120 likes | 339 Views
Announcements. H5CPE2 module homepage will soon change http://hermes.eee.nott.ac.uk/teaching/h5cpe2 Simulation packages: PSIM: http://www.powersimtech.com/download.html PSPICE: http://hermes.eee.nott.ac.uk/teaching/h54pqe/ LIVE Power Electronics Course (prof. Mohan)
E N D
Announcements H5CPE2 module homepage will soon change http://hermes.eee.nott.ac.uk/teaching/h5cpe2 Simulation packages: PSIM: http://www.powersimtech.com/download.html PSPICE: http://hermes.eee.nott.ac.uk/teaching/h54pqe/ LIVE Power Electronics Course (prof. Mohan) http://www.ece.umn.edu/groups/PowerElectronics_Drives/ User: klumpner@ieee.org Password: powerelectronics Free Power Electronics Literature on Internet Power Modules - Application Manual (Semikron) http://www.semikron.com/skcweb/e/applica/applica_help.html Switched-Mode Power Supplies (Philips) http://www.semiconductors.philips.com/acrobat/applicationnotes/APPCHP2.pdf Power Control with Thyristors and Triacs (Philips) http://www.semiconductors.philips.com/acrobat/applicationnotes/APPCHP6.pdf
Ideal Diode Rectifier (1) Line A Id=const A x Phase voltage D1 D3 D5 Neutral Vxy 0 D4 D6 D2 C y B Highly inductive load (excitation of synchronous generator) IA D1, D3, D5 select the most positive input line potential to x D4, D6, D2 select the most negative input line potential to y Vx 150 D3 D5 D1 30 D2 D4 D6 Vy Vxy “Six-pulse diode rectifier” IA D1 (+Id) D4 (-Id)
Ideal Diode Rectifier (2) If E= 240 then Mean (average) rectified voltage y potential= x potential mirrored Or it could be determined from integrating the line-to-line voltage over 60
Measures of Power Quality (1) Lets consider a nonlinear load that draws a nonlinear and displaced current from a sinusoidal supply VA VA IA IA=1*sin(100t)+1/3*sin(300t)+1/5*sin(500t)+1/7*sin(700t) FFT of IA 1. Extract the fundamental component IA1(Amp&angle) IA IA1 Iah= IA -IA1
Measures of Power Quality (2) VA IA1 IA1-active IA1-reactive • 2. Decompose the fundamental component into an active IA1-active in phase with VA and a reactive IA1-reactive which is 90-deg displaced to VA IA1 IA1-reactive IA1-active • Displacement Power Factor (DPF) • DPF= Cosine of phase shift between fundamental component of line current and the associated phase voltage (ie IA and VAN) • Accounts for proportion of fundamental current that does something useful. DPF = 1 is good, DPF = 0.5 is bad (50% of fundamental current does nothing useful)
Measures of Power Quality (3) RMS (root-mean square) current = the value of an equivalent DC current that produces the same amount of heat into a given resistor as the current in question • Distortion Factor (DF) • DF = RMS fundamental current/total RMS current (<1 !!) • Accounts for the proportion of RMS current that is due to harmonics. DF = 1 is good (pure sinewave). DF = 0.5 is bad (50% of RMS current is due to the harmonics) • Total Harmonic Distortion Factor (THD) This parameters is more often used to define the degree of distortion of a waveform: %
Measures of Power Quality (4) • Power Factor (PF) • PF= active power delivered/total RMS volts * total RMS amps • Tells us what proportion of the total apparent power (VA) represents useful power delivered to the load (Watts) • PF=1 is good, PF = 0.5 is bad (have to supply 1000VA to get 500W for example) • Relationship between quantities • Assume voltage is undistorted compared to current – normally reasonable for a decent power system • Then total RMS voltage = fundamental RMS voltage Power factor = Displacement factor * Distortion factor Note if the current is undistorted then DF = 1 and Power factor = Displacement factor = Cosine of phase shift (as given in 1st year notes)
Ideal Diode Rectifier (5) Analyze Input Current Waveform=general case +Id -Id 0 T/2 T Half-wave symmetric: f (t+T/2) = - f (t) Decomposition in Fourier series results in: where n=1,3,5,7,9… In=0 for n =2,4,6…because of half-wave symmetry In=0 for the situations where n* = 2 As for 3-phase diode rectifier (ideal case) = 2/3 Harmonics current component: In=0 for n=3, 9, 15 etc > Id (peak) RMS input current: Displacement factor DPF = 1 Distortion factor DF = Power factor
Non-ideal Diode Rectifier (1) L Line A Id=const A x D1 D5 Phase voltage D3 a’ Neutral b’ Vxy c’ 0 L D4 D6 D2 C y B L Effect of supply inductance = this causes overlap since current can’t transfer instantaneously from one device to another ia ib x D1 D3 a’ b’ VA VB VC ia Instant commutation between D1 and D3 is not possible !
Non-ideal Diode Rectifier (2) Outgoing diode (D1) Incoming diode (D3) Both diodes conduct simultaneously (overlap) Loss in Volt-time area!! smaller Vxy
Non-ideal Diode Rectifier (3) The missing VTA in Vxy is required to commutate the load current Id from D1 to D3 Total VTA loss from Vx in one cycle = 3*L*Id Similar for Vy hence total VTA loss in Vxy = 6*L*Id Loss in mean Vxy Causes volt-drop with Id known as regulation Effect on the line current waveform ia ib Negligible effect on the harmonics unless is very large Id 0
Non-ideal Diode Rectifier (4) L A Id=const x D1 D3 ia VBA a’ b’ ib 0 L B Detrimental effect on other users Current in L2 is distorted due to overlap Id=const A L2 L1 PCC Voltage in PCC = partially distorted 0 L1 L2 B C To other users Calculation of Easiest way to find it is by considering the following loop: VBA Area = 2 Id *L • During this time, Vba causes the change of : • ia from Id to 0 • ib from 0 to Id