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:

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

(4)

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

Rewriting the transfer function as:

$H\left(s\right)=H\left(j\omega \right)=-{\frac {\omega _{0}}{j\omega }}={\frac {1}{-j{\frac {\omega }{\omega _{0}}}}}={\frac {1}{R\left(\omega \right)+jX\left(\omega \right)}}$

(5)

We can then define the quality factor of an ideal integrator:

$Q={\frac {X\left(\omega \right)}{R\left(\omega \right)}}\rightarrow \infty$

(6)

Since $R\left(\omega \right)=0$. Fig. 5 shows a multiple-input integrator, with output voltage:

$v_{o}\left(s\right)=-{\frac {1}{s\cdot R_{1}C}}\cdot v_{1}\left(s\right)-{\frac {1}{s\cdot R_{2}C}}\cdot v_{2}\left(s\right)$

(7)

Integrator Noise

Fig. 6 shows an integrator where the output is fed back to one of its inputs, giving us:

$v_{o}\left(s\right)=-{\frac {1}{s\cdot R_{1}C}}\cdot v_{i}\left(s\right)-{\frac {1}{s\cdot R_{2}C}}\cdot v_{o}\left(s\right)$

(8)

Ignoring the noise from the amplifier, the output noise of the integrator in Fig. 6 can be expressed as:

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

(9)

The total integrated noise is then:

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

(10)