Power Electronics. Exercise: Circuit Feedback

Transcription

1 Lehrstuhl für Elektrische Antriebssysteme und Leistungselektronik Technische Universität München Prof Dr-Ing Ralph Kennel Aricsstr 21 Tel: +49 (0) D München Internet: Fax: +49 (0) Power Electronics Exercise: Circuit Feedback

2 1 Theory 11 Introduction Power electronic circuits generate strong electromagnetic interferences (EMI) in power switches and loads during operation These interferences can be conducted back to the power circuit and power grid through the lines and cable and thus influence the control circuits and contaminate the grid Interferences with lower frequencies of several khz to 30 MHz are usually propagated through conductive materials Higher frequencies in the range from 30 MHz to 10 GHz are easily radiated and propagated in free space Since power electronic circuits operate in the frequency range below 1 MHz, the interferences in fundamental and lower harmonic frequency range are mainly propagated through conductors That means, they are easily fed back from the load to the power electric circuits and further to the power grid Power electronic components and the electric machines working in switching modes are strong (EMI) sources The interference signals are usually generated through fast changes of voltages and currents 12 EMI through Voltage and Current Changes In general, fast changes of current (high di/dt value) cause over voltages because of stray inductances in the circuits; and fast changes of voltage (high du/dt value) cause leakage currents because of stray capacitances Moreover, the effect introduced by transmission line also brings high frequency oscillations in the cables Such interferences caused by fast changes of voltages and currents are normally classified as electromagnetic interferences (EMI) A transmission line (an electrical wire) is in practice not an ideal conductor It includes resistance, inductance and capacitance An infinitesimally short segment of a transmission line could be represented with a complex circuit as shown in Figure 1, where L, C, Rs and Rp are stray inductance, stray capacitance, serial resistance and parallel resistance of the line segment dl, respectively Figure 1 Equivalent circuit (b) of a short segment of a pair of transmission line (a) Therefore, a long cable will be composed of infinitely many circuits of Figure 1(b) connected in series In this case, a transmission line will be treated as a high order circuit Therefore, oscillation will occur if the impedances of the transmitter, the line and the receiver are not matched Any conductor has stray inductance For example, a one meter straight wire would have an inductance of over one uh Considering this effect, a single wire could be represented as the circuit in Figure 2 2

3 Figure 2 Simplified equivalent circuit of a straight wire with stray inductance A changing current in such wire will generate a voltage between points A and B The current magnitude is, ( 1 ) where di/dt is the changing rate of the current It is seen that the induced voltage is directly proportional to the stray inductance and the current changing speed If an electric wire is close to another conductor, stray capacitance exists between them, as shown in Figure 3 Figure 3 Simplified equivalent circuit of the stray capacitance of an electric wire With the existence of the stray capacitance, a changing voltage will generate a current between points A and B, even if A and B are insulated The value is calculated to be, ( 2 ) where du/dt is the changing rate of the voltage It is clear that the leakage current is directly proportional to the stray capacitance and the voltage changing speed Therefore, the reduction of the changing rate of voltages and currents helps to reduce the interferences However, this brings more losses in the power components 3

4 13 Harmonics The switching operation of power electronics will generate harmonics, which are also caused by voltage or current changes but have special characteristics For inductance loads, which are commonly seen in motor drive, the voltage usually has a rectangular waveform Figure 4 gives an example with the voltage in square wave and thus the current in symmetric triangle form for inductance loads Figure 4 Voltage and current curves of a inductance load with PWM operation The square and triangle waves could be decomposed in Fourier series For the voltage with square wave, it is,, ( 3 ) where U is the maximum value of the voltage For the current in symmetric triangle form, the Fourier decomposition is, where I is the maximum value of the current, ( 4 ) It is seen that higher switching frequency will push the harmonics in higher frequency range and thus reduce the magnitude of the harmonic 14 Summary Operating frequency and changing rate of voltages and currents determine the amount of interferences Meanwhile, they also influence the losses (or efficiency) of power systems The relationship between these aspects is summarized in the following table Table 1 EMI, harmonics and losses depending on frequency and voltage and current changes Increasing frequency Increasing dv/dt or di/dt EMI Increase Increase Harmonics Decrease Increase Switching Losses Increase Decrease 4

