In the following article we discuss the main op amp parameters and the related op amp basic application circuits with equations, for solving their specific component values.

Op-amps (operational amplifiers) are a specialized type of integrated circuit that include a directly-coupled, high-gain amplifier with an overall response characteristics adjusted by a feedback.

The op-amp derives its name from the fact that it can execute a wide range of mathematical calculations. Because of its response, an op-amp is also known as a linear integrated circuit and is the core component of many analogue systems.

An op amp features an extraordinarily high gain (possibly nearing infinity), which may be adjusted via a feedback. The addition of capacitors or inductors to the feedback network might result in gain that changes with frequency, affecting the integrated circuit's overall operational state.

As shown in the figure above, the fundamental op amp is a three terminal device having two inputs and one output. The input terminals are classified as "inverting" or "non-inverting."

## Op Amp Parameters

When supplied with equal input voltages, the output of the ideal operational amplifier, or "op amp," is zero, or "0 volts."

VIN 1 = VIN 2 gives VOUT = 0

Practical op-amps have an imperfectly balanced input, causing uneven bias currents to flow through the input terminals. In order to balance the op amp output, an input offset voltage must be provided between the two input terminals.

### 1) Input Bias Current

When the output is balanced, or when V_{OUT} = 0, the input bias current (I_{B}) is equal to one-half the total individual currents entering the two input connections. It is often a very tiny number; for example, I_{B} = 100 nA is a normal value.

### 2) Input offset current

The difference between each individual current reaching the input terminals is known as the input offset current (I_{io}). Again, it is often of extremely low value; for instance, a common value is I_{io} = 10 nA.

### 3) Input offset voltage

In order to keep the op amp balanced, an input offset voltage V_{io} needs to be applied across the input terminal. Usually the value of V_{io} is = 1 mV.

Values of I_{io} and V_{io} may both vary with temperature, and this variation is referred to as I_{io} drift and V_{io} drift, respectively.

### 4) Power Supply Rejection Ratio (PSRR)

The ratio of the change in input offset voltage to the corresponding change in the power supply voltage is known as the power supply rejection ratio, or PSRR. This is often in the range of 10 to 20 uV/V.

Additional parameters for op-amps that might be mentioned are:

### 5) Open-loop gain/Closed loop Gain

Open-loop gain refers to an op-amp's gain without a feedback circuit, whereas closed-loop gain refers to an op-amp's gain with a feedback circuit. It is generally represented as A_{d}.

### 6) Common-mode rejection ratio (CMRR)

This is the ratio of the difference signal to the common-mode signal and serves as a measure of a differential amplifier's performance. We use Decibels (dB) to express this ratio.

### 7) Slew Rate

Slew rate is the rate at which the output voltage of an amplifier changes under large signal conditions. It is represented using the unit V/us.

## Op Amp Basic Application Circuits

In the following paragraphs we will learn about several interesting op amp basic circuits. Each of the basic designs are explained with formulas to solve their component values and features.

### AMPLIFIER OR BUFFER

The circuit for an inverting amplifier, or an inverter, can be seen in Figure 1, above. The gain of the circuit is given by:

Av = - R2/R1

Note that the gain is negative one, indicating that the circuit operates as a phase-inverting voltage follower, if the two resistances are equal (i.e., R1 = R2). The output would be identical to the input, with the polarity reversed.

In reality, the resistors may be removed for unity gain and substituted by direct jumper wires, as shown in Fig. 2 below.

This is possible because R1 = R2 = 0 in this circuit. Typically, R3 is removed from the inverting voltage follower circuit.

The op amp output will amplify the input signal if R1 is less than R2. For instance, if R1 is 2.2 K and R1 is 22 K, the gain could be expressed as:

Av = - 22,000/2,200 = -10

The negative symbol denotes phase inversion. The input and output polarities are reversed.

By making R1 bigger than R2, the very same circuit may also attenuate (decrease the strength of) the input signal. For instance, if R1 is 120 K and R2 is 47 K, the circuit gain would be roughly:

Av = 47,000/120,000 = - 0.4

Again, the polarity of the output is the inverse of that of the input. Although R3's value is not particularly important, it should be about equal to the parallel combination of R1 and R2. Which is:

R3 = (R1 x R2)/(R1 + R2)

To demonstrate this, consider our previous example, where R1 = 2.2 K and R2 = 22 K. R3's value in this situation should be approximately:

R3 = (2200 x 22000)/(2200 + 22000) = 48,400,000/24,200 = 2000 Ω

We can choose the closest standard resistance value for R3 because the precise value is not necessary. A 1.8 K or a 2.2 K resistor could be used in this case.

