Power systems — solar arrays (I-V curve, power tracking), batteries (DoD, cycles), RTG: Exercises

This is a self-testing ladder. Each problem states everything you need. Try it, then open the collapsible solution. The rungs climb from L1 Recognition (name the thing) to L5 Mastery (design a whole subsystem). Everything here builds on the parent power-systems note and touches Semiconductor Physics, Power Electronics, Orbital Mechanics (Keplerian) and Spacecraft Thermal Control.

Reference curve for the whole set — this figure shows a solar cell's I-V curve (blue: current falling as voltage rises), the power curve $P = V \cdot I$ (pink: a single hill), and the three key points $I_{sc}$ (max current at $V = 0$ ), $V_{oc}$ (max voltage at $I = 0$ ), and the MPP (the peak of the power hill):

Figure — Power systems — solar arrays (I-V curve, power tracking), batteries (DoD, cycles), RTG

Level 1 — Recognition

Problem 1.1 — Read the curve

On the I-V curve above, name the three labelled operating points and state, in one phrase each, what is special about them.

Recall Solution

==Short-circuit current $I_{sc}$ == — the point where terminal voltage $V = 0$ (wires shorted). Current is at its maximum, power is zero (because $P = V \cdot I = 0 \cdot I_{sc} = 0$ ).
==Open-circuit voltage $V_{oc}$ == — the point where current $I = 0$ (no load connected). Voltage is at its maximum, power is again zero ( $P = V_{oc} \cdot 0 = 0$ ).
Maximum power point (MPP) $(V_{m p}, I_{m p})$ — the knee of the curve where the rectangle of area $V \cdot I$ is largest. This is where we want to operate.

Problem 1.2 — Units check

A battery is rated $28 V$ , $20 Ah$ . What is its total stored energy in watt-hours?

Recall Solution

Energy $= voltage \times charge capacity$ . Amp-hours times volts gives watt-hours. Writing the stored energy as $E_{total}$ (we reserve $Q$ for charge, never energy): $E_{total} = 28 V \times 20 Ah = 560 Wh .$ Why this works (explicit unit chain): carry the units through step by step.

Convert amp-hours to coulombs (charge): $20 Ah \times 3600 \frac{s}{h} = 72000 A\cdots = 72000 C$ .
Multiply charge by volts (a volt is a joule per coulomb, so C·V = J): $72000 C \times 28 V = 2016000 J$ .
Convert joules back to watt-hours (one Wh $= 3600$ J): $2016000 J \div 3600 \frac{J}{Wh} = 560 Wh$ . The $3600$ that entered in step 1 is exactly undone in step 3, which is why the shortcut "volts × amp-hours = watt-hours" works — the two conversions cancel.

Level 2 — Application

Problem 2.1 — Open-circuit voltage

A solar cell obeys the ideal single-diode equation $I = I_{L} - I_{0} (e^{V / (n V_{T})} - 1),$ where $I$ is the output current, $V$ the terminal voltage, $I_{L}$ the photocurrent, $I_{0}$ the dark saturation current, $n$ the ideality factor, and $V_{T}$ the thermal voltage. Given $I_{L} = 5 A$ , $I_{0} = 1 0^{- 12} A$ , $n = 1.5$ , $V_{T} = 0.026 V$ , find the open-circuit voltage $V_{oc}$ (the voltage at which $I = 0$ ).

Recall Solution

$V_{oc}$ is the voltage where current vanishes. Setting $I = 0$ in the ideal cell equation $I = I_{L} - I_{0} (e^{V / n V_{T}} - 1)$ and solving: $V_{oc} = n V_{T} ln (\frac{I _{L}}{I _{0}} + 1) .$ Why the logarithm? The diode current grows exponentially with voltage; to cancel a fixed photocurrent we must undo that exponential — the inverse of $e^{x}$ is $ln x$ . Plugging in: $V_{oc} = 1.5 \times 0.026 \times ln (\frac{5}{1 0 ^{- 12}} + 1) = 0.039 \times ln (5 \times 1 0^{12}) \approx 0.039 \times 29.25 \approx 1.14 V .$

Problem 2.2 — Fill factor

For the same cell, the numerically found MPP is $V_{m p} = 1.00 V$ , $I_{m p} = 4.90 A$ . Compute the fill factor $F F$ and $P_{ma x}$ .

