ee707 advanced electrical engineering lab manual (2006 scheme)

EE 707

ADVANCED ELECTRICAL ENGINEERING LAB

2006 Scheme

July 15, 2011

ACKNOWLEDGEMENT

Any suggestions for the improvement of this manual will be acknowledged and appre-ciated.

ANITH KRISHNAN

INTRODUCTION

EE 707 - Advanced Electrical Engineering Lab is a part of the course study of the studentsof Electrical Engineering registered under CUSAT.

The students are advised to refer standard textbooks along with this laboratory manualto have a thorough understanding of the concepts and the principles involved.

INSTRUCTIONS

• Discipline has to be maintained throughout the lab session.

• All students should wear shoes.

• Male students should tuck in their shirts.

• Students should not lean on the work bench or on the computer table.

• Do not switch on the supply before the circuit is verified by the concerned lab faculty.

• Shutdown the computer before leaving the lab.

• Use of mobile phone inside the laboratory is strictly prohibited.

• Entry will be allowed only to students who have their rough record and fair record(completed) in custody.

• Use of USB drives, CDs, DVDs or any other portable data storage device withoutthe permission of the lab in-charge is forbidden.

All the students are to strictly abide by the instructions given here and any additionalinstruction given during the lab session, failing to which he/she will have to face disciplinaryaction.

Contents

Familiarization of MATLAB I. 1

Familiarization of MATLAB II. 3

Familiarization of MATLAB III. 5

Familiarization of SIMULINK. 7

Modelling and Simulation of Mechanical Translational System. 10

Lead Compensator. 12

Lag Compensator. 14

Comparison of Lag Compensated and Lead Compensated System. 16

PID Control System. 17

Synchro. 20

Steady-State Stability Analysis. 21

Bus Admittance Matrix. 24

LVDT. 28

Step Response of Second Order System. 29

Diode Rectifier circuits. 30

Lab Manual: Advanced Electrical Engineering Lab (EE 707) 2006 Scheme

Experiment No.: 1

FAMILIARIZATION OF MATLAB I

Aim:

To familiarize with some basic commands in MATLAB.

Main Window:

Some Basic Commands:

• To store a row matrix A =[1 2 3

]>> A=[1 2 3]

• To store a column matrix, A =

123

>> A=[1;2;3]

Department of Electrical and Electronics Engineering,College of Engineering, Thalassery

1


• To store a square matrix, A =

1 2 34 5 67 8 9

>> A=[1 2 3;4 5 6;7 8 9]

• To find the inverse of matrix A

>> A=[1 2;3 4]

>> inv(A)

gives the answer

[−2 11.5 −0.5

]• To find the determinant of A

>> det(A)

gives the answer −2

• Multiplication of two matrices, A =

[1 23 4

]and B =

[4 32 1

]>> A=[1 2;3 4]

>> B=[4 3;2 1]

>> A*B

gives the answer

[8 520 13

]

• Square of a matrix A2 =

[1 23 4

]2>> A=[1 2;3 4]

>> A^2

gives the answer

[7 1015 22

]Result:

Note: The students are expected to learn all the commands not given in this manual as and

when required.


2


Experiment No.: 2

FAMILIARIZATION OF MATLAB II

Aim:

To familiarize with some basic commands in MATLAB related to electrical engineering.

Commands and their usage:

• Realize a transfer function G(s) = 1s2 + 2s+ 5

>> num=[1] % coefficients of numerator

>> den=[1 2 5] % coefficients of denominator

>> G=tf(num,den)

The transfer function can also be realized directly as shown below

>> G=tf([1],[1 2 5])

• Multiplication of two transfer functions G1(s) = 1s2 + 2s+ 5

and G2(s) = 32s2 + 5

>> G1=tf([1],[1 2 5])

>> G2=tf([3],[2 0 5])

>> G1*G2

The answer will be a fourth order transfer function (as evident).