The phase inversion created by the circuit in Fig. 2 may not be acceptable in several situations. To use the op-amp as a non-inverting amplifier (or like a simple buffer), connect it as illustrated in Fig. 3 below.

The gain in this circuit is expressed as follows:

Av = 1 + R2/R1

The output and input have the same polarity and are in phase.

Keep in mind that the gain must be always at minimum 1 (unity). It is not possible to attenuate (reduce) signals using a non-inverting circuit.

The gain of the circuit will be comparatively stronger if R2 value is significantly greater than R1. For instance, if R1 = 10 K and R2 = 47 K, the gain of the op amp will be as given below:

Av = 1 + 470,000/10,000 = 1 + 47 = 48

However, if R1 is significantly bigger than R2, the gain will only be somewhat more than unity. For example, if R1 = 100 K and R2 = 22 K, the gain would be:

Av = 1 + 22,000/100,000 = 1 + 0.22 = 1.22

In case the two resistances are identical (R1 = R2), the gain would always be 2. To convince yourself of this, try the gain equation in a few scenarios.

A specific situation is when both resistances are set to 0. In other words, as seen in Fig. 4 below, direct connections are used in place of the resistors.

The gain is exactly one in this case. This conforms to the gain formula:

Av = 1 + R2/R1 = 1 + 0/0 = 1

The input and the output are identical. Applications for this non-inverting voltage follower circuit include impedance matching, isolation, and buffer.

### ADDER (Summing Amplifier)

A number of input voltages could be added using an op amp. As illustrated in Fig. 5 below, input signals V1, V2,… Vn are applied to the op amp via resistors R1, R2,… Rn.

These signals are then combined to produce the output signal, which equals the sum of the input signals. The following formula may be used to calculate the op-amp's real performance as an adder:

VOUT = - Ro ((V1/R1) + (V2/R2) . . . + (Vn/Rn))

See the negative symbol. This means that the output has been inverted (the polarity is reversed). In other words, this circuit is an inverting adder.

The circuit may be changed to function as a non-inverting adder by switching the connections to the op-amp's inverting and non-inverting inputs, as illustrated in Fig. 6 below.

The output equation could be made simpler by assuming that all of the input resistors have identical values.

VOUT = - Ro ((V1 + V2 . . . + Vn)/R)

### DIFFERENTIAL AMPLIFIER

Fig. 7 above depicts the basic circuit of a differential amplifier. The component values are set so that R1 = R2 and R3 = R4. Therefore, performance of the circuit can be calculated using the following formula:

VOUT = VIN 2 - VIN 1

Only as long as the op amp can accept that inputs 1 and 2 have different impedances (input 1 has an impedance of R1 and input 2 has an impedance of R1 plus R3).

### ADDER/SUBTRACTOR

Figure 8 above depicts the configuration for an op amp adder/subtractor circuit. In the event when R1 and R2 have the identical values and R3 and R4 are likewise set to the same values, then:

VOUT = (V3 + V4) - (V1 - V2)

In other words, Vout = V3 + V4 is the total of the V3 and V4 inputs while it is the subtraction of the V1 and V2 inputs. The values for R1, R2, R3, and R4 are selected to match the characteristics of the op amp. R5 should be equal to R3 and R4, and R6 should be equal to R1 and R2.

### MULTIPLIER

Simple multiplication operations may be done with the circuit seen in Fig. 9 above. Keep in mind that this is the same circuit as in Fig. 1. To achieve a consistent gain (and subsequently a multiplication of the input voltage in the ratio R2/R1) and precise results, precision resistors with the prescribed values for R1 and R2 should be used. Notably, the output phase is inverted by this circuit. The voltage at the output will be equal to:

VOUT = - (VIN x Av)

where Av is the gain, as determined by R1 and R2. VOUT and VIN are the output and input voltages, respectively.

As seen in Fig. 10 above, the multiplication constant can be altered if R2 is a variable resistance (potentiometer). Around the control shaft you can mount a calibration dial with marks for various common gains. The multiplication constant may be read directly from this dial using a calibrated reading.

### INTEGRATOR

An op-amp will, at the very least, theoretically function as an integrator when the inverting input is coupled with the output through a capacitor.

As indicated in Fig. 11 above, a parallel resistor must be connected across this capacitor in order to maintain DC stability. This circuit implements the following relationship to integrate input signal:

R2's value should be selected to match the op amp parameters, such that:

VOUT = R2/R1 x VIN

### DIFFERENTIATOR

The differentiator op amp circuit includes a capacitor in the input line that connects to the inverting input and a resistor that connects this input to the output. However, this circuit has clear limits, therefore a preferable setup would be to parallel the resistor and capacitor as illustrated in Fig. 12 above.