Recall Solution

Maximum power: $P_{ma x} = V_{m p} I_{m p} = 1.00 \times 4.90 = 4.90 W .$ Fill factor is how "square" the curve is — the MPP rectangle divided by the ideal $V_{oc} I_{sc}$ rectangle: $F F = \frac{V _{m p} I _{m p}}{V _{oc} I _{sc}} = \frac{4.90}{1.14 \times 5} = \frac{4.90}{5.70} \approx 0.86.$ This is in the typical GaAs-space-cell band (0.85–0.88).

Problem 2.3 — Sun-angle scaling

Sunlight hits a panel at $β = 3 0^{\circ}$ off the surface normal. The full-sun intensity is $1367 W/m^{2}$ . What intensity reaches the panel, and by roughly what fraction does $I_{sc}$ drop?

Recall Solution

Only the component of light along the surface normal delivers power, and that projection is a cosine (adjacent over hypotenuse on the light-angle triangle): $S = 1367 cos 3 0^{\circ} = 1367 \times 0.866 = 1184 W/m^{2} .$ Photocurrent is proportional to absorbed photon flux, so $I_{sc}$ scales with intensity: $\frac{I _{sc}}{I _{sc, 0}} = cos 3 0^{\circ} = 0.866,$ a 13.4% drop.

Level 3 — Analysis

Problem 3.1 — Derive the MPP condition

Starting from $P (V) = V I (V)$ , show that at the maximum power point $\frac{d I}{d V} = - \frac{I}{V}$ .

Recall Solution

WHAT: we want the top of the power hill. WHY the derivative: the slope $d P / d V$ is zero at a peak — that is the definition of a smooth maximum. $\frac{d P}{d V} = \frac{d}{d V} (V I) = I + V \frac{d I}{d V} .$ (Product rule: differentiate $V$ leaving $I$ , plus $V$ times the derivative of $I$ .) Set it to zero at the MPP: $I + V \frac{d I}{d V} = 0 ⟹ \frac{d I}{d V} = - \frac{I}{V} .$ What it means: the incremental conductance $d I / d V$ exactly cancels the bulk conductance $I / V$ . This is the identity the incremental-conductance MPPT controller watches for. See Power Electronics.

Problem 3.2 — Cycle-life scaling

Li-ion cycle life follows $N \approx N_{0} (DoD / DoD_{0})^{- k}$ with $N_{0} = 5000$ cycles at $DoD_{0} = 50%$ and $k = 1.5$ . How many cycles at $80%$ DoD? At $30%$ DoD?

Recall Solution

What the formula says: deeper discharge ( $↑$ DoD) means fewer cycles — the negative exponent flips the ratio. At $80%$ : $N = 5000 (\frac{80}{50})^{- 1.5} = 5000 \times (1.6)^{- 1.5} = 5000 \times 0.4941 \approx 2470 cycles .$ At $30%$ (shallower — should give more cycles): $N = 5000 (\frac{30}{50})^{- 1.5} = 5000 \times (0.6)^{- 1.5} = 5000 \times 2.152 \approx 10760 cycles .$ The trend is right: shallower cycling more than doubles life.

Problem 3.3 — Temperature shift of MPP

A cell string has 40 series cells. Each cell's $V_{oc}$ drifts by $- 2.2 mV/^{\circ} C$ , and each cell's short-circuit current rises slightly with heat at $+ 0.05% /^{\circ} C$ of $I_{sc}$ . Each cell starts at $V_{oc} = 1.14 V$ . The panel warms from $- 1 0^{\circ} C$ to $+ 6 0^{\circ} C$ . How much does the string $V_{oc}$ move, what is the per-cell percentage voltage drop, how much does $I_{sc}$ change, and which way must MPPT push the operating voltage?

