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

![[op-amp-basic-schematic-symbol.svg]]

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|400]]

**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

```circuitjs
$ 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 $V_o$

```circuitjs
$ 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

```circuitjs
$ 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

```circuitjs
$ 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
```

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