Operational amplifier applications

In order for a particular device to be used in an application, it must satisfy certain requirements. The operational amplifier must

• have large open-loop signal gain (voltage gain of 200,000 is obtained in early integrated circuit exemplars), and
• have input impedance large with respect to values present in the feedback network.

With these requirements satisfied, the op-amp is considered ideal, and one can use the method of virtual ground to quickly and intuitively grasp the 'behavior' of any of the op-amp circuits below.

Resistors used in practical solid-state op-amp circuits are typically in the kΩ range. Resistors much greater than 1 MΩ cause excessive thermal noise and make the circuit operation susceptible to significant errors due to bias or leakage currents.

Practical operational amplifiers draw a small current from each of their inputs due to bias requirements (in the case of bipolar junction transistor-based inputs) or leakage (in the case of MOSFET-based inputs).

These currents flow through the resistances connected to the inputs and produce small voltage drops across those resistances. Appropriate design of the feedback network can alleviate problems associated with input bias currents and common-mode gain, as explained below. The heuristic rule is to ensure that the impedance "looking out" of each input terminal is identical.

To the extent that the input bias currents do not match, there will be an effective input offset voltage present, which can lead to problems in circuit performance. Many commercial op-amp offerings provide a method for tuning the operational amplifier to balance the inputs (e.g., "offset null" or "balance" pins that can interact with an external voltage source attached to a potentiometer). Alternatively, a tunable external voltage can be added to one of the inputs in order to balance out the offset effect. In cases where a design calls for one input to be short-circuited to ground, that short circuit can be replaced with a variable resistance that can be tuned to mitigate the offset problem.

Operational amplifiers using MOSFET-based input stages have input leakage currents that will be, in many designs, negligible.

Although power supplies are not indicated in the (simplified) operational amplifier designs below, they are nonetheless present and can be critical in operational amplifier circuit design.

Power supply imperfections (e.g., power signal ripple, non-zero source impedance) may lead to noticeable deviations from ideal operational amplifier behavior. For example, operational amplifiers have a specified power supply rejection ratio that indicates how well the output can reject signals that appear on the power supply inputs. Power supply inputs are often noisy in large designs because the power supply is used by nearly every component in the design, and inductance effects prevent current from being instantaneously delivered to every component at once. As a consequence, when a component requires large injections of current (e.g., a digital component that is frequently switching from one state to another), nearby components can experience sagging at their connection to the power supply. This problem can be mitigated with appropriate use of bypass capacitors connected across each power supply pin and ground. When bursts of current are required by a component, the component can bypass the power supply by receiving the current directly from the nearby capacitor (which is then slowly recharged by the power supply).

Additionally, current drawn into the operational amplifier from the power supply can be used as inputs to external circuitry that augment the capabilities of the operational amplifier. For example, an operational amplifier may not be fit for a particular high-gain application because its output would be required to generate signals outside of the safe range generated by the amplifier. In this case, an external push–pull amplifier can be controlled by the current into and out of the operational amplifier. Thus, the operational amplifier may itself operate within its factory specified bounds while still allowing the negative feedback path to include a large output signal well outside of those bounds.[1]

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[nb 1] 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

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

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.[2]

As the negative input of the op-amp acts as a virtual ground, the input impedance of this circuit is equal to R_in.

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:

${\displaystyle Z_{\text{in}}=(1+A_{\text{OL}}B)Z_{\text{dif}}}$

where Z_dif is the op-amp's input impedance to differential signals, and A_OL is the open-loop voltage gain of the op-amp (which varies with frequency), and B is the feedback factor (the fraction of the output signal that returns to the input).[3][4] In the case of the ideal op-amp, with A_OL infinite and Z_dif infinite, the input impedance is also 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.

The feedback loop similarly decreases the output impedance:

${\displaystyle Z_{\text{out}}={\frac {Z_{\text{OL}}}{1+A_{\text{OL}}B}}}$

where Z_out is the output impedance with feedback, and Z_OL is the open-loop output impedance.[4]

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), multiplied by the open-loop gain of the op-amp)

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.

The input and output impedance are affected by the feedback loop in the same way as the non-inverting amplifier, with B=1.[3][4]

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)

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.

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

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

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

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:

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

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.

In a practical application one encounters a significant difficulty: unless the capacitor C is periodically discharged, the output voltage will eventually drift outside of the operational amplifier's operating range. This can be due to any combination of:

• The input V_in has a non-zero DC component,
• Input bias current is non-zero,
• Input offset voltage is non-zero.[6]

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 ||:

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

Op amp integrator § Practical circuit explains the output drift adds a small finite DC error voltage:

${\displaystyle V_{\text{error}}=\left({\frac {R_{\text{f}}}{R_{\text{i}}}}+1\right)\left(V_{\text{OS}}+I_{\text{B-}}\left(R_{\text{f}}\parallel R_{\text{i}}\right)\right).}$

The circuit functions now more like a low-pass filter with flat response up to the cutoff frequency ${\displaystyle {\frac {1}{2\pi C_{f}R_{f}}}}$ and only significantly above that does it function as an integrator, with equation:

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

Differentiates the (inverted) signal over time:

${\displaystyle V_{\text{out}}=-RC{\frac {dV_{\text{in}}}{dt}},}$

where ${\displaystyle V_{\text{in}}}$ and ${\displaystyle V_{\text{out}}}$ are functions of time.

The transfer function of the inverting differentiator has a single zero in the origin (i.e., where angular frequency ${\displaystyle \omega =0}$). The high-pass characteristics of a differentiating amplifier 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.

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 geometrically smaller than the inductor it simulates, and its capacitance is less subject to changes in value due to environmental changes. Applications where this circuit may be superior to a physical inductor are simulating a variable inductance or simulating a very large inductance.

This circuit is of limited use in applications relying on the back EMF property of an inductor, as this effect will be limited in a gyrator circuit to the voltage supplies of the op-amp.

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.

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.

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 and the op-amp inputs draw no current. So the current flowing from the source through the resistor and diode 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

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

When the voltage across the diode V_D is greater than zero, this current 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.

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, 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

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