Design and Implementation of On-board Electrical Power ... - OUFTI-1
Design and Implementation of On-board Electrical Power ... - OUFTI-1 Design and Implementation of On-board Electrical Power ... - OUFTI-1
Figure 5.37: Schematics of the first dissipation system.The advantage of this implementation is that nearby all the power can be dissipated ina resistor in series with the transistor. The disadvantage is that the transistor acts as aswitch which can only be open or closed. When it is closed, the voltage of the batteries busmay suddenly drop and the control opens the switch. The circuit then enters in oscillation.To avoid this, a feed-back resistor is inserted on the comparator, introducing an hysteresis.This fixes a problem to bring another one: the dissipation system will partially discharge thebatteries each time it is activated.The circuit was tested and failed: the field effect transistor had a too high gate thresholdvoltage. Moreover, the oscillations occurred (test report: [1]).It was discarded and a second version was designed.5.4.3 Second versionThis version is closer to the example of figure 5.36. A TL1431 is used instead of the Zenerdiode, for a better precision in the regulation. The TL1431 has three PIN: the cathode, theanode and the reference. The functional block diagram of the TL1431 is shown on figure 5.38.Figure 5.38: Functional block diagram of TL1431.It looks similar to the first dissipation circuit, but it uses a operational amplifier insteadof the comparator and a bipolar transistor instead of the FET. This is an analogical circuit.When it is activated, the circuit will find an equilibrium, and the amount of power dissipated inthe shunt will exactly be the amount of exceeding power. The batteries will not be dischargedby the circuit.80
The PNP transistor is needed because the current in the TL1431 should not be higherthan 100mA. A shunt resistor is inserted in the shunt branch but attention must be paid tothe polarization of the transistor.Figure 5.39: Schematics of the shunt regulator.5.4.4 Components designThe solar panels output power may reach 5.5W. The shunt regulator must be able to dissipatethis amount of power. 5.5W corresponds to a shunt current of around 1.3A at 4.2V. The powerwill be dissipated in the transistor and in the shunt resistor. The interest of the shunt resistoris that it can be located outside the EPS card. The planned solution is to place the shuntresistor on the face with no solar cells.The shunt regulator is a critical system. A failure would cause the loss of the mainpower buss and thus the loss of the satellite. The TL1431 is available in space model. If theprocurement of the space model was too difficult, military and automotive models may besuitable. A flight model PNP transistor has been provided by Thales ETCA: the Q2N5153.R KRThe cathode current of the TL1431 (I KA ) must be inferior to 100mA. I KA = I B + I KR ,where I B is the base current in the transistor and I KR the current in the KR resistor. Theminimal β of the transistor is 35, the maximal V EB is 1.45V (for I C = 2.5A) and the maximumI C is 1.3A, thusI KA = I B + I KR < 100mA,I Cβ + V EBR KR< 100mA,81
- 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 52 and 53: The power losses in the inductor ar
- 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
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- 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 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
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- Page 112 and 113: Appendix DSchematics of the enginee
- Page 114 and 115: 876543213V3 CONVERTER AND INPUT FIL
- Page 116: 87654321ANTENNA DEPLOYMENT CIRCUITB
The PNP transistor is needed because the current in the TL1431 should not be higherthan 100mA. A shunt resistor is inserted in the shunt branch but attention must be paid tothe polarization <strong>of</strong> the transistor.Figure 5.39: Schematics <strong>of</strong> the shunt regulator.5.4.4 Components designThe solar panels output power may reach 5.5W. The shunt regulator must be able to dissipatethis amount <strong>of</strong> power. 5.5W corresponds to a shunt current <strong>of</strong> around 1.3A at 4.2V. The powerwill be dissipated in the transistor <strong>and</strong> in the shunt resistor. The interest <strong>of</strong> the shunt resistoris that it can be located outside the EPS card. The planned solution is to place the shuntresistor on the face with no solar cells.The shunt regulator is a critical system. A failure would cause the loss <strong>of</strong> the mainpower buss <strong>and</strong> thus the loss <strong>of</strong> the satellite. The TL1431 is available in space model. If theprocurement <strong>of</strong> the space model was too difficult, military <strong>and</strong> automotive models may besuitable. A flight model PNP transistor has been provided by Thales ETCA: the Q2N5153.R KRThe cathode current <strong>of</strong> the TL1431 (I KA ) must be inferior to 100mA. I KA = I B + I KR ,where I B is the base current in the transistor <strong>and</strong> I KR the current in the KR resistor. Theminimal β <strong>of</strong> the transistor is 35, the maximal V EB is 1.45V (for I C = 2.5A) <strong>and</strong> the maximumI C is 1.3A, thusI KA = I B + I KR < 100mA,I Cβ + V EBR KR< 100mA,81
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