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I've seen time again the formula for an inverting integrator. I was wondering if I could swap the configurations on the inverting and non inverting inputs of the op amp to make the circuit a non-inverting integrator. I did the math, but when I searched online to make sure my math was correct, I found nothing supporting my derivation. So I'm wondering if my math is incorrect or if I have a conceptual gap with the functionality of an op amp. Here's my work:

I called the non-inverting input to the op amp \$v_{+}\$ the inverting input \$v_{-}\$, the amplification \$A\$, the output voltage \$v_{out}\$, and the input voltage \$v_{in}\$. $$ v_{out} = A\times(v_+ - v_-) $$ Because I grounded \$v_-\$, $$v_{out} = Av_+$$ $$v_+ = \frac{v_{out}}{A}$$ The current going through resistor \$I_R\$ is the same current going through the capacitor, \$I_C\$. Where \$I_R=\frac{v_{in}-v_+}{R}\$, and \$I_C = \frac{dv}{dt}C\$. Therefore, by equating these equations we can solve: $$I_R = I_C$$ $$\frac{v_{in}-v_+}{R}=\frac{dv}{dt}C$$ $$\frac{1}{RC}(v_{in}-v_+)=\frac{dv}{dt}$$ $$\frac{1}{RC}(v_{in}-\frac{v_{out}}{A})=\frac{dv}{dt}$$ $$\lim_{A\rightarrow\infty}\frac{1}{RC}(v_{in}-\frac{v_{out}}{A}) = \frac{1}{RC}v_{in}=\frac{dv}{dt}$$ $$\frac{1}{RC}v_{in}dt=dv$$ $$\frac{1}{RC}\int v_{in}dt=v(t)-v(0)$$ $$\therefore v(t)=\frac{1}{RC}\int v_{in}dt + v(0)$$ The inverting integrator follows the same formula, except it has a negative tacked on infront of the integral: $$v(t)=-\frac{1}{RC}\int v_{in}dt + v(0)$$

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  • \$\begingroup\$ What do you propose this circuit looks like? \$\endgroup\$ – Matt Young Nov 16 '19 at 3:06
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I think your math is correct. However, real op amps will not work with this configuration, because real op amps have finite frequency response, or a time delay, or they shift the phase of a sine wave: all of which are about the same thing.

So, when you try this with a real op amp, the circuit will be unstable.

You can see this mathematically by adding a pole to the gain of the op amp... perhaps writing it is A/(s+a), or A/s. (s being the Laplace variable, "a" a small number), then doing the math in the Laplace domain.

This just shows that using an ideal model can lead to the wrong conclusions sometimes.

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  • \$\begingroup\$ No - neither a finite frequency response nor time delay or unwanted phase shift are a problem (all these effects are effective for working circuits as well). The most severe fact is the missing NEGATIVE feedback (because you have grounded the inverting terminal). Please note that each working opamp circuit needs negative DC feedback for a stable operating point. This even applies to the classical (inverting) MILLER integrator . \$\endgroup\$ – LvW Nov 16 '19 at 9:26
  • \$\begingroup\$ @User69795..The error in your calculation appears in the third line: While writing V+=Vout/A you are assuming linear amplifier operation for pure positive feedback ...and this is not possible, in reality. \$\endgroup\$ – LvW Nov 16 '19 at 9:31
  • \$\begingroup\$ @LvW: Yes, modeling the op amp as an infinite gain with no dynamics leads allows one to create a circuit, (using positive feedback), with a well defined, finite gain. The math works out great. Mr. Expresso shows that with this model you can also make an ideal, noninverting integrator. The model is not good enough to reflect real op amps, which will not work in these ciircuits. But, there is nothing wrong with the math. \$\endgroup\$ – user69795 Nov 17 '19 at 4:45
  • \$\begingroup\$ @user69795....at first, I did not say that the math would be wrong!. But the hardware will not follow the (idealized) calculation. Consider the following mechanical example:Theoretically, a stable system of two superimposed spheres is conceivable - and the mathematical treatment shows that then both centers of gravity must lie exactly on top of each other and no disturbance affects the system. This is the mechanical analogon to positive feedback. However - everyone knows that the calculation is correct, but the prerequisites for the calculation cannot be fulfilled. \$\endgroup\$ – LvW Nov 17 '19 at 10:26

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