1 / 44

Advanced electronics

Advanced electronics. Definitions. EMF Electromotive "force" is not considered a force measured in newtons, but a potential, or energy per unit of charge, measured in volts PD

signa
Download Presentation

Advanced electronics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Advanced electronics

  2. Definitions • EMF • Electromotive "force" is not considered a force measured in newtons, but a potential, or energy per unit of charge, measured in volts • PD • Potential difference measured between two points (eg across a component) if a measure of the energy of electric charge between the two points also measured in volts

  3. Definitions • Current The flow of electric charge • Resistance The resistance to current • Capacitor Store charge in circuit

  4. Definitions • A circuit with a number of ‘elements’ or ‘branches’ is called a NETWORK A network which has one or more sources of EMF is said to be an ACTIVE circuit A network with no source of EMF is said to be PASSIVE

  5. Definitions • Active: Those devices or components which produce energy in the form of Voltage or Current are called as Active Components

  6. Definitions • Passive: Those devices or components which do not produce energy are known as Passive. Some components, which may may store or maintain Energy in the form of Voltage or Current are known as Passive Components

  7. Double subscript notation IAB A B VBC EDA D C

  8. Kirchoff’s first law

  9. Kirchoff’s first law The total current is shared by the components in a parallel circuit A1 = A4 = A2 + A3

  10. Kirchoff’ second law The sum of all the pd’s around the circuit are equal to the EMF of the supply In this example we are ignoring the internal resistance of the battery 12 V I =2A 2Ω 4Ω IxR = 4V IxR = 8V

  11. Kirchoff’s second law This time we are taking the internal resistance of the battery into consideration The sum of the pd’s across the two resistors does not equal the EMF of the cell. The current has dropped to 1.5 A 12 V I =1.5A 2Ω 4Ω I x R = 3V I x R = 6V

  12. Kirchoff’s second law • Task • Using what you know about Kirchoff’s second law work out the internal resistance of the battery,

  13. Kirchoff’s second law This time we are taking the internal resistance of the battery into consideration The sum of the pd’s across the two resistors does not equal the EMF of the cell. There is a 3 V across over the internal resistance r= 2Ω 12 V I =1.5A 2Ω 4Ω I x R = 3V I x R = 6V

  14. Kirchoff’ second law U s i n g K i r c h h o f f ’ s s e c o n d l a w e r 1 A D I I R B C I e I I = R + r f o r l o o p A B C D 1

  15. Kirchoff’s second law Example: A circuit consists of a cell of emf 1.6 V in series with a resistance 2.0 Ω connected to a resistor of resistance 3.0 Ω in parallel with a resistor of resistance 6.0 Ω. Determine the total current drawn from the cell, the potential difference across the 3.0 Ω resistor and the current I1

  16. Solution Total resistance of the parallel resistors = (R1 x R2)/R1 +R2 (3 x 6)/3 + 6 18/9 = 2Ω This is in series with the 2Ω internal resistance Total resistance 4Ω

  17. Total current = V/R = 1.6/4 = 0.4A Pd from cell (1.6v) = Pd across parallel set + Pd across internal resistance = (total current x Rparallel set) + (total current x R int) = (total current x 2) + (total current x 2) = 1.6v (total current x Rparallel set) = 0.8v Pd across the 3Ω = 0.8v

  18. Current through the 3 resistor • = V/R • = 0.8/3 = 0.267A

  19. Kirchoff’s second law Current directions IAB, IAD and IBD Develop expressions for the following meshes ABC, supply voltage A ABDA BDCB The resistance of monitoring device M = 80Ω B 20 Ω X A C M 40 Ω 30 Ω D 4 V

  20. Kirchoff’s second law 1, 4 = 20IAB + X(IAB – IBD) 2, 0 = 20IAB + 80IBD – 40IAD 3, 0 = 80IBD + 30(IBD+IAD) – X(IAB-IBD)80BD +30IBD + 30AD – XIAB + XIBD = 110IBD + XIBD + 30 IAD - XIAB

  21. If X = 60Ω calculate the current flowing through the monitoring device 1, 4 = 20IAB + 60(IAB – IBD) 4 = 20IAB + 60IAB – 60IBD 4 = 80IAB – 60IBD 2, 0 = 20IAB + 80IBD – 40I AD 3, 0 =110IBD + 60IBD + 30 IAD – 60IAB 0 =170IBD + 30 IAD – 60IAB Multiply 2, by 3 and 3, by 4

  22. Kirchoff’s second law 0 = 60IAB + 240IBD – 120I AD Add them together 0 =920IBD – 180IAB 920IBD = 180IAB 5.1IBD = IAB Substitute for IAB in 1,

  23. Kirchoff’s second law 4 = 408IBD – 60IBD 4 = 348IBD IBD = 0.0115A Current through monitoring device 11.5 mA

  24. Kirchoff’s second law example 2 Current directions IAB, IAD and IBD Develop expressions for the following meshes ABC, supply voltage A ABDA BDCB Find the current through the 50Ω resistor B 10 Ω 25Ω A C 50Ω 15 Ω 30 Ω D 6 V

