KIGALI INSTITUTE OF SCIENCE AND TECHNOLOGY

KIST

FACULTY OF ENGINEERING

DEPARTMENT OF ELECTRICAL AND ELECTRONICS

PROGRAM OF ELECTRONICS AND TELECOMMUNICATION

YEAR 3, SEMESTER II

ACADEMIC YEAR 2011

COURSE: LINEAR INTEGRATED CIRCUITS AND ITS APPLICATIONS (ETE 3321)

LAB REPORT

DONE BY: Odifax MUNYANEZA

- GS20090178

EXPERIMENT No.2: A.SUMMING AMPLIFIER

AIM:1. To demonstrate the operation of inverting amplifier using 741 Op-Amp as a summing amplifier.2. To measure the slew rate of 741 Op-Amp IC.

EQUIPEMENTS REQUIRED:

(a)LT SPICE(b)COMPUTER

COMPONENTS REQUIRED:

LT SPICE COMPONENTS LIKE: Resistors, Op-Amp 741

SPICE SIMULATION R EQUIREMENTS AND PROCEDURESUse Resistors, the Op-Amp IC 741 designed in last LAB, and connect them as in the circuits below, then simulate them.

THEORY:

Summing Amplifier:Op-amp may be used to design a circuit whose output is the sum of several inputSignals. Such a circuit is called a Summing Amplifier.

Inverting Summing Amplifier:A typical summing amplifier with three input voltages V1, V2 and V3, three inputResistors R1, R2 and a feedback resistor Rf is shown in fig. The following analysis isCarried out assuming that the op-amp is an ideal one, that is, AOL = ∞and Ri =∞. Since, the input bias current is assumed to be zero, there is no voltage drop across the resistor Rcomp and hence the non-inverting input terminal is at ground potential.The voltage at node ‘a’ is zero as the no inverting input terminal isGrounded. The nodal equation by KCL at node ‘a’ is

V1/R1V2/R2V3/R3Vo/Rf0 OR Vo -(Rf.V1/R1Rf.V2/R2Rf.V3/R3)

Thus, the output is an inverted, weighted sum of the inputs. In the special case, whenR1 = R2 = R3 = Rf, we have

Vo -(V1+V2+V3)` In which case the output Vo is the inverted sum of the inputs signals. We may also set

R1 = R2 = R3 = 3Rf

In this case Vo V1 V2 V3)/3Thus, the output is the average of the input signals (inverted)

PROCEDURES1. Construct the circuit in LTSPICE2. Apply the given inputs and observe the output on CRO.3. Calculate the amplitude & frequency of the output and capture it in your report.4. Vary the values of inputs to vary the gain and observe the output.5. Calculate the gain for various values of inputs.6. Plot the inputs and outputs waveforms for various inputs on the graph paper.

.CIRCUIT DIAGRAM:

Fig.1: Summing Amplifier Where R1=R2=R3=Rf

Fig.2: InputsOutput Waveforms

CIRCUIT DIAGAM:Where R1=R2=R3=3Rf

InputsOutput Waveforms

CIRCUIT DIAGRAM: Where R1, R2, R3 and Rf are different

InputsOutput Waveforms

B.SLEW RATEThe slew rate is defined as the maximum rate of change of output voltage caused by a

Step input voltage and is usually specified in V/μs. For example, a 1 V/μs slew rateMeans that the output rises or falls by 1V in one microsecond. An ideal slew rate isInfinite which means that op-amp’s output voltage should change instantaneously inResponse to input step voltage. Practical IC op-amps have specified slew rates from 0.1V/μs to well above 1000 V/μs.

PROCEDURE:1. Construct the circuit in LTSPICE2. Apply the sine input and increase the frequency until the output gets distorted.3. Calculate the slew rate by using SR= (2fπVm/106 ) V/μsWhere Vm=Peak output amplitude in volts F=frequency in Hz4. Vary the inputs (square & triangular wave) and observe the frequency & amplitude toCalculate the slew rate.

CIRCUIT DIAGRAM: Voltage Follower to measure slew rate

Input Waveform of a voltage follower

Output Waveform of Voltage Follower

Observation table:

For Summing Amplifier:

S No. Resistors Input Output

Voltage Frequency Voltage

Frequency

R1 R2 R3 Rf V1

V2 V3

F1 F2 F3 Vo F

1.R1=R2=R3=Rf

4k 4k 4k 4k 1v 3v 5v

1k 1k 1k -9v 1kHz

2.R1=R2=R3=3Rf

12k

12k 12k 4k 1v 3v 5v

1k 1k 1k -3v 1kHz

3.R1,R2,R3 and Rf are different

3k 4k 5k 6k 1v 3v 5v

1k 1k 1k -12.5v 1kHz

CALCULATIONS:

