Pin Configuration

For a standard 8-pin, 741 type Operational Amplifier (IC), the following pin configuration has to be implemented for its operation to be successful:

Pin	Useage
1	Offset Null
2	Inverted Input
3	Non-Inverted Input
4	-V Supply
5	No use
6	Output
7	+V Supply
8	No use

Symbol

The standard symbol for an op amp is shown below:

Inverting input v_-

Non-inverting input v₊

Output v_o

Operation

${\displaystyle v_{o}=A(v_{+}-v_{-})}$

Non-inverting Amplifier

Consider next this circuit:

Using the idealized op amp assumptions, we see that

${\displaystyle v^{-}=v^{+}=v_{i}.}$

This means that current flowing from right to left through ${\displaystyle R_{1}}$ is

${\displaystyle I_{R_{1}}={\frac {v_{i}}{R_{1}}}}$

and all of this current must also flow from right to left through ${\displaystyle R_{2}}$ since the op amp takes in no current at its input terminals. So we can find

${\displaystyle {\begin{aligned}v_{o}&=v_{-}+I_{R_{2}}R_{2}\\&=v_{i}+{\frac {v_{i}}{R_{1}}}R_{2}\\&=\left(1+{\frac {R_{2}}{R_{1}}}\right)v_{i}\\G&={\frac {v_{o}}{v_{i}}}=1+{\frac {R_{2}}{R_{1}}}.\end{aligned}}}$

In this configuration the gain is positive and at least as great as one. This configuration provides a non-inverting amplifier.

Voltage Follower (Current Buffer)

The op amp in this configuration:

is identical to the non-inverting amplifier configuration if we pick ${\displaystyle R_{1}=\infty }$ and ${\displaystyle R_{2}=0}$. The output, therefore, is

${\displaystyle v_{o}=v_{i}.}$

Because the ideal op amp draws no current from the source ${\displaystyle v_{i}}$ and because it has very low output impedance, this circuit provides a replica of the input without loading the circuit that provides the input voltage. Many sensors, such as some microphones, have a high output impedance. Drawing current from them would result in a reduction of the voltage sensed by the amplifier. This circuit prevents that, leading to the alternate name current buffer for this circuit.

Inverting Amplifier

Noting that ${\displaystyle v^{+}=0}$ and solving the second of these equations for ${\displaystyle v^{-}}$ gives

${\displaystyle v^{-}={\frac {v_{o}}{A}}.}$

Now we can substitute this into the first of our two equations:

${\displaystyle {\frac {v_{o}}{A}}={\frac {R_{2}}{R_{1}+R_{2}}}v_{i}+{\frac {R_{1}}{R_{1}+R_{2}}}v_{o}.}$

Subtracting ${\displaystyle {\frac {R_{1}}{R_{1}+R_{2}}}v_{o}}$ from both sides and factoring out ${\displaystyle v_{o}}$ on the left hand side yields

${\displaystyle \left({\frac {1}{A}}-{\frac {R_{1}}{R_{1}+R_{2}}}\right)v_{o}={\frac {R_{2}}{R_{1}+R_{2}}}v_{i}.}$

Dividing boths sides by ${\displaystyle {\frac {1}{A}}-{\frac {R_{1}}{R_{1}+R_{2}}}}$ and by ${\displaystyle v_{i}}$ yields the gain

${\displaystyle G={\frac {v_{o}}{v_{i}}}={\frac {\frac {R_{2}}{R_{1}+R_{2}}}{{\frac {1}{A}}-{\frac {R_{1}}{R_{1}+R_{2}}}}}}$

This can be rewritten more simply if both numerator and denominator are multiplied by ${\displaystyle R_{1}+R_{2}}$:

${\displaystyle G={\frac {R_{2}}{{\frac {R_{1}+R_{2}}{A}}-R_{1}}}.}$

Now if ${\displaystyle A}$ is very large, the first term in the denominator can be neglected and the gain is

${\displaystyle G={\frac {v_{o}}{v_{i}}}\approx -{\frac {R_{2}}{R_{1}}}.}$

This shows that the amplification constant ${\displaystyle A}$ of the op amp vanishes from the overall expression for the gain because it is very large. The gain of the circuit is instead determined solely by the external components. This means that the high variability of ${\displaystyle A}$ in manufacture does not affect the behaviour of the circuit in which the op amp is embedded. As long as ${\displaystyle A}$ is sufficiently large, the approximation is reasonable.

