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% Operational Amplifiers % Diego Trapero

The Operational Amplifier

$$ \begin{circuitikz}[scale = 1] \draw (0,0) nodeop amp {} (opamp.+) node[left] {$v_+$} (opamp.-) node[left] {$v_-$} (opamp.out) node[right] {$v_0$} (opamp.down) -- (-0.1,-1) node[ocirc] {$V_{S+}$} (opamp.up) -- (-0.1,1) node[ocirc] {$V_{S-}$} ; \end{circuitikz} $$

Op-amp terminals The terminals of an op-amp are

  • Positive terminal, $+$, non-inverting terminal.
  • Negative terminal, $-$, inverting terminal
  • Positive power supply, $V_{S+}$
  • Negative power supply, $V_{S-}$
  • Output terminal, $v_o$

Op-amp equivalent circuit Some of the non-ideal resistive characteristics of an op-amp can be modelled by the following circuit

$$ % Operational amplifier model, by Diego Trapero \ctikzset{bipoles/length=0.8cm} \begin{circuitikz}[scale = 1] \draw (0,0) node[left] {$v_-$} (0,0) toshort, o- (2,2) toR, l=$R_i$, v=$v_d$ (0,2) node[left] {$v_+$} (0,2) toshort, o- (4,0) node[ground] {} (4,0) tocontrolled voltage source, l_=$G v_d$ (4,2) toR, l=$R_o$ ,-o (6,2) node[right] {$v_o$} ; \end{circuitikz} $$

Ideal Operational Amplifier

$$ \begin{circuitikz}[scale =1] \draw (0,0) nodeop amp {} (opamp.+) node[left] {$v_+$} (opamp.-) node[left] {$v_-$} (opamp.out) node[right] {$v_0$} ; \end{circuitikz} $$

The ideal operational amplifier is a model of the real amplifier, where the equivalent circuit has

  • Infinite open-loop gain, $G \rightarrow \infty$.
  • Zero output impedance, $R_{out} = 0$.
  • Infinite input impedance, $R_{in} \rightarrow \infty$.
  • None of the other non-ideal behaviours: CMRR, noise, offset, bandwith...

Ideal op-amp transfer function, op-amp formula. With the simplifications of the ideal op-amp mode, the output voltage is

$$v_o = G (v_+ - v_-)$$

where $G$ is a very large number, and can be considered infinite. Nevertheless, we should take into account saturation, so $v_o$ must be in the following interval:

$$v_o \in [v_{S-}, v_{S+}]$$

Ideal op-amp golden rules

  1. In a negative feedback configuration, the output attempts to do whatever is necessary to make the voltage difference between the inputs zero.
  2. The inputs draw no current.

[The Art of Electronics, page 177]

Op-amp configurations

Comparator

$$ \begin{circuitikz}[scale = 1.2] \draw (0,0) nodeop amp {} (opamp.+) node[left] {$v_+$} (opamp.-) node[left] {$v_-$} (opamp.out) node[right] {$v_0$} (opamp.down) -- (-0.05,-1) node[ocirc] {$V_{S+}$} (opamp.up) -- (-0.05,1) node[ocirc] {$V_{S-}$} ; \end{circuitikz} $$

$$ v_o = \begin{cases} v_{S+} & \text{if } v_+ > v_- \\ v_{S-} & \text{if } v_- > v_+ \end{cases} $$

Voltage follower

$$ % Voltage follower, by Diego Trapero \ctikzset{bipoles/length=0.8cm} \begin{circuitikz}[scale =1] \draw (2.5,2.5) nodeop amp {} (0,2) toshort, o- (opamp.-) toshort (1.3,4) toshort (opamp.out) toshort (opamp.out) toshort, -o ; \end{circuitikz} $$

$$v_o = v_i$$

Circuit analysis

a) With the op-amp formula

$$ v_o = G (v_i - v_o) $$
$$ v_o = Gv_i - Gv_o $$
$$ v_o + Gv_o = Gv_i $$
$$ \frac{v_o}{G} + v_o = v_i \xrightarrow{G \rightarrow \infty} v_o = v_i$$

b) With the golden rules: Since it is a feedback configuration, we can apply the first rule. We obtain that $v_+ = v_-$, or $v_i = v_o$, which is the transfer function of the voltage follower.

