Operational Amplifiers Operational Amplifier was originally used for D.C. Amplifiers which perform mathematical operations such as summation, subtraction,

Operational AmplifiersOperational AmplifiersOperational Amplifier was originally used for D.C. Amplifiers which perform mathematical operations such as summation, subtraction, integration, and differentiation in analog computers.

Op-Amp is a high gain amplifier. Output is taken from only one terminal. Vin small, Vo high. So Vo/ Vin = high

Used to amplify both A.C and D.C signals.

Its applications are,A.C. to D.C. conversion, Analog-to-Digital and Digital-to-Analog conversion, Signal conditioning, Active filters, and Many other applications.www.bookspar.com | Website for

Students | VTU NOTES | QUESTION PAPERS

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Operational AmplifiersOperational AmplifiersThe Ideal Operational Amplifier:

It has two input terminals 1(V1) and 2(V2) and output terminal 3(V0).

Since the Op-amp is an active device, it requires a D.C. power supply for its operation. As shown in figure a positive supply +VCC is connected to terminal 4 and a negative supply –VEE is connected to terminal 5 (Gnd ???).

The Op-amp is also known as a differential amplifier. This means that the Op-amp, senses the difference between the input signals applied at terminals 1 and 2 and amplifies the difference by an amount A.www.bookspar.com | Website for


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Operational AmplifiersOperational AmplifiersThe output Vo is therefore given by

Vo = A(V2 – V1)

Where V1 and V2 are the input signals and the constant ‘A’ is known as the open-loop gain (There is no feedback loop connecting the output terminal to the input terminals).

Properties of an ideal Op-Amp:1.Infinite Input Impedance: The ideal Op-amp does not draw any current from the voltage sources connected to its input terminals. This implies that the input impedance of an Op-amp is infinity.

2.Zero Output Impedance: The voltage at the output terminal is independent of the current drawn from it i.e. output impedance is zero. Hence the Op-amp can drive an infinite number of devices.www.bookspar.com | Website for


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Operational AmplifiersOperational Amplifiers3. Infinite Bandwidth: This implies that the amplifier can amplify

any frequency from zero to infinity without attenuation. In other words, the ideal Op-amp will amplify signals of any frequency with equal gain.

4. Infinite Voltage Gain: The open-loop voltage gain of an ideal Op-amp is very large, i.e., infinity.

5. Perfect Balance: The output voltage is zero when equal voltages are present at the two input terminals.

6. Infinite CMRR : This means that the output common-mode noise voltage is zero.

7. Infinite Slew Rate: Slew rate indicates the rapidity with which the output of an Op-amp changes in response to the changes in input frequency. (how fast the output of op-amp is going to respond for any change in input)

8. Temperature: The characteristics do not change with temperature.www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

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Operational AmplifiersOperational AmplifiersIn the figure, two signals V1 & V2

are fed to the input of an operational amplifier. One of the properties of an ideal Op-amp is that no current should enter the terminals 1, & 2, i.e., il and i2 are both zero as shown in figure.

It is observed that in the expression for the output V0, the input signal at terminal 1 appears with a negative sign attached to it. On the other hand, the input signal at terminal 2 appears without alteration in its sign.

Terminal 1 is therefore known as the Inverting Terminal while Terminal 2 is called the Non-inverting Terminal. www.bookspar.com | Website for


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Operational AmplifiersOperational AmplifiersSaturable Property of an Op-Amp:‘A’ is a constant for a particular Operational Amplifier and is known as the open-loop gain since there is no feedback loop connecting the output terminal to the input terminal, which is the basic circuit of an op-amp.

This op-amp senses the difference between the input signals and amplifies the difference between the input signals.

The output Vo is given by

Vo = A*(V2 - V1) ---(i)

Hence, if V1= 0, Vo = A*V2 ----------(ii)

and if V2= 0, Vo = - A*V1 --------(iii)www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

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Operational AmplifiersOperational AmplifiersThe open-loop voltage gain of an ideal op-amp is infinity; then,

substituting A = ∞ in eq’s (ii) & (iii) above, the output Vo of the op-amp should range from +∞ to -∞.

However, in practice, Vo is limited by the magnitudes of the power supply voltages. If the supply voltage are ± 15V, (Terminals 4 & 5) V0 would be about ±10 V.

