Integrators

The Ideal Integrator

The ideal integrator, shown in Fig. 1, with symbol shown in Fig. 2, makes use of an ideal operational amplifier with $A_{v}\rightarrow \infty$, $R_{i}\rightarrow \infty$, and $R_{o}=0$. The current through the resistor, $i_{R}$, can be expressed as:

$i_{R}={\frac {v_{i}}{R}}=i_{C}=-C\cdot {\frac {\partial v_{o}}{\partial t}}$

(1)

Thus, we can write the integrator output voltage, $v_{o}$, as:

$v_{o}=-{\frac {1}{RC}}\int v_{i}\cdot dt$

(2)

In the Laplace domain:

${\frac {v_{i}\left(s\right)}{R}}=-sC\cdot v_{o}\left(s\right)$

(3)

Or equivalently:

${\frac {v_{o}\left(s\right)}{v_{i}\left(s\right)}}=-{\frac {1}{s\cdot RC}}=-{\frac {1}{s\cdot \tau }}=-{\frac {\omega _{0}}{s}}$

(4)

The magnitude and phase response of an ideal integrator is shown in Figs. 3 and 4.

Fig. 5 shows a multiple-input integrator, and Fig. 6 shows an integrator where the output is fed back to one of its inputs.

Integrator Noise

Ignoring the noise from the amplifier, the output noise can be expressed as:

${\begin{aligned}{\overline {v_{o}^{2}}}&=\left|H_{1}\left(s\right)\right|^{2}\cdot {\overline {v_{n1}^{2}}}+\left|H_{2}\left(s\right)\right|^{2}\cdot {\overline {v_{n2}^{2}}}\\&=\left|{\frac {1}{1+sRC}}\right|^{2}\cdot {\overline {v_{n1}^{2}}}+\left|{\frac {1}{1+sRC}}\right|^{2}\cdot {\overline {v_{n2}^{2}}}\\&={\frac {1}{1+\omega ^{2}R^{2}C^{2}}}\cdot {\overline {v_{n1}^{2}}}+{\frac {1}{1+\omega ^{2}R^{2}C^{2}}}\cdot {\overline {v_{n2}^{2}}}\\&={\frac {1}{1+\omega ^{2}R^{2}C^{2}}}\cdot 4kTR\cdot \Delta f+{\frac {1}{1+\omega ^{2}R^{2}C^{2}}}\cdot 4kTR\cdot \Delta f\\\end{aligned}}$

(5)

The total integrated noise is then:

${\begin{aligned}{\overline {v_{o,{\text{T}}}^{2}}}&=\int _{0}^{\infty }\left|H_{1}\left(s\right)\right|^{2}\cdot {\overline {v_{n1}^{2}}}df+\int _{0}^{\infty }\left|H_{2}\left(s\right)\right|^{2}\cdot {\overline {v_{n2}^{2}}}df\\&=\int _{0}^{\infty }{\frac {1}{1+\omega ^{2}R^{2}C^{2}}}\cdot 4kTR\cdot df+\int _{0}^{\infty }{\frac {1}{1+\omega ^{2}R^{2}C^{2}}}\cdot 4kTR\cdot df\\&={\frac {kT}{C}}+{\frac {kT}{C}}=2\cdot {\frac {kT}{C}}\\\end{aligned}}$

(6)