Inverting amplifier

$$ % Inverting amplifier, by Diego Trapero \ctikzset{bipoles/length=0.8cm} \begin{circuitikz}[scale =1] \draw (3.5,1.5) nodeop amp {} (0,2) toR, l=$R_i$, o- (opamp.-) toshort (2.3,3) toR, l=$R_f$ (opamp.out) toshort (opamp.out) toshort, -o (opamp.+) toshort (2.3,0) node[ground]{} ; \end{circuitikz} $$

Circuit analysis

a) With the golden rules

a.1) From the golden rules we know

	* $i_{opamp} = 0$

	* $v_- = v_+ = 0$

a.2) KCL in the negative terminal
$$ i_i = i_o $$
$$ \frac{v_i - v_-}{R_i} = \frac{v_- - v_o}{R_f}$$

a.3) $v_- = 0$
$$ \frac{v_i}{R_i} = - \frac{v_o}{R_f}$$

a.4) Obtain $v_o$
$$ v_o = - \frac{R_f}{R_i} v_i $$

Noninverting amplifier

$$ % Noninverting amplifier, by Diego Trapero \ctikzset{bipoles/length=0.8cm} \begin{circuitikz}[scale =1] \draw (3.5,2.5) nodeop amp {} (0,3) toR, l=$R_i$ (0,3) node[ground] {} (2.3,2) node[ocirc] {} (2.3,2) node[left] {$v_i$} (opamp.-) toshort (2.3,4) toR, l=$R_f$ (opamp.out) toshort (opamp.out) toshort, -o (5,2.5) node[right] {$v_o$} ; \end{circuitikz} $$

Circuit analysis

a) With the op-amp formula

a.1) Op-amp formula
$$ v_o = G (v_i - v_-) $$

a.2) Calculate $v_-$ with the voltage divider formula:
$$ v_- = \frac{R_i}{R_i+R_f} v_o $$

a.3) Substitute $v_-$ in the op-amp formula
$$ v_o = G (v_i - \frac{R_i}{R_i+R_f} v_o) $$

a.4) Take all $v_o$ to the left
$$ v_o + G \frac{R_i}{R_i+R_f} v_o = G v_i $$

a.5) Divide all by $G$ and take the limit when $G$ tends to infinity
$$ \frac{v_o}{G} + \frac{R_i}{R_i+R_f} v_o = v_i \xrightarrow{G \rightarrow \infty} \frac{R_i}{R_i + R_f} v_o = v_i $$

a.6) The transfer function is
$$ v_o = \frac{R_i + R_f}{R_i} v_i $$
$$ v_o = ( 1 + \frac{R_f}{R_i} ) v_i $$

Noninverting amplifier circuit analysis with the golden rules

Differential amplifier

$$ % Differential amplifier, by Diego Trapero \ctikzset{bipoles/length=0.8cm} \begin{circuitikz}[scale =1] \draw (3.5,2.5) nodeop amp {} (0,2) toR, l=$R_3$, o- (0,3) toR, l=$R_1$, o- (opamp.-) toshort (2.3,4) toR, l=$R_2$ (opamp.out) toshort (opamp.out) toshort, -o (opamp.+) toR, l=$R_4$ (2.3,0) node[ground]{} ; \end{circuitikz} $$

$$v_o = \frac{R_4}{R_3 + R_4}(1 + \frac{R_2}{R_1}) v_2 - \frac{R_2}{R_1} v_1 $$

Integrator

$$ % Integrator, by Diego Trapero \ctikzset{bipoles/length=0.8cm} \begin{circuitikz}[scale =1] \draw (3.5,1.5) nodeop amp {} (0,2) toR, l=$R_i$, o- (opamp.-) toshort (2.3,3) toC, l=$C$ (opamp.out) toshort (opamp.out) toshort, -o (opamp.+) toshort (2.3,0) node[ground]{} ; \end{circuitikz} $$

Differentiator

$$ % Differentiator, by Diego Trapero \ctikzset{bipoles/length=0.8cm} \begin{circuitikz}[scale =1] \draw (3.5,1.5) nodeop amp {} (0,2) toC, l=$C$, o- (opamp.-) toshort (2.3,3) toR, l=$R_f$ (opamp.out) toshort (opamp.out) toshort, -o (opamp.+) toshort (2.3,0) node[ground]{} ; \end{circuitikz} $$