The following equation determine how well this circuit performs:

VOUT = - (R2 x C1) dVIN/dt

### LOG AMPLIFIERS

The fundamental circuit (Fig. 13 above) employs an NPN transistor and an op-amp to generate an output proportional to the input's log:

VOUT = (- k log_{10}) VIN/RI_{o}

The "inverted" circuit, working as a fundamental anti-log amplifier, is depicted in the lower diagram. Typically, the capacitor is of low value (e.g., 20 pF).

### AUDIO AMP

An op amp, is essentially a dc amplifier but may also be applied for ac applications. A straightforward audio amplifier is shown in Figure 14 above.

### AUDIO MIXER

A modification of the audio amplifier is shown in this circuit (Fig. 15 above). You can see how it resembles the adder circuit in Fig. 5. The different input signals are blended or merged. Each input signal's input potentiometer allows for level adjustment. The relative proportions of the different input signals in the output can thus be adjusted by the user.

### SIGNAL SPLITTER

The signal splitter circuit seen in Fig. 16 above is just the opposite of a mixer. A single output signal is divided into several identical outputs that feed various inputs. The multiple signal lines are separated from one another using this circuit. To adjust the required level, each output line includes a separate potentiometer.

### VOLTAGE TO CURRENT CONVERTER

The circuit presented in Fig. 17 above will cause the load impedance R2 and R1 to experience the same current flow.

This current's value would be proportional to the input signal voltage and independent of the load.

However, due to the high input resistance provided by the non-inverting terminal, the current will be of relatively low value. This current has a value that is directly proportional to VIN/R1.

### CURRENT TO VOLTAGE CONVERTER

If the output voltage is equal to IIN x R2 and the design (Fig. 18 above) is used, the input signal current can flow straight via the feedback resistor R2.

To put it another way, input current is transformed into a proportionate output voltage.

The bias circuit created at the inverting input sets a lower limit on current flow, which prevents any current from passing through R2. To eliminate "noise," a capacitor can be added to this circuit as illustrated in the figure.

### CURRENT SOURCE

The above figure 19 shows how an op amp can be used like a current source. The resistor values can be calculated using the following equations:

R1 = R2

R3 = R4 + R5

The output current can be evaluated using the following formula:

Iout = (R3 x VIN) / (R1 x R5)

### MULTIVIBRATOR

You may adapt an op amp to use as a multivibrator. Fig. 20 above displays two fundamental circuits. The design at the top left is a free running (astable) multivibrator, whose frequency is controlled by:

A monostable multivibrator circuit that can be activated by a square wave pulse input could be seen in the lower right diagram. The component values provided are for a CA741 op amp.

### SQUARE WAVE GENERATOR

Fig. 21 above depicts a functional square wave generator circuit centered around an op amp. This square wave generator circuit could possibly be the most straightforward one. Just three external resistors and one capacitor are needed in addition to the op amp itself.

The two main elements that determine the circuit's time constant (output frequency) are the resistor R1 and the capacitor C1. However the R2 and R3-based positive feedback connection also has an impact on the output frequency. Although equations are often somewhat complicated, they can be made simpler for particular R3/R2 ratios. For illustration:

If R3/R2 ≈ 1.0 then F ≈ 0.5/(R1/C1)

or,

If R3/R2 ≈ 10 then F ≈ 5/(R1/C1)

The most practical method is to employ one of these standard ratios and change the values of R1 and C1 to achieve the required frequency. For R2 and R3, conventional values may be employed. For instance, the R3/R2 ratio will be 10 if R2 = 10K and R3 = 100K, thus:

F = 5/(R1/C1)

In most cases, we will already be aware of the required frequency, and we will only need to choose the appropriate component values. The simplest method is to first choose a C1 value that seems reasonable, and then rearrange the equation to find R1:

R1 = 5/(F x C1)

Let's look at a typical example of 1200 Hz frequency we're looking for. If C1 is connected to a 0.22uF capacitor, then R1 should have the value as depicted in the following formula:

R1 = 5/(1200 x 0.00000022) = 5/0.000264 = 18,940 Ω

A typical 18K resistor might be employed in the majority of applications. A potentiometer may be added in series with R1 to increase the usefulness and adaptability of this circuit, as illustrated in Fig. 22 below. This makes it possible to manually adjust the output frequency.

For this circuit, the very same calculations are used, however R1's value is changed to match the series combination of the fixed resistor R1a and the adjusted value of the potentiometer R1b:

R1 = R1a + R1b

The fixed resistor is inserted to ensure that the value of R1 never drops to zero. The range of output frequencies is determined by the fixed value of R1a and the highest resistance of R1b.

