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