EEE123

Power efficiency when maximum power transfer

50%

Current Source

Current supplied is always same regardless of voltage

Step 4 to show maximum power transfer with internal resistance

Differentiate denominator wrt Rl and set equal to 0 to find minimum (minimum denominator gives largest power)

Step 2 for charging capacitor through a resistor (current)

Differentiate equation wrt time and set differential of Vs to 0 as source voltage is constant

Peak to peak detector

Diode clamp attached to peak detector Since diode clamp causes Vopeak to become 2Vipeak, when a peak detector is used on Vo it gives 2Vipeak which is the peak to peak voltage of Vi

Flemmings right hand rule

Direction of induced emf

Transformer

Electrical energy is transferred from one device to another without any physical connection between the two

Reactive Power

Energy flowing between source and load that is not dissipated

Reluctance

Magnetic Resistance (H^-1)

Root Mean Square Voltage

RMS is the DC value that would give the same heating effect as the AC waveform is Vrms = Vp/sqrt(2)

Step 5 to show maximum power transfer with interal resistance

Rearrange to give Rint = Rl for maximum power transfer

Conductor

Material which allows electric charge to flow freely through it when a voltage is applied across it

Insulator

Material which does not allow electric charge to flow through it

Step 2 to working out diode conduction state (fixed sources)

Replace diode with a voltage source of its forward drop voltage (usually 0.7V) in its reverse bias direction

Step 3 for current growth in an inductor through a resistor (current)

Replace term with I0, separate the variables and integrate

Variation of resistance with temperature (Insulators)

Resistance decreases

Variation of resistance with temperature (Conductors)

Resistance increases

Kirchoff's 1st law (current law)

The sum of currents entering a node is 0 at all times (taking into account directions)

Temperature coefficient of resistance

Symbol - alpha subscript 0 Ratio of change in temperature per degree Celsius with reference to a definite temperature (0 degrees celcius)

Conductivity and units

Symbol - lower case sigma Reciprocal of resistivity Siemens per metre (S/m))

Flux Density

Symbol: B Units: Tesla, T Flux per metre squared

Time constant for resistor inductor circuit

T = L/R

Time constant for resistor capacitor circuit

T = RC

Kirchoffs 2nd law (voltage law)

The sum of voltages around a loop is 0 at all times (taking into account directions)

Units VA

Volt-Amps (Voltage x Current)

Units VAr

Volt-Amps Reactive

Power factor

cos(phi) 1 if purely resistive, 0 if purely inductive or capacitive, so no power dissipated in purely inductive or capacitive circuits

Electric to magnetic analogies

emf -> mmf current -> flux Resistance -> reluctance V = IR -> F(mmf) = flux X reluctance

Wire moving perpendicular to magnetic field

emf = BLv

Forward voltage drop/Diode drop

Voltage required in the forward bias direction to cause a diode to start conducting

Voltage Source

Voltage supplied always same regardless of current

Voltage divider equation

Vout = Vin(R2/R1 + R2)

Step 1 for charging capacitor through a resistor (current)

Vs = Vr + Vc Sub in equations for Vr and Vc

Current divider equation

i1 = it(R2/R1+R2)

Loop

Closed path formed by connecting branches

Norton's Theorem Step 3

Create equivalent circuit with current flowing through path as a single current source and the total resistance as a single parallel resistor

Step 3 to show maximum power transfer with internal resistance

Divide through by Rl

Back emf

Due to self inductance when there is a change of current. Acts to oppose the change in current

Step 2 to working out voltage at which a diode will begin to conduct (varying sources)

Label voltage across diode as 0.7V and current in branch as 0A

Step 5 for current growth in an inductor through a resistor (current)

Sub constant back in, use log laws to get just one ln on RHS, take exponential of both sides, rearrange to get I by itself on LHS

Voltage across capacitor when charging

Sub in Q=CV to time varying current equation, sub in current for I in the integral and solve integral. Vs=I0R so sub this in afterwards

