S Power Electronics Exam

Transkriptio

1 S Power Electronics Exam Answer all five questions (in English, Finnish or Swedish). Questions in Finnish are on the reverse side. 1. Figure below shows a setup using two pulse-width modulated converters to supply a three-phase load, in this case a AC motor, from a three-phase AC power system. Explain shortly the basic operating principle of the converter. The converter on the left hand side can be replaced by a diode rectifier. Compare the advantages and disadvantages of these two cases. 2. A three-phase thyristor rectifier is connected to a 50 Hz supply with an rms line-to-line voltage of 400 V. Assume that the thyristors are ideal and that the load is an ideal DC current source of 50 A. a) Draw the input and output voltage and current waveforms when the control angle is 60. b) Derive equations for and calculate the average value of the DC voltage, the total harmonic distortion of the line current and c) the active power consumed by the rectifier. As there is no line inductance on the ac-side the commutation time can be ignored. 3. The dc input voltage of a buck (step-down) converter is 24 V and the desired DC output voltage is 16.8 V. Assume the converter to be lossless and omit the ripple in the output voltage (i.e., large filter capacitance). a) Draw the circuit diagram of the converter, when the switching power-pole is implemented with an IGBT and a diode. b) Draw the waveforms of the inductor voltage, the inductor current, and the input current below each other. c) Calculate the IGBT duty ratio, the average input current, the average output current and the peak-to-peak ripple in the inductor current. The inductance is 0.1 mh, the output power is 28 W, and the switching frequency is 200 khz. 4. The simplified schematic of a single-phase voltage source inverter is shown below. Draw the implementation of this circuit with semiconductor switches. The switching functions of switches A and B, as well as the output current i o are shown in the right. Draw the output voltage waveform u o. Define the conducting semiconductor devices for different time intervals. 5. Efficiency in power electronics. What are the elements having an effect on the efficiency of converters and why efficiency is important in power electronics? 1 Fourier-series f ( t) a0 ahcos h t bhsin h t 2 h 1 ah ( )cos, 0,, ( )sin, π f t h t d t h b 0 h f t n t d t π 0 h 1,,

2 S PowerElectronics Tentti Vastaa kaikkiin viiteen kysymykseen joko suomeksi, ruotsiksi tai englanniksi. Kysymykset löytyvät englanniksi paperin toiselta puolelta. 1. Alla olevassa kuvassa on esitetty suuntaaja, jossa käytetään kahta pulssinlevysmoduloitua suuntaajaa syöttämään vaihtovirtakonetta kolmivaiheisesta sähköverkosta. Selitä lyhyesti kytkennän toimintaperiaate. Vasemmanpuoleinen suuntaaja voidaan korvata diodisillalla. Vertaa näiden kahden eri vaihtoehdon etuja ja haittoja keskenään. 2. Kolmivaiheinen tyristoritasasuuntaaja on kytketty 50 Hz verkkoon, jonka pääjännitteen tehollisarvo on 400 V. Tyristorit oletetaan ideaalisiksi ja kuorma ideaaliseksi 50 A tasavirtalähteeksi. a) Piirrä tulo- ja lähtöpuolen jännitteiden ja virtojen käyrämuodot kun ohjauskulma on 60. b) Johda yhtälöt ja laske lukuarvot tasajännitteen keskiarvolle ja verkkovirran särökertoimelle (THD) sekä c) tasasuuntaajan kuluttamalle pätöteholle. Koska sähköverkossa ei ole induktanssia, kommutoimiskulman vaikutusta ei tarvitse ottaa huomioon. 3. Jännitettä laskevan katkojan (buck) syöttöjännite on 24 V ja haluttu lähtöjännite on 16.8 V. Oleta suuntaaja häviöttömäksi ja lähtöjännite ideaaliseksi (eli suuri suodatuskondensaattori). a) Piirrä suuntaajan kytkentäkaavio kun vaihtokytkin on toteutettu IGBTllä ja diodilla. b) Piirrä induktanssin jännitteen ja virran sekä tulovirran käyrämuodot allekkain. c) Laske IGBTn ohjaussuhteen arvo, tulo- ja lähtövirran keskiarvot sekä induktanssin virran huipusta huippuun vaihtelu. Induktanssi on 0.1 mh, lähtöteho 28 W ja käytettävä kytkentätaajuus 200 khz. 4. Yksivaiheisen jännitevälipiirillisen vaihtosuuntaajan periaatekuva on esitetty alla. Piirrä kytkennän toteutus tehopuolijohdekomponenttien avulla. Oikeanpuoleisessa kuvassa on A ja B kytkimien kytkemisfunktioiden kuvaajat sekä lähtövirran i o käyrämuoto. Piirrä lähtöjännitteen u o käyrämuoto. Määrittele kullakin ajan hetkellä johtavat tehopuolijohdekomponentit. 5. Tehoelektroniikan hyötysuhde. Mitkä asiat vaikuttavat suuntaajien hyötysuhteeseen ja miksi hyötysuhteella on suuri merkitys tehoelektroniikassa? 1 Fourier-sarja f ( t) a0 ahcos h t bhsin h t 2 h 1 ah ( )cos, 0,, ( )sin, 1,, π f t h t d t h b 0 h f t n t d t h π 0