• To obtain the step response of the system G1(s) = 1s2 + 2s+ 5

>> G1=tf([1],[1 2 5])

>> step(G1)

0 1 2 3 4 5 60

0.05

0.1

0.15

0.2

0.25

Step Response

Time (sec)

Am

plitu

de


3


• To obtain the bode plot of the system G1(s) = 1s2 + 2s+ 5

>> G1=tf([1],[1 2 5])

>> bode(G1)

−80

−70

−60

−50

−40

−30

−20

−10

Mag

nitu

de (

dB)

10−1

100

101

102

−180

−135

−90

−45

0

Pha

se (

deg)

Bode Diagram

Frequency (rad/sec)

• To obtain the root locus of the system G1(s) = 1s3 + 3s2 + 5s+ 4

>> G1=tf([1],[1 3 5 4])

>> rlocus(G1)

−5 −4 −3 −2 −1 0 1−4

−3

−2

−1

0

1

2

3

4

Root Locus

Real Axis

Imag

inar

y A

xis

Result:


4


Experiment No.: 3

FAMILIARIZATION OF MATLAB III

Aim:

To familiarize the usage of M-File.

Procedure:

1. From the main window, choose to open a new Blank M-File from the File menu.

2. In the new window, the commands can be entered in the sequence in which they aremeant to be executed.

3. After all the necessary commands are entered, the file is saved using .m extension.

4. The file (all the commands in sequence) can be executed by clicking on the runbutton on the toolbar or by pressing F5.

5. The commands will be executed, if there was no error in the code. The commandprompt will display all the errors (if any).


5


Excercise:

Obtain the unit-impulse response of the following system:[x1

x2

]=

[0 1−1 −1

] [x1

x2

]+

[10

]u

y =[0 1

] [x1

x2

]+ [0]u

Result:

0 2 4 6 8 10 12−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

Unit Impulse Response

Time (sec)

Am

plitu

de


6


Experiment No.: 4

FAMILIARIZATION OF SIMULINK

Aim:

To familiarize with SIMULINK.

Introduction:

SIMULINK is a Graphical User’s Interface (GUI) software which works directly with theblock diagram of a control system (rather than differential equations, or transfer functions)to produce a simulation of the system’s response to arbitrary inputs and initial conditions.The basic entity in SIMULINK is a block, which can be selected from a library of commonlyused blocks. Alternatively, a user can device special blocks out of the common blocks, M-files, MEX files, C, or Java codes through the S-function facility.

Procedure:

1. To open the SIMULINK library browser, double click on the SIMULINK icon on theMATLAB toolbar, or issue the following command at the MATLAB prompt

>> simulink

2. Click on the create a new model icon on the SIMULINK toolbar. A window for thenew model will open.


7


3. Open the subsystem library in the general SIMULINK library browser by double-clicking on the appropriate icon. The subsystems are: Continuous, Discontinuous,Sinks, Sources etc.

4. Select the required blocks from the subsystems libraries and drag them individuallyto the open new model window.

5. Once the required blocks are dragged and placed in the new model window, the in-ports and out-ports of the adjacent blocks can be joined to create a block diagram asdesired. Double clicking on each block will open a dialog blox, in which the block’sparameters can be set.

6. Once the block diagram is ready, it is saved into a location on the hard disk drive.

7. In order to simulate the block diagram, click on the play button on the toolbar.If the model created was correct, the simulation will start and the results can beviewed using any of the sink blocks in the model. However, if there was an error,SIMULINK prompts with a diagnostics dialog box which describes what went wrongin the simulation and also what has to be modified in the model for a successfulsimulation.

8. The simulation can be further refined by adjusting the simulation parameters.SIMULINK provides several simulation parameters that can be adjusted to achieve adesired accuracy in a simulation. A user can select from a number of time-integrationschemes, such as Runge-Kutta, Euler, etc and as well as refine the tolerances and timestep sizes used for performing the simulation.


8


Excercise:

Simulate the following block diagram using SIMULINK.

