Editing Operational amplifier (section)

==Operation==
[[Image:Op-amp open-loop 1.svg|thumb|An op amp without negative feedback (a comparator)]]
The amplifier's differential inputs consist of a non-inverting input (+) with voltage ''V<sub>+</sub>'' and an inverting input (&minus;) with voltage ''V<sub>&minus;</sub>''; ideally the op amp amplifies only the difference in voltage between the two, which is called the ''differential input voltage''. The output voltage of the op amp ''V''<sub>out</sub> is given by the equation
:<math>V_\text{out} = A_\text{OL} (V_+ - V_-),</math>
where ''A''<sub>OL</sub> is the [[Electronic feedback loops|open-loop]] gain of the amplifier (the term "open-loop" refers to the absence of an external feedback loop from the output to the input).

=== Open-loop amplifier ===
The magnitude of ''A''<sub>OL</sub> is typically very large (100,000 or more for integrated circuit op amps, corresponding to +100&nbsp;[[Decibel|dB]]). Thus, even small microvolts of difference between ''V''<sub>+</sub> and ''V''<sub>&minus;</sub> may drive the amplifier into [[Clipping (signal processing)|clipping]] or [[Saturation current|saturation]]. The magnitude of ''A''<sub>OL</sub> is not well controlled by the manufacturing process, and so it is impractical to use an open-loop amplifier as a stand-alone [[differential amplifier]].

Without [[negative feedback amplifier|negative feedback]], and optionally [[positive feedback]] for [[Regenerative circuit|regeneration]], an ''[[Electronic feedback loops|open-loop]]'' op amp acts as a [[comparator]], although comparator ICs are better suited.<ref>{{Cite web |last=Bryant |first=James |date=2011 |title=Application Note AN-849: Using Op Amps as Comparators |url=https://www.analog.com/media/en/technical-documentation/application-notes/AN-849.pdf |url-status=live |archive-url=https://web.archive.org/web/20230202080015/https://www.analog.com/media/en/technical-documentation/application-notes/AN-849.pdf |archive-date=2023-02-02}}</ref> If the inverting input is held at ground (0&nbsp;V), and the input voltage ''V''<sub>in</sub> applied to the non-inverting input is positive, the output will be maximum positive; if ''V''<sub>in</sub> is negative, the output will be maximum negative. 

=== Closed-loop amplifier ===
[[Image:Operational amplifier noninverting.svg|thumb|An op amp with negative feedback (a non-inverting amplifier)]]

If predictable operation is desired, negative feedback is used, by applying a portion of the output voltage to the inverting input. The ''closed-loop'' feedback greatly reduces the gain of the circuit. When negative feedback is used, the circuit's overall gain and response is determined primarily by the feedback network, rather than by the op-amp characteristics. If the feedback network is made of components with values small relative to the op amp's input impedance, the value of the op amp's open-loop response ''A''<sub>OL</sub> does not seriously affect the circuit's performance. In this context, high input [[Electrical impedance|impedance]] at the input terminals and low output impedance at the output terminal(s) are particularly useful features of an op amp. 

The response of the op-amp circuit with its input, output, and feedback circuits to an input is characterized mathematically by a [[transfer function]]; designing an op-amp circuit to have a desired transfer function is in the realm of [[electrical engineering]]. The transfer functions are important in most applications of op amps, such as in [[analog computers]].

In the non-inverting amplifier on the right, the presence of negative feedback via the [[voltage divider]] ''R''<sub>f</sub>, ''R''<sub>g</sub> determines the ''closed-loop gain'' ''A''<sub>CL</sub>&nbsp;= {{nowrap|''V''<sub>out</sub> / ''V''<sub>in</sub>}}. Equilibrium will be established when ''V''<sub>out</sub> is just sufficient to pull the inverting input to the same voltage as ''V''<sub>in</sub>. The voltage gain of the entire circuit is thus {{nowrap|1 + ''R''<sub>f</sub> / ''R''<sub>g</sub>}}. As a simple example, if ''V''<sub>in</sub>&nbsp;= 1&nbsp;V and R<sub>f</sub>&nbsp;= ''R''<sub>g</sub>, ''V''<sub>out</sub> will be 2&nbsp;V, exactly the amount required to keep ''V''<sub>&minus;</sub> at 1&nbsp;V. Because of the feedback provided by the ''R''<sub>f</sub>, ''R''<sub>g</sub> network, this is a ''closed-loop'' circuit.

Another way to analyze this circuit proceeds by making the following (usually valid) assumptions:<ref>{{cite book |first=Jacob |last=Millman |title=Microelectronics: Digital and Analog Circuits and Systems |publisher=McGraw-Hill |date=1979 |isbn=0-07-042327-X |pages=[https://archive.org/details/microelectronics00mill_0/page/523 523–527] |url=https://archive.org/details/microelectronics00mill_0/page/523 }}</ref>
# When an op amp operates in linear (i.e., not saturated) mode, the difference in voltage between the non-inverting (+) and inverting (&minus;) pins is negligibly small.
# The input impedance of the (+) and (&minus;) pins is much larger than other resistances in the circuit.

The input signal ''V''<sub>in</sub> appears at both (+) and (&minus;) pins per assumption 1, resulting in a current ''i'' through ''R''<sub>g</sub> equal to {{nowrap|''V''<sub>in</sub> / ''R''<sub>g</sub>}}:
<math display=block>i = \frac{V_\text{in}}{R_\text{g}}</math>

Since Kirchhoff's current law states that the same current must leave a node as enter it, and since the impedance into the (&minus;) pin is near infinity per assumption 2, we can assume practically all of the same current ''i'' flows through ''R''<sub>f</sub>, creating an output voltage
<math display=block>V_\text{out} = V_\text{in} + iR_\text{f} = V_\text{in} + \left(\frac{V_\text{in}}{R_\text{g}} R_\text{f}\right) = V_\text{in} + \frac{V_\text{in}R_\text{f}} {R_\text{g}} = V_\text{in} \left(1 + \frac{R_\text{f}}{R_\text{g}}\right)</math>

By combining terms, we determine the closed-loop gain ''A''<sub>CL</sub>:
<math display=block>A_\text{CL} = \frac{V_\text{out}}{V_\text{in}} = 1 + \frac{R_\text{f}}{R_\text{g}}</math>