### VARIABLE PULSE WIDTH GENERATOR

A square wave is totally symmetrical. The square wave signal's duty cycle is defined as the ratio of high level time to total cycle time. Square waves have a 1:2 duty cycle by definition.

With just two more components, the square wave generator from the previous section may be transformed into a rectangle wave generator. Fig. 23 above depicts the updated circuit.

Diode D1 restricts the passage of current via R4 on negative half cycles. R1 and C1 make up the time constant as expressed in the following equation:

T1 = 5/(2C1 x R1)

However, on positive half-cycles, the diode is allowed to conducts, and the parallel combination of R1 and R4 along with C1 defines the time constant, as shown in the following calculation:

T2 = 5/(2C1 ((R1 R4)/(R1 + R4)))

The overall cycle length is just the total of the two half-cycle time constants:

Tt = T1 + T2

The output frequency is the inverse of the total time constant of the whole cycle:

F = 1/Tt

Here the duty cycle will not equal 1:2 because the time constant for the high and low level sections of the cycle will differ. Asymmetrical waveforms will be produced as a result. It is possible to make R1 or R4 adjustable, or even both of them, but be aware that doing so would change both the output frequency and duty cycle.

### SINE WAVE OSCILLATOR

The sine wave, which is shown in Fig. 24 below, is the most basic of all ac signals.

There is absolutely no harmonic content in this extremely pure signal. There is just one fundamental frequency in a sine wave. Actually, creating a completely pure, distortion-free sine wave is rather difficult. Thankfully, using an oscillator circuit built around an op-amp, we can get pretty near to an optimal waveform.

Fig. 25 above depicts a conventional sine wave oscillator circuit incorporating an op-amp. A twin-T circuit serving as a band-reject (or notch) filter serves as the feedback network. The capacitor C1 and the resistors R1 and R2 make up one T. C2, C3, R3, and R4 make up the other T. The schematic has it reversed. The component values must have the following relationships for this circuit to work properly:

The following formula determines the output frequency:

F = 1/(6.28 x R1 x C2)

By changing R4's value, the twin-T feedback network tuning could be tweaked somewhat. Typically, this could be a tiny trimmer potentiometer. The potentiometer is set to its highest resistance and then gradually reduced until the circuit just hovers on the verge of oscillation. The output sine wave might get corrupted if the resistance is adjusted too low.

### SCHMITT TRIGGER

Technically speaking, a Schmitt trigger may be referred to as a regenerative comparator. Its primary function is to transform an input voltage that is slowly changing into an output signal, at a particular input voltage.

To put it another way, it has a "backlash" property called hysteresis which functions like a voltage "trigger." The op amp becomes the basic building block for the Schmitt trigger operation (see Fig. 26 above). The following factors determine the triggering or trip voltage:

V_{trip} = (V_{out} x R1) / (-R1 + R2)

In this type of a circuit, the hysteresis is double the trip voltage.

In Fig. 27 below, another Schmitt trigger circuit is depicted. In this circuit, the output is said to be "triggered" when the dc input hits about one-fifth of the supply voltage.

The supply voltage may be anywhere between 6 and 15 volts, therefore depending on the supply voltage chosen, the trigger can be set to operate at 1.2 to 3 volts. If necessary, the actual triggering point could also be altered by modifying the value of R4.

The output will be the same as the supply voltage as soon as it is triggered. If the output is attached to an incandescent bulb or LED (through a series ballast resistor), the lamp (or LED) will illuminate once the input voltage hits the triggering value, indicating that this precise voltage level has been achieved at the input.

### BUFFERED VOLTAGE REFERENCE GENERATOR

This useful little circuit compels the output of the opamp to be set at a value defined by the supply voltage and the voltage divider made up of RI and R2.

By using this functionality, you could provide other circuits a constant bias voltage reference. In the example, the output voltage is equal to 50% of the input voltage.

The 50% or 1/2 the supply voltage value at the output is due to the use of identical resistors for R1 and R2.

You could change the R1 and R2 values to get any other desired output reference voltage. The output is fully buffered by the op amp.

### BURGLAR ALARM

This circuit uses an op amp to configure a closed loop alarm system with a buzzer. The inverting input of the op amp is rigged at 1/2 the supply voltage, while the non-inverting input is connected to the ground line through a "normally-closed" push button switch. This switch works like a intruder detector.

In the above situation while the switch is depressed and closed, the output of the op amp is low or at zero volts.

