vce physics unit 3 - topic 2 electronics & photonics
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VCE Physics Unit 3 - Topic 2
1.0 Unit Outline apply the concepts of current, resistance, potential difference (voltage drop), power to the operation of electronic circuits comprising diodes, resistors, thermistors, and photonic transducers including light dependent resistors (LDR), photodiodes and light emitting diodes (LED); V = IR, P = VIcalculate the effective resistance of circuits comprising parallel and series resistance and unloaded voltage dividers;describe energy transfers and transformations in opto-electronic devicesdescribe the transfer of information in analogue form (not including the technical aspects of modulation and demodulation) usingLight intensity modulation i.e. changing the intensity of the carrier wave to replicate the amplitude variation of the information signal so that the signal may propagate more efficientlyDemodulation i.e. the separation of the information signal from the carrier wave design, investigate and analyse circuits for particular purposes using technical specifications related to potential difference (voltage drop), current, resistance, power, temperature, and illumination for electronic components such as diodes, resistors, thermistors, light dependent resistors (LDR), photodiodes and light emitting diodes (LED); analyse voltage characteristics of amplifiers including linear voltage gain (VOUT/VIN) and clipping; identify safe and responsible practices when conducting investigations involving electrical, electronic and photonic equipment
Chapter 1Topics covered:Electric Charge.Electric Current.Voltage.Electromotive Force.Electrical Energy.Electric Power.
1.0 Electric ChargeThe fundamental unit of electrical charge is that carried by the electron (& the proton).This is the smallest discrete charge known to exist independently and is called the ELEMENTARY CHARGE.Electric Charge (symbol Q) is measured in units called COULOMBS (C).The electron carries - 1.6 x 10-19 C.The proton carries +1.6 x 10-19 C.
If 1 electron carries 1.6 x 10-19 CThen the number of electrons in 1 Coulomb of Charge
1.1 Flowing ChargesWhen electric charges (in particular electrons) are made to move or flow, an Electric Current (symbol I) is said to exist.The SIZE of this current depends upon the NUMBER OF COULOMBS of charge passing a given point in a given TIME.Section of Current Carrying WireMathematically: I = Q/t where: I = Current in Amperes (A) Q = Charge in Coulombs (C) t = Time in Seconds (s)then 6.25 x 1018 electrons pass here every second.
1.2 Electric CurrentElectric CURRENTS usually flow along wires made from some kind of CONDUCTING MATERIAL, usually, but not always, a METAL.Currents can also flow through a Liquid (electrolysis), through a Vacuum (old style radio valves), or through a Semiconductor (Modern Diodes or Transistors).A Current can only flow around a COMPLETE CIRCUIT. A break ANYWHERE in the circuit means the current stops flowing EVERYWHERE, IMMEDIATLY.The current does not get weaker as it flows around the circuit, BUT REMAINS CONSTANT. It is the ENERGY possessed by the electrons (obtained from the battery or power supply) which gets used up as the electrons move around the circuit.In circuits, currents are measured with AMMETERS, which are connected in series with the power supply. Typical Electric Circuit
1.3 Conventional Current vs Electron CurrentConventional vs Electron CurrentWell before the discovery of the electron, electric currents were known to exist.It was thought that these currents were made up of a stream of positive particles and their direction of movement constituted the direction of current flow around a circuit.This meant that in a Direct Current (D.C.) circuit, the current would flow out of the POSITIVE terminal of the power supply and into the NEGATIVE terminal.Currents of this kind are called Conventional Currents, and ALL CURRENTS SHOWN ON ALL CIRCUIT DIAGRAMS EVERYWHERE are shown as Conventional Current, as opposed to the real or ELECTRON CURRENT.
1.4 VoltageTo make a current flow around a circuit, a DRIVING FORCE is required.This driving force is the DIFFERENCE in VOLTAGE (Voltage Drop or Potential Difference) between the start and the end of the circuit.The larger the current needed, the larger the voltage required to drive that current.VOLTAGE is DEFINED as the ENERGY SUPPLIED TO THE CHARGE CARRIERS FOR THEM TO DO THEIR JOB ie.TRAVEL ONCE AROUND THE CIRCUIT.