Voltage across resistor when capacitor charging through it

Sub in current for capacitor into V=IR, and Vs=I0R so sub this in afterwards

Superposition Theorem Step 2

Superimpose (sum) currents in each branch (or branch of interest) taking care of directions. Result is current through that branch

Conductance and units

Symbol - G Reciprocal of resistance Siemens (S)

Resistivity Units

Ohm-metres

Semi-Conductor

A material who's electrical conductivity is dependent on the voltage supplied

Single phase half wave rectifier

A single diode is used so that only voltage in the positive direction above 0.7V appears across the load. When the source provides a voltage that is in reverse bias for the diode, no current flows and hence the load voltage is 0V during this period

Peak detectors

AC to DC converter Peak detector when used for signals, rectifier when used for AC to DC power conversion When signal in forward bias direction, capacitor is charged up to Vs-0.7V and Vo=Vs-0.7V, when signal is in reverse bias direction, diode does not conduct and Vo=Vc as Vc decreases due to it discharging through R If R was removed, there is no way for C to discharge so once charged, Vo = Vi-0.7V forever

Units A-h

Amp Hour (Amps x Hours)

Step 1 to working out diode conduction state (fixed sources)

Assume diode is conducting

Step 1 to working out voltage at which a diode will begin to conduct (varying sources)

Assume diode is on point of conducting (0.7V across diode but 0A through it)

Parallel internal resistance

Current Source

Energy stored in capacitor

By virtue of the electric field between the plates

Energy stored in inductor

By virtue of the magnetic field produced

Thevenin's Theorem Step 1

Calculate voltage across terminals that will be connected to the load

Single phase half wave rectifier with smoothing

Capacitor put in parallel with load. Charged once per input cycle

Single phase full wave rectifier with smoothing

Capacitor put in parallel with load. Charged twice per input cycle

Diode clamp

Clamps the negative peak of an AC signal to -0.7V When signal is first positive, C cannot charge as R too big, and diode does not conduct, so Vo=Vi. When signal becomes lower than -0.7V, diode conducts and C charges to Vipeak-0.7V. R is very large so C doesn't discharge, so minimum Vo can only be -Vipeak + Vipeak-0.7V = -0.7V. When signal goes positive again, total Vo is Vc + Vi (Vc=Vipeak) therefore Vo=Vipeak+Vi. When signal next goes negative, small amount of charge has been lost from C through R, so recharged.

Magnetic Circuit

Closed loop path followed by magnetic flux lines

Shunt regulator

Connected in parallel with the load and acts with a resistor that is in series with the load. Alters the current it draws, and therefore alters the voltage dropped across the resistor, so that a constant Vo is maintained

Series regulator

Connected in series with the load and acts as a valve by restricting the flow of current from input to output so that a constant Vo can be maintained

Clipping circuit

Consist of a signal source, a series resistor, and a diode with a voltage source connected to it. Limits extremes of a voltage waveform by clipping them off

Thevenin's Theorem Step 3

Create equivalent circuit with voltage across terminals as a single voltage source and the total resistance as a single series resistor

Step 4 for current growth in an inductor through a resistor (current)

Find constant of integration by using initial conditions (so t = 0, I = 0) therefore c = ln(I0)

Step 4 for charging capacitor through a resistor (current)

Find constant of integration by using initial conditions (so t = 0, I = I0) therefore c = ln(I0)

Step 1 to show maximum power transfer with internal resistance

Find current in circuit I = E/(Rint + Rl)

Superposition Theorem Step 1

Find currents in each branch of the circuit when supplied by each source in turn when all other voltage sources are short circuited and current sources open circuited. Only their internal resistances considered

Step 2 to show maximum power transfer with internal resistance

Find power dissipated in load P = I^2Rl (sub in I from step 1) and expand

Norton to Thevenin

Find voltage across terminals and use that voltage as a voltage supply and place the resistor in series

Step 4 to working out diode conduction state (fixed sources)

If current flows in forward bias direction, diode is conducting and assumption was right. If flows in reverse bias direction then diode is not conducting and assumption was wrong