When the output attains this level, it does not increase any further even if the input voltages are increased. This op-amp is now said to be saturated.

www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

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Operational AmplifiersOperational AmplifiersOp-Amp as Comparator:A comparator is a device which compares a signal voltage with a reference voltage. An Op-Amp comparator is an open loop Op-Amp. The reference voltage is applied to one of its input terminals, and the signal to be compared is applied to the other input terminal. Depending upon which of the two voltages is greater, the output voltage is held at the positive or negative saturation voltage.


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Operational AmplifiersOperational AmplifiersA time-varying signal Vin is applied to the non-inverting terminal through a resistor R, and the reference voltage Vref of 1 V is applied to the inverting terminal through resistor R1.

When Vin < Vref the voltage at the inverting (-) terminal is greater than the voltage at the non-inverting (+) terminal and hence Vo = -Vsat it being approximately equal to -VEE.

When Vin > Vref the voltage at the non-inverting (+) terminal is greater than the voltage at the inverting (-) terminal and hence Vo = + Vsat it being approximately equal to +Vcc Therefore, when Vin crosses Vref the output voltage V0 changes instantaneously from one saturation level i.e. from +Vcc to -VEE or from -VEE to +Vcc to other and waveforms are shown in the figure.www.bookspar.com | Website for


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Operational AmplifiersOperational Amplifiers


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Operational AmplifiersOperational AmplifiersThe Inverting Op-Amp Circuit:An Inverting Op-Amp circuit is one whose output is out of phase by 180o with respect to the input.

The point G is called the ‘virtual ground’. since G is not directly connected to the ground, no current from G flows to the ground. Also, the voltage drop across R1 makes the potential V at G to be nearly zero. This is the reason G is referred to as Virtual Ground (not the true ground)


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Operational AmplifiersOperational AmplifiersBecause of the infinite input impedance of the Op-Amp, no current enters the Op-Amp. Hence the same current flows through the feedback resistor Rf.

where Vo is the output voltage

The ratio Vo/Vi is the closed loop voltage gain VCL of the circuit. As there is a feedback loop, the gain is known as the closed loop gain. We see that it differs from A, the open-loop gain and depends only on the values of the external resistances R1and Rf. The minus sign (-) indicates that the output is inverted with respect to the input. Thus the above circuit represents an inverting amplifier.www.bookspar.com | Website for


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Operational AmplifiersOperational AmplifiersThe Non-Inverting Op-Amp Circuit:A non-inverting Op-Amp circuit is one whose output is in phase with the input. Figure shows an Op-Amp circuit with a small input Vi applied to the non-inverting terminal. The inverting terminal is connected through a resistor R1 to the ground. Rf is the feedback resistor.

Because of infinite gain of Op-Amp, practically no voltage drop exists between the input terminals. Hence the potential at G is also taken to be Vi. Further, assuming no current flows through the Op-amp, the same current ‘i’ flows through R1and Rf. Thus we havewww.bookspar.com | Website for


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Since the gain is positive, no phase change occurs for the signal.Thus the above circuit represents a non-inverting amplifier.


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Operational AmplifiersOperational AmplifiersProperties of an Practical Op-Amp:1.Input Impedance: This is the ratio of the input voltage change to input current change, measured at one input terminal. Input impedance may be in the range of 10 KΩ to 100 MΩ.2.Output Impedance: This is the ratio of the output voltage change to output current change. Output impedance could be in the range of tens to hundreds of ohms. (75 Ω for µA-742 IC).3.Gain: Though an ideal Op-amp has infinite gain, a practical Op-amp has a gain of the order of 5 x 104.4.Common Mode Rejection Ratio (CMRR): The relative sensitivity of an Op-amp to a difference signal as compared to a common mode signal is called the common-mode rejection ratio, and is given by


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For commercial Op-amps the CMRR lies in the range 60 to 100 dB.

5. Output Voltage Swing: This is the peak output voltage with respect to zero available at the output without distortion. It is a function of the supply voltage. Thus, in an amplifier operating between supply voltages +6 and -6 V, the peak-to-peak undistorted output swing might be from -4 V to +4V (in practice the allowable voltage swing is not always symmetrical).6. Input Common-mode Voltage Swing: This is the maximum range of input voltage that can be simultaneously applied to both inputs without causing cut off or saturation of amplifier stages.