  25. Pd from A – C = 6V Pd across the 10 Ω resistor = current x resistance = 10x IAB (10IAB) Pd across the 25Ω resistor = 25(IAB – IBD) = 25IAB –25IBD The pd across both the resistors (ABC) is 10IAB + 25IAB –25IBD = 35IAB –25IBD = 6V Kirchoff’s second law example 2 B 10 Ω 25Ω A C 50Ω 15 Ω 30 Ω D 6 V

  26. Pd from ABDA= 0V Pd across the 50Ω resistor = current x resistance = 50x IBD (30IBD) Pd across the 15Ω resistor = -15IAD The pd across both the 10Ω resistor is 10IAB PD of the mesh ABDA 50IBD +10IAB -15IAD = 0V Kirchoff’s second law B 10 Ω 25Ω A C 50Ω 15 Ω 30 Ω D 6 V

  27. =105IBD + 30IAD - 25IABPd from BDCB= 0V Pd across the 50Ω resistor = current x resistance =50 x IBD (50IBD) The pd across both the 30Ω resistor is 30IAD + 30IBD Pd across the 25Ω resistor = -(25IAB –25IBD) PD of the mesh BDCD 50IBD +30IAD +30IBD -25IAB + 25IBD = 0 Kirchoff’s second law B 10 Ω 25Ω A C 50Ω 15 Ω 30 Ω D 6 V

  28. We now have 3 equations 1, 35IAB –25IBD = 6V 2, 50IBD +10IAB -15IAD = 0V 3,105IBD + 30IAD – 25AB =0V Multiply equation 2, by2 and call it 4, 4,100IBD +20IAB -30IAD = 0V add 3, and 4, 5, 205IBD - 5IAB = 0V 6, 205IBD = 5IAB

  29. 6, 205IBD = 5IAB IAB = 41IBD go back to equation 1and substitute 35IAB –25IBD = 6 1435I BD –25IBD = 6 1410I BD = 6 IBD =.000425amps • 0.4mA

  30. Thevenin’s theorem • Thevenin’s theorem • "Any linear circuit containing several voltages and resistances can be replaced by just a Single Voltage in series with a Single Resistor". In other words, it is possible to simplify any "Linear" circuit, no matter how complex, to an equivalent circuit with just a single voltage source in series with a resistance connected to a load as shown below. Thevenins Theorem is especially useful in analysing power or battery systems and other interconnected circuits where it will have an effect on the adjoining part of the circuit.

  31. Thevenin’s theorem R1 R3 A R LOAD R2 V1 B Consider a circuit consisting of a power source and resistors

  32. Thevenin’s theorem R1 R3 A VTH R2 V1 B Thevenin’s voltage (VTH) is the open circuit voltage

  33. R1 and R2 act as a potential divider Thevenin’s theorem R1 R3 A VTH R2 V1 B Thevenin’s voltage (VTH) = V1xR2/(R1 + R2). (R3 has no affect because there is no current through it)

  34. Thevenin’s theorem R1 R3 A R2 V1 B Thevenin’s resistance (r) = R3 + (R1 xR2/(R1 + R2)). (All voltage sources are short circuited and all current sources open circuited)

  35. Thevenin’s theorem Example 2Ω 3Ω A 10Ω 4Ω 4 Volts B Calculate VTH ,r and the current through the 10Ω load

  36. Thevenin’s theorem Example 2Ω 3Ω A 10Ω 4Ω 4 Volts B VTH = V1xR2/(R1 + R2). = 4 x 4/6 = 2.6 Volts

  37. Thevenin’s theorem Example 2Ω 3Ω A 10Ω 4Ω 4 Volts B r = 3 + (8/6) = 4.3Ω

  38. Thevenin’s theorem Example 2Ω 3Ω A 10Ω 4Ω 4 Volts B Current through the load = V/I = 2.6/(4.3 +10) = 0.18A

  39. Thevenin’s theorem with two power sources A Load E1 E2 B E1 = 8V with internal resistance 4Ω E2 = 6V with internal resistance 6Ω Load = 12 Ω Use Thevenin’s Theorem to find the current through the load

  40. Thevenin’s theorem with two power sources A Load E1 E2 B Using the theorem with the load disconnected the current circulating E1 and E2 IE1E2 = (E1 – E2) / ( r1 +r2) = (8-6) / (4+6) = 0.2A

  41. Thevenin’s theorem with two power sources A Load E1 E2 B Hence equivalent emf of sources =E1 – I1r1 = 8 – ( 0.2 x 4) = 7.2 V

  42. Thevenin’s theorem with two power sources Load E1 E2 Total internal resistance of sources in parallel 1/R= ¼ + 1/6 =3/12 + 2/12 = 5/12 R = 12/5 = 2.4Ω

  43. Thevenin’s theorem with two power sources Load E1 E2 IL = 7.2 / (2.4 + 12) =7.2/ 14.4 = 0.5 A

  44. Hence equivalent emf of sources • =E1 – I1r1 = 8 – ( 0.2 x 4) = 7.2 V • Total internal resistance of sources in parallel 1/R= ¼ + 1/6 =3/12 + 2/12 = 5/12 • IL = 7.2 / (2.4 + 12) =7.2/ 14.4= 0.5 A

More Related