The configuration we have analyzed here is that of an inverting amplifier. It is inverted because of the negative sign. The magnitude of the gain is determined by the ratio of ${\displaystyle R_{1}/R_{2}}$ and can be either greater than or less than one.

Zero / Nonzero Crossing Detector

Differentiator / Integrator

Comparator

The amplification of the operational amplifier depends purely upon the ratio of a feedback resistor to the input resistor. And now, let us assume that the feedback resistor is missing, and the input is connected directly to the op amp IC without and input resistors.

If the input resistor is missing, the gain loop is open. In another word we can say that the Feedback Resistance is infinitely high, and the gain of the operational amplifier is infinitely high. However, what does determine the sign (+/-) of the gain. This is the difference between negative and positive inputs.

• In non-inverting amplifier circuit the inverting input is grounded (through resistors), and the input goes to positive.
• In the inverting amplifier circuit, the non-inverting input is grounded, and the input goes to the inverting input.
• In both cases the voltage on the non-in-use input is close to ZERO. Therefore the incoming signal is absolute.

In comparator, a reference voltage is supplied to the positive input, and the negative input measures the input voltage. The reference voltage is obtained by a simple voltage divider.

So, if the input voltage will be less then reference voltage, the output will bi “infinitely negative” And if the input voltage will be higher then reference voltage, the output voltage will be “infinitely positive”.

However, as in any amplifier, the highest value of the output is dictated by the power voltage. So the word “infinite” should be changed by “maximal” or “as high as power voltage”.

SO

  If (input_voltage > reference voltage)
      Output Voltage = NEGATIVE
  If (input_voltage < reference voltage)
      Output voltage = POSITIVE.

For the further analysis, let us consider the output of the op amp just as a voltage source. Along with a power source circuit, and the LED (Light Emitting Diode) they form a circuit which highs up in case when a current flows through the circuit.

When the comparator output is high, the output circuit is similar to that one, on the LEFT drawing. “batteries” are coupled so, that their net voltage is 0, and if we assume that the voltage in the circuit is 0, the LED will NOT light.

When the comparator input is low, we can assume the circuit on the RIGHT figure. IT has voltage of 18 Volts, so the LED will LIGHT UP.

Bargraph Voltmeter

Analog /Digital Converter

As a component of 555 timer

Differential amplifier (difference amplifier)

Amplifies the difference in voltage between its inputs.

The name "differential amplifier" must not be confused with the "differentiator," which is also shown on this page.

The "instrumentation amplifier," which is also shown on this page, is a modification of the differential amplifier that also provides high input impedance.

The circuit shown computes the difference of two voltages, multiplied by some gain factor. The output voltage:

${\displaystyle V_{\text{out}}={\frac {\left(R_{\text{f}}+R_{1}\right)R_{\text{g}}}{\left(R_{\text{g}}+R_{2}\right)R_{1}}}V_{2}-{\frac {R_{\text{f}}}{R_{1}}}V_{1}=\left({\frac {R_{1}+R_{\text{f}}}{R_{1}}}\right)\cdot \left({\frac {R_{\text{g}}}{R_{\text{g}}+R_{2}}}\right)V_{2}-{\frac {R_{\text{f}}}{R_{1}}}V_{1}.}$

Or, expressed as a function of the common mode input V_com and difference input V_dif

${\displaystyle V_{\text{com}}=(V_{1}+V_{2})/2;V_{\text{dif}}=V_{2}-V_{1}\,,}$

the output voltage is

${\displaystyle V_{\text{out}}{\frac {R_{1}}{R_{\text{f}}}}=V_{\text{com}}{\frac {R_{1}/R_{\text{f}}-R_{2}/R_{\text{g}}}{1+R_{2}/R_{\text{g}}}}+V_{\text{dif}}{\frac {1+(R_{2}/R_{\text{g}}+R_{1}/R_{\text{f}})/2}{1+R_{2}/R_{\text{g}}}}.}$

