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I am working on MatLab simulation software to plot a Bode plot in the Matlab Environment. The open loop transfer function is $$a(s) = \frac{a_0}{(1+s/\omega_1)(1+s/\omega_2)}$$

Where \$\omega_1\$ and \$\omega_2\$ are pole frequencies (on the assumption that the op amp has 2 pole) and \$a_0\$ is the open loop DC gain of the op-amp.

To plot a bode plot for general purpose op-amp 741 we know that \$a_0=2\times 10^5\$.

But what are the pole frequencies (\$\omega\$) for op-amp 741?

Datasheet is attached in this link.

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Looking only at the datasheet you provide, the line you should be referring is the "bandwidth". The value is 437 kHz (minimum), 1.5 MHz (typical). The assumption they make here is that only one pole is present, or rather meaningful to the designer.

If you want more information, search for other datasheets that have a gain vs frequency plot, and measure the pole(s) frequency(ies) from the plot.

Finally, if there is no particular reason for which you are referring to the uA741, you might want to consider switching op amp. The 741 is extremely old, and the info you can find on it is a bit more limited with respect to a modern op amp.

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  • \$\begingroup\$ What is the reason behind referring to the line "Bandwidth" to know about poles? \$\endgroup\$ – VKJ Feb 11 '18 at 12:08
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1) The 741 is ancient, even though your work might be purely theoretical I suggest forgetting about the 741 and to choose a more modern opamp, also read Reasons no use a 741 Op-amp. I'd choose an opamp where there are some gain over frequency plots in the datasheets so that I can determine the approximate pole locations from there.

2) Even if you'd know the exact pole locations, an experienced analog circuit designer (like myself) will never rely on their values. That is because the pole locations are not very well controlled and vary over almost anything you can think of including temperature and what manufacturer actually made that opamp chip.

3) because the pole locations are so unpredictable circuit designers use a feedback network around the opamp, that can fix the poles (of the opamp + feedback) into well determined positions assuming the feedback network includes frequency compensation. Then as long as the opamp has the gain over frequency behavior which the feedback network requires from it, the outcome will be predictable. Example: if the opamp has the 2nd pole at 1 MHz, we cannot expect much gain at 10 MHz.

4) To learn more about opamps read Opamps for everyone. There are also some parts in that ebook dealing with open loop gain and feedback.

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  • \$\begingroup\$ Bimpelrekkie, you wrote: "feedback...can fix the poles...into well determined positions." For my opinion, this is not correct. I rather think that feedback allows a well-determined closed-loop gain, however, the bandwidth of the closed-loop (that is the closed-loop pole) depends on the open-loop properties which are not well-known. However. fortunately this closed-loop pole plays no important role - otherwise we could use the closed-loop circuit as a lowpass with a well-designed corner frequency (which not the case). \$\endgroup\$ – LvW Feb 11 '18 at 13:06
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    \$\begingroup\$ @LvW You're right assuming there's no frequency compensation in the feedback network. I was (silently) assuming there was, updated my answer to be more clear. \$\endgroup\$ – Bimpelrekkie Feb 11 '18 at 14:02

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