Power Electronics

Prof. Dr. Ing. Joachim Böcker Power Electronics 3.09.06 Last Name: Student Number: First Name: Study Program: Professional Examination Performance Proof Task: (Credits) (0) (0) 3 (0) 4 (0) Total (80) Mark Duration: 0 minutes Permitted auxiliaries: Nonprogrammable calculator without graphic display Drawing material (circle, triangle, ruler, pens ) Please note the following remarks: You may only attend the exam if you have registered for it in the system PAUL. If you participate without registration your generated results will not be evaluated and accepted. Please hold your student identity card with photograph in readiness! Please label every blank sheet of paper with your name and your student number. For every task please use a new blank sheet of paper. Please do not use pencils and red pens. Every numerical calculation has to be furnished with units. Non-compliance will lead to point deduction. All approaches have to be documented in a comprehensible way! Specifying a single value as final result without a recognizable approach will not be evaluated. Good Luck! 3.09.06 Power Electronics Page of 9

Task : Buck Converter (0 Credits) For an automotive application a buck converter is used. The output capacitor C is very large so that the output voltage ripple can be neglected. All components are assumed to be ideal. Please assume steady state for the complete task. Figure.: Circuit diagram of buck converter with resistive load The converter operates in discontinuous conduction mode (DCM). The chosen and adjusted values are given as follows: Input voltage: U = V Load resistor: R = 0 Ω Duty cycle: D DCM = 0. Switching frequency f S = 50 khz Furthermore the plot of the input current i over two switching periods is shown in Figure.: Figure.: Input current i over two switching periods. Calculate the average value i of the input current i. Determine the power P = P = P which has to be provided by the DC voltage source.. Calculate the values of the output voltage u = U and the output current i = I..3 Determine the value of the inductance L..4 Determine the current conducting duration T off of the diode. Draw the inductor current i L in the diagram in Figure. and calculate its average value i L. 3.09.06 Power Electronics Page of 9

.5 Now the buck converter should operate in boundary conduction mode (BCM). Determine the minimum possible duty cycle D BCM that has to be adjusted for that operation status. How large are the current ripple i L and the average current i L?.6 Insert the known points of DCM and BCM in the diagram in Figure.3 and sketch roughly the characteristic of the average inductor current i L over the duty cycle D. Figure.3: Average inductor current i L over duty cycle D 3.09.06 Power Electronics Page 3 of 9

Task : Boost Converter (0 Credits) A boost converter is used in a PC for voltage adjustment. The circuit diagram of the converter is shown in Figure.. Figure.: Topology of boost converter The following values of the converter are known: Input voltage: U = 5 V Output voltage: u = V Inductance: L = 00 µh Switching frequency: f S = 00 khz The boost converter can be considered as ideal and operates in steady state. The input DC voltage U is assumed constant. The output voltage ripple u can be neglected.. Derive the ideal control characteristic U /U = f(d) of the boost converter. Set up the equations for i L (t = DT S ) and i L (t = T S ) as functions of U, U, D and T S in order to solve for the asked voltage ratio U /U. Assume that the boost converter operates in continuous conduction mode (CCM). Calculate the duty cycle D.. Determine the current ripple i L of the inductor current i L (t)..3 The boost converter operates in CCM. The load resistor R should now be adjusted so that the converter operates in boundary conduction mode (BCM). Determine the required resistance value R BCM. Now the load resistor is further increased (R > R BCM ) to force the converter to operate in discontinuous conduction mode (DCM). In order to keep the output voltage constant the duty cycle is changed to D = 0.35..4 Draw the transistor voltage u T (t), the inductor current i L (t) and the diode current i D (t) over two periods in the diagrams in Figure.. Determine the current conducting time span t D of the diode (after the transistor has been switched off). 3.09.06 Power Electronics Page 4 of 9

Figure.: Transistor voltage, inductor and diode current.5 Calculate the average values i L and i D of the inductor current i L (t) and the diode current i D (t). Determine the value of the load resistance R. 3.09.06 Power Electronics Page 5 of 9

