Op-Amps

Operational Amplifiers are fundamental to modern electronics

$$V_{out}=A(V_2-V_1)$$

Properties of an ideal Op-Amp:

• $A\to \infty$
• $R_{in}\to \infty$
  • $I_{in}\to 0$
• $R_{out}\to 0$
• Slew rate $\to \infty$
  • Output can change as fast as we want
• Common Mode Rejection Ratio (CMRR) $\to \infty$
  • Signals where $V_2=V_1$ are rejected and not amplified
• Power supply rejection ratio $\to \infty$
• Bandwith $\to \infty$

A large gain $A$ drives the differential input to zero $(V_2-V_1)\to 0$, as the op-amp always tries to keep the two inputs the same.

Buffers

A buffer provides unity gain while acting as a signal buffer.

$$\begin{gathered} V_{out}=V_{in} \\ {A}_v=1 \end{gathered}$$

As $R_{in}$ is high and $R_{out}$ is low, no current flows in and there is no impedance to current flowing out, meaning the buffer acts to isolate stages of a circuit.

Active Filters

(they aren't really active, according to Ryan.)

$$\frac{V_{out}}{V_{in}}=A(j\omega )=-\frac{Z_2}{Z_1}$$

$Z_1$ and $Z_2$ are generalised impedances and can take any value. If $Z_2=R_2||C$, for example, then:

$$A(j\omega )=-\frac{R_2/R_1}{1+j\omega C{R}_2}$$

A limits test shows that this would make a low pass filter:

• As $f\to 0$, $A(j\omega )\to -R_2/R_1$
• As $f\to \infty$, $A(j\omega )\to 0$
• The mid-band is where $A(j\omega )=R_2/R_1$

Cutoff frequency is where $A(j\omega )=1+j$, which is $\frac{1}{2\pi R_{2}C}$Hz

The other way round, where $Z_1=C+R_1$ and $Z_2=R_2$ is a high pass filter:

$$A(j\omega )=-\frac{j\omega C{R}_1}{1+j\omega C{R}_1}$$

This gives a cutoff frequency of f_c = \frac{1}{2 \pi R_1 C, where the max gain as $f\to \infty$ is $R_2/R_1$