Gain Calculation:Inverting Summing Amplifier:

CASE1: Where R1, R2, R3 and Rf are different

Vo Rf.V1/R1Rf.V2/R2Rf.V3/R3)

R1=3k V1=1v R2=4k V2=3vR3=5k V3=5vRf=6k

Vo=-(6/3+6.3/4k+6.5/5k)=-12.5v

G=-Rf/R1G=12.5/9

G=1.38CASE2:R1=R2=R3=Rf=4k

Vo V1 V2 V3) Vo=-(1+3+5) =-9

G=9/9

G=1

CASE3: Where R1=R2=R3=3Rf

R1=R2=R3=12kRf=4k

Vo=-(V1+V2+V3)/3

Vo=-(1+3+5)/3=-3

G=3/9G=O.3

THEORITICAL GAIN PRACTICAL GAINGAIN1 1.38 1.37GAIN2 1 1.05GAIN3 0.3 0.3

Slew Rate:

SR=(2fπVm/106 ) V/μsWhere,Vm=Peak output amplitude in voltsf=frequency in Hz.

f=45kHzVm=2.45k.π.2/106

Vm=565.2mS

CONCLUSION:

The gain of the summing amplifier calculated practically and theoretically areAs the Same, and the input & output waveforms are plotted on the graph paper.The slew rate calculated value is 565.2mS

REFERENCE:

1. Op Amps & Linear Integrated circuits by Ramakant Gayakwad , Chapter 3, pp102-108

ANSWERED QUESTIONS:Q.1 what is slew rate? What will be the value of slew rate for ideal Op-Amp?

The maximum rate at which the op amp is able to change to a different voltage, the slew rate value will be SR=2πfV Q.2.What do you mean by CMRR & PSRR?

CMRR stands for Common Mode Rejection Ratio, and it is a measure of how well the amplifier rejects signals that appear on both leads. The idea is that an amplifier should amplify the (Differential Mode) signal, but not any noise (Common Mode) that might appear on the lines, perhaps due to induction from nearby AC power sources. Since induction will show up on both leads, a high CMRR amplifier will have a greater signal to noise ratio overallThe PSRR is defined as the ratio of the change in supply voltage to the equivalent (differential) input voltage it produces in the op-amp, often expressed in decibels. An ideal op-amp would have infinite PSRR. The output voltage will depend on the feedback circuit, as is the case of regular input offset voltages. But testing is not confined to DC (zero frequency); often an operational amplifier will also have its PSRR given at various frequencies (in which case the ratio is one of RMS amplitudes of sine waves present at a power supply compared with the output, with gain taken into account).

Q.3 what is the operating temperature of 741 IC for Military & Industrial Applications?

A: - For Military Applications, the operating temperature is −55 °C to 125 °C (sometimes -65 °C to 175 °C).- For Industrial Applications, the operating temperature is −40 °C to 85 °C (sometimes −25 °C to 85 °C

Q.4 What are the different applications of summing amplifier?

A: One of the most common applications for a summing amplifier is to algebraically add two (or more) signals or voltages to form the sum of those signals. Practically Summing Amplifier are used in audio mixers and digital to analogue converters.

Q.5 What is the practical value of Slew Rate for 741 IC?For practical implementation, a slew rate of 0.5 v/us is suitable for the LM 741IC.

EXPERIMENT No: 3AIM:Design an Op-Amp differentiator that will differentiate an input signal for sine waveUsing 741 Op-Amp for multiple inputs

EQUIPMENTS REQUIRED1. LT SPICE2. COMPUTER

COMPONENTS REQUIRED:LTSPICE COMPONENTS LIKE: Resistors, Op-Amp 741,

SPICE SIMULATION R EQUIREMENTS AND PROCEDURES

Use Resistors, the Op-Amp IC 741 designed in last LAB, and connect them as in the circuits, below, then simulate them.

THEORY:

Differentiator:One of the simplest of the op-amp circuits that contain capacitor is theDifferentiating Amplifier or Differentiator. As the name suggests, the circuit performsThe mathematical operation of differentiation, that is, the output waveform is theDerivative of input waveform. A differentiator circuit is shown in fig.The node N is at virtual ground potential i.e., Vn = 0. The current iC the capacitor is,iC = C1 d/dt (Vi - Vn) = C1 dVi/dt ---------------(1)The current if through the feedback resistor is Vo/Rf and there is no current into the opamp.Therefore, the nodal equation at node N is,C1 (dVi/dt) + Vo/Vf = 0From which we haveVo = - Rf C1 (dVi/dt) ------------------ (2)Thus the output voltage Vo is a constant (-RfC1) times the derivative of the input voltageVi and the circuit is a differentiator. The sign indicates a 180o phase shift of the outputWaveform V0 with respect to the input signal.The phasor equivalent of Eq. (2) is, Vo(s) = - RfC1 s Vi(s) where Vo and Vi is the