Inverting amplifier with offset adjustment

$$ % Inverting amplifier with offset adjustment, by Diego Trapero \ctikzset{bipoles/length=0.8cm} \begin{circuitikz}[scale =1] \draw (3.5,1.5) nodeop amp {} (0,2) toR, l=$R_i$, o- (0,2) node[left] {$v_1$} (opamp.-) toshort (2.3,3) toR, l=$R_f$ (opamp.out) toshort (opamp.out) toshort, -o (5,1.5) node[right] {$v_o$} (opamp.+) node[ocirc] {} (opamp.+) node[left] {$v_2$} ; \end{circuitikz} $$

$$v_o = \left( 1 + \frac{R_2}{R_1} \right) v_2 - \frac{R_2}{R_1} v_1 $$

Circuit analysis

a) Using the differential amplifier formula

a.1) The differential amplifier formula is

$$v_o = \frac{R_4}{R_3 + R_4} \left( 1 + \frac{R_2}{R_1} \right) v_2  - \frac{R_2}{R_1} v_1 $$

a.2) This op-amp circuit is a special case where $R_4 \rightarrow \infty$

$$ \text{if } R_4 \rightarrow \infty : v_o = \cancelto{1}{ \frac{R_4}{R_3 + R_4} } (1 + \frac{R_2}{R_1}) v_2  - \frac{R_2}{R_1} v_1 $$

$$v_o = \left( 1 + \frac{R_2}{R_1} \right) v_2 - \frac{R_2}{R_1} v_1 $$

b) Using superposition

Summing amplifier

$$ % Summing amplifier, by Diego Trapero \ctikzset{bipoles/length=0.8cm} \begin{circuitikz}[scale =1] \draw (3.5,1.5) nodeop amp {} (-0.5,5) toR, l=$R_1$, o- (-0.5,5) node[left] {$v_1$} (-0.5,4) toR, l=$R_2$, o- (-0.5,4) node[left] {$v_2$} (-0.5,2) toR, l=$R_n$, o- (-0.5,2) node[left] {$v_n$} (1.8,5) toshort (1.8,2)toshort (opamp.-) toshort (2.3,3) toR, l=$R_f$ (opamp.out) toshort (opamp.out) toshort, -o (5,1.5) node[right] {$v_o$} (opamp.+) toshort (2.3,0) node[ground]{} ; \end{circuitikz} $$

$$v_o = - \frac{R_f}{R_1} v_1 - \frac{R_f}{R_2} v_2 - ... - \frac{R_f}{R_n} v_n$$

$$v_o = - R_f \sum{ \frac{v_i}{R_i} } $$

\newpage

Op Amp table

\tabulinesep=2mm

\begin{longtabu} to\linewidth{|X[1,m,c]|X[2,m,c]|X[2,m,c]|}

\hline
Op Amp & Circuit & Formula \\
\hline

Comparator
&
\begin{circuitikz}[scale = 0.6, transform shape]
	\draw (0,0) node[op amp](opamp) {}
		(opamp.+) node[left] {$v_+$}
		(opamp.-) node[left] {$v_-$}
		(opamp.out) node[right] {$v_o$}
	;
\end{circuitikz}
&
$$
v_o =
  \begin{cases}
   v_{S+} & \text{if } v_+ > v_- \\
   v_{S-} & \text{if } v_- > v_+
  \end{cases}
$$
\\
\hline

Voltage follower
&
\begin{circuitikz}[scale = 0.6, transform shape]
	\draw
	(2.5,2.5) node[op amp](opamp) {}
	(0,2) to[short, o-](opamp.+)
	(opamp.-) to[short](1.3,4)
	(1.3,4) to[short](3.7,4)
	(opamp.out) to[short](3.7,4)
	(opamp.out) to[short, -o](4,2.5)
	;
\end{circuitikz}
&
$$v_o = v_i$$
\\
\hline

Inverting amplifier
&
\begin{circuitikz}[scale = 0.6, transform shape]
	\draw
	(3.5,1.5) node[op amp](opamp) {}
	(0,2) to[R, l=$R_i$, o-](opamp.-)
	(opamp.-) to[short](2.3,3)
	(2.3,3) to[R, l=$R_f$](4.7,3)
	(opamp.out) to[short](4.7,3)
	(opamp.out) to[short, -o](5,1.5)
	(opamp.+) to[short](2.3,0)
	(2.3,0) node[ground]{}
	;
\end{circuitikz}
&
$$ v_o = - \frac{R_f}{R_i} v_i $$
\\
\hline