C

D

A

ACMRR 10(log) log20


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Operational AmplifiersOperational Amplifiers7. Input Offset Current (Iio): This is the difference of the currents into

the input terminals with the output at zero volts.

8. Input Offset Voltage: Whenever both the input terminals of the Op-amp are grounded, ideally, the output voltage should be zero. However, in this condition, the practical Op-amp shows a small non-zero output voltage. To make this output voltage zero, a small voltage in milli volts (1-4) is required to be applied to one of the input terminals. Such a voltage makes the output exactly zero. This d.c. voltage, which makes the output voltage zero, when the other terminal is grounded is called input offset voltage denoted as Vio.

9. Input Bias Current: Input bias current can be defined as the current flowing into each of the two input terminals when they are biased to the same voltage level, i.e., when the Op-amp is balanced.


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Operational AmplifiersOperational Amplifiers10. Power-supply Voltage-rejection ratio: A change in supply voltage

ΔVCC will produce a change ΔVo in the amplifier output. The power-supply voltage-rejection ratio ΔVo/ΔVCC is usually specified for the condition that the difference voltage input Vi = 0. Typically this ratio is in the range 10-5 to 7 x 10-5.

11. Frequency Response: Op-amps are usually low-frequency devices. The open-loop (without feed back) 3 dB bandwidth of general-purpose Op-amps will be in the region of 100 Hz or less. Bandwidths as low as 10Hz are possible.

12. Slew Rate: Slew Rate is one of the most important specifications of an Op-amp because it limits the size of the output voltage at higher frequencies.www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

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Operational AmplifiersOperational AmplifiersThe charging current in a capacitor is given by i = C dV/dt

OR i/C = dV/dtThe above equation says that the rate of change of voltage is equal to the charging current divided by the capacitance. The larger the charging current, the faster the capacitor charges. If the charging current is limited to a maximum value, the rate of change of voltage is also limited to a maximum value.


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Operational AmplifiersOperational AmplifiersThe rate of change of voltage with respect to time is

1. If Imax = 80 µA and Cc = 40 pF the maximum rate of voltage change is

This shows that the output voltage across the capacitor changes at a maximum rate of 2V/µS. The voltage cannot change faster than this unless Imax is increased or Cc is decreased.Slew Rate is defined as the maximum rate of output voltage change.

Slew rate = Imax/CCwww.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

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Operational AmplifiersOperational AmplifiersModes of Operation: (Ideal)1.Single-ended mode: Single-ended input operation results when the input signal is connected to one input with the other input connected to ground.

Figure-1 shows the input is applied to the plus input (with minus input at ground), which results in an output having the same polarity. as the applied input signal. Figure-2 shows an input signal applied to the minus input, the output then being opposite in phase to the applied signal.www.bookspar.com | Website for


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Operational AmplifiersOperational Amplifiers2. Differential mode:Op-Amp has the ability to greatly amplify signals that are opposite at the two inputs, while only slightly amplifying signals that are common to both inputs.

In differential mode, the two input signals are equal but have opposite polarity at every instant of time. Thus referring to figure, Vi1 = -Vi2. So Differential input VDM = (Vi1 - Vi2)


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Operational AmplifiersOperational Amplifiers3. Common Mode: In this case, the two input signals are identical both in amplitude & phase at every instant of time, and the circuit is said to be operating in the common mode. The input signals are called common mode signals.

Therefore, VCM = ½ * (Vi1 + Vi2) i.e. either Vi1 or Vi2 (or average of both).


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Operational AmplifiersOperational AmplifiersSince amplification of the opposite input signals is much greater than that of the common input signals, the circuit provides a common mode rejection as described by a numerical value called the Common-Mode Rejection Ratio (CMRR).

The overall operation being to amplify the difference signal while rejecting the common signal at the two inputs. Since noise (any unwanted input signal) is generally common to both inputs, the differential connection attenuates this unwanted input while providing an amplified output of the difference signal applied to the inputs.