Task 3: Four-Quadrant Converter (0 Credits) A suburban train is equipped with a four-quadrant converter (4QC) to rectify the AC line voltage and to provide a nearly constant DC voltage for the traction drives and the on-board electrical systems. The system is shown in Figure 3.: Figure 3.: Transformer, four-quadrant converter and DC-link between grid and load The following physical quantities are known: Grid voltage: U AC = 5 kv Grid frequency: f = 6.667 Hz Secondary-side leakage inductance: L σs = 6.5 mh DC-link voltage: U DC =.5 kv The four-quadrant converter can be considered as ideal (no losses, no interlocking times). The primary leakage inductance and the winding resistances of the transformer can be neglected. In the above diagram L σs is the transformer s leakage inductance. So, the transformer circuit symbol is considered as ideal transformer. 3. Draw the structure of the PWM unit for interleaved switching mode in Figure 3.. Input is the normalized reference voltage, output are the two switching signals for the two legs of the 4QC. Figure 3.: PWM unit for interleaved switching mode 3. What value for the switching frequency f S has to be selected so that the maximum occurring inductor current ripple i max does not exceed 00 A for interleaved switching mode? Determine the frequency of the current ripple that results from interleaved switching mode. For which values of s ref (t) does the maximum current ripple appear? 3.09.06 Power Electronics Page 6 of 9

In Figure 3.3 the root mean square value of the secondary current I and the power P drawn from the grid were recorded during the ride of the train from one station to the next. The ride required a total amount of energy of W = 3.5 kwh. Figure 3.3: Diagrams showing secondary side current (top), active power (centre) and reactive power (bottom) 3.3 Calculate the transformer turns ratio ü with help of the given diagrams above. Determine the duration t total of the train ride. 3.4 Draw the grid-side phasor diagrams for the times t = t (end of acceleration phase), t < t < t (constant power phase) and t = t (start of the decelerating phase). Calculate the phase angles φ (angle between grid voltage and transformed converter voltage) for the three cases. It can be assumed that the controller operates in power factor correction mode (cos(φ ui ) = ) leading solely to active power drawn from or fed into the grid. Use the following scales: 0 kv 5 cm; 00 A 5 cm. 3.5 Calculate the reactive power Q which is required by the transformer and delivered by the 4QC for the three cases of subtask 3.4), insert the calculated values in the power diagram in Figure 3.3 and sketch roughly its course. Task 4: Line-Commutated Rectifier (0 Credits) 3.09.06 Power Electronics Page 7 of 9

A 3-pulse thyristor-controlled converter is connected to the 3-phase grid via a Dy-transformer configuration and supplies a resistive load. The complete configuration is shown in Figure 4.: Figure 4.: Line commutated rectifier in M3 configuration The transformer and converter can be considered as ideal (no losses, no magnetic leakage). 4. Sketch the output voltage u d (t) in the diagrams in Figure 4. for the control angles α = 0, 45 and 35. What is the control angle limit value α L that represents the boundary control angle between CCM and DCM operation? Figure 4.: Normalized output voltage u d /u for different control angles 4. Calculate and draw in Figure 4.3 the control characteristic U d /U dmax = f(α) of the converter for the continuous conduction mode (α < α L ) and discontinuous conduction mode (α > α L ) range. 3.09.06 Power Electronics Page 8 of 9

Figure 4.3: Control characteristic of converter for CCM and DCM 4.3 Determine the active power P R which is consumed by the load resistance R as a function of the control angle α for the continuous conduction mode (CCM). 4.4 The converter supplies a resistive load. Describe in full sentences why reactive power appears on the mains side as well? Assign the different types of losses to its causes. The following mathematical relations might be needed for some of the subtasks: sin(x ± y) = sin (x) cos (y) ± cos (x) sin (y) cos(x ± y) = cos (x) cos (y) sin (x) sin (y) cos (ax)dx = x + sin (ax) 4a sin (ax)dx = x 4a sin(ax) 3.09.06 Power Electronics Page 9 of 9

Solution Task ) Buck Converter [0 Credits].) Average value of input current and power drawn from the DC source: i = D DCM T S i max = D T S DCM i max = 0, 0.876 A = 0.0876 A P = U i = V 0.0876 A =.05 W [3 Credits].) Output voltage and output current: U = P R =.05 W 0 Ω = 3.4 V I = U R [ Credits] = 3.4 V 0 Ω = 0.34 A.3) Value of inductance: + i L = U U L [ Credits] D DCM T S L = U U D i DCM T S = L V 3.4 V 0.876 A 0, 0.0000 s = 40 µh.4) Off-time, inductor current and average value of inductor current: + i Lmin = i Lmax U T L off = 0 T off = Li Lmax 0.00004 H 0.876 A = = 0.8 µs U 3.4 V i L = (D DCM T S +T off ) i max T S = (0. 0 µs+0.8 µs) 0,876 A = 0.34 A = I 0 µs [4 Credits].5) Duty cycle for BCM, inductor current ripple and average inductor current: 3.09.06 Power Electronics Page 0 of 9