Non-inverting amplifier
&
\begin{circuitikz}[scale = 0.6, transform shape]
	\draw
	(3.5,2.5) node[op amp](opamp) {}
	(0,3) to[R, l=$R_i$](opamp.-)
	(0,3) node[ground] {}
	(2.3,2) node[ocirc] {}
	(2.3,2) node[left] {$v_i$}
	(opamp.-) to[short](2.3,4)
	(2.3,4) to[R, l=$R_f$](4.7,4)
	(opamp.out) to[short](4.7,4)
	(opamp.out) to[short, -o](5,2.5)
	(5,2.5) node[right] {$v_o$}
	;
\end{circuitikz}
&
$$ v_o = ( 1 + \frac{R_f}{R_i} ) v_i $$
\\
\hline

Differential amplifier
&
\begin{circuitikz}[scale = 0.6, transform shape]
	\draw
	(3.5,2.5) node[op amp](opamp) {}
	(0,2) to[R, l=$R_3$, o-](opamp.+)
	(0,3) to[R, l=$R_1$, o-](opamp.-)
	(opamp.-) to[short](2.3,4)
	(2.3,4) to[R, l=$R_2$](4.7,4)
	(opamp.out) to[short](4.7,4)
	(opamp.out) to[short, -o](5,2.5)
	(opamp.+) to[R, l=$R_4$](2.3,0)
	(2.3,0) node[ground]{}
	;
\end{circuitikz}
&
$$v_o = \frac{R_4}{R_3 + R_4}(1 + \frac{R_2}{R_1}) v_2  - \frac{R_2}{R_1} v_1 $$
\\
\hline

Integrator
&
\begin{circuitikz}[scale = 0.6, transform shape]
	\draw
	(3.5,1.5) node[op amp](opamp) {}
	(0,2) to[R, l=$R_i$, o-](opamp.-)
	(opamp.-) to[short](2.3,3)
	(2.3,3) to[C, l=$C$](4.7,3)
	(opamp.out) to[short](4.7,3)
	(opamp.out) to[short, -o](5,1.5)
	(opamp.+) to[short](2.3,0)
	(2.3,0) node[ground]{}
	;
\end{circuitikz}
&
$$?$$
\\
\hline

Differentiator
&
\begin{circuitikz}[scale = 0.6, transform shape]
	\draw
	(3.5,1.5) node[op amp](opamp) {}
	(0,2) to[C, l=$C$, o-](opamp.-)
	(opamp.-) to[short](2.3,3)
	(2.3,3) to[R, l=$R_f$](4.7,3)
	(opamp.out) to[short](4.7,3)
	(opamp.out) to[short, -o](5,1.5)
	(opamp.+) to[short](2.3,0)
	(2.3,0) node[ground]{}
	;
\end{circuitikz}
&
$$?$$
\\
\hline

Inverting amplifier with offset adjustment
&
\begin{circuitikz}[scale = 0.6, transform shape]
	\draw
	(3.5,1.5) node[op amp](opamp) {}
	(0,2) to[R, l=$R_i$, o-](opamp.-)
	(0,2) node[left] {$v_1$}
	(opamp.-) to[short](2.3,3)
	(2.3,3) to[R, l=$R_f$](4.7,3)
	(opamp.out) to[short](4.7,3)
	(opamp.out) to[short, -o](5,1.5)
	(5,1.5) node[right] {$v_o$}
	(opamp.+) node[ocirc] {}
	(opamp.+) node[left] {$v_2$}
	;
\end{circuitikz}
&
$$v_o = \left( 1 + \frac{R_2}{R_1} \right) v_2 - \frac{R_2}{R_1} v_1 $$
\\
\hline

Summing amplifier
&
\begin{circuitikz}[scale = 0.6, transform shape]
	\draw
	(3.5,1.5) node[op amp](opamp) {}
	(-0.5,5) to[R, l=$R_1$, o-](1.8,5)
	(-0.5,5) node[left] {$v_1$}
	(-0.5,4) to[R, l=$R_2$, o-](1.8,4)
	(-0.5,4) node[left] {$v_2$}
	(-0.5,2) to[R, l=$R_n$, o-](1.8,2)
	(-0.5,2) node[left] {$v_n$}
	(1.8,5) to[short](1.8,2)
	(1.8,2)to[short](opamp.-)
	(opamp.-) to[short](2.3,3)
	(2.3,3) to[R, l=$R_f$](4.7,3)
	(opamp.out) to[short](4.7,3)
	(opamp.out) to[short, -o](5,1.5)
	(5,1.5) node[right] {$v_o$}
	(opamp.+) to[short](2.3,0)
	(2.3,0) node[ground]{}
	;
\end{circuitikz}
&
$$v_o = - R_f \sum{ \frac{v_i}{R_i} } $$
\\

\end{longtabu}