Exercise 2 solution. 2cm. B sat 0.3 T K 0.5 mm 2. i mt 10 3 T. = mm l e2. = 25 mm 2 A e2. = mm 2 A w2. = 1.96.

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1 Exercise solution Problem a.5 cm a cm a 3 cm µr 5 µr µr 3 Ipk.4 A Ipk 4A L 4mH L 6 µh f sw khz J 4.5 A B sat.3 T K.5 mm µ 4. π. 7 H m i.. 3 mt 3 T The magnetic core parameters are: r l e π. in r out r e π. l ei. π a i a.. i l e = 47.4 mm l e = mm l e3 = mm A ei a i A e = 5 mm A e = mm A e3 = 4 mm A wi π. a i A w = mm A w = mm A w3 = mm AP i A. ei A wi AP = mm 4 AP = mm 4 AP 3 = mm 4 Vol i A. ei l ei Vol =.8. 3 mm 3 Vol = mm 3 Vol 3 = mm 3 For small I (. Ipk) Irms approx. equals to Iaverage. Skin depth is: 76 S. mm Hz S =.4 mm f sw

2 . Area Product needed for the first inductor (taking B Ipk JK.. B.5 T): = mm 4 => the smallest core a =.5cm Check the imum number of turns in the window of chosen toroid: I Irms Iav Ipk Irms.9. Ipk.9. Ipk Needed winding area for inductor RMS current : Irms =.36 A Irms J =.8 mm strands ceil No need for a litz wire. π. S strands = 4A. cu π. strands =.39 mm Winding: one strand of copper wire =.39 mm Maximum number of turns: n A w π.. strands K n = Choose µr such that n < n and B <.5T : n Ll. e µr. µ. A e n i round l e µr. i µ. A n = 346 n = 45 n 3 = 73 e Ipk B i A. e n B i =.85 T B =.6 T B 3 =.37 T

3 B B is acceptable and µr = 5, number of turns n The following core is choosen: a =.5 cm µr = 5 B B. I B.. Ipk B.. B B = 37 mt. Area Product needed for the second inductor (taking B.5 T): Ipk JK.. B = mm 4 => the smallest core a = a = cm Check the imum number of turns in the window of chosen toroid: I Irms Iav Ipk Irms.9. Ipk.9. Ipk Needed winding area for inductor RMS current : Irms = 3.6 A Irms J =.8 mm strands ceil π. S strands = 5 Litz wire needed! 4A. cu π. strands =.45 mm Winding: ten strands of copper wire.5 mm Maximum number of turns: n A w π.. strands Choose µr such that n < n and B <.5T : K n = 6

4 n Ll. e µr. µ. A e n i round l e µr. i µ. A n = 95 n = 67 n 3 = 47 e Ipk B i A. e n B i =.53 T B =.358 T B 3 =.5 T The flux density is too big, the core with a = a = cmis not suitable. Next try - core with a = a 3 = c Maximum number of turns: n A w3 π.. strands K n = 64 Choose µr such that n < n and B <.5T : n Ll. e µr. µ. A e n i round l e3 µr. i µ. A n = 67 n = 47 n 3 = 34 e3 Ipk B i A. e3 n B i =.9 T B =.8 T B 3 =.76 T B B 3 is acceptable and µr = 5 number of turns n 39 The following core is choosen: a 3 B B. I B.. Ipk B.. B = cmµr 3 = (this core is too big, but the only possible B = 35.3 mt

5 Problem. V in V V out 48 V I out Af sw 5 khz J 4.5 A mm B. T K.5 V out Vin D D V in D =.75 V out I Lav I out D I.. I Lav I Lav = 4A I =.4 A V. in D L I. f sw I I pk I Lav I I rms I Lav L = H I pk = 4. A I rms = 4A. S 76 mm. Hz S = m f sw First find area product to choose the core LI.. pk I rms AP JK.. B AP = m 4 The core that satisfies the needed area product is E4/7/. A e 49 mm A w 4 mm A. e A w = mm 4 Number of turns: N LI. pk A. e B N = Rounding turns number to N 64 I rms J =.889 mm strands ceil π. S strands = 3

6 4A. cu π. strands =.64 mm Winding: Three strands of copper wire Window area occupied by the winding:.5 mm Window Air gap: π.. strands. N K Window = mm µ. N. A e l g L l g =.74 mm Air gap l g =.74 mm in the middle leg of E core, or two gaps l g =.85 mm in the side legs. A. cu N = mm µ. N. A e l g L l g =.74 mm

7 f s = 5 khz; P = W; V in = ±5 V; V out = ±5 V; ΔB =.5 T; k =.6; J = 4 6 A/m ; A P = A e A w = { V, } t on I (RMS) ΔB J k P I(RMS) = = =.333A ; Vin 5 T S = μs; t on = μs; {V,ton} = μ 5 =.5m פתרון שאלה מס' 3 : A P נתונים: נתחיל בנוסחת נחשב את הפרמטרים החסרים: A P.5m.333 = =.n [m ] הגוף המתאים ETD34 עם הפרמטרים: A e [mm ] l e [mm] A w [mm ] V e [mm 3 ] A P [m 4 ] ETD n נחזור ונחשב את מספר הליפופים ע"פ ה- A e האמיתי של הגוף :ETD34 { V, } t on.5m [V sec] n = = 3 [Turns] ΔB A 6 e.5 [T] 97. [m ] V n out = n = n 35 [Turns] Vin 3 ועכשיו נקבע את עובי החוט: I (RMS) =.333A; I (RMS) = = 4A; A.333 [A] 4 [A] =.33 [mm ] ; A [mm ] Cu = = ; 4 [A/mm ] 4 [A/mm ] Cu = δ[mm] f 7 s [Hz] = 7 5k =.3 [mm] לא נשכח להתחשב בתופעת ה- effect :Skin שטח חטך וקוטר יעיל מרבי של חוט עגול: S m S = πr = πδ.36 [mm ]; d = =.644[mm] ; π ACu.33 ACu = = ; m = = 3 ; S.36 S.36 נשתמש בחוט בעל קוטר הקרוב ביותר ל- d וזה חוט מהסוג: SWG- או 8 mil או בקוטר של.7. mm בראשוני נצטרך חוט יחיד ובמשני חוט ליצה המורכב משלושה חוטים מבודדים.

8 Problem 4 ) Buck transfer function, duty cycle is a parameter:..5 V.5V.V V(Load)/V(in) V_V ) Boost transfer function, duty cycle is a parameter: V.5V.V V(Load)/V(in) V_V

9 ) Buck a) Input voltage change of %: V 9V 8V 7V s ms 4ms 6ms 8ms ms V(out) b) Duty Cycle change of.: 4V V V 8V c) Load change of 5%: 6V s ms 4ms 6ms 8ms ms V(out) 8.8V 8.4V 8.V 7.6V 7.V s ms 4ms 6ms 8ms ms V(Load)

10 ) Boost a) Input voltage change of %: V 8V b) Duty Cycle change of.: 6V s 5ms ms 5ms ms V(Load) 4V V 5V V c) Load change of 5%: V s 5ms ms 5ms ms V(Load) 8V 6V 4V s 5ms ms 5ms ms V(Load) 3