- Muchas notas - Fran Acién

20210224 - Calibration Cubesat - Power Budget

References:


Expected energy generation

ISS orbit is 92.7 minutes with 36 minutes in eclipse. That is 140º out of 360 of eclipse.

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Cubesat standard photovoltaic triple junction cell has a maximum electrical power of 2400 mW. Howover, MPPT has a efficient of 90%.

Se define la irradiancia como la densidad superficial de potencia: $$ B_0 = 1367 \ W/m^2 $$

Solo el 70% de la radiación solar termina por llegar al suelo y ser absorbida.

I am going to supose that we will use Triple Junction GaAs Solar Cell: $$ V_{op} = 2350 mV \\ I_{op} = 417 mA \\ \text{Effiency} = 29.5 \% \\ \text{MPPT Effiency} = 0.9 \\ \text{Absorbed Solar radiation after atmosphere} = 0.7 \\ 1350 \cdot 0.3 \cdot 0.7 \cdot 0.9 = 255.15 [W/m^2] \\ \text{End of life} = 2 \% \text{per year} \\ 255.15 \cdot 0.98 = 250 [W/m^2] \\ \text{Para generar 10W necesitarmos} = 0.04 \ m^2 \\ \text{Cell Area} = 26.51 \ cm^2 \\ \text{total cell} = 16 \\ \text{Utilizaremos } 18 \text{ cells} $$

6 x 3 celulas.

Células de 4 x 7. Nos caben dos células por cada pared.

18 celdas, 2 en cada panel… 9 paneles.

Calcular cual es la mejor forma, si desplegado o .

Diferentes disposiciones:

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Diposición número 1

Un única cara con 18 celdas. Su valor máximo de potencia es de 11.9 W. Su media de potencia será de 3.68 W.

5eefa4ba9b85199b2db2b2cf9e525970.png

Teniendo en cuenta que tendremos aproximadamente 140º de eclipse, el peor escenario es el siguiente:

1d2096adb1752a33ada40adcbb11c4d3.png

Dónde la potencia media generada con eclipse será de 2.19 W. Esto es que la mínima energía generada será de 2.19 W x 1.54 h = 3.3726 Wh.

Disposición número dos

La disposición en la que una cara siempre está mirando a tierra y 3 caras tienen 6 celdas. La potencia máxima es de 5.6 W. Su potencia media será de 3.68 W.

aef63f78ff3df1d6a878266c0af6aa2b.png

Con eclipse se quedaría de la siguiente forma:

826b080aa0886ae19d106d47543de998.png

En el que la potencia media en este escenario será de 2.19 W. Esto es que la mínima energía generada será de 2.19 W x 1.54 h = 3.3726 Wh.

Expected Power Consumption

Module Peak [mW] Average [mW] Normal Mode [mW] Safe Mode [mW] Note
EPS 200 75 75 75 1
Battery Heaters 200 200 40 40 2
ADCS 1500 300 60 60 3
Magnetorquers 775 775 39 0 4
Reaction Wheels 1500 750 156 60 5
OBC 500 320 64 60 6
COMMS RX 400 300 60 30 7
COMMS TX 2000 750 83 60 8
Payload 1 [Ka-band] 1000 375 42 60 9
Payload 2 [Ku-band] 1000 375 42 60 10
Payload 3 [Star-tracker] 1000 250 150 60 11
Total 10075 4470 811 565 12
Energy losses 4324 1918 348 243 13
Total + Losses 14399 6388 1159 808 14

Notes:

  • 3: Considering the peak power of the GPS.
  • 2: Temporaly value. Taken from Upsat
  • 4: Temporaly value. Taken from Lisboa TFM
  • 6: CC3200 have several modes for power saving. I am taking a mean value.
  • 10: Only emitting when there is LOS from tenerife (maximum of 60 minutes per day)
  • 11: Only emitting when there is LOS from tenerife (maximum of 60 minutes per day)
  • 13: Considering the 70% of the input power

Expected a minimum availabe power coming from the solar panels of 4.5W.

Energy Budget

Considering ISS Orbit of 92.5 (1.54h) minutes, where 36 minutes is eclipse.

Energy Figures CubeSat in Safe Mode CubeSat in Normal Mode
Energy Generated / Orbit [mWh] 3370 3370
Consumed Energy without Losses / Orbit [mWh] 870 1249
Energy Losses / Orbit [mWh] 373 536
Consumed Energy / Orbit [mWh] 1243 1785

Power Management during Phase

LOS Tenerife

Module Peak [mW] Safe Mode [mW] Note
EPS 200 75 1
Battery Heaters 200 40 2
ADCS 1500 60 3
Magnetorquers 775 0 4
Reaction Wheels 1500 60 5
OBC 500 60 6
COMMS RX 0 30 7
COMMS TX 0 60 8
Payload 1 [Ka-band] 1000 60 9
Payload 2 [Ku-band] 1000 60 10
Payload 3 [Star-tracker] 1000 60 11
Total 10075 565 12
Energy losses 4324 243 13
Total + Losses 14399 808 14

Batteries

TODO: Revisar cálculos

Assuing we will use APR18650 batteries. That is 1.1Ah, 3.3V.

The solar panel bus is 12V. So, we need 3 18650 in series.

36 minuttes of clipse.

$$ \frac{1159 \cdot 10^{-3}}{12} = 100 \ mA $$

That is way less than 1.1 Ah.