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CIRCUIT SIMULATION -(E) LABORATORY REPORT
2021-2022
DEPARTMENT OF ELECTRICAL ENGINEERING
SUBMITTED BY:
ANSHUMAN SINGH
5th SEM
ROLL NUMBER: 20UELE6006
1
S.NO TITLE DATE PAGE NO. REMARKS
1.
INTRODUCTION OF PS-PICE AND ITS
LIBRARY FUNCTIONS.
2. TO VERIFY KCL & KVL IN A GIVEN
ELECTRICAL DC NETWORK USING PS-PICE
SIMULATIONS.
3.
TO VERIFY SUPERPOSITION THEOREM FOR
INDEPENDENT SOURCE DC NETWORK
USING PS-PICE.
TO VERIFY THEVENIN’S
THEOREM FOR INDEPENDENT
SOURCE DC NETWORK USING PS-
PICE.
4.
TO VERIFY MAXIMUM POWER TRANSFER
THEOREM IN A GIVEN ELECTRICAL DC
NETWORK USING PS-PICE SIMULATION.
5.
2
EXPERIMENT - 1
AIM: INTRODUCTION OF PS-PICE AND ITS LIBRARY FUNCTIONS.
THEORY:
• SPICE WAS FIRST DEVELOPED AT THE UNIVERSITY OF CALIFORNIA, BERKELEY, IN THE EARLY 1970S.
SUBSEQUENTLY AN IMPROVED VERSION SPICE 2 WAS AVAILABLE IN THE MID-1970S ESPECIALLY TO SUPPORT
COMPUTER AIDED DESIGN.
• PSPICE WAS RELEASED IN JANUARY 1984, AND WAS THE FIRST VERSION OF UC BERKELEY SPICE AVAILABLE ON
AN IBM PERSONAL COMPUTER. PSPICE LATER INCLUDED A WAVEFORM VIEWER AND ANALYSER PROGRAM
CALLED PROBE. SUBSEQUENT VERSIONS IMPROVED ON PERFORMANCE AND MOVED TO DEC/VAX
MINICOMPUTERS, SUN WORKSTATIONS, APPLE MACINTOSH, AND MICROSOFT WINDOWS. VERSION 3.06 WAS
RELEASED IN 1988, AND HAD A "STUDENT VERSION" AVAILABLE WHICH WOULD ALLOW A MAXIMUM OF UP TO
TEN TRANSISTORS TO BE INSERTED.
• ORCAD EE PSPICE IS A SPICE CIRCUIT SIMULATOR APPLICATION FOR SIMULATION AND VERIFICATION OF
ANALOG AND MIXED-SIGNAL CIRCUITS.[17] PSPICE IS AN ACRONYM FOR PERSONAL SIMULATION PROGRAM
WITH INTEGRATED CIRCUIT EMPHASIS.
• PSPICE WAS A MODIFIED VERSION OF THE ACADEMICALLY DEVELOPED SPICE, AND WAS COMMERCIALIZED BY
MICROSIM IN 1984. MICROSIM WAS PURCHASED BY ORCAD A DECADE LATER IN 1998.
ORCAD PSPICE DESIGNER IS AVAILABLE IN TWO OPTIONS: PSPICE DESIGNER AND PSPICE DESIGNER PLUS.
3
GENERAL GUIDELINE ON HOW TO USE PSPICE:
THE GENERAL PROCEDURE FOR USING PSPICE CONSISTS OF 3 BASIC STPES.
STEP 1
THE USER DRAWS THE CIRCUIT IN SCHEMATIC FORM WHICH HE WANTS TO SIMULATE.
STEP 2
THE USER SPECIFIES THE TYPE OF ANALYSIS DESIRED, AND DIRECTS PSPICE TO PERFORM THAT
ANALYSIS. THIS CAN, FOR INSTANCE, BE DC ANALYSIS, AC ANALYSIS, TRANSIENT ANALYSIS...
STEP 3
THE USER INSTRUCTS THE COMPUTER TO PRINT OR PLOT THE RESULTS OF THE ANALYSIS. IN THIS STEP,
THE USER SEES THE GRAPHICAL RESULTS OF THE ANALYSIS DONE. FOR EXAMPLE, HE CAN SEE THE
GRAPH OF THE OUTPUT VOLTAGE VS. OUTPUT CURRENT (V VS. I), OR ANY DATA WHICH HE WANTS TO
ANALYZE.
4
EXPERIMENT- 2
AIM: TO VERIFY KCL & KVL IN A GIVEN ELECTRICAL DC NETWORK USING PS-
PICE SIMULATIONS.
COMPONENTS USED:
S.NO COMPONENT NOTATION QUANTITY VALUES
1.
2.
3.
4.
RESISTORS
VOLTAGE SOURCE
CURRENT SOURCE
GROUND
R1,R2,R3,R4,R5
V1,V2,V3
I3
GND
5
3
1
1
10Ω,8Ω,7Ω,17Ω,100Ω
10V,10V,15V
15A
0V
5
THEORY: IN 1845, A GERMAN PHYSICIST, GUSTAV KIRCHHOFF DEVELOPED A PAIR OR SET OF RULES OR LAWS
WHICH DEAL WITH THE CONSERVATION OF CURRENT AND ENERGY WITHIN ELECTRICAL CIRCUITS. THESE TWO
RULES ARE COMMONLY KNOWN AS: KIRCHHOFFS CIRCUIT LAWS WITH ONE OF KIRCHHOFFS LAWS DEALING WITH
THE CURRENT FLOWING AROUND A CLOSED CIRCUIT, KIRCHHOFFS CURRENT LAW, (KCL) WHILE THE OTHER LAW
DEALS WITH THE VOLTAGE SOURCES PRESENT IN A CLOSED CIRCUIT, KIRCHHOFFS VOLTAGE LAW, (KVL).
KIRCHHOFFS FIRST LAW – THE CURRENT LAW, (KCL):
KIRCHHOFFS CURRENT LAW OR KCL, STATES THAT THE “TOTAL CURRENT OR CHARGE ENTERING A JUNCTION OR
NODE IS EXACTLY EQUAL TO THE CHARGE LEAVING THE NODE AS IT HAS NO OTHER PLACE TO GO EXCEPT TO LEAVE,
AS NO CHARGE IS LOST WITHIN THE NODE“. IN OTHER WORDS THE ALGEBRAIC SUM OF ALL THE CURRENTS
ENTERING AND LEAVING A NODE MUST BE EQUAL TO ZERO, I(EXITING) + I(ENTERING) = 0. THIS IDEA BY KIRCHHOFF IS
COMMONLY KNOWN AS THE CONSERVATION OF CHARGE.
