# Is this Non-Inverting integrator correct?

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: $$v(t)=-\frac{1}{RC}\int v_{in}dt + v(0)$$

• What do you propose this circuit looks like? – Matt Young Nov 16 '19 at 3:06