Result:

0 5 10 150

0.2

0.4

0.6

0.8

1

1.2

Time (s)

Am

plitu

de

Unit Step Response


9


Experiment No.: 5

MODELLING AND SIMULATION OF MECHANICALTRANSLATIONAL SYSTEM

Aim:

To model a mechanical translational system and analyse the system using computer sim-ulation.

Modelling:

With the system initially at rest, a force of 25 Newtons is applied at time t = 0. Assumethat the mass M = 1 kg, frictional coefficient B = 5N/m/s, and the spring constantK = 25N/m.

Applying Newton’s law of motion, the force equation of the system is

Md2x

dt2+B

dx

dt+Kx = f(t)

Let

x1 = x

x2 =dx

dt

then

x1 = x2

x2 =1

M[f(t)−Bx2 −Kx1]

The system can be analyzed by simulating the above system to a step input.


10


Result:

Exercise:

Determine the behaviour of the following system using SIMULINK.

2d3y

dt3+ 4

d2y

dt2+ 8

dy

dt+ 10y = 10u(t)


11


Experiment No.: 6

LEAD COMPENSATOR

Aim:

To design a lead compensator for a system whose open-loop transfer function is given by

G(s) =5

s(s+ 2)

so that the static velocity error constant kv is 20s−1, the phase margin is atleast 55◦ andthe gain margin is atleast 12dB.

Theory:

Students are expected to complete this section by themselves.

Procedure:


Result:

G(s)Gc(s) =20(0.26s+ 1)

s(0.5s+ 1)(0.04s+ 1)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

Time (s)

Am

plitu

de

Ramp Response

InputUn−compensated SystemCompensated System


12


3.95 4 4.05 4.1 4.15 4.2 4.25 4.3 4.35 4.4 4.454

4.05

4.1

4.15

4.2

4.25

4.3

Time (s)

Am

plitu

de

Ramp Response


0.05

−80

−60

−40

−20

0

20

40

60

Mag

nitu

de (

dB)

10−1

100

101

102

103

−180

−135

−90

Pha

se (

deg)

Bode Diagram of Compensated SystemGm = Inf dB (at Inf rad/sec) , Pm = 58.3 deg (at 10.1 rad/sec)

Frequency (rad/sec)


13


Experiment No.: 7

LAG COMPENSATOR

Aim:

To design a lag compensator for a system whose open-loop transfer function is given by

G(s) =5

s(s+ 2)

so that the static velocity error constant kv is 20s−1, the phase margin is atleast 55◦ andthe gain margin is atleast 12dB.

Theory:


Procedure:


Result:

G(s)Gc(s) =20(8.7s+ 1)

(166.7s+ 1)s(0.5s+ 1)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

Time (s)

Am

plitu

de

Ramp Response



14


4.95 5 5.05 5.1 5.15 5.2 5.25 5.3 5.35 5.45

5.05

5.1

5.15

5.2

5.25

5.3

Time (s)

Am

plitu

de

Ramp Response


0.05

−100

−50

0

50

100

150

Mag

nitu

de (

dB)

10−4

10−3

10−2

10−1

100

101

102

−180

−135

−90

Pha

se (

deg)

Bode Diagram of Compensated SystemGm = Inf dB (at Inf rad/sec) , Pm = 58.1 deg (at 0.95 rad/sec)

Frequency (rad/sec)


15


Experiment No.: 8

COMPARISON OF LAG COMPENSATED AND LEADCOMPENSATED SYSTEM

Aim:

To analyse the performance of a system when it is lag compensated and when it is leadcompensated.


16


Experiment No.: 9

PID CONTROL SYSTEM

Aim:

To familiarize with P, PI, PD and PID control using PID control trainer kit.

Apparatus Required:

The students are expected to complete this section before coming to the lab.

Theory:


Procedure:

Proportional (P) Control System:

1. With the help of jumper links, connect the test points 2 and 4 to ground.

2. Set all pots (P1 to P5) in the minimum position.

3. Switch on the unit and see that the supply indicator glows.

4. With the help of DMM, observe the set value (P1) and simulated value (P2). Boththese values are 0V.

5. Observe the error signal at the output of the adder, where the set value and thesimulated value are subtracted. Also check the final output. All these values will beat 0V.