When a potential burglar or an intruder steps over the switch, the switch opens. As soon as this happens, the non-inverting input pin#3 of the IC goes higher than the pin#2 potential. Due to this the output of the op amp turns high equal to the supply DC level and it sounds the alarm or the connected buzzer.

A large value capacitor could be connected across the base/emitter of the 2N2222 to ensure that the buzzer keeps sounding even after the push switch is returned to its original condition.

### PEAK VOLTAGE DETECT AND HOLD

Our next basic op amp circuit will detect the peak value of a varying input voltage, and then hold it so that the peak level can be analyzed with a meter.

Opamp IC1a is wired as a high impedance voltage follower circuit. This is because its positive input pin 3 is linked to the modulating input signal.

When the input voltage reaches its maximum value, the peak voltage is applied across C1, charging it to the maximum.

Diode D1 prevents the voltage from discharging back into IC1a by allowing only the positive peak voltage to reach C1.

Opamp IC1b is another voltage follower and generates the maximum output at pin #7. This may be evaluated using a digital or analogue voltmeter.

Now, as the input level begins dropping below the peak value, C1 begins to discharge gradually. To obtain the most precise peak reading, the output measurement must be carefully monitored.

### Basic Timer Circuit using Op amps

Our next design is a basic op amp based timer circuit. A couple of op amps IC1a and IC1b from the IC LM358 can be seen in the circuit.

IC1a is configured as a voltage follower and the basic timer circuit. IC1b is configured in a detector/comparator arrangement.

When the switch S1 in the OFF position, the pin#3 of IC1a remains grounded or at zero volts. IC1a being a voltage follower replicates its input zero volt at its output pin#1. Thus pin#1 also stays at 0V.

This causes pin#6 of IC2 to remains at 0V, causing the non-inverting pin#5 to remain at a higher potential than pin#6.

Since IC1b is rigged as a comparator, a higher potential at pin#5 causes its output to go high.

This in turn allows the transistor and the relay to remain switched ON. In this situation the load also remains switched ON.

To start the timer action we have to now move the S1 to the "start timer" position.

As soon as S1 is moved to "start timer" position, C1 begins charging via R1. The potential at pin#3 of IC1a now slowly starts rising and the same is replicated on its pin#1 and pin#6 of IC1b.

Resistor R2 and R3 being identical ensures that the potential at pin#5 of IC1b is fixed 1/2 of the supply voltage. Meaning, the potential at pin#5 will be 6 V.

Now, as C1 charges the potential at pin#6 also rises slowly, until it exceeds the 6 V mark. When the pin#6 potential exceeds the pin#5 potential, the output of IC1b pin#7 instantly turns low or 0V.

When this happens the transistor Q1 is switched OFF. This likewise switches of the relay and the connected load.

The value of R1 and C1 can be tweaked as desired to change the timing period from a few seconds to many minutes.

### MOTOR SPEED AND DIRECTION CONTROLLER USING A SINGLE OP AMP

The next design above is a very fundamental op amp circuit which not only regulates the speed of a motor, but also determines its direction of rotation.

The op-amp here is yet again configured in a voltage follower mode. Its positive input at pin #3 hooked up with the pot R3 which serves as a motor speed controller and also as a direction controller.

The output of the op-amp will be close to zero whenever the wiper of the pot R3 is in the middle of its range. In this position neither Q1 nor Q2 is switched on.

The op amp output turns positive as soon as the pot wiper is moved toward the positive side. When this happens and Q1 begins transferring current to the motor.

The output of the op-amp swings to a negative voltage when the potentiometer is adjusted toward the negative supply, shutting off Q1 and turning on Q2. This action also changes the rotating direction of the motor.

The speed of the motor rises in the direction it is spinning, depending on how far the R3 wiper is pulled toward either end of the potentiometer. To find the maximum DC voltage range acceptable for the selected motor, you may have to monitor the voltage variation on emitters of Q1 and Q2.

### SIMPLE MICROPHONE AMPLIFIER

Our next basic op amp circuit can be seen in the above figure. It is an electret condenser microphone amplifier circuit. In this design, the op amp is operated in an inverted adjustable gain amplifier circuit with a low potential electret microphone feeding the input.

Take note that this amplifier circuit is powered by a single 9-volt source (which also provides a necessary level of DC via the microphone).

Using potentiometer R4, you may change the gain of the circuit from a low gain of 10 to a high gain of 110 (when R4 is adjusted to its maximum resistance of 100k -ohm).

#### Wrapping up

So these were a few op amp basic circuits with their parameters explained. Hope you have understood all the characteristics and formulas related to an op amp.

If you have any other basic op amp circuit design that you think needs to be included in the above article, please feel free to mention them through your comments below.

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