Recall Solution

Temperature change: $Δ T = 60 - (- 10) = 7 0^{\circ} C$ . Voltage side (large, negative): per-cell drop $- 2.2 mV/^{\circ} C \times 70 = - 154 mV$ . Across 40 series cells the voltages add: $Δ V_{oc, string} = 40 \times (- 0.154 V) = - 6.16 V .$ Per-cell percentage drop (relative to the $1.14 V$ start): $\frac{- 0.154}{1.14} \times 100% = - 13.5% per cell.$ Current side (tiny, positive): heat mildly narrows the bandgap so a few more photons are absorbed, nudging $I_{sc}$ up: $\frac{Δ I _{sc}}{I _{sc}} = + 0.05% /^{\circ} C \times 70 = + 3.5%.$ Compare the two effects: voltage falls by $13.5%$ per cell while current rises only $+ 3.5%$ . The voltage loss dominates, so as the panel heats $V_{m p}$ falls and MPPT must lower the operating voltage to stay on the peak. (Hotter carriers → more diode leakage $I_{0}$ → lower voltage; see Semiconductor Physics and Spacecraft Thermal Control.)

Level 4 — Synthesis

Problem 4.1 — Eclipse energy and battery sizing

A LEO spacecraft has a $90$ -minute orbit with a $35$ -minute eclipse. Eclipse-load is $500 W$ . The battery is $28 V$ , $20 Ah$ . If mission rules cap DoD at $80%$ , does the battery cover one eclipse, and with what margin?

Recall Solution

Step 1 — energy needed: convert eclipse minutes to hours: $35 min = 0.583 h$ . $E_{need} = 500 W \times 0.583 h = 291.7 Wh .$ Step 2 — usable energy at 80% DoD: $E_{usable} = (28 \times 20) \times 0.80 = 560 \times 0.80 = 448 Wh .$ Step 3 — compare. $448 > 291.7$ , so yes, one eclipse is covered. Margin: $448 - 291.7 = 156.3 Wh$ , i.e. $\frac{156.3}{291.7} \times 100% \approx 53.6% spare.$ Equivalently the eclipse only uses $291.7/560 = 52.1%$ DoD, comfortably under the 80% cap.

Problem 4.2 — Mission lifetime in eclipse cycles

Using the actual DoD from 4.1 ( $52.1%$ ) and the cycle-life law from 3.2 ( $N_{0} = 5000$ at $50%$ , $k = 1.5$ ), estimate the battery's cycle life and the mission years, given one eclipse per $90$ -min orbit.

Recall Solution

Cycle life at 52.1% DoD: $N = 5000 (\frac{52.1}{50})^{- 1.5} = 5000 \times (1.042)^{- 1.5} = 5000 \times 0.9403 \approx 4701 cycles .$ Orbits per year: one cycle per orbit; a year is $365.25 \times 24 \times 60 = 525960$ minutes, so orbits per year $= 525 960/90 = 5844$ . Mission years: $\frac{4701}{5844} \approx 0.80 years .$ Interpretation: at ~52% DoD the battery is the life-limiter — under a year of eclipse cycling. To reach multi-year life you must lower DoD (bigger battery) or accept scheduled capacity fade. This is the classic tension in LEO power design.

Level 5 — Mastery

Problem 5.1 — RTG power budget over mission life

A radioisotope thermoelectric generator uses $^{238} Pu$ (half-life $T_{1/2} = 87.7 yr$ ). Its thermal output at launch is $4400 W_{th}$ ; thermoelectric conversion efficiency is $6.5%$ . Let $t$ be the mission elapsed time measured in years since launch. (a) Electrical power at launch? (b) Electrical power after $t = 14$ years, treating both fuel decay and a $0.8%$ /year converter degradation? Assume converter efficiency multiplies as $(1 - 0.008)^{t}$ with $t$ in years.