6. Now adjust the set value with pot P1 to 1V and simulated process value at 0V withpot P2 and record it in the table.

7. Observe the error signal. The error signal will be +1V.

8. Observe the final output of proportional block for different positions of pot P5 andnote down the same in the table.

9. Find the gain Kp of the proportional block for different positions of pot P5.

10. Now adjust the set value with pot P1 to 1V and the simulated process value withpot P2 to 2V respectively.

11. Observe the error signal, the error signal will be -1V.

12. Observe the final output of the proportional block for different positions of pot P5and note down the same in the table.

13. Find the gain Kp of the proportional block for different positions of pot P5.

14. Repeat the above procedure for different values of set value and simulated processvalue and note down the same in the table.

15. Switch off the unit.


17


Proportional Integral (PI) Control System:

1. Connect test point 1 and 2 and the test point 4 to ground using jumper links.

2. Set all pots P1 to P5 to minimum position.





7. Observe the error signal and the final output (the final output rises immediately toabout +9V i.e. the saturation level).

8. Press the first switch in the discharge block and observe the final output on DMM(the final output is same as the error signal when the switch is in pressed positionand the output goes to positive saturation as soon as the switch is released).

9. Now keep pot P3 in midway position and again press the discharge switch momen-tarily (when the discharge switch is released, the final output starts rising towards+ve saturation taking few seconds to reach the same).

10. Repeat the above procedure with a -ve error signal.

Proportional Derivative (PD) Control System:

1. Connect test point 3 and 4 and test point 2 to ground using jumper links.

2. Set all pots P1 to P5 to minimum position.





7. Observe the error signal and the final output (the final output is same as the errorsignal).

8. Press the second switch in the discharge block and observe the final output on DMM(the final output goes to +ve saturation level when the switch is pressed and it comesback to the error signal level as soon as the switch is released).

9. Now keep pot P4 in midway position and again press the discharge switch momen-tarily (the final output is at +ve saturation when the discharge switch is pressed anddischarges to the error signal level when the switch is released taking a longer timeas compared to the earlier case).


18


10. Repeat the above step with a -ve error signal.

Proportional Integral Derivative (PID) Control System:

1. Connect test point 1 to 2 and test point 3 to 4 using jumper links to select the PIDmode.

2. Set all pots (P1 to P5) to minimum position.

3. Switch on the supply (the supply indicator glows).

4. Connect one DMM to observe the error signal and the other DMM at the final outputof the proportional block.

5. Adjust the set value with pot P1 to 1V and the simulated process value with pot P2to 0V.

6. Observe the error signal and the final output (the final output will rise immediatelyto +9V).

7. Press both the discharge switches momentarily (simultaneously) and observe thefinal output (the output remains at +ve saturation when switch is pressed, and whenswitch is released, it tries to discharge to the error signal level due to derivativeaction and simultaneously due to integral action, since error is +ve, it again rises tosaturation level).

8. Now keep pot P3 and P4 in midway position and observe the final output when boththe discharge switches are pressed momentarily.

9. Now adjust pots P1 and P2 such that the error signal is -1V and repeat the aboveprocedure.

10. Observe the effect of changing the error signal from 1V to 3V and then 3V to 1V andobserve the final output. Analyse your results.

11. Similarly, observe the effect of changing the error signal from -1V to -3V and the -3Vto -1V and analyse your results.

Result:


19


Experiment No.: 10

A.C. POSITION CONTROL SYSTEM: A.C. SYNCHRO

Aim:

To study the operation of Synchro Transmitter and Receiver.

