notes/Areas/electricity/active-components/op-amp.md

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The operational amplifier has a very high input impedance which makes it very good for amplifying low voltage signals.

Basically the OpAmp is a function like this:

\displaystyle Y = A_v (X_1 - X_2)

Where:

\begin{flalign}
&Y = \text{Output Voltage}&\\\
&A_v = \text{Open Loop Gain}\\
&X_1 = \text{Input V1 (Non Inverting Input)}\\
&X_2 = \text{Input V2 (Inverting Input)}\\
\end{flalign}

Regions

Op Amps functions in different regions, just like diodes, and transistors.

!op-amp-regions.png

Linear Region This is how the Op-Amp normally functions.

Saturation Region When the output of the op-amp would be higher than +V_{CC} or lower than -V_{CC} the output value is clamped to those values.

In real life OpAmps have A_V values as high as 10^8 or 10^9 due to this even very small input voltages would quickly leave the linear region. That is why we need

Negative Feedback To use negative feedback we connect the output of the OpAmp to one of its inputs. This connection is modified by a feedback factor (\beta) which can be in the range 0 \le \beta \le 1.

Due to this feedback the new formula for the output V_O is now:

\begin{flalign}
&V_o = A_V * V_\Delta&\\\
\\
&V_- = \beta * V_o\\
&\text{Now we can say that }V_\Delta \text{is equal to:}\\
&V_\Delta = V_+ - \beta *V_o &| \textit{ Solve for }V_o \\
&V_o = \frac{V_+ - V_\Delta}{\beta}
\end{flalign}

Non-Inverting Amplifier

$ 64 0.000005 1.0312258501325766 50 5 50 5e-11
a 192 240 304 240 9 15 -15 1000000 4.9999000019999595 5 100000
r 192 320 192 400 0 1000
r 304 320 192 320 0 1000
w 192 320 192 256 0
w 304 240 304 320 0
O 304 240 368 240 1 0
g 192 400 192 432 0 0
v 96 352 96 224 0 0 40 5 0 0 0.5
w 96 224 192 224 2
g 96 352 96 432 0 0
b 144 288 289 401 0
x 264 386 278 389 4 24 β
x 240 345 252 348 4 12 Rf
x 160 363 176 366 4 12 Rg

What is the closed loop gain of this circuit?

\begin{flalign}
&V_- = V_+ = V_s&\\\
&V_- \text{is the output of a voltage divider}\\
\end{flalign}

Because V_- is equal to V_+ and

$V_- = V_s = V_o (\frac{R_g}{R_G+R_F})$ t If we solve that equation for \frac{V_O}{V_s} we get the following formula:

\displaystyle Gain =\frac{V_o}{V_s} = 1+\frac{R_F}{R_G}

Buffer (Voltage-Follow)

This circuit is usefull, because the output always replicates the voltage at the input. For example if you connect the output of a voltage divider you can drive a load wth V_o and the impedance in the Load will not change the V_o

$ 64 0.000005 1.0312258501325766 50 5 50 5e-11
a 192 240 304 240 9 15 -15 1000000 4.999950000499995 5 100000
w 192 320 192 256 0
w 304 240 304 320 0
O 304 240 368 240 1 0
v 96 304 96 224 0 0 40 5 0 0 0.5
w 96 224 192 224 2
g 96 304 96 352 0 0
w 192 320 304 320 0

Inverting Amplifier

$ 64 0.000005 1.0312258501325766 50 5 50 5e-11
v 48 304 48 192 0 0 40 5 0 0 0.5
r 176 192 96 192 0 1000
w 176 224 176 304 0
g 176 304 176 352 0 0
g 48 304 48 352 0 0
r 176 112 336 112 0 1000
w 336 112 336 208 0
w 176 192 176 112 0
w 336 208 416 208 0
g 416 320 416 352 0 0
p 416 208 416 320 1 0 0
x 249 86 261 89 4 12 Rf
x 129 166 141 169 4 12 Ri
a 176 208 336 208 8 15 -15 1000000 0.00004999900001999959 0 100000
w 96 192 48 192 2

\begin{flalign}
&\frac{V_o}{V_S} = -\frac{R_F}{R_I}&\\\
\end{flalign}

Difference Amplifier

$ 64 0.000005 1.0312258501325766 50 5 50 5e-11
R 144 192 96 192 0 0 40 5 0 0 0.5
R 144 224 96 224 0 0 40 4 0 0 0.5
r 144 192 224 192 0 1000
r 144 224 224 224 0 1000
a 224 208 352 208 8 15 -15 1000000 2.000009999800004 2 100000
r 224 128 352 128 0 1000
w 352 128 352 208 0
w 224 192 224 128 0
w 352 208 400 208 0
p 400 208 400 320 1 0 0
g 400 320 400 352 0 0
r 224 224 224 320 0 1000
g 224 320 224 352 0 0
x 176 171 192 174 4 12 R1
x 176 244 192 247 4 12 R1
x 199 279 215 282 4 12 R2
x 279 148 295 151 4 12 R2

\displaystyle V_O = \frac{R2}{R1}(V_2-V_1)