V1 = (+Vin) + N, V2 = (-Vin) + N, Vo = V1 – V2 = [(+Vin) + N – ((-Vin) + N)] = 2Vin

+Vin

-Vin


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Operational AmplifiersOperational AmplifiersHere Common mode signals is ‘N’ i.e. Noise.Differential mode signals are +Vin and –Vin.So Common mode signals are cancelled out and differential mode signals are amplified.

1. VDM = (Vi1 - Vi2)

Vo = AD VDM

2. VCM = ½ * (Vi1 + Vi2)

Vo = AC VCM

Since any signals applied to an op-amp in general have both in-phase and out-of phase components, the resulting output can be expressed as

Total Vo = (ADVDM + ACVCM)

Where VDM = difference mode input voltage VCM = common mode input voltage AD = differential mode gain of the amplifier AC = common-mode gain of the amplifierwww.bookspar.com | Website for


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Operational AmplifiersOperational AmplifiersOpposite Polarity Inputs:If opposite polarity inputs applied to an op-amp are ideally opposite signals, Vi1 = -Vi2 = Vs, the resulting difference voltage is VDM = Vi1 -Vi2 = Vs - (-Vs) = 2Vs

While the resulting common voltage isVCM = ½ * (Vi1 + Vi2) = ½ * [Vs + (-Vs)] = 0

so that the resulting output voltage is

This shows that when the inputs are an ideal opposite signal (no common element), the output is the differential gain times twice the input signal applied to one of the inputs.

Vo = ADVDM + ACVCM = AD 2Vs + AC 0 = 2 ADVs


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Operational AmplifiersOperational AmplifiersSame Polarity Inputs:If same polarity inputs applied to an op-amp Vi1 = Vi2 = Vs,

the resulting difference voltage is VDM = Vi1 -Vi2 = Vs - (Vs) = 0

While the resulting common voltage isVCM = ½ * (Vi1 + Vi2) = ½ * [Vs + Vs] = Vs

so that the resulting output voltage is

This shows that when the inputs are ideal in-phase signals (no difference signal), the output is the common-mode gain times the input signal Vs, which shows that only common-mode operation occurs.

Vo = ADVDM + ACVCM = AD 0 + AC Vs = AcVs


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Operational AmplifiersOperational AmplifiersCommon-Mode Rejection:The solutions above provide the relationships that can be used to measure AD and AC in op-amp circuits.

To measure AD: Set Vi1 = -Vi2 = Vs = 0.5 V, so that

VD = (Vi1 - Vi2) = [0.5 V - (-0.5 V)] = 1V

VC = ½ * (Vi1 + Vi2) = ½ [0.5 V + (-0.5 V)] = 0V

Under these conditions the output voltage isVo = ADVD + ACVC = AD(1 V) + AC(0) = AD

Thus, setting the input voltages Vi1 = -Vi2 = 0.5V results in an output voltage numerically equal to the value of AD.

(i.e. AD = 2*Vs = 2*Vi1 or 2*Vi2)


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Operational AmplifiersOperational AmplifiersTo measure AC: Set Vi1 = Vi2 = Vs = 1V, so that

VD = (Vi1 - Vi2) = [1V – 1V] = 0V

VC = ½ * (Vi1 + Vi2) = ½ [1V + 1V] = 1V

Under these conditions the output voltage isVo = ADVD + ACVC = AD(0 V) + AC(1 V) = AC

Thus, setting the input voltages Vi1 = Vi2 = 1V results in an output voltage numerically equal to the value of AC. (i.e. AC = Vs = Vi1 or Vi2)

C

D

C

D

A

ACMRR

A

ACMRR

10log20(log)www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

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Operational AmplifiersOperational AmplifiersOp-Amp Applications:1.Comparator:2.Voltage follower:The input voltage vi is applied to the non-inverting input, and the inverting input is grounded through a resistor R. The input and output are shorted, as shown in figure.

We can show that this Op-Amp gives an output voltage which is equal to the input voltage, and is in phase with it.


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Operational AmplifiersOperational AmplifiersAs the Op-Amp has practically infinite input impedance and zero current through the virtual short at N, the potential at N is the same as the potential at the non-inverting input. But the voltage (with respect to ground) of this terminal (i.e., non-inverting input) is Vi

Vo = Vi

Hence the output voltage is equal to the input voltage. Also, it is in phase with the input voltage. As the output voltage follows the input voltage at all instants, this Op-Amp is called Voltage Follower.