Apparatus Required:


Theory:


Procedure:

1. Arrange the power supply, synchro transmission and synchro receiver near to eachother.

2. Connect the power supply output to R1-R2 terminals of the transmitter and receiver.

3. Short S1-S1, S2-S2, S3-S3 windings of the transmitter and receiver with the help ofpatch chords.

4. Switch on the unit (the supply indicator will glow).

5. As the power is switched on, the transmitter and receiver shaft will come to the sameposition on the dial.

6. Vary the shaft position of the transmitter and observe the corresponding change inthe shaft position of the receiver (the receiver shaft will move by an equal amount asthat of the transmitter).

Result:


20


Experiment No.: 11

STEADY-STATE STABILITY ANALYSIS

Aim:

To analyse the steady-state performance of a power system using MATLAB.

System Description:

3.0' =dX

'E2.0=tX 3.012 =X

3.012 =X

0.1=V

∞

A 60Hz synchronous generator having inertia constantH = 9.94MJ/MV A and a transientreactance X ′

d = 0.3 per unit is connected to an infinite bus through a purely reactive circuitas shown in the figure. Reactances are marked on the diagram on a common system base.The generator is delivering real power of 0.6 per unit, 0.8 power factor lagging to an infinitebus at a voltage of V = 1 per unit. Assume the per unit damping power coefficient isD = 0.138. The generator is operating in the steady state at δ0 = 16.79◦. The input poweris increased by a small amount ∆P = 0.2 per unit. The new steady-state behaviour canbe studied by analyzing the step response of the rotor angle and the generator frequency.

Modelling:

It is given that the input power is increased by a small amount ∆P . The linearized swingequation is thus

H

πf0

d2∆δ

dt2+D

d∆δ

dt+ Ps∆δ = ∆P

ord2∆δ

dt2+

πf0H

Dd∆δ

dt+

πf0H

Ps∆δ =πf0H

∆P

or in terms of the standard second-order differential equation,

d2∆δ

dt2+ 2ζωn

d∆δ

dt+ ω2

n∆δ = ∆u


21


where

∆u =πf0H

∆P

ωn =

√πf0H

Ps

ζ =D

2

√πf0HPs

The synchronizing coefficient Ps is given by

Ps = Pmax cos δ0

where Pmax (steady-state stability limit) is given by

Pmax =|E ′| |V |

X

X = 0.3 + 0.2 +0.3

2= 0.65

E ′ is computed fromE ′ = V + jXI

where current I is

I =S∗

V ∗ = 0.756 − 36.87◦

The per unit apparant power

S =0.6

0.86 cos−1 0.8 = 0.756 36.87◦

Transforming to the state variable form by taking x1 = ∆δ and x2 = ∆ω = ∆δ, thestate-space representation is given by[

x1

x2

]=

[0 1

−ω2n −2ζωn

] [x1

x2

]+

[01

]∆u

and

y(t) =

[1 00 1

] [x1

x2

]Matlab Code:

A=[0 1;-37.705 -2.617];

Du=3.79;

B=[0;1]*Du;

C=[1 0;0 1];


22


D=[0;0];

t=0:0.01:5;

[y,x]=step(A,B,C,D,1,t);

Dd=x(:,1);

Dw=x(:,2);

d=d0+Dd)*180/pi;

f=f0+Dw/(2*pi);

subplot(2,1,1);

plot(t,d);

grid on;

subplot(2,1,2);

plot(t,f);

grid on;

Simulink Block Diagram:

Result:

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 515

20

25

30

δ 0 (de

g.)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 559.95

60

60.05

60.1

Time (s)

f (H

z)


23


Experiment No.: 12

CALCULATION OF BUS ADMITTANCE MATRIX

Aim:

To find the bus admittance matrix of a simple power system from its impedance diagram.

System Description:

The emfs shown in the figure are E1 = 1.16 0◦ and E2 = 1.0 6 0◦.