6
HERE, THE THREE CURRENTS ENTERING THE NODE, I1, I2, I3 ARE ALL POSITIVE IN VALUE AND THE TWO CURRENTS LEAVING THE
NODE, I4 AND I5 ARE NEGATIVE IN VALUE. THEN THIS MEANS WE CAN ALSO REWRITE THE EQUATION AS;
I1 + I2 + I3 – I4 – I5 = 0
THE TERM NODE IN AN ELECTRICAL CIRCUIT GENERALLY REFERS TO A CONNECTION OR JUNCTION OF TWO OR MORE CURRENT
CARRYING PATHS OR ELEMENTS SUCH AS CABLES AND COMPONENTS. ALSO FOR CURRENT TO FLOW EITHER IN OR OUT OF A
NODE A CLOSED CIRCUIT PATH MUST EXIST. WE CAN USE KIRCHHOFF’S CURRENT LAW WHEN ANALYSING PARALLEL CIRCUITS.
KIRCHHOFFS SECOND LAW – THE VOLTAGE LAW, (KVL)
KIRCHHOFFS VOLTAGE LAW OR KVL, STATES THAT “IN ANY CLOSED LOOP NETWORK, THE TOTAL VOLTAGE AROUND
THE LOOP IS EQUAL TO THE SUM OF ALL THE VOLTAGE DROPS WITHIN THE SAME LOOP” WHICH IS ALSO EQUAL TO
ZERO. IN OTHER WORDS THE ALGEBRAIC SUM OF ALL VOLTAGES WITHIN THE LOOP MUST BE EQUAL TO ZERO. THIS
IDEA BY KIRCHHOFF IS KNOWN AS THE CONSERVATION OF ENERGY.
7
STARTING AT ANY POINT IN THE LOOP CONTINUE IN THE SAME DIRECTION NOTING THE DIRECTION OF ALL THE
VOLTAGE DROPS, EITHER POSITIVE OR NEGATIVE, AND RETURNING BACK TO THE SAME STARTING POINT. IT IS
IMPORTANT TO MAINTAIN THE SAME DIRECTION EITHER CLOCKWISE OR ANTI-CLOCKWISE OR THE FINAL VOLTAGE
SUM WILL NOT BE EQUAL TO ZERO. WE CAN USE KIRCHHOFF’S VOLTAGE LAW WHEN ANALYSING SERIES CIRCUITS.
CIRCUIT DIAGRAM:
8
METHODOLOGY:
1. SELECT THE ‘GET NEW PART’ BUTTON THEN COLLECT ALL THE REQUIRED COMPNENTS FOR YOUR
CIRCUIT.
2. CONNECT ALL THE COMPONENTS WITH THE HELP OF CONNECTING WIRES.
3. NOW NAME THE COMPONENTS AND GIVE RESPECTIVE VALUES TO THE COMPONENTS.
4. FIRST SIMULATE THE CIRCUIT AND NOTE CURRENT IN EACH BRANCH AND ANALYSE CURRENT AT A
PARTICULAR NODE A.
5. NOW ADD A BRANCH IN CIRCUIT WITH RESISTOR R5.
6. NOW SIMULATE THE CIRCUIT AND NOTE CURRENT IN EACH BRANCH AND ANALYSE CURRENT AT
PREVIOUS NODE.
7. OBSERVE IF THEIR IS ANY DIFFERENCE IN BOTH CASES.
8. SIMILARLY,OBSERVE VOLTAGE DROP IN ANY CLOSE LOOP IN BOTH CASES AND OBSERVE THE
DIFFERENCE.
ANALTYTICAL SOLUTION AND OBSERVATION:
NEXT PAGE----->
9
10
RESULT AND DISCUSSION:
WE HAVE VERIFIED KVL AND KCL AND SEEN THAT THERE IS SOME AMOUNT OF ERROR
BETWEEN ANALYTICAL VALUE AND SIMULATED VALUE THAT MAYBE BECAUSE OF
APPROXIMATION AT SOME STEPS.
ALSO AFTER ADDING ONE MORE BRANCH WITH RESISTOR R5 KCL AND KVL IS STILL VALID
BUT THE VALUES OF CURRENT CHANGES BECAUSE SOME AMOUNT OF CURRENT GOES
INTO THE R5 BRANCH AND ALSO VOLTAGE AT NODE CHANGES.
*******
11
EXPERIMENT - 3
AIM: TO VERIFY SUPERPOSITION THEOREM FOR DEPENDENT AND INDEPENDENT
SOURCE DC NETWORK USING PS-PICE.
S.NO COMPONENT NOTATION QUANTITY VALUES
1.
2.
3.
4.
RESISTORS
VOLTAGE SOURCE
CURRENT SOURCE
GROUND
R1,RX,R3,R4,R5
V1,V2,V3
I3
GND
5
3
1
1
10V,10V,15V
15A
0V
COMPONENTS USED :
70Ω,40Ω,30Ω,30Ω,50Ω
12
THEORY :
SUPERPOSITION THEOREM STATES THAT IN ANY LINEAR, BILATERAL NETWORK WHERE MORE THAN ONE SOURCE IS PRESENT,
THE RESPONSE ACROSS ANY ELEMENT IN THE CIRCUIT, IS THE SUM OF THE RESPONSES OBTAINED FROM EACH SOURCE
CONSIDERED SEPARATELY WHILE ALL OTHER SOURCES ARE REPLACED BY THEIR INTERNAL RESISTANCE. SUPERPOSITION
THEOREM IS A CIRCUIT ANALYSIS THEOREM THAT IS USED TO SOLVE THE NETWORK WHERE TWO OR MORE SOURCES ARE
PRESENT AND CONNECTED.
TO CALCULATE THE INDIVIDUAL CONTRIBUTION OF EACH SOURCE IN A CIRCUIT, THE OTHER SOURCE MUST BE REPLACED OR
REMOVED WITHOUT AFFECTING THE FINAL RESULT. WHILE REMOVING A VOLTAGE SOURCE, ITS VALUE IS SET TO ZERO. THIS IS
DONE BY REPLACING THE VOLTAGE SOURCE WITH A SHORT CIRCUIT. WHEN REMOVING A CURRENT SOURCE, ITS VALUE IS SET
TO ZERO. THIS IS DONE BY REPLACING THE CURRENT SOURCE WITH AN OPEN CIRCUIT.
THE SUPERPOSITION THEOREM IS VERY IMPORTANT IN CIRCUIT ANALYSIS BECAUSE IT CONVERTS A COMPLEX CIRCUIT INTO A
NORTON OR THEVENIN EQUIVALENT CIRCUIT.
GUIDELINES TO KEEP IN MIND WHILE USING THE SUPERPOSITION THEOREM
•WHEN YOU SUM THE INDIVIDUAL CONTRIBUTIONS OF EACH SOURCE, YOU SHOULD BE CAREFUL WHILE ASSIGNING SIGNS TO
THE QUANTITIES. IT IS SUGGESTED TO ASSIGN A REFERENCE DIRECTION TO EACH UNKNOWN QUANTITY. IF A CONTRIBUTION
FROM A SOURCE HAS THE SAME DIRECTION AS THE REFERENCE DIRECTION, IT HAS A POSITIVE SIGN IN THE SUM; IF IT HAS THE
OPPOSITE DIRECTION, THEN A NEGATIVE SIGN.
