Designing an Integrator Circuit with Given Absolute Gain

[PhysicsTruth]

Homework Statement

Keeping in mind the corner frequency at 3dB, design the R,C values for an integrator to maintain a gain of magnitude 10 throughout the band pass.

Relevant Equations

$f_{cutoff} = \frac{1}{2 \pi R_{f}C}$
$f_{0dB} = \frac{1}{2 \pi R_{1}C}$
$DC Gain = -\frac{R_{f}}{R_{1}}$
$AC Gain = -\frac{X_{C}}{R1}$

For the integrator circuit, I can design the cutoff frequency and the 0dB frequency as required. Using Laplace transforms, the gain is around $-\frac{1}{sR1C}$, where s is the complex impedance parameter. But, how do I maintain the absolute value of this gain at 10 for the whole band pass for this integrator? I need to find out the R1, Rf and C values, given the cutoff frequency and this absolute gain at 10 for the band pass, but I don't know how to utilise this absolute gain in deducing the necessary information in finding out the values. Can someone help me out in this?

 

[Tom.G]

Putting on my thinking cap, I wiggled my fingers and invoked the knower of all things...
https://www.google.com/search?&q=gain+of+integrator+op+amp

 

[DaveE]

What? This question doesn't make sense to me. Integrators don't have passbands (according to me) the gain is essentially infinite at DC and constantly decreases (inversely proportional to frequency) forever. Of course there are some real issues in the real world, like a low frequency pole, but theoretically integrators are simple.

So that makes me think that either:
1) Your instructor is an idiot, or too busy to write a good HW question.
2) You have left out part of the problem description.

So, next step: Draw a schematic of the circuit, take a picture of it with your phone and post it AND tell us the exact wording of the question. For example, you refer to R₁ and R_f, but I have no idea what those are (OK, I could guess, but YOU NEED TO TELL US).

BTW, it sounds like you are describing what I would call a low pass filter. A 1st order LPF will have a passband at low frequencies (i.e. constant gain), and will act like an integrator at higher frequencies (gain proportional to 1/f).

 

[Tom.G]

Perhaps both of you could read:
the section

The AC or Continuous Op-amp Integrator​

At:
https://www.electronics-tutorials.ws/opamp/opamp_6.html

And:
http://www.visionics.a.se/html/curriculum/Experiments/Opamp integrator/Opamp Integrator1.html

 

[alan123hk]

The name AC Op-amp Integrator with DC Gain Control is indeed a bit cumbersome, but it can reduce misunderstandings when used.

 

[DaveE]

Perhaps both of you could read:
the section

The AC or Continuous Op-amp Integrator​

At:
https://www.electronics-tutorials.ws/opamp/opamp_6.html

And:
http://www.visionics.a.se/html/curriculum/Experiments/Opamp integrator/Opamp Integrator1.html

Thanks for the referral, but I already know what an integrator is. Better than the person that wrote that second link. That is a 1st order LPF. If you remove R2, it's an integrator. This is assuming we ignore the fact that op-amps aren't actually ideal. Fortunately, the first link has it correctly.

Here's a hint: Integrators have $H(s)=\frac{1}{(\frac{s}{\omega_o})}$; LPF have $H(s)=\frac{1}{(1+ \frac{s}{\omega_o})}$. Now I admit that at higher frequencies they behave the same. But, they are topologically different and that is worth recognizing. Fortunately the textbooks get this right; the web links, well that depends...