Theory and Modelling:

The circuit has been redrawn in terms of admittances and transformation to currentsources. With source transformation, the equivalent current sources are

I1 =1.1

j1.0= −j1.1 pu

I2 =1.0

j0.8= −j1.25 pu


24


Node 0 (which is normally ground) is taken as reference. Applying KCL to the inde-pendent nodes 1 through 4 reslts in

I1 = y10V1 + y12(V1 − V2) + y13(V1 − V3)

I2 = y20V2 + y12(V2 − V1) + y23(V2 − V3)

0 = y23(V3 − V2) + y13(V3 − V1) + y34(V3 − V4)

0 = y34(V4 − V3)

Rearranging these equations yields

I1 = (y10 + y12 + y13)V1 − y12V2 − y13V3

I2 = −y12V1 + (y20 + y12 + y23)V2 − y23V3

0 = −y13V1 − y23V2 + (y13 + y23 + y34)V3 − y34V4

0 = −y34V3 + y34V4

Let

Y11 = y10 + y12 + y13

Y22 = y20 + y12 + y23

Y33 = y13 + y23 + y34

Y44 = y34

Y12 = Y21 = −y12

Y13 = Y31 = −y13

Y23 = Y32 = −y23

Y34 = Y43 = −y34

The node equation reduces to

I1 = Y11V1 + Y12V2 + Y13V3 + Y14V4

I2 = Y21V1 + Y22V2 + Y23V3 + Y24V4

I3 = Y31V1 + Y32V2 + Y33V3 + Y34V4

I4 = Y41V1 + Y42V2 + Y43V3 + Y44V4

In the above network, since there is no connection between bus 1 and 4, Y14 = Y41 = 0;similarly Y24 = Y42 = 0.

Extending the above relation to an n bus system, the node-voltage equation in matrixform is

I1I2...In

=

Y11 Y12 · · · Y1n

Y21 Y22 · · · Y2n...

......

Yn1 Yn2 · · · Ynn

V1

V2...Vn


25


or

Ibus = YbusVbus

where Ibus is the vector of the injected bus currents (i.e. external current sources). Thecurrent is positive when flowing towards the bus, and it is negative if flowing away fromthe bus. Vbus is the vector of bus voltages measured from the reference node (i.e. nodevoltages). Ybus is known as the bus admittance matrix. The diagonal element of each nodeis the sum of admittances connected to it. It is known as the self-admittance or drivingpoint admittance, i.e.,

Yii =n∑

j=0

yij j 6= i

The off-diagonal element is equal to the negative of the admittance between the nodes. Itis known as the mutual admittance or transfer admittance, i.e.,

Yij = Yji = −yij

The inverse of bus admittance matrix is known as the bus impedance matrix Zbus.

MATLAB function ybus:

A function called Y=ybus(zdata) is written for the formation of the bus admittancematrix. zdata is the line data input and contains four columns. The first two colmns arethe line bus numbers and the remaining columns contain the line resistance and reactanceper unit. The function returns the bus admittance matrix.

In the program, the line impedances are first converted to admittances. Y is theinitialized to zero. In the first loop, the line data is searched to find the elements connectedto a bus, and the diagonal elements are thus formed.

The following is the program for building the bus admittance matrix:

function[Y]=ybus(zdata)

nl=zdata(:,1);

nr=zdata(:,2);

R=zdata(:,3);

X=zdata(:,4);

nbr=length(zdata(:,1));

nbus=max(max(nl),max(nr));

Z=R+j*X;

y=ones(nbr,1)./Z;

Y=zeros(nbus,nbus);

for k=1:nbr;

if nl(k)>0 & nr(k)>0

Y(nl(k),nr(k))=Y(nl(k),nr(k))-y(k);

Y(nr(k),nl(k))=Y(nl(k),nr(k));


26


end

end

for n=1:nbus

for k=1:nbr

if nl(k)==n | nr(k)==n

Y(n,n)=Y(n,n)+y(k);

else, end

end

end

Matlab Code:

% From To R X

z = [0 1 0 1.0

0 2 0 0.8

1 2 0 0.4

1 3 0 0.2

2 3 0 0.2

3 4 0 0.08];

Y=ybus(z)

Ibus=[-j*1.1; -j*1.25; 0; 0];

Zbus=inv(Y)

Vbus=Zbus*Ibus

Result:


27


Experiment No.: 13

DISPLACEMENT MEASUREMENT USING LVDT

Aim:

To measure the displacement using linear variable differential transducer (LVDT).

