Operational Amplifier

A voltage amplifier (multiplies a voltage)
Can create 4 types of Linear Amplifier:

name	input	output
Voltage amplifier	$V$	$V$
Current amplifier	$I$	$I$
Transresistance amplifier	$I$	$V$
Transconductance amplifier	$V$	$I$

\usepackage{circuitikz}
\begin{document}
\begin{circuitikz}[american]
	\node[op amp, yscale=-1] at (4.69, 5.51){};
	\node[circ](N1) at (2.5, 6){} node[anchor=south] at (N1.north){$V+$};
	\node[circ](N2) at (2.5, 5){} node[anchor=north] at (N2.south){$V-$};
	\node[circ](N3) at (4.607, 7){} node[anchor=south] at (N3.north){$V_{DD}$};
	\node[circ](N4) at (4.607, 4){} node[anchor=north] at (N4.south){$V_{SS}$};
	\node[circ](N5) at (6.5, 5.5){} node[anchor=west] at (N5.east){$V_{out}$};
	\draw (3.5, 6) -- (2.5, 6);
	\draw (3.5, 5.02) |- (2.5, 5);
	\draw (4.607, 6.049) |- (4.6, 6.998);
	\draw (4.607, 4.971) -| (4.613, 3.991);
	\draw (5.88, 5.51) -| (6.5, 5.5);

\end{circuitikz}
\end{document}

Kinda like a Voltage controlled voltage source
Ideally, 0 current flows through the inputs, and the open loop voltage gain is infinite
Open loop voltage gain: $A$
% error of $A_{f}$ is $\frac{| A_{f} - A_{f, i d e a l} |}{| A_{f, i d e a l} |}$ , where $A_{f}$ is the gain of one of the following circuits
practically it cannot supply more than Vdd

Inverting Voltage Amplifier

Op amp is in a closed loop
- There is a path from the output back to one of the inputs
Closed loop negative feedback (it connects to the negative input)
the voltage gain is $\frac{V_{o}}{V_{i}}$ or $- \frac{R_{2}}{R_{1}}$
In the ideal case, $V + = V - = 0$
We call this a virtual short circuit across the input terminals
$V -$ is the virtual ground

\usepackage{circuitikz}
\begin{document}
\begin{circuitikz}[american]
\draw
(0, 0) node [op amp] (opamp) {}
(opamp.-) to[short, *-] ++(0, 1.5) coordinate (leftR)
to[R=$R_{2}$] (leftR -| opamp.out)
to[short, -*] (opamp.out)
(opamp.out) to[short, -*] ++(1, 0) node [circ, label=below:$V_{o}$] {}
(opamp.-) to[R, l_=$R_{1}$] ++(-2, 0)
to[V, l_=$V_{i}$] ++(0, -2) node [ground] {}
(opamp.+) node [ground] {}
;
\end{circuitikz}
\end{document}

Finding

A_{f}

begin with Kirchhoff's Laws#Current at $V -$ :
$(\frac{V_{i} - V_{-}}{R_{1}}) - (\frac{V_{-} - V_{o}}{R_{2}}) = 0$ (recall that no current enters - of the op amp)
Also recall that $V_{-} = 0 V$
so $\frac{V_{o}}{R_{2}} = - \frac{V_{i}}{R_{1}}$
$A_{f} = \frac{V_{o}}{V_{i}} = - \frac{R_{2}}{R_{1}}$

Adding a load resistor does not affect the gain
You cannot find output current directly, instead, find output voltage

Summing Amplifier

Take an inverting amplifier
If you put another voltage source + resistor into $V -$ , they add before being scaled

Difference Amplifier

Take an inverting amplifier
If you put another voltage source + resistor into $V +$ , they subtract before being scaled

