Design and Implementation of On-board Electrical Power ... - OUFTI-1

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The power losses in the inductor are equal to I 2 avR L , where R L is the series resistance ofthe inductor. They represent a loss of 1% efficiency ifR L < 0.01 V outI out= 0.01R. (5.10)We now have all the elements necessary to design the inductor.follows:The procedure is as• The inductance is given by Eq. 5.9, with maximum V in and maximum I out .• The peak current in the inductor is given by Eq. 5.7, with maximum V in and maximumI out .• An acceptable inductor series resistance is given by Eq. 5.10.Capacitor designFrom the integration of i C = C dv Cdtis found to beand Fig. 5.3, the voltage ripple on the capacitor ∆v CUsing the value of ∆i L given by Eq. 5.6, we have∆v C = ∆i LT s8C . (5.11)∆v C = V outηV inV in − V out8f 2 s LC , (5.12)C = V outηV inV in − V out8f 2 s L∆v C. (5.13)This formula gives the value of the capacitor C corresponding to a chosen voltage ripple∆v C .”Boost” converter [16]PrincipleThe power stage of a boost converter includes the same components than a Buck converter,but the position of the switch and the inductor are inverted (Fig. 5.4). The switch is inposition 1 during a time interval DT s (phase 1) and in position 2 during an interval D ′ T s(phase 2), with D ′ = 1 − D.In phase 1, the voltage v L across the inductor is equal to the input voltage V in . Theinductor is being charged of magnetic energy. The current in the inductor is increasing andthe slope of the waveform is given by v L (t) = L di L(t)dt. The variation of i L over the phase 1 is52
Figure 5.4: Power stage of a boost converter.∆i L,1 = V inDT s. (5.14)LIn the phase 2, the magnetic energy of the inductor is released. The voltage across theinductor is inverted. V s = V in + v L and V out = V s . As a result, C is charged at a voltagehigher than V i n. The current in the inductor i L is decreasing and the variation over the phase2 is∆i L,2 = (V in − V out )D ′ T s. (5.15)LIn steady-state (Fig. 5.5), the net variation of i L equals zero.Figure 5.5: Evolution of voltages and currents in a boost converter (from Wikipedia).As a consequence, from Eq. 5.14 and Eq. 5.15, we have53
Page 2 and 3: AcknowledgementsI want to thank the
Page 4 and 5: Contents1 Introduction 91.1 Cubesat
Page 6 and 7: 4.4.2 Mean case . . . . . . . . . .
Page 8 and 9: B Power budget worksheet 106C Pictu
Page 10 and 11: design and the tests are delegated
Page 12 and 13: Chapter 2Requirements of the EPS2.1
Page 14 and 15: • The Single-Event Upset (SEU)Thi
Page 16 and 17: the P-POD. RBF pins must fit within
Page 18 and 19: Figure 2.6: Top view of the PC104 c
Page 20 and 21: Chapter 3Design of EPS architecture
Page 22 and 23: • Voltage (4) and current (5) at
Page 24 and 25: Figure 3.6: The equivalent circuit
Page 26 and 27: of our Lithium-Polymer batteries va
Page 28 and 29: Figure 3.12: I-V curve of a solar p
Page 30 and 31: 3.3.3 CapacityA important value to
Page 32 and 33: Parameter SLPB723870H4 SLPB554374HN
Page 34 and 35: of the batteries is kept between -2
Page 36 and 37: Over Charge Prohibition 4.275 ± 0.
Page 38 and 39: supplied in 5V. The circuit will be
Page 40 and 41: Chapter 4The Power Budget4.1 Introd
Page 42 and 43: Figure 4.1: P-V curve of a solar pa
Page 44 and 45: 4.3.2 Efficiency of convertersTo at
Page 46 and 47: Figure 4.3: Consumptions in % in me
Page 48 and 49: Chapter 5Electrical Design of EPS5.
Page 50 and 51: V outV in= D. (5.1)Since D ≤ 1, t
Page 54 and 55: ∆i L,1 + ∆i L,2 = 0, (5.16)V in
Page 56 and 57: Using the value of ∆i L given by
Page 58 and 59: There is no data about the case to
Page 60 and 61: Capacitor selectionFour 10µF ceram
Page 62 and 63: • Output voltage: 5V.• Maximum
Page 64 and 65: Figure 5.12: Burst mode operation (
Page 66 and 67: Figure 5.14: Simplified schematics
Page 68 and 69: Figure 5.15: Worksheet for 3.3V con
Page 70 and 71: sequently, the k was chosen above 0
Page 72 and 73: where G 1 is the initial control-to
Page 74 and 75: Figure 5.21: Measured Bode diagram
Page 76 and 77: Figure 5.26: Equivalence between th
Page 78 and 79: C f =12πf f R 0f,L f = R 2 0f C f
Page 80 and 81: Figure 5.37: Schematics of the firs
Page 82 and 83: R KR >1.45V100mA − 1.3A35= 23.07
Page 84 and 85: The schematics is shown on figure 5
Page 86 and 87: A commercial model meets all requir
Page 88 and 89: Figure 5.45: Schematics of the heat
Page 90 and 91: PrefixX7X5Y5Z5SuffixTemperature ran
Page 92 and 93: 6.2.1 The second dissipation system
Page 94 and 95: • The antenna deployment system.
Page 96 and 97: 6.3.3 TestsThe engineering model of
Page 98 and 99: 7.3 ActivitiesAs OUFTI-1 is designe
Page 100 and 101: 8.1.2 DesignA model of Li-Po batter
Page 102 and 103:
[15] Fabien Jordan, Phase B Electri
Page 104 and 105:
TaK = 273 + TaC;%Photo-current ther
Page 106 and 107:
Appendix BPower budget worksheetIn
Page 108 and 109:
108
Page 110 and 111:
Appendix CPictures of the prototype
Page 112 and 113:
Appendix DSchematics of the enginee
Page 114 and 115:
876543213V3 CONVERTER AND INPUT FIL
Page 116:
87654321ANTENNA DEPLOYMENT CIRCUITB
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