So, in passing through a Voltage of 1 Volt, 1 Coulomb of Charge picks up 1 Joule of Electrical Energy. OR A 12 Volt battery will supply each Coulomb of Charge passing through it with 12 J of Energy.Mathematically; V = W/Q where: V = Voltage (Volts) W = Electrical Energy (Joules) Q = Charge (Coulombs)
1.5 E.M.F. Voltage is measured with a VOLTMETER.The term EMF (ELECTROMOTIVE FORCE) describes a particular type of voltage.It is the VOLTAGE of a battery or power supply when NO CURRENT is being drawn.
This is called the Open Circuit Voltage of the battery or supplyWith S closed, a current begins to flow and V drops and now measures voltage available to drive the current through the external circuitVoltmeters are placed in PARALLEL with the device whose voltage is being measured.Voltmeters have a very high internal resistance, so they have little or no effect the operation of the circuit to which they are attached.
*Electronics & Photonics RevisionQuestion Type:Q1: Which one of the following statements (A to D) concerning the voltage across the resistor in Figure 1 is true?
A. The potential at point A is higher than at point B.B. The potential at point A is the same as at point B.C. The potential at point A is lower than at point B.D. The potential at point A varies in sign with time compared to that at point B.Potential Difference
1.6 Electrical EnergyThe conversion of Electrical Energy when a current passes through a circuit element (a computer) is shown below.Mathematically W = VQ 1, where: W = Electrical energy (Joule) V = Voltage (Volts) Q = Charge (Coulomb)Current and Charge are related through: Q = It. substituting for Q, in equation 1 we get: W = VItElectrical Energy (W) is defined as the product of the Voltage (V) across, times the Charge (Q), passing through a circuit element (eg. a light globe).
Electronics & Photonics RevisionQuestion Type: Q2: Determine the electrical energy dissipated in the 100 resistor of Figure 1 in 1 second. In your answer provide the unit.Electrical EnergyA: Electrical energy W = VQ = VIt = (4.0)(40 x 10-3)(1) = 0.16 Joule
1.7 Electrical PowerElectrical Power is DEFINED as the Time Rate of Energy Transfer: P = W/t where P = Power (Watts, W) W = Electrical Energy (Joule) t = Time (sec)From W = VI t we get:P = VI From Ohms Law (V = IR) [see next chapter] we get:P = VI = I2R = V2/R where: I = Current (Amps) R = Resistance (Ohms) V = Voltage (Volts)Electrical Power is sold to consumers in units of Kilowatt-Hours. (kW.h)
A 1000 W (1kW) fan heater operating for 1 Hour consumes 1kWh of electrical power.
Since P = W/t or W = P x t, we can say:1 Joule = 1 Watt.secso 1000 J = 1kW.secso3,600,000 J = 1 kW.houror 3.6 MJ = 1 kW.h
1.8 A.C. ElectricityThere are two basic types of current electricity:(a) D.C. (Direct Current) electricity where the current flows in one direction only.(b) A.C. (Alternating Current) where the current changes direction in a regular and periodic fashion.The Electricity Grid supplies domestic and industrial users with A.C. electricity. A.C. is favoured because: it is cheap and easy to generate it can be transformed; its voltage can be raised or lowered at will by passage through a transformer.The only large scale use of high voltage D.C. electricity is in public transport, ie. trams and trains. A.C. ELECTRICITY - PROPERTIESVPtoP = Peak to Peak Voltagefor Domestic Supply VPtoP = 678 VT = Periodfor Domestic Supply T = 0.02 secVP = Peak Voltage for Domestic Supply VP = 339 V
1.9 R.M.S. Voltage and CurrentGRAPHICAL DEVELOPMENT OF THE RMS VOLTAGE FROM AN A.C. VOLTAGEWith an A.C. supply, the average values for both voltage and current = 0, so Vav and Iav cannot be used by the Power Companies to calculate the amount of electric power consumed by its customers.To get around this problem R.M.S. or Root Mean Square values for AC voltage and current were developed. RMS values are DEFINED as: The AC Voltage/Current which delivers the same voltage/current to an electrical device as a numerically equal D.C. supply would deliver.An AC source operating at 240 V RMS delivers the same power to a device as a DC source of 240 V. Yet, AC circuits do consume power, so a method of calculating it had to be found.
1.10 Peak versus RMS ValuesIn AC supplies, the Peak and RMS values are related through simple formulae:For Voltage:VRMS = VP/2For Current:IRMS = IP/2In Australia Domestic Electricity is supplied at 240 V, 50 HzThe Voltage quoted is the RMS value for the AC supply.Thus the Peak value for voltage is VP = VRMS x 2 = 240 x 1.414 = 339 V