3. Summing Amplifier: (Adder)The three inputs to be added are connected to the inverting input as shown. The non-inverting terminal is connected to the ground. This op-amp summing amplifier is a feedback amplifier in which the feedback is provided through Rf. Point G is at virtual ground.


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Since G is at virtual ground,if = il + i2 + i3 ---- (1)

where i1 i2 and i3 are the currents through input resistors R1, R2 and R3 respectively.


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Operational AmplifiersOperational AmplifiersSubstituting these above values in eq (i) we get,

An averaging circuit is a special case of a summing amplifier.Suppose R1 = R2 = R3 = R,


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Operational AmplifiersOperational Amplifiers4. Subtractor:The subtraction of 2 input voltages is possible by using an Op-amp circuit called the subtractor. Its operation is similar to that of a summing amplifier.

In order to determine the output of the subtractor, the principle of superposition is used. (Analyze one input by grounding the other input).

Vo = (V2 – V1)


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Operational AmplifiersOperational AmplifiersFirst V1 will be analyzed by grounding V2.So Op-Amp becomes inverting amplifier. V01 = (-Rf/R1) * V1

If Rf = R1 then V01 = -V1

Second V2 will be analyzed by grounding V1.So Op-Amp becomes Non-inverting amplifier.At point ‘N’ the potential will be (Voltage divider theorem)

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2

22

2

VV

RRif

VRR

RV

N

f

fN


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Operational AmplifiersOperational AmplifiersFor Non-inverting amplifier,V02 = [1 + (Rf/R2)] * VN (Here Rf = R2)V02 = 2 VN = 2 (V2/2) = V2

So total Vo = V01 + V02 Vo = -V1 + V2

i.e. Vo = (V2 – V1) (Subtractor)

The output voltage is equal to the voltage V2 applied to the Non-inverting terminal minus the voltage V1 applied to the inverting terminal. Hence the circuit is called “Subtractor”.


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Operational AmplifiersOperational Amplifiers5. Integrating Circuit: (Integration amplifier)An integrator circuit is one whose output is the integral of the input. It is obtained by using a basic inverting amplifier configuration if the feedback resistor ‘Rf’ is replaced by a capacitor ‘Cf’.

The charge on the capacitor is Q = C * VQ = C (0 - Vo) where ‘0’ is the potential at G (virtual ground).


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Operational AmplifiersOperational AmplifiersEquate the two equations

But i = Vi/R, where ‘Vi’ is the input voltage and ‘R’ resistor through which the input is fed.

Thus we see that the output voltage is the integral of the input voltage and hence the circuit functions as an integrator.


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Operational AmplifiersOperational Amplifiers6. Differentiating Circuit:A differentiator circuit is one whose output is the differential coefficient of the input. Figure shows a differentiator circuit using an Op-Amp. ‘G’ is the virtual ground.

The input voltage ‘Vi’ is applied to the inverting terminal of the Op-Amp through the capacitor. The non-inverting terminal is connected to the ground.


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Operational AmplifiersOperational AmplifiersIf ‘q’ is the charge acquired by the capacitor then,Vi = q/CDifferentiating equation w.r.t. ‘t’ we get

The charging current of the capacitor is

but i = dq/dt


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Operational AmplifiersOperational AmplifiersEquating these two equations

but

Substitute for dq/dt

Output = (Constant) x (Differential coefficient of the Input)


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Operational AmplifiersOperational Amplifiers1. An Inverting Amplifier has R1 = 20 KΩ and Rf = 100 KΩ. Find the Output Voltage, the Input Resistance and the Input Current for an input voltage of 1V.

2. Design an inverting amplifier with an input resistance of 30 KΩ and a closed loop gain of -10.

ACL = -10, R1 = 30 KΩ

Rf = 300 KΩwww.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

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Operational AmplifiersOperational Amplifiers3. An inverting amplifier of figure has a load resistor of 50 KΩ connected to its output. If the input voltage is 0.5 V, find the load current, output voltage and input current.