3 S Power Electronics Solutions, Exam Question 1, This is a voltage sourced converter, which means that we have a large dc capacitor between the two converters. As shown in the figure below. Conv2 is a normal PWM dc/ac converter supplying the load with balanced three phase voltages. By using PWM both the amplitude and frequency can be controlled according to the needs of the load. In the first case conv1 is similar converter as conv2 by its layout. However, the control is bit different. Converter 1 keeps the DC bus voltage constant, this means that the power taken from the utility is adjusted to be the same as supplied to the load. Otherwise dc bus voltage either drops or increases, i.e. difference in the two powers is affecting the energy of the capacitor. Converter 1 uses PWM too as conv2 but now the utility voltage is nearly constant as well as the frequency is. This means that conv1 voltage is only adjusted slightly around the voltage of the utility. Most importantly, the PWM voltage of conv1 is synchronized with the phase of the utility voltage so that there is wanted power flow from utility to the converter or vice versa. This is highlighted in the third figure below. If conv1 is PWM converter power can flow in both directions and we can also have nearly sinusoidal line currents in the utility side. If conv1 is diode rectifier power can only flow from utility to the converter and line currents are far from sinusoidal. Of course PWM converters are more expensive than diode rectifiers. Many gave comments on the losses of these two approaches. It is true that diodes have conduction losses but so do IGBTs or other switching devices. It is good to remember that there are also switching losses if conv1 uses PWM. Therefore, the losses of diode-bridge are supposed to be lower than those of PWM rectifier. Figure 1.16 from the textbook Figure a) and c) from the textbook

4 Question 2 DC voltage follows one of the line-to-line voltages and it repeats similar six times during a period of the line voltage. When control angle is used the turning of the thyristors is delayed the same amount. Therefore, average of the DC voltage can be calculated as: ( ( ) ( )) ( ) This can be calculated as shown on Pages of the textbook too. In order to find the total harmonic distortion (THD) of the line current the rms values of the line current and its fundamental are necessary. Line current is positive for 120 degrees, zero 60 degrees, negative 120 degrees and again zero 60 degrees during one line cycle. This means that it differs from zero 2/3rds of the line cycle. The line current s rms value is When using the same place of origo as in calculating the rms value and by using the Fourier series, the rms value of the line current s 1st harmonic is Then by definition the THD is: ( ) As the rectifier can be considered ideal (no losses) power can be calculated from the AC side too: ( ) The phase shift between voltage and fundamental component of current is equal to the control angle.

5 Question 3

6 Question 4 Solutions are shown below, the switches S xx must be turn-off devices like IGBTs or MOSFETs. As can be seen from the voltage can current waveforms, load is assumed to be inductive, i.e. current increases with positive voltage, doesn t change when voltage is zero and decreases with negative voltage. The polarities of voltage and current are defining the conducting components. It is good to note that there must always be exactly two components conducting, not more or less. Note that components S xx are turned on and off according to h A and h B but it doesn t meant that they would be conducting current. Question 5 Topics that should be discussed are: 1) Definition of the efficiency in general, output power/input power. 2) Sources of the losses, switching and conduction losses of power semiconductor devices and losses in other components too (capacitors, inductors, wires). 3) Why a high efficiency is important, price of the energy, need for a cooling system, high temperature reduces life time of components => has an impact on the size of the equipment and both construction and life time costs.