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

This page is the drill hall for Power Systems (3.6.22). The parent note built the tools: the I–V curve, the maximum-power point, depth-of-discharge, cycle life. Here we throw every kind of input at those tools — normal cases, zero cases, extreme cases, and one nasty exam twist — so that when a real mission problem lands on your desk, you have already seen its shape.

Before we start, one promise: every symbol used below was defined in the parent. If you have not met $I_L$, $I_0$, $V_T$, $V_{oc}$, $I_{sc}$, DoD, or $N_{\text{cycles}}$ yet, read the parent first — this page assumes them and drills them.

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The scenario matrix

Think of a power-system problem as a machine with knobs. Each knob can be set to a normal value, a zero/degenerate value, or an extreme/limiting value. The table below lists every knob-setting class this topic can throw at you, and which worked example covers it.

#	Case class	What is special about it	Covered by
C1	Normal operating point	Cell working mid-curve, find $P$, $V_{mp}$	Ex 1
C2	Zero input: $V=0$	Short circuit — current only	Ex 2
C3	Zero input: $I=0$	Open circuit — voltage only	Ex 2
C4	Limiting: full darkness ($I_L\to 0$)	Eclipse — cell becomes a plain diode	Ex 3
C5	Limiting: high temperature	$V_{oc}$ falls, MPP shifts	Ex 4
C6	Off-normal sun angle ($\cos \beta$)	Geometry reduces intensity	Ex 5 (figure)
C7	Series/parallel scaling	Many cells → array voltage & current	Ex 6
C8	Battery sizing (real word problem)	Eclipse energy vs. capacity vs. DoD	Ex 7
C9	Cycle-life / mission-lifetime	Non-linear DoD penalty	Ex 8
C10	Exam twist: which converter?	Battery voltage ≠ $V_{mp}$	Ex 9

Nine examples cover all ten cells (Ex 2 covers both C2 and C3 — they are the two ends of one curve).

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Example 1 — Normal operating point (C1)

Forecast: Guess before computing — will the current be close to $I_L=0.50\text{ A}$, or nearly zero? (Hint: is $2.00\text{ V}$ close to the open-circuit voltage?)

1. Compute the thermal-voltage denominator $n{V}_T=1.5\times 0.026=0.039\text{ V}$. Why this step? The diode's exponential is $e^{V/(n{V}_T)}$; every voltage in this problem is measured in units of this $0.039\text{ V}$. We need the ruler before we measure.

2. Evaluate the diode exponent $\frac{V}{n{V}_T}=\frac{2.00}{0.039}=51.28$. Why this step? This dimensionless number tells us how deep into "diode turns on" we are. A value near $50$ means the exponential is enormous.

3. Diode current $I_0(e^{51.28}-1)=10^{-12}\times (1.86\times 10^{22})\approx 1.86\times 10^{10}\text{ A}$?? Why this step? That is absurdly larger than $I_L$. This is the check that catches you: at $V=2.00\text{ V}$ the diode has already swallowed everything — meaning $2.00\text{ V}$ is past the open-circuit voltage. The physical current would be hugely negative, which the ideal equation permits mathematically but a real illuminated cell never delivers.

4. Find where the cell actually stops: open circuit is where $I=0$: $V_{oc}=n{V}_T\ln \left(\frac{I_L}{I_0}+1\right)=0.039\ln \left(\frac{0.50}{10^{-12}}\right).$ $\ln (5\times 10^{11})=26.94$, so $V_{oc}=0.039\times 26.94=1.05\text{ V}$. Why this step? The cell can never exceed $V_{oc}$ under load. Our chosen $2.00\text{ V}$ is off the usable curve.

5. Redo at a valid operating voltage $V=1.00\text{ V}$ (below $V_{oc}$): $I=I_L-I_0(e^{1.00/0.039}-1)=0.50-10^{-12}(e^{25.64}-1).$ Here $e^{25.64}=1.37\times 10^{11}$, which is about $10^{11}$ times larger than the $1$ we subtract, so $e^{25.64}-1\approx e^{25.64}$ to eleven significant figures. Why this step? Dropping the $-1$ is not sloppiness — it is a controlled approximation. The Shockley "$-1$" only matters at low voltage (reverse bias); once the exponent exceeds about $10$, $e^x$ dwarfs $1$ and the $-1$ is invisible at our precision. So diode current $=10^{-12}\times 1.37\times 10^{11}=0.137\text{ A}$, giving $I=0.50-0.137=0.363\text{ A}$.