•TO USE THE SUPERPOSITION THEOREM WITH CIRCUIT CURRENTS AND VOLTAGES, ALL THE COMPONENTS MUST BE LINEAR.
•IT SHOULD BE NOTED THAT THE SUPERPOSITION THEOREM DOES NOT APPLY TO POWER, AS POWER IS NOT A LINEAR
QUANTITY. 13
METHODOLOGY:
1. SELECT THE ‘GET NEW PART’ BUTTON THEN COLLECT ALL THE REQUIRED COMPNENTS FOR YOUR
CIRCUIT.
2. CONNECT ALL THE COMPONENTS WITH THE HELP OF CONNECTING WIRES.
3. NOW NAME THE COMPONENTS FOR YOUR CONVENIENCE WE HAVE HERE NAMED THE TARGETED
RESISTOR RX ACROSS WHICH WE HAVE TO FIND VOLTAGE DROP.
4. FIRST SIMULATE THE ORIGINAL CIRCUIT AND NOTE THE VOLTAGE ACROSS RX AND CONSIDER IT AS VX.
5. FIRST REPLACE V1 VOLTAGE SOURCE FROM IT’S INTERNAL RESISTANCE i.e SHORT THE TERMINALS.
6. NOW SIMULATE THE CIRCUIT AND NOTE THE VOLTAGE ACROSS RX AND CONSIDER IT AS VX1.
7. NOW REPLACE SECOND ACTIVE SOURCE WITH IT’S INTERNAL RESISTANCE i.e CURRENT SOURCE IN OUR
CASE.
8. NOW SIMULATE THE CIRCUIT AND NOTE THE VOLTAGE ACROSS RX AND CONSIDER IT AS VX2.
9. DO ALGEBRAIC SUM OF VX1 AND VX2 TO GET THE REQUIRED VOLTAGE ACROSS RX.
10. COMPARE THIS VOLTAGE WITH ORIGINAL CIRCUIT AND LOOK FOR AN ERROR.
ANALYTICAL SOLUTION AND OBSERVATION :
NEXT PAGE------>
14
15
16
CIRCUIT DIAGRAM :
17
RESULT AND DISCUSSION:
WE HAVE SUCCESSFULLY VERIFIED SUPERPOSITION THEOREM AND SEEN THAT THERE IS SOME
AMOUNT OF ERROR IN VOLTAGE ACROSS RX MAYBE BECAUSE OF APPROXIMATION AT SOME
STEPS.
******
18
EXPERIMENT N0 - 4
AIM: TO VERIFY THEVENIN’S THEOREM FOR INDEPENDENT SOURCE DC NETWORK
USING PS-PICE.
S.NO COMPONENT NOTATION QUANTITY VALUES
1.
2.
3.
4.
RESISTORS
VOLTAGE SOURCE
CONNECTING WIRES
GROUND
R1,R2,R3,RL
V1
-
GND
4
1
1
1
15V
-
0V
50Ω,60Ω,70Ω,100Ω
COMPONENTS USED:
19
THEORY: THEVENIN’S THEOREM STATES THAT IT IS POSSIBLE TO SIMPLIFY ANY LINEAR
CIRCUIT, IRRESPECTIVE OF HOW COMPLEX IT IS, TO AN EQUIVALENT CIRCUIT WITH A SINGLE
VOLTAGE SOURCE AND A SERIES RESISTANCE.
THEVENIN THEOREM APPLICATIONS
•THEVENIN’S THEOREM IS USED IN THE ANALYSIS OF POWER SYSTEMS.
•THEVENIN’S THEOREM IS USED IN SOURCE MODELLING AND RESISTANCE MEASUREMENT USING THE
WHEATSTONE BRIDGE.
THEVENIN THEOREM LIMITATIONS
•THEVENIN’S THEOREM IS USED ONLY IN THE ANALYSIS OF LINEAR CIRCUITS.
•THE POWER DISSIPATION OF THE THEVENIN EQUIVALENT IS NOT IDENTICAL TO THE POWER DISSIPATION
OF THE REAL SYSTEM. 20
THEVENIN’S THEOREM EXAMPLE
STEP 1: FOR THE ANALYSIS OF THE ABOVE CIRCUIT USING THEVENIN’S
THEOREM, FIRSTLY REMOVE THE LOAD RESISTANCE AT THE CENTRE, IN
THIS CASE, 40 Ω.
STEP 2: REMOVE THE VOLTAGE SOURCES’ INTERNAL RESISTANCE BY
SHORTING ALL THE VOLTAGE SOURCES CONNECTED TO THE CIRCUIT, I.E.
V = 0. IF CURRENT SOURCES ARE PRESENT IN THE CIRCUIT, THEN
REMOVE THE INTERNAL RESISTANCE BY OPEN CIRCUITING THE
SOURCES. THIS STEP IS DONE TO HAVE AN IDEAL VOLTAGE SOURCE OR
AN IDEAL CURRENT SOURCE FOR THE ANALYSIS.
STEP 3: FIND THE EQUIVALENT RESISTANCE. IN THE EXAMPLE, THE
EQUIVALENT RESISTANCE OF THE CIRCUIT IS CALCULATED AS FOLLOWS:
WITH THE LOAD RESISTANCE REMOVED AND THE VOLTAGE SOURCE
SHORTED, THE EQUIVALENT RESISTANCE OF THE CIRCUIT IS
CALCULATED AS FOLLOWS:
THE RESISTOR 10 Ω IS PARALLEL TO 20 Ω, THEREFORE THE EQUIVALENT
RESISTANCE OF THE CIRCUIT IS:
RT=(R1×R2)/(R1+R2)=(20×10)/(20+10)=6.67Ω
STEP 4: FIND THE EQUIVALENT VOLTAGE.
21
TO CALCULATE THE EQUIVALENT VOLTAGE, RECONNECT THE
VOLTAGE SOURCES BACK INTO THE CIRCUIT. VS = VAB, THEREFORE
THE CURRENT FLOWING AROUND THE LOOP IS CALCULATED AS
FOLLOWS:
I=VR=(20V−10V)/(20Ω+10Ω)=0.33A
THE CALCULATED CURRENT IS COMMON TO BOTH RESISTORS, SO
THE VOLTAGE DROP ACROSS THE RESISTORS CAN BE CALCULATED
AS FOLLOWS:
VAB = 20 – (20 Ω X 0.33 A) = 13.33 V
OR,
VAB = 10 + (10 Ω X 0.33 A) = 13.33 V
THE VOLTAGE DROP ACROSS BOTH RESISTORS IS THE SAME.
