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