6. Power $P=V\cdot I=1.00\times 0.363=0.363\text{ W}$.

Verify: At $V=V_{oc}=1.05\text{ V}$ the current must be $\approx 0$. Plug in: $0.50-10^{-12}{e}^{1.05/0.039}=0.50-10^{-12}{e}^{26.9}\approx 0.50-0.50=0$. ✓ Units: $\text{V}\times \text{A}=\text{W}$. ✓

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Example 2 — The two zero-input ends: short & open circuit (C2, C3)

Forecast: Both endpoints deliver zero power — do you see why before reading on?

1. Short circuit ($V=0$): exponent $=0$, so $e^0-1=0$. Then $I=I_L-0=0.50\text{ A}=I_{sc}$. Why this step? With terminals shorted there is no voltage to drive the diode, so the diode "hides" and all photocurrent flows out. This is the top-left corner of the I–V curve.

2. Power at short circuit: $P=V\cdot I=0\times 0.50=0\text{ W}$. Why this step? Maximum current but zero volts → zero watts. A short delivers no useful energy.

3. Open circuit ($I=0$): solve $0=I_L-I_0(e^{V/n{V}_T}-1)$, giving $V_{oc}=n{V}_T\ln (I_L/I_0+1)=1.05\text{ V}$ (from Ex 1). Why this step? No load means nothing draws current; the photocurrent internally recirculates through the diode until the voltage self-limits.

4. Power at open circuit: $P=1.05\times 0=0\text{ W}$.

Verify: $I_{sc}=0.50\text{ A}$ ✓ (equals $I_L$). $V_{oc}=1.05\text{ V}$ ✓ (matches Ex 1). Both powers $=0\text{ W}$ ✓.

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Example 3 — Limiting case: full eclipse, cell in the dark (C4)

Forecast: With no light, is this still a solar cell — or has it become an ordinary diode?

1. Set $I_L=0$ in the cell equation: $I=-I_0(e^{V/n{V}_T}-1)$. Why this step? The parent's single-diode model is $I=I_L-I_0(e^{V/n{V}_T}-1)$. Removing the light term leaves the bare Shockley diode. A dark solar cell is a diode.

2. Evaluate at $V=0.5\text{ V}$: exponent $0.5/0.039=12.82$, $e^{12.82}=3.7\times 10^5$. $I=-10^{-12}(3.7\times 10^5)=-3.7\times 10^{-7}\text{ A}$. Why this step? The negative sign says current flows into the cell — it absorbs power now.

3. Power $P=V\cdot I=0.5\times (-3.7\times 10^{-7})=-1.85\times 10^{-7}\text{ W}$. Why this step? Negative power = the shaded cell is a small load, heated by the other cells. This is why bypass diodes matter in real arrays.

Verify: $I<0$ (consuming) ✓. Sign: forward-biased diode with no light must draw current, matching the Shockley curve. Units $\text{V}\cdot \text{A}=\text{W}$ ✓.

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Example 4 — Limiting case: hot cell, MPP shifts (C5)

Forecast: Hot solar panels — more power or less? (Most people guess "more sun-heat = more power." Wrong.)

1. Temperature rise $\Delta T=75-25=50{}^{\circ }\text{C}$. Why this step? The coefficient multiplies the change in temperature, not the absolute temperature.

2. Voltage drop $\Delta V_{oc}=-2.2\text{ mV}{/}^{\circ }\text{C}\times 50{}^{\circ }\text{C}=-110\text{ mV}=-0.110\text{ V}$. Why this step? Higher $T$ raises $I_0$ (dark leakage) exponentially, and since $V_{oc}\propto \ln (I_L/I_0)$, a bigger $I_0$ lowers the log. Heat is the enemy of voltage.

3. New open-circuit voltage $V_{oc}^{\prime }=1.05-0.110=0.94\text{ V}$.

4. Fractional power change (current ~constant): $\frac{\Delta P}{P}\approx \frac{\Delta V_{oc}}{V_{oc}}=\frac{-0.110}{1.05}=-0.105=-10.5\%$. Why this step? If $I$ barely changes, power tracks voltage, so a $10.5\%$ voltage loss is roughly a $10.5\%$ power loss.

Verify: $V_{oc}^{\prime }=0.94\text{ V}$ ✓, $\Delta P/P=-10.5\%$ ✓. Sign negative (power falls) ✓.