STEP 5: DRAW THE THEVENIN’S EQUIVALENT CIRCUIT. THE
THEVENIN’S EQUIVALENT CIRCUIT CONSISTS OF A SERIES
RESISTANCE OF 6.67 Ω AND A VOLTAGE SOURCE OF 13.33 V.
THE CURRENT FLOWING IN THE CIRCUIT IS CALCULATED USING THE
FORMULA BELOW:
I=V/R=13.33V/(6.67Ω+40Ω)=0.286A
22
THEVENIN’S THEOREM CAN BE APPLIED TO BOTH AC AND DC CIRCUITS. BUT IT SHOULD BE NOTED THAT THIS
METHOD CAN ONLY BE APPLIED TO AC CIRCUITS CONSISTING OF LINEAR ELEMENTS LIKE RESISTORS,
INDUCTORS, CAPACITORS. LIKE THEVENIN’S EQUIVALENT RESISTANCE, EQUIVALENT THEVENIN’S IMPEDANCE IS
OBTAINED BY REPLACING ALL VOLTAGE SOURCES WITH THEIR INTERNAL IMPEDANCES.
CIRCUIT DIAGRAM:
ORIGINAL CIRCUIT
23
THEVENIN’S EQUIVALENT CIRCUIT
ANALYTICAL SOLUTION AND OBSERVATION :
NEXT PAGE ----->
24
25
METHODOLOGY:
1. SELECT THE ‘GET NEW PART’ BUTTON THEN COLLECT ALL THE REQUIRED COMPNENTS FOR YOUR
CIRCUIT.
2. CONNECT ALL THE COMPONENTS WITH THE HELP OF CONNECTING WIRES.
3. FIRST SIMULATED THE ORIGINAL CIRCUIT AND NOTE THE CURRENT IN RL.
4. NOW OPEN THE TERMINALS ACROSS RL IN THE CIRCUIT AND SIMULATE.
5. NOTE THE VOLTAGE ACROSS THE OPEN CIRCUITED BRANCH CONSIDER IT AS VTH.
6. NOW FOR RTH SHORT THE RL TERMINALS AND SIMULATE THE CIRCUIT.
7. NOTE THE CURRENT IN THE SHORT CIRCUITED BRANCH AND NAME IT AS ISC.
8. FOR RTH DIVIDE VTH BY ISC.
9. NOW DRAW THE THEVENIN’S EQUIVALENT CIRCUIT AND SIMULATE IT.
10. OBSERVE THE READING IN EQUIVALENT AND ORIGINAL CIRCUIT.
11. IF READINGS ARE CORRECT DO ERROR OBSERVATION BETWEEN SIMULATED AND ANALYTICAL VALUES.
26
RESULT AND DISCUSSION:
WE HAVE SUCCESSFULLY VERIFIED THEVENIN THEOREM AND THEIR IS NO ERROR IN VTH BUT THERE IS
-.13% ERROR IN ISC AND SIMILAR ERROR IN RTH i.e. .13% ERROR.THESE ERRORS ARE MAYBE BECAUSE OF
APPROXIMATION AT SOME STEPS.
27
*******
EXPERIMENT NO - 5
AIM: TO VERIFY MAXIMUM POWER TRANSFER THEOREM IN A GIVEN ELECTRICAL DC NETWORK USING
PS-PICE SIMULATION.
S.NO COMPONENT NOTATION QUANTITY VALUES
1.
2.
3.
4.
RESISTORS
VOLTAGE SOURCE
CONNECTING
WIRES
GROUND
R2,R4,R5,R6,R7
V1
-
GND
5
3
-
1
100V
-
0V
COMPONENTS:
28
Rx,100Ω,90Ω,150Ω,300Ω
THEORY: THE MAXIMUM POWER TRANSFER THEOREM IS NOT SO MUCH A MEANS OF ANALYSIS AS IT IS AN
AID TO SYSTEM DESIGN. SIMPLY STATED, THE MAXIMUM AMOUNT OF POWER WILL BE DISSIPATED BY A LOAD
RESISTANCE WHEN THAT LOAD RESISTANCE IS EQUAL TO THE THEVENIN/NORTON RESISTANCE OF THE
NETWORK SUPPLYING THE POWER. IF THE LOAD RESISTANCE IS LOWER OR HIGHER THAN THE
THEVENIN/NORTON RESISTANCE OF THE SOURCE NETWORK, ITS DISSIPATED POWER WILL BE LESS THAN THE
MAXIMUM.
THIS IS ESSENTIALLY WHAT IS AIMED FOR IN RADIO TRANSMITTER DESIGN, WHERE THE ANTENNA OR
TRANSMISSION LINE “IMPEDANCE” IS MATCHED TO FINAL POWER AMPLIFIER “IMPEDANCE” FOR MAXIMUM
RADIO FREQUENCY POWER OUTPUT. IMPEDANCE, THE OVERALL OPPOSITION TO AC AND DC CURRENT, IS
VERY SIMILAR TO RESISTANCE AND MUST BE EQUAL BETWEEN SOURCE AND LOAD FOR THE GREATEST
AMOUNT OF POWER TO BE TRANSFERRED TO THE LOAD. A LOAD IMPEDANCE THAT IS TOO HIGH WILL
RESULT IN LOW POWER OUTPUT. A LOAD IMPEDANCE THAT IS TOO LOW WILL NOT ONLY RESULT IN LOW
POWER OUTPUT BUT POSSIBLY OVERHEATING OF THE AMPLIFIER DUE TO THE POWER DISSIPATED IN ITS
INTERNAL (THEVENIN OR NORTON) IMPEDANCE.
MAXIMUM POWER DOESN’T MEAN MAXIMUM EFFICIENCY
MAXIMUM POWER TRANSFER DOES NOT COINCIDE WITH MAXIMUM EFFICIENCY. APPLICATION OF THE MAXIMUM
POWER TRANSFER THEOREM TO AC POWER DISTRIBUTION WILL NOT RESULT IN MAXIMUM OR EVEN HIGH
EFFICIENCY. THE GOAL OF HIGH EFFICIENCY IS MORE IMPORTANT FOR AC POWER DISTRIBUTION, WHICH DICTATES A
RELATIVELY LOW GENERATOR IMPEDANCE COMPARED TO THE LOAD IMPEDANCE.
29
CIRCUIT DIAGRAM:
30
METHODOLOGYAND ANALYTICAL SOLUTION:
1. SELECT THE ‘GET NEW PART’ BUTTON THEN
COLLECT ALL THE REQUIRED COMPNENTS FOR
YOUR CIRCUIT.
2. CONNECT ALL THE COMPONENTS WITH THE HELP OF
CONNECTING WIRES.
3. NOW TAKE R2 AS VARIABLE PARAMETER AND
SIMULATE THE CIRCUIT.
4. GRAPH WILL BE SHOWN IN OTHER WINDOW.
31
32
RESULT:
WE HAVE SUCCESSFULLY VERIFIED MAXIMUM POWER TRANSFER THEOREM AND
RESISTANCE AT WHICH MAXIMUM POWER CAME IS 360Ω. i.e. RTH=RX .