In order for this circuit to produce a signal proportional to the voltage difference of the input terminals, the coefficient of the V_com term (the common-mode gain) must be zero, or

${\displaystyle R_{1}/R_{\text{f}}=R_{2}/R_{\text{g}}}$

With this constraint If you think of the left-hand side of the relation as the closed-loop gain of the inverting input, and the right-hand side as the gain of the non-inverting input, then matching these two quantities provides an output insensitive to the common-mode voltage of ${\displaystyle V_{1}}$ and ${\displaystyle V_{2}}$.</ref> in place, the common-mode rejection ratio of this circuit is infinitely large, and the output

${\displaystyle V_{\text{out}}={\frac {R_{\text{f}}}{R_{1}}}V_{\text{dif}}={\frac {R_{\text{f}}}{R_{1}}}\left(V_{2}-V_{1}\right).}$

where the simple expression R_f / R₁ represents the closed-loop gain of the differential amplifier.

The special case when the closed-loop gain is unity is a differential follower, with:

${\displaystyle V_{\text{out}}=V_{2}-V_{1}.\,}$

Inverting amplifier

An inverting amplifier is a special case of the differential amplifier in which that circuit's non-inverting input V₂ is grounded, and inverting input V₁ is identified with V_in above. The closed-loop gain is R_f / R_in, hence

${\displaystyle V_{\text{out}}=-{\frac {R_{\text{f}}}{R_{\text{in}}}}V_{\text{in}}\!\,}$.

The simplified circuit above is like the differential amplifier in the limit of R₂ and R_g very small. In this case, though, the circuit will be susceptible to input bias current drift because of the mismatch between R_f and R_in.

To intuitively see the gain equation above, calculate the current in R_in:

${\displaystyle i_{\text{in}}={\frac {V_{\text{in}}}{R_{\text{in}}}}}$

then recall that this same current must be passing through R_f, therefore (because V₋ = V₊ = 0):

${\displaystyle V_{\text{out}}=-i_{\text{in}}R_{\text{f}}=-V_{\text{in}}{\frac {R_{\text{f}}}{R_{\text{in}}}}}$

A mechanical analogy is a seesaw, with the V₋ node (between R_in and R_f) as the fulcrum, at ground potential. V_in is at a length R_in from the fulcrum; V_out is at a length R_f. When V_in descends "below ground", the output V_out rises proportionately to balance the seesaw, and vice versa.

Non-inverting amplifier

A non-inverting amplifier is a special case of the differential amplifier in which that circuit's inverting input V₁ is grounded, and non-inverting input V₂ is identified with V_in above, with R₁ ≫ R₂. Referring to the circuit immediately above,

${\displaystyle V_{\text{out}}=\left(1+{\frac {R_{\text{2}}}{R_{\text{1}}}}\right)V_{\text{in}}\!\,}$.

To intuitively see this gain equation, use the virtual ground technique to calculate the current in resistor R₁:

${\displaystyle i_{1}={\frac {V_{\text{in}}}{R_{1}}}\,,}$

then recall that this same current must be passing through R₂, therefore:

${\displaystyle V_{\text{out}}=V_{\text{in}}+i_{1}R_{2}=V_{\text{in}}\left(1+{\frac {R_{2}}{R_{1}}}\right)}$

Unlike the inverting amplifier, a non-inverting amplifier cannot have a gain of less than 1.

A mechanical analogy is a class-2 lever, with one terminal of R₁ as the fulcrum, at ground potential. V_in is at a length R₁ from the fulcrum; V_out is at a length R₂ further along. When V_in ascends "above ground", the output V_out rises proportionately with the lever.

The input impedance of the simplified non-inverting amplifier is high, of order R_dif × A_OL times the closed-loop gain, where R_dif is the op amp's input impedance to differential signals, and A_OL is the open-loop voltage gain of the op amp; in the case of the ideal op amp, with A_OL infinite and R_dif infinite, the input impedance is infinite. In this case, though, the circuit will be susceptible to input bias current drift because of the mismatch between the impedances driving the V₊ and V₋ op amp inputs.

Voltage follower (unity buffer amplifier)

Used as a buffer amplifier to eliminate loading effects (e.g., connecting a device with a high source impedance to a device with a low input impedance).

