[FrontPage]

Basic Op Amp Circuits

Ideal Op Amp and Equivalent Circuit

An ideal op amp has very high input impedance (for derivations often assumed to be infinite) and low output impedance (assumed to be zero). The difference between the voltages at the non-inverting $V_+$ and inverting $V_-$ input terminals is amplified by the so-called open-loop gain $G$, thus the characteristic equation $V_o = G(V_+ - V_-)$.

 \begin{circuitikz} \draw  (0,0) node[op amp] (opamp) {}  (opamp.+) node[left] {$V_+$}  (opamp.-) node[left] {$V_-$}  (opamp.out) node[right] {$V_o$}; \end{circuitikz}

The two supply terminals are usually not shown and are assumed to be connected to the positive and negative rail of a split power supply (in synthesizers typically $\pm 12$V or $\pm 15$V). In applications with a single power supply like guitar stomp boxes (typically 9V) a so-called virtual ground configuration is used in which the negative supply terminal of the op amp is connected to ground, the positive terminal to $V_{cc}$ (9V) and instead of ground a bias voltage of $\frac{1}{2} V_{cc}$ (4.5V) is used.

Equivalent circuit:

 \begin{circuitikz}  \draw  (0,-1) to [short, o-o] (6,-1)  (0,0) to [short,o-,l_=$V_-$] (2,0)  (0,2) to [short,o-,l=$V_+$] (2,2)  (2,0) to [R, l=$R_{in}$] (2,2)  (3,0) to [cV, v_=$G(V_+ + V_-)$] (3,2)  (3,2) to [R, l=$R_{out}$] (5,2) to [short, -o,l=$V_{o}$] (6, 2)  (3,0) to [short, -o, l] (3,-1)  ;\end{circuitikz}

Inverting Amplifier

 \begin{circuitikz}  \draw  (5,2) node[op amp] (opamp) {}  (0,2.5) to [R, l=$R_{in}$] (opamp.-)  (0,2.5) to [short, o-] (0,2.5)  (0,0.5) to [short, o-o] (7,0.5)  (3.8,0.5) to [short, *-] (opamp.+)  (3.8,3.5) to [R, l=$R_{f}$] (6.2,3.5)  (3.8,3.5) to [short, -*] (opamp.-)  (6.2,3.5) to [short, -*] (opamp.out)  (opamp.out) to [short, -o] (7, 2);  \draw[->] (0.25,2.2) to node[right] {$V_{in}$} (0.25,0.7);  \draw[->] (6.75,1.8) to node[right] {$V_{out}$} (6.75,0.7);\end{circuitikz}

$V_{out} = -\frac{R_f}{R_{in}} V_{in}$

Non-Inverting Amplifier

 \begin{circuitikz}  \draw  (3,4) node[op amp, yscale=-1] (opamp) {}  (0,4.5) to [short, o-] (opamp.+)  (0,0) to [short, o-o] (6,0)  (4.2,4) to [R, l=$R_2$] (4.2,2)  (4.2,2) to [R, l=$R_1$] (4.2,0)  (4.2,0) to [short, *-] (4.2,0)  (opamp.-) to [short] (1.8,2)  (1.8,2) to [short, -*] (4.2,2)  (4.2,4) to [short, *-o] (6,4);  \draw[->] (0.25,4.3) to node[right] {$V_{in}$} (0.25,0.2);  \draw[->] (5.75,3.8) to node[right] {$V_{out}$} (5.75,0.2);\end{circuitikz}

$V_{out} = V_{in} \left(1+\frac{R_2}{R_1}\right)$

Integrator

 \begin{circuitikz}  \draw  (5,2) node[op amp] (opamp) {}  (0,2.5) to [R, l=$R$] (opamp.-)  (0,2.5) to [short, o-] (0,2.5)  (0,0.5) to [short, o-o] (7,0.5)  (3.8,0.5) to [short, *-] (opamp.+)  (3.8,3.5) to [C, l=$C$] (6.2,3.5)  (3.8,3.5) to [short, -*] (opamp.-)  (6.2,3.5) to [short, -*] (opamp.out)  (opamp.out) to [short, -o] (7, 2);  \draw[->] (0.25,2.2) to node[right] {$V_{in}$} (0.25,0.7);  \draw[->] (6.75,1.8) to node[right] {$V_{out}$} (6.75,0.7);\end{circuitikz}

$\displaystyle V_{out} = - \int_0^t \frac{V_{in}}{RC} \mathrm{d}t + V_0$ where $V_0$ is the output voltage at time $t=0$.