In BCM, value of average inductor current is half of the inductor current ripple: i L = i L = D( D)U T S L i L = U = DU R R Equating the two equations gives: D( D)U T S L = DU R ( D)T S L = R D = L T S R D = L 40 µh = = 0.4 = 0.6 T S R 0 µs 0 Ω i L = D( D)U T S L = i L = i L =.44 A = 0.7 A [5 Credits] 0.6( 0.6) V 0 µs 40 µh =.44 A.6) Average inductor current over duty cycle: + [4 Credits] Task ) Boost Converter [0 Credits] 3.09.06 Power Electronics Page of 9

.) Control characteristic and duty cycle: U U = f(d) =? Setting up the two formulas for the inductor current (steady state is assumed): i L (t = DT S ) = U L DT S + i L (t = 0) i L (T S ) = i L (t = DT S ) + U U ( D)T L S = i L (t = 0) Solving for the searched voltage ratio: U DT L S + i L (t = 0) + U U L U DT L S + U U L U D + U U L L ( D)T S = 0 ( D) = 0 U D + (U U )( D) = 0 U D + U U U D + U D = 0 U U + U D = 0 U + U D = U U U D = U U ( D) = U ( D)T S = i L (t = 0) U = U D Adjusted duty cycle: D = U U D = U = U U V 5 V = = 0.583 U U V [6 Credits] 3.09.06 Power Electronics Page of 9

.) Inductor current ripple: i L = U L DT S = 5 V 00 µh 0.583 0 µs = 0.95 A [ Credit].3) Resistance value for boundary conduction mode: i D = i R = U = i R L ( D) = i L ( D) max U = i L ( D) R R BCM BCM = U = V = 97.44 Ω i L ( D) 0.95 A ( 0.583) [3 Credits].4) Current conduction time span of diode and diagrams: ++ i Lmax = U L DT S = 5 V 00 µh 0.35 0 µs = 0.75 A i L (t = t Lück ) = U U t L D + i Lmax = 0 A t D = (0 A i Lmax )L U U [7 Credits] = L i Lmax 00 µh 0.75 A = =.5 µs U +U 5 V+ V 3.09.06 Power Electronics Page 3 of 9

.5) Average value of inductor and diode current and value of resistor: i L = i D = i Lmax 0.6T S = i T S 4 L max = 0.055 A i Lmax 0.5T S = i T S 8 L max = 0.087 A R = U = U = V = 548.57 Ω i R i D 0.087 A [3 Credits] 3.09.06 Power Electronics Page 4 of 9

Task 3) Four-Quadrant Converter [0 Credits] 3.) PWM unit structure for interleaved operation mode of 4QC: [3 Credits] 3.) Leakage inductance for requested maximum current ripple: i _INTERLEAVED = s ( s ) U DC f S L σs Maximum current ripple for s ref (t) = ±0.5: i _max_interleaved = U DC 8f S L σs f S = Pulse frequency: f p = f S = 500 Hz = 000 Hz [4 Credits] U DC i _max _IL 8 L σs =.5 kv 00 A 8 6.5 mh = 500 Hz 3.3) Transformer turns ratio and duration of train ride: ü = I = I 000 A = = 0 I AC P/U AC 750 kw/5 kv W = P(t = t ) t + P(t < t < t ) (t t ) + P(t = t ) (t total t ) W = P(t = t ) t + P(t < t < t ) 3t + P(t = t ) t t = W = [ P(t = t ) + P(t < t < t ) 3 + P(t = t )] t W P(t=t )+P(t <t<t ) 3+ = 3.5 kwh P(t=t ) 750 kw+50 kw 3 600 kw t = 3.5 kwh 3600 s h 55 kw = 4 s 3.09.06 Power Electronics Page 5 of 9

t total = 5t = 0 s [4 Credits] 3.4) Grid-side phasor diagrams for three cases: Case : End of acceleration phase (t = t ) P = U AC I AC cos(φ ui ) I AC = 750 kw 5 kv cos (0 ) = 50 A I AC U AC j L s ü² I AC üu φ = arctan ( ω L σs ü I AC U AC ) = arctan ( π 6.667 Hz 0.0065 H 0 50 A ) = 4. 5 kv Case : Constant power phase (t < t < t ) 0.5 P = U AC I AC cos(φ ui ) I AC = 50 kw 5 kv cos (0 ) = 0 A I AC U AC üu j L s ü² I AC 0.5 0.5 φ = arctan ( ω L σs ü I AC U AC ) = arctan ( π 6.667 Hz 0.0065 H 0 0 A ) = 9.90 5 kv Case 3: Start of deceleration phase (t = t ) 3.09.06 Power Electronics Page 6 of 9