Phasor representation of Vo and Vi. In steady state, put s = jω. We may now write theMagnitude of gain A of the differentiator as,A = Vo/Vi = - jωRf C1 = ωRf C1 ---------------- (3)From Eq. (3), one can draw the frequency response of the op-amp differentiator. Equation (3) may be rewritten as

A = f / faWhere, fa = 1/2π Rf C1

f=operating frequency

PROCEDURE:1. Construct the circuit in LTSPICE2. Apply the given inputs and observe the output on CRO.3. Calculate the amplitude & frequency of the output.4. Vary the inputs to and observe the output.5. Plot the inputs and outputs waveforms for various inputs and capture it in your report

CIRCUIT DIAGRAM:

OBSERVATION TABLE:

S No

Inputs Outputs

Voltage frequency Voltage Frequency1. 3v 0.5Hz -9v 0.5Hz2. 4v 0.17Hz -2v 0.17Hz

CALCULATION:

Calculate the output for the input using,Vo = - Rf C1 (dVi/dt)

CONCLUSION:

The output and input waveforms of the differentiated outputs with the various inputs are plotted on the graph paper.

REFERENCE:1. Op Amps & Linear Integrated circuits by Ramakant Gayakwad, Chapter 6, 250-255

ANSWERED QUESTIONS:

1. The difference between practical differentiator and ideal differentiator is Practical operational amplifiers draw a small current from each of their inputs wile the ideal has some limitations2. Application of differentiator is that it works as high pass filter.3. The frequency at which the gain is 0 dB is given by 1Hz4. The condition at which the input will be differentiated properly is that. Capacitors will not oppose the voltage change by creating current in the circuit: in this condition, but we can say that the output voltage will be constant. Another condition is to differentiate a signal representing the data given

EXPERIMENT No. 04AIM:Design an Integrator to demonstrate the operation of an integrator using 741 Op- Amp for multiple inputs.

EQUIPEMENTS REQUIRED:1. LT SPICE2. COMPUTER

COMPONENTS REQUIRED:LTSPICE COMPONENTS LIKE: Resistors, Capacitor, Supplies, Op-Amp 741.

SPICE SIMULATION R EQUIREMENTS AND PROCEDURESUse Resistors, the Op-Amp IC 741 designed in last LAB, and connect them as in the circuits below, then simulate them.

THEORY:

Integrator:If we interchange the resistor and capacitor of the differentiator, we have the circuitof an integrator. The nodal equation at node N is,Vi/R + Cf.dVo/dt = 0Or, (dVo/dt) = - (1/R1Cf.)Vi

Integrating both sides, we get

∫0

t

dVo=−1/ R 1C 1∫0

t

Vidt=

Vo(t)=-(1/R1Cf)∫0

t

Vi (t)dt + Vo(0)

Where Vo (0) is the initial output voltage.

The circuit, thus provides an output voltage which is proportional to the time integral of the input and R1Cf is the time constant of the integrator. It may be noted that there is a negative sign in the output voltage, and therefore, this integrator is also known as an Inverting integrator. A simple low pass RC circuit can also work as an integrator when time constant is very large. This requires very large values of R & C. The components R& C cannot be made infinitely large because of practical limitations.

PROCEDURE:

1. Construct the circuit in LTSPICE.2. Apply the given inputs and observe the output on CRO.3. Calculate the amplitude & frequency of the output and capture it in your report.4. Vary the inputs to and observe the output.5. Plot the inputs and outputs waveforms for various inputs and capture it in report

CIRCUIT DIAGRAM:

OBSERVATION TABLE:

S No

Inputs OutputsVoltage Frequency Voltage Frequenc

y

1. 6v 0.5Hz -3.5v 0.5Hz2. 2v 0.125Hz -4v 0.125Hz

CALCULATIONS:

Calculate the output for the input using,`

Vo( t)=−( 1R 1Cf

)∫0

t

Vi ( t ) dt+Vo (0)

CONCLUSION:

The output and input waveforms of the integrated output with the various inputs arePlotted on the graph paper

REFERENCE:

1. Op Amps & Linear Integrated circuits by Ramakant Gayakwad, Chapter 6, pp-245-249

ANSWERED QUESTIONS:

1. Roll-off is a term commonly used to describe the steepness of a transmission function with frequency, particularly in electrical network analysis.2. Practical operational amplifiers draw a small current from each of their inputs , may lead to noticeable deviations from ideal operational amplifier behaviour. .... Bias currents will have no impact on the difference between the two inputs. .... In effect, this resistor reduces the DC gain of the "integrator" – it 3. Integrators are used in low pass filters4. Capacitors oppose voltage change by creating current in the circuit: in this condition, but we can say that the output voltage will be constant.