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Example 5 — Off-normal sun angle (C6) — geometric

Figure: A blue solar panel is drawn tilted so its green "panel normal" arrow points $\beta =60^{\circ }$ away from the red "to Sun" arrow. Parallel orange sunlight rays arrive horizontally; the gray arc marks the angle $\beta$ between the Sun direction and the panel normal. The picture shows that only the component of the beam along the normal lands on the cells — the geometric reason the effective intensity is $S_0\cos \beta$, not the full $S_0$.

Forecast: At $60^{\circ }$ off-normal, do we lose "$60\%$" of the sunlight, or something else?

1. Projected intensity $S=S_0\cos \beta$. Why this step? Look at the red arrow in the figure: only the component of the sunbeam along the panel's normal delivers energy. The spreading of a beam over a tilted surface is a pure $\cos$ geometry — the same rule as Orbital Mechanics (Keplerian) solar-flux projections.

2. Evaluate at $\beta =60^{\circ }$: $\cos 60^{\circ }=0.5$, so $S=1367\times 0.5=683.5{\text{ W/m}}^2$. Why this step? Not $40\%$ — it is exactly half. The cosine is what earns the number.

3. $I_{sc}$ falls proportionally to intensity: factor $=0.5$, i.e. $I_{sc}$ halves. Why this step? Photocurrent $I_L\propto$ absorbed photon flux $\propto S$. Since $I_{sc}=I_L$, the current tracks intensity linearly (voltage would drop only logarithmically — see parent).

4. Extreme $\beta =0^{\circ }$ (face-on): $\cos 0^{\circ }=1$, full $1367{\text{ W/m}}^2$, $I_{sc}$ unchanged. Extreme $\beta =90^{\circ }$ (edge-on): $\cos 90^{\circ }=0$, $S=0$, $I_{sc}=0$ — the panel is a knife-edge to the Sun and collects nothing. Why this step? Covering both endpoints proves the formula behaves sanely at the limits — no sudden jumps.

Verify: $S(60^{\circ })=683.5{\text{ W/m}}^2$ ✓, factor $0.5$ ✓, $S(0^{\circ })=1367$ ✓, $S(90^{\circ })=0$ ✓.

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Example 6 — Series & parallel scaling to an array (C7)

Forecast: Which knob (series or parallel) sets the voltage, and which sets the current?

1. Series stacks voltage: $N_s=\frac{28}{0.90}=31.1\Rightarrow N_s=32$ cells. Why this step? Cells in series add their voltages (like batteries stacked end-to-end) but share the same current. Round up so the string can still reach the bus voltage after ageing/heat losses.

2. Array string voltage $V=32\times 0.90=28.8\text{ V}$ (slightly above bus, headroom for the MPPT converter).

3. Parallel stacks current: $N_p=\frac{4.5}{0.45}=10$ strings. Why this step? Parallel strings add currents at the same voltage.

4. Total cells $=N_s\times N_p=32\times 10=320$.

5. Array max power $P=(N_s{V}_{mp})\times (N_p{I}_{mp})=28.8\times 4.5=129.6\text{ W}$. Why this step? Power scales with the total number of cells: $P_{array}=N_s{N}_p{V}_{mp}{I}_{mp}=320\times 0.405\text{ W}=129.6\text{ W}$.

Verify: $N_s=32$, $N_p=10$, $V=28.8\text{ V}$, $P=129.6\text{ W}$. Cross-check: $320\times (0.90\times 0.45)=320\times 0.405=129.6\text{ W}$ ✓.

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Example 7 — Battery sizing, the real word problem (C8)

Forecast: With a $560\text{ Wh}$ battery and a $\sim 290\text{ Wh}$ demand, will we be near or far from the $80\%$ limit?

1. Eclipse duration in hours $=35/60=0.583\text{ hr}$. Why this step? Energy is power × time, so time must be in hours to pair with watt-hours.

2. Energy needed $E_{\text{ecl}}=500\text{ W}\times 0.583\text{ hr}=291.7\text{ Wh}$. Why this step? This is what the battery must deliver every single eclipse.

3. Battery total energy $E_{\text{batt}}=28\text{ V}\times 20\text{ Ah}=560\text{ Wh}$. Why this step? Volts × amp-hours = watt-hours; this is the full tank. (We write energy as $E_{\text{batt}}$ in watt-hours, not $Q$: in this chapter $Q$ means electric charge in amp-hours, so calling energy $Q$ would clash. Keep charge and energy in separate symbols.)

