# Op Amp Circuits

concept

The op amp is often used in one of several common configurations. We've already seen the inverting and non-inverting amplifiers as well as the summing amplifier.
In this topic we'll cover the unity gain buffer, the active rectifier (or ideal diode) and the differentiator and integrator circuits.
Together these circuit fragments are the most common ways to use op amps in real world circuits.

fact

The unity gain buffer is:
Whose output is given by:
$$ V_o = V_{in} $$

From the circuit we can clearly see that \(V_o = V_{in}\) follows from the fact that \(V_+ = V_-\)
Why would we want a circuit that does nothing to the input voltage? A few reasons:

- Your input signal has a very small current but the op amp can output a larger one
- If the output impedance of the input circuit is very large the unity buffer can effectively lower it so that we don't load the next section of the circuit

fact

The ideal diode (or active rectifier) circuit looks like:
Whose output is given by:
$$ V_o = \begin{cases}
0 & V_{in}\leq 0 \\
V_{in} & V_{in} \gt 0
\end{cases} $$

The ideal diode's op amp outputs \(V_{in} + 0.7\)V so that \(V_o = V_{in}\) (assuming the diode drops 700mV).
And since the diode performs a rectifying function the resistor pulls \(V_o\) to 0 when \(V_{in} \lt 0\)
This allows us to rectify input signals without the usual voltage drop of 700mV. This is very useful for weak input signals. Say we want to amplify a wireless signal, we don't want to just lose all of our information if we're far away from the radio tower and only receiving a wave that peaks at 500mV.
The precision rectifier is, however, considerably larger and more expensive than a simple diode so its use is usually restricted to situations where the 700mV drop is unacceptable.

fact

The differentiator amplifier approximates the differential function:
Whose output is given by:
$$ V_o = -R\cdot C\frac{dV_{in}}{dt} $$

derivation

Since \(V_+ = 0\)V we know that \(V_- = 0\)V too.
So the current running through the capacitor is \(I = C\frac{dV_{in}}{dt}\)
Now \(V_+ = 0\) and we know the current running through the capacitor, all of which runs through the resistor which means that:
$$\begin{align}
V_o & = -R\cdot I \\
& = -R \cdot C \frac{dV_{in}}{dt}
\end{align}$$
Which is a scaled derivative of the input voltage.

fact

The integrator amplifier approximates the integral of the input voltage:
Whose output is given by:
$$ V_o = -\frac{1}{R\cdot C}\int_0^tV_{in}dt $$

derivation

Starting out as usual with \(V_- = V_+ = 0\)V we see that the current running through the resistor is:
$$ I = \frac{V_{in}}{R} $$
Now all of that current runs through the capacitor and \(V_-\) is 0V so:
$$\begin{align}
V_o & = 0 - \int_0^t \frac{I}{C}dt \\
& = -\frac{1}{R\cdot C}\int_0^tV_{in}dt
\end{align}$$

practice problems