Centre Tap Transformer

If defined as 12-0-12, then from centre tap to V+ is 12V and from centre tap to V- is -12V

Design of a zener diode regulator

Iz must be greater than 0 at all times, and the conditions that are most likely to stop the diode from conducting are when Vi is a minimum and IL is a maximum. Ir = IL + Iz Ir = (Vimin-Vo)/R = ILmax + Izmin Rearranging the above for R gives the largest value of R that can be used

Active network

Network has a source of emf

Passive network

Network has no source of emf

Direction of magnetic field

North pole to south pole

Branch

Part of a circuit connecting two nodes

Components in series

Same current flows through them all but different voltages

Power Factor Correction

Same power can be supplied with minimal current if power factor = 1 For more inductive circuits, there is a lagging power factor, and a capacitor is used to bring the power factor near to 1 For more capacitive circuits, there is a leading power factor, and an inductor is used to bring the power factor near to 1 Corrective inductors and capacitors added in parallel so as to not affect the voltage across the load

Components in parallel

Same voltage across each branch but different currents

Step 3 for charging capacitor through a resistor (current)

Separate the variables for integration and integrate both sides

Step 2 for current growth in an inductor through a resistor (current)

Set differential equal to 0 and rearrange to find I0

Thevenin to Norton

Short circuit terminals, find current and then create norton circuit using that current as the source and place resistor in parallel

Norton's Theorem Step 1

Short circuit the nodes and calculate total current flowing through path

Norton's Theorem Step 2

Short circuit voltage sources, open circuit current sources (leaving only internal resistances) and calculate total resistance as if voltage supply from nodes

Thevenin's Theorem Step 2

Short circuit voltage sources, open circuit current sources (leaving only internal resistances) and calculate total resistance as if voltage supply from nodes

Step 5 for charging capacitor through a resistor (current)

Sub constant back in, put on RHS of equation to leave just ln(I) on LHS and take exponential of both sides to get it in exponential form as shown in equation sheet

Voltage across inductor during current growth

Sub in current for voltage in an inductor equation, differentiate and cancel terms to get in version on equation sheet, Vs=RI0 so sub this in

Reluctances in Series

St = S1 + S2 + S3 + ... + Sn

Node

The point where two or more branches meet

Breakdown voltage

The reverse bias voltage at which current will be able to flow the wrong way through the diode

Assumptions made when choosing a capacitor to meet a ripple specification

The transformer and power source are ideal Diode forward voltage drop is negligible Load current is constant Discharge occurs for the whole interval between charging peaks

Changes the ripple voltage in a peak detector

Time constant (RC) of the capacitor resistor combination. As RC increases, less ripple meaning that when diode is not conducting, less charge is lost in the capacitor

Series internal resistance

Voltage Source

Open circuit voltage/EMF

Voltage across terminals if load disconnected

Internal Resistance

Voltage at terminals will be a function of time Power dissipated in source

Step 1 for current growth in an inductor through a resistor (current)

Vs = Vr + Vl Sub in equations for Vr and Vl

Zener diode regulator

When IL is to be kept constant and Vi changes, a reduction in Vi gives the same reduction in Vr, and so the current reduction through the diode is change in Vi/R. The reduction in Ir is compensated by an equal reduction in Iz, and a small decrease in Vo (due to the steepness of the zener diode slope this is very small) When Vi is kept constant but IL changes, a reduction in IL causes a reduction in Vo, however a small drop in Vo causes a large drop in Iz, hence increasing IL again so there is no change in IL overall. Vo and Ir are slightly changed but not much

Step 3 to working out diode conduction state (fixed sources)

Work out the current through the source representing the diode

Step 3 to working out voltage at which a diode will begin to conduct (varying sources)

Work out value of source that gives conditions of step 2

Step 4 to working out voltage at which a diode will begin to conduct (varying sources)

Work out whether an increase in source voltage will increase current through diode whilst voltage across it remains at 0.7V. If it does then the diode conducts for variable sources larger than the point of conduction voltage calculated, if not then the diode conducts for voltages smaller than the point of conduction voltage calculated