********

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CSL (1).pptx

  • 1. CIRCUIT SIMULATION -(E) LABORATORY REPORT 2021-2022 DEPARTMENT OF ELECTRICAL ENGINEERING SUBMITTED BY: ANSHUMAN SINGH 5th SEM ROLL NUMBER: 20UELE6006 1
  • 2. S.NO TITLE DATE PAGE NO. REMARKS 1. INTRODUCTION OF PS-PICE AND ITS LIBRARY FUNCTIONS. 2. TO VERIFY KCL & KVL IN A GIVEN ELECTRICAL DC NETWORK USING PS-PICE SIMULATIONS. 3. TO VERIFY SUPERPOSITION THEOREM FOR INDEPENDENT SOURCE DC NETWORK USING PS-PICE. TO VERIFY THEVENIN’S THEOREM FOR INDEPENDENT SOURCE DC NETWORK USING PS- PICE. 4. TO VERIFY MAXIMUM POWER TRANSFER THEOREM IN A GIVEN ELECTRICAL DC NETWORK USING PS-PICE SIMULATION. 5. 2
  • 3. EXPERIMENT - 1 AIM: INTRODUCTION OF PS-PICE AND ITS LIBRARY FUNCTIONS. THEORY: • SPICE WAS FIRST DEVELOPED AT THE UNIVERSITY OF CALIFORNIA, BERKELEY, IN THE EARLY 1970S. SUBSEQUENTLY AN IMPROVED VERSION SPICE 2 WAS AVAILABLE IN THE MID-1970S ESPECIALLY TO SUPPORT COMPUTER AIDED DESIGN. • PSPICE WAS RELEASED IN JANUARY 1984, AND WAS THE FIRST VERSION OF UC BERKELEY SPICE AVAILABLE ON AN IBM PERSONAL COMPUTER. PSPICE LATER INCLUDED A WAVEFORM VIEWER AND ANALYSER PROGRAM CALLED PROBE. SUBSEQUENT VERSIONS IMPROVED ON PERFORMANCE AND MOVED TO DEC/VAX MINICOMPUTERS, SUN WORKSTATIONS, APPLE MACINTOSH, AND MICROSOFT WINDOWS. VERSION 3.06 WAS RELEASED IN 1988, AND HAD A "STUDENT VERSION" AVAILABLE WHICH WOULD ALLOW A MAXIMUM OF UP TO TEN TRANSISTORS TO BE INSERTED. • ORCAD EE PSPICE IS A SPICE CIRCUIT SIMULATOR APPLICATION FOR SIMULATION AND VERIFICATION OF ANALOG AND MIXED-SIGNAL CIRCUITS.[17] PSPICE IS AN ACRONYM FOR PERSONAL SIMULATION PROGRAM WITH INTEGRATED CIRCUIT EMPHASIS. • PSPICE WAS A MODIFIED VERSION OF THE ACADEMICALLY DEVELOPED SPICE, AND WAS COMMERCIALIZED BY MICROSIM IN 1984. MICROSIM WAS PURCHASED BY ORCAD A DECADE LATER IN 1998. ORCAD PSPICE DESIGNER IS AVAILABLE IN TWO OPTIONS: PSPICE DESIGNER AND PSPICE DESIGNER PLUS. 3
  • 4. GENERAL GUIDELINE ON HOW TO USE PSPICE: THE GENERAL PROCEDURE FOR USING PSPICE CONSISTS OF 3 BASIC STPES. STEP 1 THE USER DRAWS THE CIRCUIT IN SCHEMATIC FORM WHICH HE WANTS TO SIMULATE. STEP 2 THE USER SPECIFIES THE TYPE OF ANALYSIS DESIRED, AND DIRECTS PSPICE TO PERFORM THAT ANALYSIS. THIS CAN, FOR INSTANCE, BE DC ANALYSIS, AC ANALYSIS, TRANSIENT ANALYSIS... STEP 3 THE USER INSTRUCTS THE COMPUTER TO PRINT OR PLOT THE RESULTS OF THE ANALYSIS. IN THIS STEP, THE USER SEES THE GRAPHICAL RESULTS OF THE ANALYSIS DONE. FOR EXAMPLE, HE CAN SEE THE GRAPH OF THE OUTPUT VOLTAGE VS. OUTPUT CURRENT (V VS. I), OR ANY DATA WHICH HE WANTS TO ANALYZE. 4
  • 5. EXPERIMENT- 2 AIM: TO VERIFY KCL & KVL IN A GIVEN ELECTRICAL DC NETWORK USING PS- PICE SIMULATIONS. COMPONENTS USED: S.NO COMPONENT NOTATION QUANTITY VALUES 1. 2. 3. 4. RESISTORS VOLTAGE SOURCE CURRENT SOURCE GROUND R1,R2,R3,R4,R5 V1,V2,V3 I3 GND 5 3 1 1 10Ω,8Ω,7Ω,17Ω,100Ω 10V,10V,15V 15A 0V 5
  • 6. THEORY: IN 1845, A GERMAN PHYSICIST, GUSTAV KIRCHHOFF DEVELOPED A PAIR OR SET OF RULES OR LAWS WHICH DEAL WITH THE CONSERVATION OF CURRENT AND ENERGY WITHIN ELECTRICAL CIRCUITS. THESE TWO RULES ARE COMMONLY KNOWN AS: KIRCHHOFFS CIRCUIT LAWS WITH ONE OF KIRCHHOFFS LAWS DEALING WITH THE CURRENT FLOWING AROUND A CLOSED CIRCUIT, KIRCHHOFFS CURRENT LAW, (KCL) WHILE THE OTHER LAW DEALS WITH THE VOLTAGE SOURCES PRESENT IN A CLOSED CIRCUIT, KIRCHHOFFS VOLTAGE LAW, (KVL). KIRCHHOFFS FIRST LAW – THE CURRENT LAW, (KCL): KIRCHHOFFS CURRENT LAW OR KCL, STATES THAT THE “TOTAL CURRENT OR CHARGE ENTERING A JUNCTION OR NODE IS EXACTLY EQUAL TO THE CHARGE LEAVING THE NODE AS IT HAS NO OTHER PLACE TO GO EXCEPT TO LEAVE, AS NO CHARGE IS LOST WITHIN THE NODE“. IN OTHER WORDS THE ALGEBRAIC SUM OF ALL THE CURRENTS ENTERING AND LEAVING A NODE MUST BE EQUAL TO ZERO, I(EXITING) + I(ENTERING) = 0. THIS IDEA BY KIRCHHOFF IS COMMONLY KNOWN AS THE CONSERVATION OF CHARGE. 6
