# Circuit Theory/Operational Amplifiers

!22 transistors making up an operational
amplifier
An operational amplifier is a circuit of about 22 transistors organized
with the goal of producing an amplifier.

$$V_{\!\text{out}} = A_{OL} \, (V_{\!+} - V_{\!-})$$

Ideally the open loop gain (A~OL~) is infinite. In reality the output
will \"rail\" to either the positive or negative supply.

Open loop means that there are no devices connecting the output to the
inverting input (- or negative input). In practice most op-amps are used
in \"closed-loop\", \"negative feedback\" configuration. Closed loop
designs (in the first approximation) require remembering two \"golden
rules\":

:   I. The output attempts to do whatever is necessary to make the
    voltage difference between the inputs zero.
:   II\. The inputs draw no current.

  Op-amp                                                                        Pins                                                                        Symbol
  ----------------------------------------------------------------------------- --------------------------------------------------------------------------- -------------------------------------------------------
  ![](Schematics_voltampopamp.jpg "Schematics_voltampopamp.jpg"){width="100"}   ![](Generic_741_pinout_top.png "Generic_741_pinout_top.png"){width="300"}   ![](Opamppinouts.png "Opamppinouts.png"){width="200"}

## Subtraction

!Subtraction or \"Differential
Amplifier\"
Under the condition that the *R*~*f*~ /*R*~1~ = *R*~*g*~ /*R*~2~, the
output expression becomes:

$$V_{\text{out}} = \frac{ R_{\text{f}} }{ R_1 } (V_{\text{2}} - V_{\text{1}})\,$$
\"A\" is the *differential gain* of the circuit.

## Operating Point

Op Amps are balanced:

:   

    :   The supplies are +V and -V, not V_s and ground. This is why many
        power supplies have both +15 and -15 volts or +5 and -5 volts.
    :   They are made with almost equal numbers of PNP and NPN
        transistors.
    :   They are best operated where the output is around 0 volts.

Don\'t try to fit them into a digital circuit.

## Typical Applications

Op Amps are typically
used to amplify.

!Non-Inverting
!Inverting Vacuum tubes
(light bulbs with a grid and three wires) (1907 by Lee De
Forest) solved the existing need for
amplification of a weak data signal. But vaccuum tubes broke in less
than a year. Many famous scientist/engineer\'s first experience with
technology was radio
repair
including Feynman and
Armstrong.

Transistors came next in 1951 created by
Shockley. They enabled the transistor
radio \... that could be held in the palm of one\'s hand. Transistors
replaced vacuum tubes in almost all devices.

Operational Amplifiers emerged out of math, not inventor work shops.
They emerged from the needs that MatLab fills today, not a simple
calculator. Attempts were made to build them with vaccum tubes and every
known transistor. The 741 op amp took until transistors matured .. 1968.
The 741 is still in production. Op amps are found where the difference
between two devices sensing the environment are needed. They are used to
replace
inductors
in circuits because they are cheaper and lighter.

## Inverting or Not?

Op amps are drawn with the non-inverting input on top if the design goal
is an output that same polarity as the input. Why isn\'t it just called
\"positive input\"? Why isn\'t positive always on top?

The reason is that most op amps involve negative feedback. Most designs
require circuitry to be connected to the \"inverting\" input. If
possible the \"non-inverting\" input is grounded. This means that the
sweet middle operating point is more plausible.

But the most important reason comes from multiple op amp designs. The
negative or positive nature of the output can be more easily seen if the
inverting and non-inverting inputs flip based upon the immediate output
design intention. Negative output means that eventually a final op amp
may be needed to flip the output positive.

## Local Feedback

Op amp circuits are easy to design as long as the feedback is kept local
\... meaning the feedback loops back over only one local op amp. The
feedback does not span outputs. Only outputs connected to inputs connect
two op amps.

## Buffer

!buffer
configuraton
Buffers are needed to re-invigorate a signal.
Fanout is when an output is connected to too many
inputs. Buffers can restore the power needed to drive the rest of the
circuit.

## Adder

!Adder
configuration A
adder sums several (weighted) voltages:

$$V_{\text{out}} = -R_{\text{f}} \left( \frac{V_1}{R_1} + \frac{V_2}{R_2} + \cdots + \frac{V_n}{R_n} \right)$$

When $R_1 = R_2 = \cdots = R_n$, and $R_{\text{f}}$ independent

$$V_{\text{out}} = -\frac{R_{\text{f}}}{R_1} ( V_1 + V_2 + \cdots + V_n ) \!\$$

When $R_1 = R_2 = \cdots = R_n = R_{\text{f}}$

$$V_{\text{out}} = -( V_1 + V_2 + \cdots + V_n ) \!\$$

## Integrator

!integrator

$$V_{\text{out}} = -\int_0^t \frac{ V_{\text{in}} }{RC} \, \operatorname{d}t + V_{\text{initial}}\,$$

(where $V_{\text{in}}$ and $V_{\text{out}}$ are functions of time,
$V_{\text{initial}}$ is the output voltage of the integrator at time *t*
= 0.)

There are several potential problems with this circuit.

-   It is usually assumed that the input $V_{\text{in}}$ has zero DC
    component (i.e., has a zero average value). Otherwise, unless the
    capacitor is periodically discharged, the output will drift outside
    of the operational amplifier\'s operating range.
-   Even when $V_{\text{in}}$ has no offset, the leakage or bias
    currents into the operational amplifier inputs can add an unexpected
    offset voltage to $V_{\text{in}}$ that causes the output to drift.
    Balancing input currents **and** replacing the non-inverting ($+$)
    short-circuit to ground with a resistor with resistance $R$ can
    reduce the severity of this problem.
-   Because this circuit provides no DC feedback (i.e., the capacitor
    appears like an open circuit to signals with $\omega = 0$), the
    offset of the output may not agree with expectations (i.e.,
    $V_{\text{initial}}$ may be out of the designer\'s control with the
    present circuit).
-   The initial voltage is typically enforced by a power supply in
    parallel with the capacitor that is switched out of the circuit at
    t=0.

Many of these problems can be made less severe by adding a *large*
resistor $R_F$ in parallel with the feedback capacitor. At significantly
high frequencies, this resistor will have negligible effect. However, at
low frequencies where there are drift and offset problems, the resistor
provides the necessary feedback to hold the output steady at the correct
value. In effect, this resistor reduces the DC gain of the
\"integrator\" -- it goes from infinite to some finite value $R_F/R$.

## Differentiator

!Differentiator

This circuit differentiates the (inverted) signal over time.

$$V_{\text{out}} = -RC \,\frac{\operatorname{d}V_{\text{in}} }{ \operatorname{d}t}$$

## Inductor Replacement

!Gyrator .. replaces an
inductor

Simulates an inductor. The circuit exploits the fact that the current
flowing through a capacitor behaves through time as the voltage across
an inductor. The capacitor used in this circuit is smaller than the
inductor it simulates and its capacitance is less subject to changes in
value due to environmental changes.

This circuit is unsuitable for applications relying on the input
impedance seen left to right and indicated by the red $Z_{in}$ symbol is
the same in both circuits. The output on the left could tell the
difference between the op amp circuit and an inductor.

Of course there would not be a magnetic field (for a motor or
transformer) and any application relying on back
EMF would not work. 
