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University of Alberta *

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401

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Electrical Engineering

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Feb 20, 2024

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Question 1: A standard full-wave 4 diode rectifier, C dc filter has the following circuit parameters: Parameters: V s = 115 V, f s = 60Hz C = 2mF P load = 1,000 W V Don = 0.8 V With C known but θ c , V r unknown, V dc & I dc cannot be determine directly, but an iteration technique can be used. See section 3, chapter 11, example 4 page 26. Use 3 iteration stages where you are using an estimate of . Determine: (a) diode conduction angle, capacitor pk-to-pk ripple voltage, the average load current & voltage, (b) peak and rms of both the diode rectifier output current and capacitor current. (a) diode conduction angle, capacitor pk-to-pk ripple voltage, the average load current & voltage θ c θ c = 0 (0) first pass (1) first iteration using Iteration using (b) peak and rms of the diode rectifier output current and capacitor current (2) second iteration ECE 401 Assignment #3 Solutions V m = 2 V s = 2 × 115 = 162.62 V 2 f s C = 0.24 V r = I dc 2 f s C = 6.2098 0.24 = 28.8744 V V dc = V pk = V m − 2 V Don = 162.62 − 1.6 = 161.035 V I dc = P V dc = 1000 161.035 = 6.2098 A V dc = V pk − V r 2 = 161.035 − 28.8744 2 = 148.097 V I dc = P V dc = 1000 148.097 = 6.7523 A θ c = 2 × arccos ( 1 − V r V m ) = 2 × arccos ( 1 − 3.971 162.62 ) = 1.1437 rad . / sec . = 65.528 o θ c = 2 × arccos ( 1 − V r V m ) = 2 × arccos ( 1 − 17.8923 162.62 ) = 0.9470 rad . / sec . = 54.257 o (3) third iteration θ c = 2 × arccos ( 1 − V r V m ) = 2 × arccos ( 1 − 19.1383 162.62 ) = 0.9800 rad . / sec . = 56.1521 o V r = I dc 2 f s C ( 1 − θ c 2 ) = 6.7523 0.24 ( 1 − 1.1437 2 ) = 17.8923 V V dc = V pk − V r 2 = 161.035 − 17.8923 2 = 152.088 V I dc = P V dc = 1000 152.088 = 6.5751 A V r = I dc 2 f s C ( 1 − θ c 2 ) = 6.5751 0.24 ( 1 − 0.9470 2 ) = 19.1383 V V dc = V pk − V r 2 = 161.035 − 19.1383 2 = 151.465 V I dc = P V dc = 1000 151.465 = 6.6022 A V r = I dc 2 f s C ( 1 − θ c 2 ) = 6.6022 0.24 ( 1 − 0.9800 2 ) = 18.927 V V dc = V pk − V r 2 = 161.035 − 18.927 2 = 151.509 V I dc = P V dc = 1000 151.509 = 6.5976 A I m = I dc π 2 2 θ c = 6.5976 × π 2 2 × 0.980 = 33.33 A I 0, rms = I dc π 3 8 θ c = 6.5976 π 3 8 × 0.980 = 13.1205 A I o , rms = I dc π 3 8 θ c − 1 = 6.5976 π 3 8 × 0.980 − 1 = 11.3408 A I C , pk = I m − I dc = 6 = 33.33 − 6.5976 = 26.62 A

Question 2: A 1 phase full-bridge diode rectifier, V s =115 V f s = 60Hz V Don = 0 V, is to be operated at an input power level of 1.250 kW using an ac inductor input filter and an output dc capacitor filter (who’s voltage may be considered constant and ripple free). Using the performance curves on page 20 chapter 11. (a) Determine the ac inductance required to minimize the ac supply current as illustrated in class. You decide this inductance is too large, so you note the circuit breaker is rated at 15 A, so you decided to consider a smaller inductance but keep the current to no more than 14.8 A. (b) What inductance can you choose? Determine how this change in inductance has on: (c) the supply rms current and input power factor; (d) the diode rectifier average output voltage and current. (e) the harmonic content of the supply current and its resultant THD F . (a) ac inductance in mH required to minimize the supply rms current. (d) 1he diode rectifier average output voltage and current (c) the supply rms current and input power factor (e) the harmonic content of the supply current and its resultant THD F . from the plots at I pu = 1.362 L pu = 0.04 p.u. V s = 115 V f s = 60 Hz P in = 2 kW L base = V 2 s 2 π f s P base = 115 2 2 π × 60 × 1250 = 28.06 m H L mh ,0.04 p . u . = 0.04 * 28.06 = 1.123 m H I rms ,0.11 = 1.31 × 10.87 = 14.24 A I rms ,0.04 = 1.36 × 10.87 = 14.8 A L mh ,0.11 p . u . = 0.11 * 28.064 = 3.087 m H (b) What inductance can you choose? I base = P Vs = 1250 115 = 10.87 A I pu = 14.8 10.87 = 1.362 PF 0.11 = 0.76 PF 0.04 = 0.74 I dc ,0.11 = 0.9 × 10.87 = 9.78 A I dc ,0.04 = 0.8 × 10.87 = 8.70 A V dc ,0.11 = 1.12 × 115 = 128.8 A V dc ,0.04 = 1.25 × 115 = 143.75 A I H ,0.11 = 0.59 × 10.87 = 6.41 A I dc ,0.04 = 0.81 × 10.87 = 8.80 A THDF 0.11 = 50 % THDF 0.04 = 75 %

Question 3: Power Quality of 1-phase diode rectifier: A 1-phase diode rectifier draws a current of 11 A rms at a power level of 2 kW from a 230 V ac supply. If the fundamental current lags the supply voltage by 30°, determine the CDF, DPF, PF and THD f associated with the rectifier input current. Question 4: Power Quality of a 3-phase diode rectifier Not knowing what filters are being used, a rectifier input electrical operating parameters are: P in = 5 kW V LL = 208 V @ 60Hz I s = 16 A I 1 = 15 A Determine: (i) total rms of the harmonic currents excluding the fundamental (I H ), (ii) input PF, CDF and THD F , (iii) phase angle displacement of the fundamental current ( ɸ 1 ) (i) I H (iii) ɸ 1 (ii) PF, CDF, THD F V s = 230 V f s = 60 Hz P in = 2 kW I s = 12 A lag ϕ 1 = 30 o I 1 = P s V s cos ϕ 1 = 2000 230 cos 30 o = 10.04 A DPF = cos ϕ 1 = cos 30 o = 0.866 CDF = I 1 I s = 10.04 11 = 0.9128 PF = CDF × DPF = 0.9128 × 0.866 = 0.7905 I H = I 2 s − I 2 1 = 11 2 − 10.041 2 = 4.492 A THD F = 100 × I H I 1 = 100 × 4.492 10.041 = 44.74 % I H = I 2 s − I 2 1 = 16 2 − 15 2 = 5.568 A CDF = I 1 I s = 15 16 = 0.9375 DPF = PF CDF = 0.8674 0.9375 = 0.9252 THD F = 100 × I H I 1 = 100 × 5.568 15 = 37.12 % PF = P S = 5000 3 V LL × I s = 5000 3 × 208 × 16 = 5000 5764.3 = 0.8674 ϕ 1 = acos ( DPF ) = acos (0.9252) = 22.30 o = 0.3892 rad . / sec .

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