  • 7. HERE, THE THREE CURRENTS ENTERING THE NODE, I1, I2, I3 ARE ALL POSITIVE IN VALUE AND THE TWO CURRENTS LEAVING THE NODE, I4 AND I5 ARE NEGATIVE IN VALUE. THEN THIS MEANS WE CAN ALSO REWRITE THE EQUATION AS; I1 + I2 + I3 – I4 – I5 = 0 THE TERM NODE IN AN ELECTRICAL CIRCUIT GENERALLY REFERS TO A CONNECTION OR JUNCTION OF TWO OR MORE CURRENT CARRYING PATHS OR ELEMENTS SUCH AS CABLES AND COMPONENTS. ALSO FOR CURRENT TO FLOW EITHER IN OR OUT OF A NODE A CLOSED CIRCUIT PATH MUST EXIST. WE CAN USE KIRCHHOFF’S CURRENT LAW WHEN ANALYSING PARALLEL CIRCUITS. KIRCHHOFFS SECOND LAW – THE VOLTAGE LAW, (KVL) KIRCHHOFFS VOLTAGE LAW OR KVL, STATES THAT “IN ANY CLOSED LOOP NETWORK, THE TOTAL VOLTAGE AROUND THE LOOP IS EQUAL TO THE SUM OF ALL THE VOLTAGE DROPS WITHIN THE SAME LOOP” WHICH IS ALSO EQUAL TO ZERO. IN OTHER WORDS THE ALGEBRAIC SUM OF ALL VOLTAGES WITHIN THE LOOP MUST BE EQUAL TO ZERO. THIS IDEA BY KIRCHHOFF IS KNOWN AS THE CONSERVATION OF ENERGY. 7
  • 8. STARTING AT ANY POINT IN THE LOOP CONTINUE IN THE SAME DIRECTION NOTING THE DIRECTION OF ALL THE VOLTAGE DROPS, EITHER POSITIVE OR NEGATIVE, AND RETURNING BACK TO THE SAME STARTING POINT. IT IS IMPORTANT TO MAINTAIN THE SAME DIRECTION EITHER CLOCKWISE OR ANTI-CLOCKWISE OR THE FINAL VOLTAGE SUM WILL NOT BE EQUAL TO ZERO. WE CAN USE KIRCHHOFF’S VOLTAGE LAW WHEN ANALYSING SERIES CIRCUITS. CIRCUIT DIAGRAM: 8
  • 9. METHODOLOGY: 1. SELECT THE ‘GET NEW PART’ BUTTON THEN COLLECT ALL THE REQUIRED COMPNENTS FOR YOUR CIRCUIT. 2. CONNECT ALL THE COMPONENTS WITH THE HELP OF CONNECTING WIRES. 3. NOW NAME THE COMPONENTS AND GIVE RESPECTIVE VALUES TO THE COMPONENTS. 4. FIRST SIMULATE THE CIRCUIT AND NOTE CURRENT IN EACH BRANCH AND ANALYSE CURRENT AT A PARTICULAR NODE A. 5. NOW ADD A BRANCH IN CIRCUIT WITH RESISTOR R5. 6. NOW SIMULATE THE CIRCUIT AND NOTE CURRENT IN EACH BRANCH AND ANALYSE CURRENT AT PREVIOUS NODE. 7. OBSERVE IF THEIR IS ANY DIFFERENCE IN BOTH CASES. 8. SIMILARLY,OBSERVE VOLTAGE DROP IN ANY CLOSE LOOP IN BOTH CASES AND OBSERVE THE DIFFERENCE. ANALTYTICAL SOLUTION AND OBSERVATION: NEXT PAGE-----> 9
  • 10. 10
  • 11. RESULT AND DISCUSSION: WE HAVE VERIFIED KVL AND KCL AND SEEN THAT THERE IS SOME AMOUNT OF ERROR BETWEEN ANALYTICAL VALUE AND SIMULATED VALUE THAT MAYBE BECAUSE OF APPROXIMATION AT SOME STEPS. ALSO AFTER ADDING ONE MORE BRANCH WITH RESISTOR R5 KCL AND KVL IS STILL VALID BUT THE VALUES OF CURRENT CHANGES BECAUSE SOME AMOUNT OF CURRENT GOES INTO THE R5 BRANCH AND ALSO VOLTAGE AT NODE CHANGES. ******* 11
  • 12. EXPERIMENT - 3 AIM: TO VERIFY SUPERPOSITION THEOREM FOR DEPENDENT AND INDEPENDENT SOURCE DC NETWORK USING PS-PICE. S.NO COMPONENT NOTATION QUANTITY VALUES 1. 2. 3. 4. RESISTORS VOLTAGE SOURCE CURRENT SOURCE GROUND R1,RX,R3,R4,R5 V1,V2,V3 I3 GND 5 3 1 1 10V,10V,15V 15A 0V COMPONENTS USED : 70Ω,40Ω,30Ω,30Ω,50Ω 12
  • 13. THEORY : SUPERPOSITION THEOREM STATES THAT IN ANY LINEAR, BILATERAL NETWORK WHERE MORE THAN ONE SOURCE IS PRESENT, THE RESPONSE ACROSS ANY ELEMENT IN THE CIRCUIT, IS THE SUM OF THE RESPONSES OBTAINED FROM EACH SOURCE CONSIDERED SEPARATELY WHILE ALL OTHER SOURCES ARE REPLACED BY THEIR INTERNAL RESISTANCE. SUPERPOSITION THEOREM IS A CIRCUIT ANALYSIS THEOREM THAT IS USED TO SOLVE THE NETWORK WHERE TWO OR MORE SOURCES ARE PRESENT AND CONNECTED. TO CALCULATE THE INDIVIDUAL CONTRIBUTION OF EACH SOURCE IN A CIRCUIT, THE OTHER SOURCE MUST BE REPLACED OR REMOVED WITHOUT AFFECTING THE FINAL RESULT. WHILE REMOVING A VOLTAGE SOURCE, ITS VALUE IS SET TO ZERO. THIS IS DONE BY REPLACING THE VOLTAGE SOURCE WITH A SHORT CIRCUIT. WHEN REMOVING A CURRENT SOURCE, ITS VALUE IS SET TO ZERO. THIS IS DONE BY REPLACING THE CURRENT SOURCE WITH AN OPEN CIRCUIT. THE SUPERPOSITION THEOREM IS VERY IMPORTANT IN CIRCUIT ANALYSIS BECAUSE IT CONVERTS A COMPLEX CIRCUIT INTO A NORTON OR THEVENIN EQUIVALENT CIRCUIT. GUIDELINES TO KEEP IN MIND WHILE USING THE SUPERPOSITION THEOREM •WHEN YOU SUM THE INDIVIDUAL CONTRIBUTIONS OF EACH SOURCE, YOU SHOULD BE CAREFUL WHILE ASSIGNING SIGNS TO THE QUANTITIES. IT IS SUGGESTED TO ASSIGN A REFERENCE DIRECTION TO EACH UNKNOWN QUANTITY. IF A CONTRIBUTION FROM A SOURCE HAS THE SAME DIRECTION AS THE REFERENCE DIRECTION, IT HAS A POSITIVE SIGN IN THE SUM; IF IT HAS THE OPPOSITE DIRECTION, THEN A NEGATIVE SIGN. •TO USE THE SUPERPOSITION THEOREM WITH CIRCUIT CURRENTS AND VOLTAGES, ALL THE COMPONENTS MUST BE LINEAR. •IT SHOULD BE NOTED THAT THE SUPERPOSITION THEOREM DOES NOT APPLY TO POWER, AS POWER IS NOT A LINEAR QUANTITY. 13