\usepackage{circuitikz}
\begin{document}
\begin{circuitikz}[american]
\draw
(0, 0) node [op amp] (opamp) {}
(opamp.-) to[short, *-] ++(0, 1.5) coordinate (leftR)
to[R=$R_{f}$] (leftR -| opamp.out)
to[short, -*] (opamp.out)
(opamp.out) to[short, -*] ++(1, 0) node [circ, label=below:$V_{o}$] {}
(opamp.-) to[R, l_=$R$] ++(-3, 0)
to[V, l_=$V_{1}$] ++(0, -2) node [ground] {}
(opamp.+) to[R, l=$R_{f}$] ++(0, -2) node [ground] {}
(opamp.+) to [R, l=$R$] ++(-2, 0) to [V, l=$V_{2}$] ++(0, -2) node [ground] {}
;
\end{circuitikz}
\end{document}

By voltage division, $V + = (\frac{R_{f}}{R + R_{f}}) V_{2} = V -$
Plug that into KCL at $V -$ to get $\frac{V_{0}}{V_{2} - V_{1}} = \frac{R_{f}}{R}$

Non-Inverting Voltage Amplifier

\usepackage{circuitikz}
\begin{document}
\begin{circuitikz}[american]
\draw
(0, 0) node [op amp] (opamp) {}
(opamp.-) to[short, *-] ++(0, 1.5) coordinate (leftR)
to[R=$R_{2}$] (leftR -| opamp.out)
to[short, -*] (opamp.out)
(opamp.out) to[short, -*] ++(1, 0) node [circ, label=below:$V_{o}$] {}
(opamp.-) to[R, l_=$R_{1}$] ++(-2, 0) node [ground] {}
(opamp.+) to[V, l_=$V_{i}$] ++(0, -2) node [ground] {}
;
\end{circuitikz}
\end{document}

There is still a virtual short circuit, with $V -$ being the virtual ground
$V + = V - = V_{i}$
- $V -$ tracks $V +$
With KCL at $V -$ , you get $\frac{V_{o}}{V_{i}} = 1 + \frac{R_{2}}{R_{1}}$
practically, $\frac{V_{o}}{V_{i}} = \frac{A}{1 + A {(1 + \frac{R_{2}}{R_{1}})}^{- 1}} = (1 + \frac{R_{2}}{R_{1}}) (\frac{A}{1 + \frac{R_{2}}{R_{1}} + A})$
- as $A \to \infty$ , this converges to $1 + \frac{R_{2}}{R_{1}}$

Voltage Follower

\usepackage{circuitikz}
\begin{document}
\begin{circuitikz}[american]
\draw
(0, 0) node [op amp] (opamp) {}
(opamp.-) to[short, *-] ++(0, 1) coordinate (leftR)
to (leftR -| opamp.out)
to[short, -*] (opamp.out)
(opamp.out) to[short, -*] ++(1, 0) node [circ, label=below:$V_{o}$] {}
(opamp.+) to[V, l_=$V_{i}$] ++(0, -2) node [ground] {}
;
\end{circuitikz}
\end{document}

Unity gain amplifier
Voltage buffer
Doesn't change voltage but might give current
Ideally, infinite power gain
$V_{o} = A V_{i} - A V_{o} ⟹ A_{f} = \frac{V_{o}}{V_{i}} = \frac{A}{1 + A}$

Practical Op Amp

Has a resistor between $V +$ and $V -$ (usually pretty big)
Has a ground to VCVS to resistor (usually pretty small) to $V_{o}$ in series
We ignore these resistors, even when deconstructing a practical op amp into its base parts
Open loop gain is usually pretty big

Integrator Circuit

\usepackage{circuitikz}
\begin{document}
\begin{circuitikz}[american]
\draw
(0, 0) node [op amp] (opamp) {}
(opamp.-) to[short, *-] ++(0, 1.5) coordinate (leftR)
to[C=$C_{f}$] (leftR -| opamp.out)
to[short, -*] (opamp.out)
(opamp.out) to[short, -*] ++(1, 0) node [circ, label=below:$V_{o}$] {}
(opamp.-) to[short] ++(-2, 0)
to[I, l_=$I_{i}$, invert] ++(0, -2) node [ground] {}
(opamp.+) node [ground] {}
;
\end{circuitikz}
\end{document}

current to voltage converter
$V_{o} = - \frac{I_{s}}{C_{f}} t$