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For my electronics course, I have been tasked with creating a circuit that will take a \$10\text{mV AC}\$ input and amplify by a factor 490 using the LM741CN op-amp with an effective bandwidth (attenuation less than 3dB) from \$1\text{ Hz}\$ to at least \$20\text{ kHz}\$. Knowing that its \$f_t = 1.5\text{ MHz}\$, I know I must split the gain into two stages to increase the bandwidth or else

$$f_{-3\text{dB}}=\frac{f_t}{A_{CL-O}}=\frac{1.5\text{ MHz}}{490}=3.06\text{ kHz.}$$

This lead me to having two stages each with the square-root of the original 490 gain (\$7\sqrt{10} \approx 22.13\$). To do this, I used two inverting op-amps configurations so that the gain would simply be \$-\frac{R_2}{R_1}\$ and set \$R_2 = 22.13 \text{ k}\Omega\$ and \$R_1 = 1 \text{ k}\Omega\$. I do this twice and I get the required gain. This also has the added benefit of increasing \$f_{-3\text{dB}}\$ to \$67.8\text{ kHz}\$.

However, I also have to account for the slew rate of the op-amp, which for the LM741CN is \$0.5\text{ V}/\mu s\$. This brings the maximum frequency to:

$$f_m = \frac{\text{SR}}{2\pi\hat{v}_o} = \frac{0.5\text{ V}/\mu s}{2\pi\times 10 \text{ mV} \times 490} = 16.1\text{ kHz}$$

This posed a problem to me so I went on to NI Multisim to simulate the bode plot for the planned circuit. Here is the circuit:

enter image description here

And here is the bode plot for the output vs input:

enter image description here

The most notable piece of information is the fact that the effective bandwidth (as I've described above, the frequency range where attenuation is less than 3dB) reaches past \$20\text{ kHz}\$, all the way above \$24\text{ kHz}\$. The regular gain (gain at 1 Hz) is 53.797 dB so at 50.797 dB, I take that to be ineffective. Yet at \$16.1\text{ kHz}\$, there is only 1.438 dB of attenuation.

My question is why is this happening? Why is the effective bandwidth larger than what my calculations tell me they should be. I can't find out what I did wrong with my calculations, and I am not sure if there are any other equations to consider.

EDIT: Some more info I was informed should be part of the post:

My actual task is described here:

enter image description here

It doesn't say anything specifically regarding 3dB being the point at which frequencies are ineffective, but it does talk about total harmonic distortion, which I assumed would be similar to the 3dB limit since my class still hasn't actually taught anything about calculating it, and I am not sure if it is possible to calculate in a theoretical POV. It also mentions the LM318N, but I got that to work fine in my calculations.

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    \$\begingroup\$ Model != datasheet specifications != reality. \$\endgroup\$
    – JRE
    Commented Sep 30, 2020 at 16:59

2 Answers 2

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My question is why is this happening?

You are probably using an AC simulation to generate the Bode plot.

The AC simulation method assumes circuit linearity, therefore it neglects the slew rate effect.

As to why you don't get the full 67.8 kHz you calculated under linear assumptions, probably there is either some loading effect due to the relatively low (for a 741) 1 kohm load on the first stage, as well as some compounding effect where the net gain of the two stages drops by more than 3 dB before either of the individual stages drops by 3 dB. The LM741 model in Multisim might also have different values for \$f_t\$ and \$A_{CL-O}\$ than you used in your calculations.

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  • \$\begingroup\$ Okay, so what would explain why it goes up to 24 kHz. My other question I guess is for my actual homework, what should I do to (in theory) get my effective bandwidth to increase to 20kHz using the LM741CN? It looks like it just can't due to the SR limitations. \$\endgroup\$ Commented Sep 30, 2020 at 17:06
  • \$\begingroup\$ @MrMineHeads, you can use a transient simulation to find the behavior including slew rate effects. Not knowing what you were asked to do on your homework, I can't say more than that. Practically, slew rate will be a fundamental limitation to what amplitude you can get out of the final stage at 20 kHz without distortion. Are you allowed to use an external transistor or two in your solution? \$\endgroup\$
    – The Photon
    Commented Sep 30, 2020 at 17:08
  • \$\begingroup\$ Here is a link to the actual question: i.imgur.com/mkyfV4B.png. It doesn't say anything specifically regarding 3dB being the point at which frequencies are ineffective, but it does talk about total harmonic distortion, which I assumed would be similar to the 3dB limit since my class still hasn't actually taught anything about calculating it, and I am not sure if it is possible to calculate in a theory POV. It also mentions the LM318N, but I got that to work fine in my calculations. \$\endgroup\$ Commented Sep 30, 2020 at 17:23
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    \$\begingroup\$ Don't bury the additional information in the comments. Edit it into your original question. \$\endgroup\$
    – Transistor
    Commented Sep 30, 2020 at 18:02
  • \$\begingroup\$ will do @Transistor \$\endgroup\$ Commented Sep 30, 2020 at 19:07
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Photon has explained why AC analysis does not give the correct answer when nonlinear effects such as slew-rate limiting (or crossover distortion) are significant.

The slew rate limitation is baked into the LM741 by the design choice of the compensation, however if you make a bridging amplifier you can achieve the design goal for typical (not worst case) slew rate as below:

schematic

simulate this circuit – Schematic created using CircuitLab

If you needed a ground-referenced output you could also build a discrete output amplifier for the output 741 and close the loop around it. It would resemble the output stage of an audio amplifier.

Of course it would be much easier to use another amplifier such as a TL081 or LM318.

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