0.5 P = U AC I AC cos(φ ui ) I AC = 600 kw 5 kv cos ( 80 ) = 40 A üu j L s ü² I AC 0.5 I AC U AC 0.5 φ = arctan ( ω L σs ü I AC U AC [6 Credits] ) = arctan ( π 6.667 Hz 0.0065 H 0 40 A ) = 34.9 5 kv 3.5) Reactive Power (required by transformer): Case : Q = π f L σs (I ) = π 6.667 Hz 0.0065 H (000 A) Q = 654.50 kva Case : Q = π f L σs (I ) = π 6.667 Hz 0.0065 H (00 A) 0.5 Q = 6.8 kva Case 3: Q = π f L σs (I ) = π 6.667 Hz 0.0065 H (800 A) 0.5 Q = 48.88 kva Reactive power has to be provided by the 4QC. The reactive power oscillates between transformer and 4QC. 3.09.06 Power Electronics Page 7 of 9

[3 Credits] 3.09.06 Power Electronics Page 8 of 9

Task 4) Line-Commutated Rectifier [0 Credits] 4.) Output voltage waveforms for three different control angles: +.5+.5 [4 Credits] 4.) Control characteristic of line-commutated rectifier: CCM (0 α < 30 ): 50 +α U d = p 3 u sin(ωt) dωt = π 30 +α U d = 3 u [ cos(50 + α) + cos(30 + α)] π π 50 +α u [ cos(ωt)] 30 +α U d = 3 u [ cos(50 )cos (α) + sin(50 ) sin(α) + cos(30 ) cos(α) sin(30 ) sin (α)] π U d = 3 u [ cos(50 )cos (α) + cos(30 ) cos(α)] π U d = 3 π u [ 3 cos(α) + 3 3 3 cos(α)] = u cos (α) π U dmax α=0 = 3 3 π u f CCM (α) = U d = U d max 3 3 u cos (α) π = cos (α) 3 3 π u DCM (30 α < 50 ): 3.09.06 Power Electronics Page 9 of 9

80 U d = p 3 u sin(ωt) dωt = π 30 +α π U d = 3 u [ cos(80 ) + cos(30 + α)] π 80 u [ cos(ωt)] 30 +α U d = 3 u [ cos (80 ) + cos(30 ) cos(α) sin(30 ) sin (α)] π U d = 3 3 u [ + cos(α) sin (α)] π U dmax α=0 = 3 3 π u (Universal reference value for CCM and DCM) 3 f DCM (α) = U d π = u [+ 3 cos(α) sin (α)] = + cos(α) sin (α) = [ + cos (α + 30 )] U 3 3 d max π u 3 3 3 Control characteristic of converter for R-load: ++ [0 Credits] 3.09.06 Power Electronics Page 0 of 9

4.3) Active power in dependency of control angle: CCM (0 α < 30 ): P R = U d RMS R = P R = 3 u R π [ωt p π 50 +α u sin (ωt)dωt 30 +α = 3 u R R π 30 +α sin(ωt) 4 P R = 3 u 5π [ 6 +α R π P R = 3 u R π [5π 50 +α 3 u ] = 30 +α sin (300 +α) 4 π 6 +α R π [50 +α sin (60 +α) + ] 4 sin(300 ) cos(α)+cos(300 ) sin (α) π 4 50 +α sin (ωt) dωt sin( 50 +α) 30 +α 4 + sin( 30 +α) ] 4 sin(60 ) cos(α)+cos(60 ) sin (α) + ] 4 P R = 3 u R π [4π 3 cos(α) 3 sin(α) + cos(α) + sin (α)] 4 4 4 4 P R = 3 u R π [π + 3 3 4 cos (α)] DCM (30 α < 50 ): P R = U d RMS R = P R = 3 u R π [ωt P R = 3 u R π [π p π 80 u sin (ωt) dωt 30 +α = 3 u 80 R R π 30 +α sin(ωt) 4 50 +α 3 u ] = 30 +α sin (360 ) π α 4 R π [80 sin( 80 ) 4 P R = 3 u R π [5π α + 3 8 cos(α) + 8 sin(α)] [4 Credits] sin (ωt)dωt 30 +α sin(60 ) cos(α)+cos (60 ) sin(α) + ] 4 + sin ( 30 +α) 4 ] 4.4) Reasons for reactive power occurring on the mains side: For each control angle α the output voltage results in a periodic but non-sinusoidal voltage that contains harmonics. These voltage harmonics lead to current harmonics (independent of the load type) which also appear in the phase currents. Distortion reactive power D In addition to that the control angle α also determines the zero crossing of the fundamental phase currents and in consequence influences the phase difference φ between grid-side voltage and current of the same phase. Fundamental reactive power Q [ Credits] 3.09.06 Power Electronics Page of 9