How to Design an Op Amp Integrator with Specific Parameters?

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Discussion Overview

The discussion revolves around designing an op-amp integrator with specific parameters, focusing on achieving a DC gain of -10 V/V and a 3dB frequency of 50 kHz. Participants explore the implications of the gain-bandwidth product of the μA741 op amp and the effects of component values on the integrator's performance.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant proposes using the gain-bandwidth product of the μA741 op amp, which is 1 MHz, to determine the appropriate capacitance for the integrator.
  • Another participant suggests modeling the op amp as a gain block and emphasizes that the resulting transfer function will be second-order, complicating the analysis compared to a first-order system.
  • It is noted that at low frequencies, the capacitive reactance (Xc) is large, resulting in a DC gain of -R2/R1, while at high frequencies, Xc becomes small, affecting the gain.
  • A participant highlights the significance of the point where Xc equals R2, indicating that the gain will be reduced to 1/√2 of the DC gain at that frequency.
  • There is a correction regarding the gain reduction, clarifying that it is 1/√2, not 1/2, when Xc equals R2.

Areas of Agreement / Disagreement

Participants express differing views on the complexity of the transfer function and the implications of the gain-bandwidth product. There is no consensus on the best approach to simplify the analysis or on the exact impact of component values on the integrator's performance.

Contextual Notes

Participants acknowledge the need to consider the gain-bandwidth product in their calculations, but the discussion does not resolve how to best incorporate this into the design process. The implications of approximating the second-order transfer function as a first-order one remain unclear.

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Homework Statement



Design an op-amp-based integrator with the dc gain equal to -10 V/V (use R1 = 200 Ω and R2 = 2 kΩ). Knowing the gain-bandwidth product of the μA741 op amp, find the proper value of a capacitance C of the integrator so that its 3dB frequency is 50 kHz.

Homework Equations



bg7QK.png
(1)

The Attempt at a Solution



I have found out that the gain-bandwidth product of the amp is 1 MHz
And this is my guess how the amplifier looks like.
zBK1s.png



I think I could use formula (1) but i would think I had to replace R2 with this formula :
dcgd6.png
(2)

because the Op amp has a capacitor.


Am I on the right track for this?
 
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The canonical approach to this is to model the op amp as a gain block ω0/s

where ω0 is your 1 MHz*2π. (So that the gain at 1 MHz = 1).

Then write the amplitude response of the combined network using your R2 and C chosen such that the new 3dB-down frequency is 50 KHz.

I might as well warn you that the result is a second-order transfer function, not just a nice
K/(Ts+1) first-order one. Of course, it's still a lowpass filter with a definite 50-KHz 3db-down cutoff frequency. 2nd-orders are just bitchier than 1st to plot & calculate, but it's straight-forward.

Your text/instructor might have given you a short-cut approximation to this problem, yielding another 1st-order transfer function. I don't know.

You could try approximating the 2nd order xfr fn by a 1st order one by factoring the 2nd-order polynomial, starting tentatively with C assuming an infinite-gain op amp, then throwing out the less significant pole.
 
Last edited:
It is also known as a low pass filter because at 'low' frquencies Xc is large and the DC(and AC) gain is given by -R2/R1.
At 'high' frequencies Xc is small and 'shorts' R2. The gain falls with increasing frequency.
An important point (from a practical point of view) is when Xc = R2. Can you see when this occurs the DC gain will have halved?
 
technician said:
It is also known as a low pass filter because at 'low' frquencies Xc is large and the DC(and AC) gain is given by -R2/R1.
At 'high' frequencies Xc is small and 'shorts' R2. The gain falls with increasing frequency.
An important point (from a practical point of view) is when Xc = R2. Can you see when this occurs the DC gain will have halved?

1. The OP is asked to include the effects of the gain-bandwidth product of the 741.
2. Gain when XC = R2 is reduced to 1/√2 of the DC gain, not 1/2.
 
Correction noted 1/√2 not 1/2
 

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