Differentiator

 \begin{circuitikz}  \draw  (5,2) node[op amp] (opamp) {}  (0,2.5) to [C, l=$C$] (opamp.-)  (0,2.5) to [short, o-] (0,2.5)  (0,0.5) to [short, o-o] (7,0.5)  (3.8,0.5) to [short, *-] (opamp.+)  (3.8,3.5) to [R, l=$R$] (6.2,3.5)  (3.8,3.5) to [short, -*] (opamp.-)  (6.2,3.5) to [short, -*] (opamp.out)  (opamp.out) to [short, -o] (7, 2);  \draw[->] (0.25,2.2) to node[right] {$V_{in}$} (0.25,0.7);  \draw[->] (6.75,1.8) to node[right] {$V_{out}$} (6.75,0.7);\end{circuitikz}

$V_{out} = -RC \frac{\mathrm{d}V_{in}}{\mathrm{d}t}$

Summing Amplifier

The summing amplifier circuit is often used in the input section of a circuit, commonly to sum control voltages or incoming audio signals from another module (either with or without adjustable gain). As can be seen in the circuit's equation the gain of the individual inputs is controlled by a resistor, thus it is amenable to easy manual control using a potentiometer.

 \begin{circuitikz} \draw  (5,2) node[op amp] (opamp) {}  (1,3.5) to [R,l=$R_{1}$] (3, 3.5)  (0,2.5) to [R,l_=$R_{2}$] (opamp.-)  (3,3.5) to [short,-*] (3,2.5)  (0,2.5) to [short, o-,l=$V_2$] (1,2.5)  (0,3.5) to [short, o-,l=$V_1$] (1,3.5)  (0,0.5) to [short, o-o] (7,0.5)  (3.8,0.5) to [short, *-] (opamp.+)  (3.8,3.5) to [R, l=$R_{f}$] (6.2,3.5)  (3.8,3.5) to [short, -*] (opamp.-)  (6.2,3.5) to [short, -*] (opamp.out)  (opamp.out) to [short, -o] (7, 2);  \draw[->] (6.75,1.8) to node[right] {$V_{out}$} (6.75,0.7);\end{circuitikz}

$V_{out} = -R_f\left(\frac{V_1}{R_1} + \frac{V_2}{R_2}\right)$

Differential Amplifier

 \begin{circuitikz}  \draw  (4,4) node[op amp] (opamp) {}  (0,3.5) to [short, o-,l=$V_2$] (1,3.5)  (0,1.5) to [short, o-o] (6,1.5)  (0,4.5) to [short, o-,l=$V_1$] (1,4.5)  (1,4.5) to [R, l=$R_1$] (2.5,4.5)  (2.5,4.5) to [short,-*] (2.8,4.5)  (2.8,3.5) to [R, l=$R_g$] (2.8,1.5)  (2.8,1.5) to [short,*-*] (2.8,1.5)  (2.8,3.5) to [short,*-*] (2.8,3.5)  (1,3.5) to [R,l_=$R_2$] (2.5,3.5)  (2.5,3.5) -- (opamp.+) (3.2,5.5) to [R, l=$R_f$] (5.2,5.5)  (5.2,5.5) to [short,-*] (opamp.out) (4.8,4) to [short,-o] (6,4)  (opamp.-) -- (2.8,5.5)  (2.8,5.5) -- (2.8,5.5) -- (3.2,5.5)  ;  \draw[->] (5.75,3.8) to node[right] {$V_{out}$} (5.75,1.7);\end{circuitikz}

$V_{out} = \frac{(R_f + R_1) R_g}{(R_g + R_2) R_1} V_2 - \frac{R_f}{R_1} V_1$

Voltage Follower

A voltage follower is useful for buffering purposes, e.g. at the output stage of a circuit.

 \begin{circuitikz}  \draw  (3,2) node[op amp,swap] (opamp) {}  (0,1.5) to [short, o-] (opamp.+)  (0,0.5) to [short, o-o] (5,0.5)  (opamp.-) -- (1.8,3.5) -- (4.2,3.5) -- (opamp.out)  (opamp.out) to [short, *-o] (5,2);  \draw[->] (0.25,1.2) to node[right] {$V_{in}$} (0.25,0.8);  \draw[->] (4.8,1.8) to node[right] {$V_{out}$} (4.8,0.8);\end{circuitikz}

$V_{out} = V_{in}$

Current-to-voltage Converter

The current-to-voltage converter (also called transimpedance amplifier) is useful in conjunction with an exponential current source to create an exponential output voltage from a linear input voltage, e.g. to drive a linear VCO with a 1V/Oct. control voltage.

 \begin{circuitikz}  \draw  (5,2) node[op amp] (opamp) {}  (2,2.5) to [short, o-, i^>=$I_{in}$] (opamp.-)  (3.8,0.5) to [short, -o] (7,0.5)  (3.8,0.5) to [short] (opamp.+)  (3.8,3.5) to [R, l=$R$] (6.2,3.5)  (3.8,3.5) to [short, -*] (opamp.-)  (6.2,3.5) to [short, -*] (opamp.out)  (opamp.out) to [short, -o] (7, 2);  \draw[->] (6.75,1.8) to node[right] {$V_{out}$} (6.75,0.7);\end{circuitikz}

 $V_{out} = -R \cdot I_{in}$


2013-12-22 15:41