${\displaystyle V_{\text{out}}=V_{\text{in}}\!\ }$

${\displaystyle Z_{\text{in}}=\infty }$ (realistically, the differential input impedance of the op-amp itself, 1 MΩ to 1 TΩ)

Due to the strong (i.e., unity gain) feedback and certain non-ideal characteristics of real operational amplifiers, this feedback system is prone to have poor stability margins. Consequently, the system may be unstable when connected to sufficiently capacitive loads. In these cases, a lag compensation network (e.g., connecting the load to the voltage follower through a resistor) can be used to restore stability. The manufacturer data sheet for the operational amplifier may provide guidance for the selection of components in external compensation networks. Alternatively, another operational amplifier can be chosen that has more appropriate internal compensation.

Summing amplifier

A summing amplifier sums several (weighted) voltages:

${\displaystyle V_{\text{out}}=-R_{\text{f}}\left({\frac {V_{1}}{R_{1}}}+{\frac {V_{2}}{R_{2}}}+\cdots +{\frac {V_{n}}{R_{n}}}\right)}$

• When ${\displaystyle R_{1}=R_{2}=\cdots =R_{n}}$, and ${\displaystyle R_{\text{f}}}$ independent

${\displaystyle V_{\text{out}}=-{\frac {R_{\text{f}}}{R_{1}}}(V_{1}+V_{2}+\cdots +V_{n})\!\ }$

• When ${\displaystyle R_{1}=R_{2}=\cdots =R_{n}=R_{\text{f}}}$

${\displaystyle V_{\text{out}}=-(V_{1}+V_{2}+\cdots +V_{n})\!\ }$

• Output is inverted
• Input impedance of the nth input is ${\displaystyle Z_{n}=R_{n}}$ (${\displaystyle V_{-}}$ is a virtual ground)

Instrumentation amplifier

Combines very high input impedance, high common-mode rejection, low DC offset, and other properties used in making very accurate, low-noise measurements

• Is made by adding a non-inverting buffer to each input of the differential amplifier to increase the input impedance.

Wien bridge oscillator

Produces a very low distortion sine wave. Uses negative temperature compensation in the form of a light bulb or diode.

Inverting integrator

The integrator is mostly used in analog computers, analog-to-digital converters and wave-shaping circuits.

Integrates (and inverts) the input signal V_in(t) over a time interval t, t₀ < t < t₁, yielding an output voltage at time t = t₁ of

${\displaystyle V_{\text{out}}(t_{1})=V_{\text{out}}(t_{0})-{\frac {1}{RC}}\int _{t_{0}}^{t_{1}}V_{\text{in}}(t)\,\operatorname {d} t}$

where V_out(t₀) represents the output voltage of the circuit at time t = t₀. This is the same as saying that the output voltage changes over time t₀ < t < t₁ by an amount proportional to the time integral of the input voltage:

${\displaystyle -{\frac {1}{RC}}\int _{t_{0}}^{t_{1}}V_{\text{in}}(t)\,\operatorname {d} t}$

This circuit can be viewed as a low-pass electronic filter, one with a single pole at DC (i.e., where ${\displaystyle \omega =0}$) and with gain.

A slightly more complex circuit can ameliorate the second two problems, and in some cases, the first as well.

Here, the feedback resistor R_f provides a discharge path for capacitor C_f, while the series resistor at the non-inverting input R_n, when of the correct value, alleviates input bias current and common-mode problems. That value is the parallel resistance of R_i and R_f, or using the shorthand notation ||:

${\displaystyle R_{\text{n}}={\frac {1}{{\frac {1}{R_{\text{i}}}}+{\frac {1}{R_{\text{f}}}}}}=R_{\text{i}}||R_{\text{f}}}$

The relationship between input signal and output signal is now:

${\displaystyle V_{\text{out}}(t_{1})=V_{\text{out}}(t_{0})-{\frac {1}{R_{i}C_{f}}}\int _{t_{0}}^{t_{1}}V_{\text{in}}(t)\,\operatorname {d} t}$

Inverting differentiator

Differentiates the (inverted) signal over time.