(i) Input current through R1 is,


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Operational AmplifiersOperational Amplifiers(ii) Output voltage

i = (0-Vo)/Rf

Vo = -iRf

Vo = 5V

(iii) Load current

4. Determine the closed loop voltage gain of the Op-Amp circuit shown in figure. Also calculate the amplitude of the output voltage if Vi = 0.2 sinωt. (Check the question)


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Operational AmplifiersOperational AmplifiersThe figure is that of an inverting amplifier

= -5 (negative sign indicates phase shift i.e. inverting configuration)

Vi =Vm sin ωt, = 0.2 sin ωt here Vm = 0.2

While determining the amplitude of the output voltage, it is not required to consider the negative sign of the gain.

Vo = 5*0.2 = 1 V

5. Design an amplifier to have a gain of + 10 using a single Op-amp.

As the gain is positive, the amplifier is a non-inverting amplifier.www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

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Operational AmplifiersOperational AmplifiersAs both R1 and Rf are external resistors, they can be selected arbitrarily.

Let R1 = 30 KΩ then, Rf = 270 KΩ

6. For the circuit shown in figure determine the output voltage.

The circuit of figure is that of an inverting adderwww.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

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7. Find the output voltage of a three-input adder circuit in which R1= R2 = R3 = 4 KΩ and the feedback resistance Rf = 6 KΩ . Given V1 = -4V, V2 = - 2V and V3 = 3V


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Operational AmplifiersOperational Amplifiers8. Design a scaling adder circuit using an Op-Amp, to give the output.

Vo = -(3Vl + 4V2 + 5V3)

For an inverting summing amplifier,


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Operational AmplifiersOperational AmplifiersLet us arbitrarily choose Rf =120KΩ

Comparing the above equation with the one given in the problem, we have,


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P – 462 M.V. Rao Q.6(b)P – 478 M.V. Rao Q.6(b)P – 496 M.V. Rao Q.6(a)P – 503 M.V. Rao Q.6(c)


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Operational AmplifiersOperational Amplifiers1. Differentiate ideal op-amp and practical op-amp. Write at least 6-8 differences.

2. Explain (a) Integrating Amplifier (b) Differentiating Amplifier

3. Find the output voltage of a three-input adder circuit in which R1= R2 = R3 = 4

KΩ and the feedback resistance Rf = 6 KΩ. Given V1 = -4V, V2 = - 2V and V3 =

3V

4. Design a scaling adder circuit using an Op-Amp, to give the output.Vo = -(3Vl + 4V2 + 5V3)

5. The gain of voltage follower is ----------A) Unity B) Zero C) Infinity D) None

6. The maximum rate at which amplifier output can change in volts per microsecond (V/µs) is calledA) Over rate B) Slew rate C) Under rate D) Nonewww.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

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Operational AmplifiersOperational Amplifiers1. Differentiate ideal op-amp and practical op-amp. Write at least 6-8

differences.

2. Explain (a) Inverting Amplifier (b) Non-Inverting Amplifier

3. Find the output voltage of a three-input adder circuit in which R1=

R2 = R3 = 4 KΩ and the feedback resistance Rf = 6 KΩ. Given V1 =

-4V, V2 = - 2V and V3 = 3V


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Operational AmplifiersOperational AmplifiersQ. Draw the non-inverting voltage amplifier circuit using an op-amp and show that the closed-loop voltage gain is given by

Where AV = open-loop voltage gain of an op-amp, β = feedback factor.

)1( V

VVf A

AA

A non-inverting Op-Amp circuit is one whose output is in phase with the input. Figure shows an Op-Amp circuit with a small input Vi applied to the non-inverting terminal. The inverting terminal is connected through a resistor R1 to the ground. Rf is the feedback resistor.


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Because of infinite gain of Op-Amp, practically no voltage drop exists between the input terminals. Hence the potential at G is also taken to be Vi Further, assuming no current flows through the Op-amp, the same current ‘i’ flows through R1 and Rf.


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Since the gain is positive, no phase change occurs for the signal. The expression in eq-(2) was derived on the assumption of infinite open-loop amplifier gain AV.


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Operational AmplifiersOperational AmplifiersFor a finite value of Av, from figure, the differential voltage amplified by the op-amp is


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Documents

Operational Amplifiers Operational Amplifier was originally used for D.C. Amplifiers which perform mathematical operations such as summation, subtraction,