5 2 Exercises 21 Exercise Question The model of transmission line (Figure 1) could be simplified with a LCR circuit as shown in the following figure Here the parallel stray resistance is omitted The calculation result can very roughly approximate the effect of the transmission line For a cable with 10 meter length, R = 01 Ohm, L = 36 uh and C = 75 pf If u changes from zero to 600 V at time t = 0, please calculate the voltage u C and current i through the capacitor The initial condition is i = 0, u C = Answer Differential equations of the equivalent circuit: Since the resistance is very small, this should be an under-damping circuit This is examined with the knowledge from the basic electronics With the known parameters, we get ( 5 ) ( 6 ) The criteria is that the circuit is under damping if u C is Using the initial condition, we have The current is And this is true in this case Then the form of ( 7 ) ( 8 ) ( 9 ) The voltage and current curves are obtained by numerical calculation They are shown in the following figures 5

6 The oscillation is represented here Because of the simplified transmission line model, the result approximates the real situation very roughly In the practice, the damping will be much faster and thus the voltage and current will get stable after only several oscillation waves 22 Exercise Problem The following figure shows one arm of an inverter At time t = 0, the current in the load coil is 30 A At this time the upper switch is turned off After 02 us the lower diode takes over the whole current from zero linearly Suppose the path between B and C of 10 cm length has a stray inductance of L s = 01 uh Please calculate the magnitude and direction of the voltage induced in path BC during the transient time Here the diode voltage drop is ignored 6

7 222 Solution When considering the stray inductance, the circuit after switching can be redrawn as following Using equation ( 1 ), we can directly get the voltage, ( 10 ) Since the current flows in the opposite direction of the given reference direction of u 1, it gets a negative value 23 Exercise Problem The following figure shows the simplified circuit of a DC-DC converter with inductive load The parameters are: L = 10 mh, R = 30 Ohm, U 0 = 600 V, diode voltage drop is zero (1) The switch is turned on and off with a frequency of f 1 = 4 khz and duty cycle of 50% Please draw the forms of voltage u and current i (2) If the switching frequency is changed to f 2 = 8 khz, Please draw the forms of voltage u and current i again 232 Solution to question 1 The voltage and current forms are shown in the following figure From the known parameters it is calculated:, The current in the rising parts (U A = 600 V) is, with α is a coefficient determined by initial value of i 1 (t = 0) which is I 1 for the rising part Therefore, 7

8 The current in the falling parts (U A = 0 V) is β is a coefficient determined by initial value of i 2 (t = 0) which is I 2 for the falling part Therefore, Within the half cycle when the switch is on i 1 rises from I 1 to I 2 Within the half cycle when the switch is off i 2 falls from I 2 to I 1 This means, So, Ripple current, The average current, 233 Solution to question 2 The voltage and current forms at doubled frequency are same as the last question The only different are the values: And using the same equations of the last questions we get the parameters of the current: Ripple current, The average current is same 8

9 234 Simpler calculation For simpler and faster calculation, we can assume the current changes are linear This is feasible because shorter than τ The half periods of the two frequencies are 0125 and ms, respectively, much Based on this consumption, the resistor voltage is roughly constant and equal to 300 V Then the voltage of the inductor is positive or negative 300 V Within a half cycle the current change can be estimated to be, and They are very close to the accurate calculation 24 Exercise Question The following figure shows a two-phase voltage rectifier using a diode full bridge The circuit parameters are: Input alternating voltage, u = 230 V (effective value), frequency f = 50 Hz; C = 3000 uf; R = 30 Ohm; diode voltage drop is zero (1) Please draw the forms of voltage and current through capacitor, C, qualitatively (2) Please draw the input current curve qualitatively (3) Please identify the circuit parts that likely generate EMI 9

10 242 Answer to questions (1) and (2) 10

11 243 Answer to question (3) The circuit parts marked with thicker lines are likely to generate EMI To reduce such EMI, these conductors should be designed as short and thick as possible 11