${\displaystyle V_{\text{out}}=-RC\,{\frac {\operatorname {d} V_{\text{in}}}{\operatorname {d} t}}\,\qquad {\text{where }}V_{\text{in}}{\text{ and }}V_{\text{out}}{\text{ are functions of time.}}}$

• This can also be viewed as a high-pass electronic filter. It is a filter with a single zero at DC (i.e., where angular frequency ${\displaystyle \omega =0}$ radians) and gain. The high-pass characteristics of a differentiating amplifier (i.e., the low-frequency zero) can lead to stability challenges when the circuit is used in an analog servo loop (e.g., in a PID controller with a significant derivative gain). In particular, as a root locus analysis would show, increasing feedback gain will drive a closed-loop pole toward marginal stability at the DC zero introduced by the differentiator.

Inductance gyrator

Simulates an inductor (i.e., provides inductance without the use of a possibly costly 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 back EMF property of an inductor as this will be limited in a gyrator circuit to the voltage supplies of the op-amp.

Negative impedance converter (NIC)

Creates a resistor having a negative value for any signal generator.

In this case, the ratio between the input voltage and the input current (thus the input resistance) is given by:

${\displaystyle R_{\text{in}}=-R_{3}{\frac {R_{1}}{R_{2}}}}$

In general, the components ${\displaystyle R_{1}}$, ${\displaystyle R_{2}}$, and ${\displaystyle R_{3}}$ need not be resistors; they can be any component that can be described with an impedance.

Precision rectifier

The voltage drop V_F across the forward biased diode in the circuit of a passive rectifier is undesired. In this active version, the problem is solved by connecting the diode in the negative feedback loop. The op-amp compares the output voltage across the load with the input voltage and increases its own output voltage with the value of V_F. As a result, the voltage drop V_F is compensated and the circuit behaves very nearly as an ideal (super) diode with V_F = 0 V.

The circuit has speed limitations at high frequency because of the slow negative feedback and due to the low slew rate of many non-ideal op-amps.

Logarithmic output

• The relationship between the input voltage V_in and the output voltage V_out is given by:

${\displaystyle V_{\text{out}}=-V_{\text{T}}\ln \left({\frac {V_{\text{in}}}{I_{\text{S}}\,R}}\right)}$

where I_S is the saturation current and V_T is the thermal voltage.

• If the operational amplifier is considered ideal, the inverting input pin is virtually grounded, so the current flowing into the resistor from the source (and thus through the diode to the output, since the op-amp inputs draw no current) is:

${\displaystyle {\frac {V_{\text{in}}}{R}}=I_{\text{R}}=I_{\text{D}}}$

where I_D is the current through the diode. As known, the relationship between the current and the voltage for a diode is:

${\displaystyle I_{\text{D}}=I_{\text{S}}\left(e^{\frac {V_{\text{D}}}{V_{\text{T}}}}-1\right).}$

This, when the voltage is greater than zero, can be approximated by:

${\displaystyle I_{\text{D}}\simeq I_{\text{S}}e^{\frac {V_{\text{D}}}{V_{\text{T}}}}.}$

Putting these two formulae together and considering that the output voltage is the negative of the voltage across the diode (V_out = −V_D), the relationship is proven.

This implementation does not consider temperature stability and other non-ideal effects.

Exponential output

• The relationship between the input voltage ${\displaystyle V_{\text{in}}}$ and the output voltage ${\displaystyle V_{\text{out}}}$ is given by:

${\displaystyle V_{\text{out}}=-RI_{\text{S}}e^{\frac {V{\text{in}}}{V_{\text{T}}}}}$

where ${\displaystyle I_{\text{S}}}$ is the saturation current and ${\displaystyle V_{\text{T}}}$ is the thermal voltage.

• Considering the operational amplifier ideal, then the negative pin is virtually grounded, so the current through the diode is given by:

${\displaystyle I_{\text{D}}=I_{\text{S}}\left(e^{\frac {V_{\text{D}}}{V_{\text{T}}}}-1\right)}$

when the voltage is greater than zero, it can be approximated by:

${\displaystyle I_{\text{D}}\simeq I_{\text{S}}e^{\frac {V_{\text{D}}}{V_{\text{T}}}}.}$

The output voltage is given by:

${\displaystyle V_{\text{out}}=-RI_{\text{D}}.\,}$