  • 14. METHODOLOGY: 1. SELECT THE ‘GET NEW PART’ BUTTON THEN COLLECT ALL THE REQUIRED COMPNENTS FOR YOUR CIRCUIT. 2. CONNECT ALL THE COMPONENTS WITH THE HELP OF CONNECTING WIRES. 3. NOW NAME THE COMPONENTS FOR YOUR CONVENIENCE WE HAVE HERE NAMED THE TARGETED RESISTOR RX ACROSS WHICH WE HAVE TO FIND VOLTAGE DROP. 4. FIRST SIMULATE THE ORIGINAL CIRCUIT AND NOTE THE VOLTAGE ACROSS RX AND CONSIDER IT AS VX. 5. FIRST REPLACE V1 VOLTAGE SOURCE FROM IT’S INTERNAL RESISTANCE i.e SHORT THE TERMINALS. 6. NOW SIMULATE THE CIRCUIT AND NOTE THE VOLTAGE ACROSS RX AND CONSIDER IT AS VX1. 7. NOW REPLACE SECOND ACTIVE SOURCE WITH IT’S INTERNAL RESISTANCE i.e CURRENT SOURCE IN OUR CASE. 8. NOW SIMULATE THE CIRCUIT AND NOTE THE VOLTAGE ACROSS RX AND CONSIDER IT AS VX2. 9. DO ALGEBRAIC SUM OF VX1 AND VX2 TO GET THE REQUIRED VOLTAGE ACROSS RX. 10. COMPARE THIS VOLTAGE WITH ORIGINAL CIRCUIT AND LOOK FOR AN ERROR. ANALYTICAL SOLUTION AND OBSERVATION : NEXT PAGE------> 14
  • 15. 15
  • 16. 16
  • 18. RESULT AND DISCUSSION: WE HAVE SUCCESSFULLY VERIFIED SUPERPOSITION THEOREM AND SEEN THAT THERE IS SOME AMOUNT OF ERROR IN VOLTAGE ACROSS RX MAYBE BECAUSE OF APPROXIMATION AT SOME STEPS. ****** 18
  • 19. EXPERIMENT N0 - 4 AIM: TO VERIFY THEVENIN’S THEOREM FOR INDEPENDENT SOURCE DC NETWORK USING PS-PICE. S.NO COMPONENT NOTATION QUANTITY VALUES 1. 2. 3. 4. RESISTORS VOLTAGE SOURCE CONNECTING WIRES GROUND R1,R2,R3,RL V1 - GND 4 1 1 1 15V - 0V 50Ω,60Ω,70Ω,100Ω COMPONENTS USED: 19
  • 20. THEORY: THEVENIN’S THEOREM STATES THAT IT IS POSSIBLE TO SIMPLIFY ANY LINEAR CIRCUIT, IRRESPECTIVE OF HOW COMPLEX IT IS, TO AN EQUIVALENT CIRCUIT WITH A SINGLE VOLTAGE SOURCE AND A SERIES RESISTANCE. THEVENIN THEOREM APPLICATIONS •THEVENIN’S THEOREM IS USED IN THE ANALYSIS OF POWER SYSTEMS. •THEVENIN’S THEOREM IS USED IN SOURCE MODELLING AND RESISTANCE MEASUREMENT USING THE WHEATSTONE BRIDGE. THEVENIN THEOREM LIMITATIONS •THEVENIN’S THEOREM IS USED ONLY IN THE ANALYSIS OF LINEAR CIRCUITS. •THE POWER DISSIPATION OF THE THEVENIN EQUIVALENT IS NOT IDENTICAL TO THE POWER DISSIPATION OF THE REAL SYSTEM. 20
  • 21. THEVENIN’S THEOREM EXAMPLE STEP 1: FOR THE ANALYSIS OF THE ABOVE CIRCUIT USING THEVENIN’S THEOREM, FIRSTLY REMOVE THE LOAD RESISTANCE AT THE CENTRE, IN THIS CASE, 40 Ω. STEP 2: REMOVE THE VOLTAGE SOURCES’ INTERNAL RESISTANCE BY SHORTING ALL THE VOLTAGE SOURCES CONNECTED TO THE CIRCUIT, I.E. V = 0. IF CURRENT SOURCES ARE PRESENT IN THE CIRCUIT, THEN REMOVE THE INTERNAL RESISTANCE BY OPEN CIRCUITING THE SOURCES. THIS STEP IS DONE TO HAVE AN IDEAL VOLTAGE SOURCE OR AN IDEAL CURRENT SOURCE FOR THE ANALYSIS. STEP 3: FIND THE EQUIVALENT RESISTANCE. IN THE EXAMPLE, THE EQUIVALENT RESISTANCE OF THE CIRCUIT IS CALCULATED AS FOLLOWS: WITH THE LOAD RESISTANCE REMOVED AND THE VOLTAGE SOURCE SHORTED, THE EQUIVALENT RESISTANCE OF THE CIRCUIT IS CALCULATED AS FOLLOWS: THE RESISTOR 10 Ω IS PARALLEL TO 20 Ω, THEREFORE THE EQUIVALENT RESISTANCE OF THE CIRCUIT IS: RT=(R1×R2)/(R1+R2)=(20×10)/(20+10)=6.67Ω STEP 4: FIND THE EQUIVALENT VOLTAGE. 21
  • 22. TO CALCULATE THE EQUIVALENT VOLTAGE, RECONNECT THE VOLTAGE SOURCES BACK INTO THE CIRCUIT. VS = VAB, THEREFORE THE CURRENT FLOWING AROUND THE LOOP IS CALCULATED AS FOLLOWS: I=VR=(20V−10V)/(20Ω+10Ω)=0.33A THE CALCULATED CURRENT IS COMMON TO BOTH RESISTORS, SO THE VOLTAGE DROP ACROSS THE RESISTORS CAN BE CALCULATED AS FOLLOWS: VAB = 20 – (20 Ω X 0.33 A) = 13.33 V OR, VAB = 10 + (10 Ω X 0.33 A) = 13.33 V THE VOLTAGE DROP ACROSS BOTH RESISTORS IS THE SAME. STEP 5: DRAW THE THEVENIN’S EQUIVALENT CIRCUIT. THE THEVENIN’S EQUIVALENT CIRCUIT CONSISTS OF A SERIES RESISTANCE OF 6.67 Ω AND A VOLTAGE SOURCE OF 13.33 V. THE CURRENT FLOWING IN THE CIRCUIT IS CALCULATED USING THE FORMULA BELOW: I=V/R=13.33V/(6.67Ω+40Ω)=0.286A 22