Recall Solution

(a) Launch electrical power: $P_{e, 0} = 4400 W_{th} \times 0.065 = 286 W_{e} .$ (b) Fuel decay follows exponential radioactive decay — activity halves every half-life, so the thermal power multiplies by $(1/2)^{t / T_{1/2}}$ where $t$ and $T_{1/2}$ are both in years (this is why we use the exponential: constant fractional decay per unit time is exactly $e^{- λ t}$ , equivalently $2^{- t / T_{1/2}}$ ). At $t = 14 yr$ : $fuel factor = (\frac{1}{2})^{14/87.7} = 2^{- 0.1596} = 0.8953.$ Converter degradation compounds yearly, with the same $t = 14 yr$ : $conv factor = (1 - 0.008)^{14} = 0.99 2^{14} = 0.8935.$ Combine (multiply the two independent losses): $P_{e} (14) = 286 \times 0.8953 \times 0.8935 \approx 286 \times 0.7999 \approx 228.8 W_{e} .$ So the RTG delivers about $229 W_{e}$ at year 14 — roughly $80%$ of launch power. Deep-space loads must be scheduled around this monotone, predictable fade (contrast with solar, which also suffers radiation-driven degradation but depends on distance and geometry).

Problem 5.2 — Choose the architecture

A probe will operate at $9.5 AU$ (near Saturn) for $12$ years, needing a steady $200 W_{e}$ . Solar intensity falls as $1/ d^{2}$ . Solar at $1 AU$ gives $1367 W/m^{2}$ ; assume $30%$ end-of-life array efficiency. Compute the required solar-array area, compare it against the RTG option from 5.1, and recommend an architecture.

Recall Solution

Step 1 — intensity at $9.5 AU$ (inverse-square: doubling distance quarters the light): $S = \frac{1367}{9. 5 ^{2}} = \frac{1367}{90.25} = 15.15 W/m^{2} .$ Step 2 — array area needed at $30%$ efficiency to make $200 W_{e}$ . Electrical power out is $P_{e} = η S A$ , so solve for area $A$ : $A = \frac{P _{e}}{η S} = \frac{200}{0.30 \times 15.15} = \frac{200}{4.545} \approx 44.0 m^{2} .$ That is a large, deployment-risky array (about a $6.6 m \times 6.6 m$ square) delivering feeble power in permanent cold and dark, worsened by long-duration radiation damage. Step 3 — RTG comparison: the year-14 RTG of 5.1 already yields $\approx 229 W_{e}$ , which exceeds the required $200 W_{e}$ for the full 12-year mission, with no dependence on distance, sun angle, or eclipse and no $44 m^{2}$ deployment. Step 4 — recommendation: choose the RTG. At $9.5 AU$ solar flux is $1367/15.15 \approx 90 \times$ weaker than at Earth, forcing an impractically huge $\sim 44 m^{2}$ array for modest power that still fails behind a planet's shadow; the RTG's fade is gentle, distance-independent, and its $\approx 229 W_{e}$ output comfortably meets the $200 W_{e}$ budget. This is exactly why outer-planet missions fly RTGs.

Recall

Which operating point has maximum power, $I_{sc}$ , $V_{oc}$ , or the MPP? ::: The MPP (the knee), because power is the product $V \cdot I$ and both endpoints have one factor equal to zero. When light dims, which changes more, $I_{sc}$ or $V_{oc}$ ? ::: $I_{sc}$ (linear in intensity); $V_{oc}$ moves only logarithmically. Why do RTGs beat solar at Saturn? ::: Solar flux falls as $1/ d^{2}$ ; at $9.5 AU$ it is $\sim 90 \times$ weaker, so solar needs a huge array while the RTG output is distance-independent.