4. Required DoD $=\frac{E_{\text{ecl}}}{E_{\text{batt}}}=\frac{291.7}{560}=0.521=52.1\%$. Why this step? DoD is the fraction of the tank we empty each orbit — here just over half.

5. Compare to the $80\%$ ceiling: $52.1\%<80\%$ → battery is comfortably sized, with margin.

Verify: $E_{\text{ecl}}=291.7\text{ Wh}$ ✓, $E_{\text{batt}}=560\text{ Wh}$ ✓, DoD $=52.1\%$ ✓ ($<80\%$). Units: $\text{Wh}/\text{Wh}$ dimensionless ✓.

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Example 8 — Cycle life and mission lifetime (C9)

Forecast: A tiny DoD increase from $50\%$ to $52\%$ — big life hit, or barely noticeable?

1. DoD ratio $\frac{52.1}{50}=1.042$. Why this step? The law only cares about the ratio of your DoD to the reference DoD.

2. Apply the exponent $N=5000\times (1.042{)}^{-1.5}$. Now $(1.042{)}^{-1.5}=0.940$, so $N=5000\times 0.940=4700\text{ cycles}$. Why this step? The negative exponent means deeper discharge → fewer cycles; the power $1.5$ makes the penalty non-linear but here mild.

3. Cycles per year: orbits per day $=\frac{1440\text{ min}}{90\text{ min}}=16$. Per year $=16\times 365.25=5844$ cycles/yr. Why this step? Each orbit is one charge/discharge cycle; count orbits to count cycles.

4. Mission life $=\frac{4700}{5844}=0.804\text{ yr}\approx 9.6\text{ months}$. Why this step? Battery cycles run out first — this is the battery-limited lifetime, a red flag for a multi-year mission.

Verify: $N=4700$ cycles ✓, orbits/yr $=5844$ ✓, life $=0.804\text{ yr}$ ✓.

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Example 9 — Exam twist: which converter? (C10)

Forecast: Direct connection is cheapest — so why does every real spacecraft add a box in between?

1. Direct-connect forces $V_{array}=V_{bus}$. When the bus sits at $24\text{ V}$, the array is dragged to $24\text{ V}$, not its $V_{mp}=28.8\text{ V}$. Why this step? A battery is a near-constant-voltage load; whatever it sits at, the array must match. That match is almost never the MPP.

2. Power lost off-MPP: at $24\text{ V}$ vs. $28.8\text{ V}$ the array sits on the steep left flank of the I–V curve, well below $P_{max}$ — you throw away watts. Why this step? The MPP is the only voltage giving max power; sitting elsewhere is strictly worse.

3. Bus goes both above and below $V_{mp}$ ($24\text{ V}<28.8\text{ V}<33.6\text{ V}$). Why this step? This is the deciding fact. Sometimes you must step the array voltage down (bus at $24\text{ V}$), sometimes up (bus at $33.6\text{ V}$).

4. Verdict — need a buck-boost MPPT converter. A pure buck (step-down) or pure boost (step-up) can only correct one direction; only a buck-boost (Power Electronics) holds the array at $28.8\text{ V}$ regardless of whether the bus is higher or lower. Why this step? The converter decouples array voltage from bus voltage, letting MPPT pin the array to $V_{mp}$ while delivering whatever the battery needs.

Verify (logic): Array $V_{mp}=28.8\text{ V}$ lies strictly between bus min $24\text{ V}$ and max $33.6\text{ V}$ → both step-up and step-down are needed → buck-boost is the unique answer. ✓

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Recall

Which endpoint of the I–V curve gives maximum current? ::: Short circuit ($V=0$), where $I=I_{sc}=I_L$. Why does a hot solar cell make less power? ::: Rising temperature increases $I_0$, which lowers $V_{oc}\propto \ln (I_L/I_0)$; current barely changes, so power falls with voltage. A cell in eclipse held at forward bias — generating or consuming? ::: Consuming — with $I_L=0$ it is a plain Shockley diode drawing current ($I<0$). Series vs parallel: which sets bus voltage? ::: Series ($N_s$) adds voltages; parallel ($N_p$) adds currents. Bus voltage swings both above and below $V_{mp}$ — which converter? ::: Buck-boost, because only it can step both up and down.