  • 23. THEVENIN’S THEOREM CAN BE APPLIED TO BOTH AC AND DC CIRCUITS. BUT IT SHOULD BE NOTED THAT THIS METHOD CAN ONLY BE APPLIED TO AC CIRCUITS CONSISTING OF LINEAR ELEMENTS LIKE RESISTORS, INDUCTORS, CAPACITORS. LIKE THEVENIN’S EQUIVALENT RESISTANCE, EQUIVALENT THEVENIN’S IMPEDANCE IS OBTAINED BY REPLACING ALL VOLTAGE SOURCES WITH THEIR INTERNAL IMPEDANCES. CIRCUIT DIAGRAM: ORIGINAL CIRCUIT 23
  • 24. THEVENIN’S EQUIVALENT CIRCUIT ANALYTICAL SOLUTION AND OBSERVATION : NEXT PAGE -----> 24
  • 25. 25
  • 26. METHODOLOGY: 1. SELECT THE ‘GET NEW PART’ BUTTON THEN COLLECT ALL THE REQUIRED COMPNENTS FOR YOUR CIRCUIT. 2. CONNECT ALL THE COMPONENTS WITH THE HELP OF CONNECTING WIRES. 3. FIRST SIMULATED THE ORIGINAL CIRCUIT AND NOTE THE CURRENT IN RL. 4. NOW OPEN THE TERMINALS ACROSS RL IN THE CIRCUIT AND SIMULATE. 5. NOTE THE VOLTAGE ACROSS THE OPEN CIRCUITED BRANCH CONSIDER IT AS VTH. 6. NOW FOR RTH SHORT THE RL TERMINALS AND SIMULATE THE CIRCUIT. 7. NOTE THE CURRENT IN THE SHORT CIRCUITED BRANCH AND NAME IT AS ISC. 8. FOR RTH DIVIDE VTH BY ISC. 9. NOW DRAW THE THEVENIN’S EQUIVALENT CIRCUIT AND SIMULATE IT. 10. OBSERVE THE READING IN EQUIVALENT AND ORIGINAL CIRCUIT. 11. IF READINGS ARE CORRECT DO ERROR OBSERVATION BETWEEN SIMULATED AND ANALYTICAL VALUES. 26
  • 27. RESULT AND DISCUSSION: WE HAVE SUCCESSFULLY VERIFIED THEVENIN THEOREM AND THEIR IS NO ERROR IN VTH BUT THERE IS -.13% ERROR IN ISC AND SIMILAR ERROR IN RTH i.e. .13% ERROR.THESE ERRORS ARE MAYBE BECAUSE OF APPROXIMATION AT SOME STEPS. 27 *******
  • 28. EXPERIMENT NO - 5 AIM: TO VERIFY MAXIMUM POWER TRANSFER THEOREM IN A GIVEN ELECTRICAL DC NETWORK USING PS-PICE SIMULATION. S.NO COMPONENT NOTATION QUANTITY VALUES 1. 2. 3. 4. RESISTORS VOLTAGE SOURCE CONNECTING WIRES GROUND R2,R4,R5,R6,R7 V1 - GND 5 3 - 1 100V - 0V COMPONENTS: 28 Rx,100Ω,90Ω,150Ω,300Ω
  • 29. THEORY: THE MAXIMUM POWER TRANSFER THEOREM IS NOT SO MUCH A MEANS OF ANALYSIS AS IT IS AN AID TO SYSTEM DESIGN. SIMPLY STATED, THE MAXIMUM AMOUNT OF POWER WILL BE DISSIPATED BY A LOAD RESISTANCE WHEN THAT LOAD RESISTANCE IS EQUAL TO THE THEVENIN/NORTON RESISTANCE OF THE NETWORK SUPPLYING THE POWER. IF THE LOAD RESISTANCE IS LOWER OR HIGHER THAN THE THEVENIN/NORTON RESISTANCE OF THE SOURCE NETWORK, ITS DISSIPATED POWER WILL BE LESS THAN THE MAXIMUM. THIS IS ESSENTIALLY WHAT IS AIMED FOR IN RADIO TRANSMITTER DESIGN, WHERE THE ANTENNA OR TRANSMISSION LINE “IMPEDANCE” IS MATCHED TO FINAL POWER AMPLIFIER “IMPEDANCE” FOR MAXIMUM RADIO FREQUENCY POWER OUTPUT. IMPEDANCE, THE OVERALL OPPOSITION TO AC AND DC CURRENT, IS VERY SIMILAR TO RESISTANCE AND MUST BE EQUAL BETWEEN SOURCE AND LOAD FOR THE GREATEST AMOUNT OF POWER TO BE TRANSFERRED TO THE LOAD. A LOAD IMPEDANCE THAT IS TOO HIGH WILL RESULT IN LOW POWER OUTPUT. A LOAD IMPEDANCE THAT IS TOO LOW WILL NOT ONLY RESULT IN LOW POWER OUTPUT BUT POSSIBLY OVERHEATING OF THE AMPLIFIER DUE TO THE POWER DISSIPATED IN ITS INTERNAL (THEVENIN OR NORTON) IMPEDANCE. MAXIMUM POWER DOESN’T MEAN MAXIMUM EFFICIENCY MAXIMUM POWER TRANSFER DOES NOT COINCIDE WITH MAXIMUM EFFICIENCY. APPLICATION OF THE MAXIMUM POWER TRANSFER THEOREM TO AC POWER DISTRIBUTION WILL NOT RESULT IN MAXIMUM OR EVEN HIGH EFFICIENCY. THE GOAL OF HIGH EFFICIENCY IS MORE IMPORTANT FOR AC POWER DISTRIBUTION, WHICH DICTATES A RELATIVELY LOW GENERATOR IMPEDANCE COMPARED TO THE LOAD IMPEDANCE. 29
  • 31. METHODOLOGYAND ANALYTICAL SOLUTION: 1. SELECT THE ‘GET NEW PART’ BUTTON THEN COLLECT ALL THE REQUIRED COMPNENTS FOR YOUR CIRCUIT. 2. CONNECT ALL THE COMPONENTS WITH THE HELP OF CONNECTING WIRES. 3. NOW TAKE R2 AS VARIABLE PARAMETER AND SIMULATE THE CIRCUIT. 4. GRAPH WILL BE SHOWN IN OTHER WINDOW. 31
  • 32. 32 RESULT: WE HAVE SUCCESSFULLY VERIFIED MAXIMUM POWER TRANSFER THEOREM AND RESISTANCE AT WHICH MAXIMUM POWER CAME IS 360Ω. i.e. RTH=RX . ********