Designing an op amp integrator

[Huumah]

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

(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.

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

(2)

because the Op amp has a capacitor.


Am I on the right track for this?

 

[rude man]

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.

 

[technician]

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?

 

[rude man]

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 X_C = R₂ is reduced to 1/√2 of the DC gain, not 1/2.

 

[technician]

Correction noted 1/√2 not 1/2