advanced electronics
DESCRIPTION
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 - PowerPoint PPT PresentationTRANSCRIPT
Advanced electronics
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
Definitions
Current The flow of electric charge Resistance The resistance to current
Capacitor Store charge in circuit
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
Definitions
Active: Those devices or components which produce energy in the form of Voltage or Current are called as Active Components
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
Definitions
Double subscript notation
A
D C
B
EDA
IAB
VBC
Kirchoff’s first law
Kirchoff’s first lawThe total current is shared by
the components in a parallel
circuitA1 = A4 = A2 + A3
Kirchoff’ second law
I =2A
12 V
2Ω 4Ω
IxR = 8VIxR = 4V
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
Kirchoff’s second law
I =1.5A
12 V
2Ω 4Ω
I x R = 6VI x R = 3V
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
Task
Using what you know about Kirchoff’s second law work out the internal resistance of the battery,
Kirchoff’s second law
Kirchoff’s second law
I =1.5A
12 V
2Ω 4Ω
I x R = 6VI x R = 3V
r= 2Ω
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
Kirchoff’ second law
A
er1
B C
D
I
I
e = I R + I r1 for loop ABCD
Using Kirchhoff’s second law
I
R
Kirchoff’s second law
I1 + I2
I1
I2
1.6 V2.0
3.0
6.0
I1 + I2
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
SolutionTotal resistance of the parallel resistors = (R1 x R2)/R1 +R2
(3 x 6)/3 + 618/9 = 2Ω
This is in series with the 2Ω internal resistanceTotal resistance 4Ω
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
Current through the 3 resistor = V/R
= 0.8/3 = 0.267A
Kirchoff’s second law
MA
B
C
D
20 Ω
40 Ω 30 Ω
X
4 V
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
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
Kirchoff’s second law
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
If X = 60Ω calculate the current flowing through the monitoring device
0 = 60IAB + 240IBD – 120I AD
Add them together
0 =920IBD – 180IAB
920IBD = 180IAB
5.1IBD = IAB
Substitute for IAB in 1,
Kirchoff’s second law
4 = 408IBD – 60IBD
4 = 348IBD
IBD = 0.0115A
Current through monitoring device
11.5 mA
Kirchoff’s second law
Kirchoff’s second law example 2
50ΩA
B
C
D
10 Ω
15 Ω 30 Ω
25Ω
6 V
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
Kirchoff’s second law example 2
50ΩA
B
C
D
10 Ω
15 Ω 30 Ω
25Ω
6 V
Pd from A – C = 6V
Pd across the 10 Ωresistor = 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
50ΩA
B
C
D
10 Ω
15 Ω 30 Ω
25Ω
6 V
Pd from ABDA= 0V
Pd across the 50Ωresistor = 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
50ΩA
B
C
D
10 Ω
15 Ω 30 Ω
25Ω
6 V
=105IBD + 30IAD - 25IABPd from
BDCB= 0V
Pd across the 50Ωresistor = 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
We now have 3 equations1, 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
6, 205IBD = 5IAB
IAB = 41IBD go back to equation 1and substitute
35IAB –25IBD = 61435I BD –25IBD = 6
1410I BD = 6IBD =.000425amps
0.4mA
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.
Thevenin’s theorem
Thevenin’s theorem
R1
R2
R3
R LOAD
A
B
V1
Consider a circuit consisting of a power source and resistors
Thevenin’s theorem
R1
R2
R3A
B
V1 VTH
Thevenin’s voltage (VTH) is the open circuit voltage
Thevenin’s theorem
R1
R2
R3A
B
V1 VTH
Thevenin’s voltage (VTH) = V1xR2/(R1 + R2). (R3 has no affect because there is no current through it)
R1 and R2 act as a potential divider
Thevenin’s theorem
R1
R2
R3 A
B
V1
Thevenin’s resistance (r) = R3 + (R1 xR2/(R1 + R2)). (All voltage sources are short circuited and all current sources
open circuited)
Thevenin’s theorem Example
2Ω
4Ω
3Ω
10Ω
A
B
4 Volts
Calculate VTH ,r and the current through the 10Ω load
Thevenin’s theorem Example
2Ω
4Ω
3Ω
10Ω
A
B
4 Volts
VTH = V1xR2/(R1 + R2). = 4 x 4/6 = 2.6 Volts
Thevenin’s theorem Example
2Ω
4Ω
3Ω
10Ω
A
B
4 Volts
r = 3 + (8/6) = 4.3Ω
Thevenin’s theorem Example
2Ω
4Ω
3Ω
10Ω
A
B
4 Volts
Current through the load = V/I = 2.6/(4.3 +10) = 0.18A
Thevenin’s theorem with two power sources
E1
E2 Load
E1 = 8V with internal resistance 4ΩE2 = 6V with internal resistance 6Ω
Load = 12 Ω Use Thevenin’s Theorem to find the current through the load
A
B
Thevenin’s theorem with two power sources
Using the theorem with the load disconnectedthe current circulating E1 and E2
IE1E2 = (E1 – E2) / ( r1 +r2) = (8-6) / (4+6) = 0.2A
E1
E2 Load
A
B
Thevenin’s theorem with two power sources
Hence equivalent emf of sources =E1 – I1r1 = 8 – ( 0.2 x 4) = 7.2 V
E1
E2 Load
A
B
Thevenin’s theorem with two power sources
Total internal resistance of sources in parallel 1/R= ¼ + 1/6 =3/12 + 2/12 = 5/12
R = 12/5 = 2.4Ω
E1
E2 Load
Thevenin’s theorem with two power sources
IL = 7.2 / (2.4 + 12) =7.2/ 14.4= 0.5 A
E1
E2 Load
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