandwidth of 32kHz. So this is pretty confusing for me. Can some one explain the question please. Thanks guys!

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  • \$\begingroup\$ The question is asking, if you take the same op-amp, and use it in an application requiring 32kHz bandwidth, what will be the highest gain you can use in that application? \$\endgroup\$ – mkeith May 14 '17 at 23:27
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    \$\begingroup\$ You're supposed to assume a constant GBW product, which is pretty good assumption for most unity-gain compensated op-amps. \$\endgroup\$ – Spehro Pefhany May 15 '17 at 0:14
  • \$\begingroup\$ I don't get it. The answer is 24 V/V. Can someone try it and explain the whole procedure to me. \$\endgroup\$ – Moeen Ahmed May 15 '17 at 0:21
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    \$\begingroup\$ Op Amps have a slope of 6dB/octave (equivalent to 20dB/dec), so if it has a -3dB gain at 8khz, it will have -9dB at 16khz and then -18dB at 32khz, use that gain value with your gain of 96 to get your final answer \$\endgroup\$ – Marcelo Espinoza Vargas May 15 '17 at 1:11
  • \$\begingroup\$ To the OP, maybe you could consider accepting one of the answers? Also, I suggest you edit the title of the question to "How do you use Gain Bandwidth Product to estimate bandwidth at different gains?" Or I can edit it, if that is OK with you. \$\endgroup\$ – mkeith May 19 '17 at 16:58
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As @SpehroPefhany mentioned, we assume that the op amp has a constant gain-bandwidth product, GBWP. That is, \$\text{GBWP}=G \cdot B\$ for any gain G and bandwidth B. From the given information, we can determine that the GBWP for this op amp must be \$96 \frac V V \cdot 8 \text{ kHz} = 768 \text{ kHz}\$. Now that we have the constant GBWP, we can solve for the gain in the second case since we know the new bandwidth, 32 kHz: \$ G = \frac {768 \text{ kHz}} {32 \text{ kHz}} = 24 \frac V V\$.

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  • \$\begingroup\$ oh I see! ty... \$\endgroup\$ – Moeen Ahmed May 16 '17 at 15:19
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This is what Signal Chain Explorer shows

enter image description here

I left the gain-set resistors at the default of 20dB. That is not the answer.

EDITING Here is BODE (with gain error curve also) for 24x gain enter image description here

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  • \$\begingroup\$ I don't know man, but the answers supplied by our instructor says that is the answer. So there must be some thing not right in your simulation. And I think the answer 24V/V is more of an estimation. But thanks anyway. And thanks to everyone who helped me understand. \$\endgroup\$ – Moeen Ahmed May 16 '17 at 15:17
  • \$\begingroup\$ I left the tool's resistors at 10:1 gain, just to illustrate the opamp performance. Had I altered the gain to 24:1, the 3dB bandwidth would have been different, as your instructor said. \$\endgroup\$ – analogsystemsrf May 16 '17 at 15:36
  • \$\begingroup\$ Added a plot with same UGBW, but 24X is closed-loop-gain, showing F3dB of 32KHz. \$\endgroup\$ – analogsystemsrf May 16 '17 at 16:26

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