Equipments Required:

The students are expected to complete this section by themselves.

Theory:


Procedure:

1. Connect the power supply and switch on the instrument by using the toggle switch.

2. Leave the instrument in the turned on condition for about 10 mins for initial warmup.

3. Rotate the micrometer till it reads 20.

4. Adjust the CAL potentiometer at the front panel so that the display reads 10.

5. Rotate the core of the micrometer till the micrometer reads 10 and adjust the ZEROpotentiometer till the display reads 0.

6. Rotate back the micrometer core upto 20 and adjust the CAL potentiometer onceagain till the display reads 10. Now the instrument has been calibrated for +/- 10mmrange. As the core of the LVDT moves, the display reads the displacement in mm.

7. Rotate the core of the micrometer in steps of 1 or 2mm and tabulate the readings. Themicrometer wil show the exact displacement and the display will show the measureddisplacement.

8. Tabulate the readings and plot the graph between actual reading and indicated read-ing.

Observation Table:

Sl. No. Micrometer Reading Indicated Reading Actual Reading Output Voltage

Result:


28


Experiment No.: 14

STEP RESPONSE OF SECOND ORDER SYSTEM

Aim:

To find the transient response of the given second order system for a unit step

Equipments Required:

The students have to complete this section by themselves.

Theory:



29


Experiment No.: 15

DIODE RECTIFIER CIRCUITS

Aim:

To familiarize with the circuit simulation tool LTSPICE and to use LTSPICE to study theworking of full-wave and half-wave rectifiers.

LTSPICE:

SPICE stands for Simulation Program with Integrated Circuit Emphasis. LTSpice is afree, full function SPICE program. The following are the main uses of LTSpice:

1. Schematic Capture

2. Circuit Simulations

3. Waveform Analysis

4. AC Analysis

5. DC Sweeps

6. Power dissipation analysis, etc.

Integrated circuits, unlike board-level designs composed of discrete parts, are impossibleto breadboard before manufacture. Further, the high costs of photolithographic masks andother manufacturing prerequisites make it essential to design the circuit to be as close toperfect as possible before the integrated circuit is first built. Simulating the circuit withSPICE is the industry-standard way to verify circuit operation at the transistor level beforecommitting to manufacturing an integrated circuit.

Board-level circuit designs can often be breadboarded for testing. Even with a bread-board, some circuit properties may not be accurate compared to the final printed wiringboard, such as parasitic resistances and capacitances. These parasitic components canoften be estimated more accurately using SPICE simulation. Also, designers may wantmore information about the circuit than is available from a single mock-up. For instance,circuit performance is affected by component manufacturing tolerances. In these cases itis common to use SPICE to perform Monte Carlo simulations of the effect of componentvariations on performance, a task which is impractical using calculations by hand for acircuit of any appreciable complexity.

Circuit simulation programs, of which SPICE and derivatives are the most prominent,take a text netlist describing the circuit elements (transistors, resistors, capacitors, etc.)and their connections, and translate this description into equations to be solved. Thegeneral equations produced are nonlinear differential algebraic equations which are solvedusing implicit integration methods, Newton’s method and sparse matrix techniques.

Procedure:



30


Circuits and Waveforms:

Half-Wave Rectifier

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04−400

−200

0

200

400

Vol

tage

(V

)

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04−40

−20

0

20

40

Time (s)

Vol

tage

(V

)

Source (primary) Voltage

Secondary VoltageVoltage across the load

Half-wave rectifier with filter


31


0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04−40

−20

0

20

40

Vol

tage

(V

)

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04−0.2

0

0.2

0.4

0.6

Time (s)

Cur

rent

(A

)

IC1

Result:


32

ee707 advanced electrical engineering lab manual (2006 scheme)

Documents