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 of the integrator in Fig. 6, for $R_{1}=R_{2}=R$, can be expressed as:

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

(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)