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

This page rebuilds the parent note's central result — the I-V curve and its maximum power point — one picture at a time. We assume you know nothing about diodes or solar cells. Every symbol is earned before it is used. By the end you will see why the curve has its shape and why there is exactly one sweet spot for power.

Parent: 3.6.22-Power-systems-—-solar-arrays-(I-V-curve,-power-tracking),-batteries-(DoD,-cycles),-

Step 1 — What "current" and "voltage" even mean here

WHAT. A solar cell has two terminals — think of them as two metal tabs. Two numbers describe what's happening at those tabs at any instant:

Current $I$ — how many charged particles flow out per second, measured in amps (A). Picture a river: current is how much water passes a point each second.
Voltage $V$ — the electrical "push" between the two tabs, measured in volts (V). Picture the height of a waterfall: bigger drop, harder push.

Power $P = V \cdot I$ is what we ultimately want (measured in watts, W) — it is push times flow, the actual energy delivered per second to the spacecraft.

WHY these two. A solar array feeds real hardware. Hardware needs power, and power is the product of exactly these two quantities. So the entire game is: for a given amount of sunlight, which pair $(V, I)$ gives the biggest product $V \cdot I$ ?

PICTURE. The cell as a black box with two tabs, an ammeter reading $I$ , a voltmeter reading $V$ .

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

Step 2 — A cell in the dark is just a diode

WHAT. Cover the cell so no light hits it. Now it behaves as a plain semiconductor diode: a one-way valve for current. Its current-voltage law is the Shockley equation:

I_{diode} = tiny leak I_{0} (e^{V / (n V_{T})} - 1)

Let's earn every symbol:

$I_{0}$ — the dark saturation current. A tiny trickle (here $\sim 1 0^{- 12} A$ ) that leaks backward even with no push. Think of a slightly leaky valve.
$V_{T} = \frac{k _{B} T}{q}$ — the thermal voltage. It sets the "scale" of voltage the diode cares about. $k_{B}$ is Boltzmann's constant (links temperature to energy), $T$ is temperature in kelvin, $q$ is the electron charge. At room temperature ( $T = 300 K$ ) this is about $26 mV$ .
$n$ — the ideality factor, a fudge number between 1 and 2 that stretches the voltage scale for real (imperfect) junctions.
$e^{(\cdot)}$ — the exponential function. Why an exponential and not, say, a straight line?

WHY subtract 1. At $V = 0$ (no push) there should be zero net current. Plugging $V = 0$ : $e^{0} - 1 = 1 - 1 = 0$ . The " $- 1$ " is there precisely to zero out the current when the push is off.

PICTURE. The classic diode curve: flat and near-zero for negative/small $V$ , then whipping upward steeply once $V$ approaches a threshold.

Step 3 — Turn on the light: a constant current source appears

WHAT. Now shine sunlight on the cell. Photons (particles of light) with energy above the material's bandgap $E_{g}$ knock electrons loose, creating a steady flow of freed charges. This is the photocurrent $I_{L}$ .

I_{L} \propto (light intensity)

$I_{L}$ — light-generated current. Double the sunlight, double $I_{L}$ . It barely cares about voltage — it is a near-constant current source, like a pump that always pushes the same amount of water regardless of the waterfall height.

WHY constant. Every second, a fixed number of photons arrive and each frees (at most) one electron-hole pair. That freeing rate is set by light, not by the terminal voltage. So $I_{L}$ is flat across the whole $V$ axis.

PICTURE. A flat horizontal line at height $I_{L}$ — the photocurrent — sitting above the (now flipped-up) diode curve from Step 2.

Step 4 — Superposition: subtract the diode from the source

WHAT. The real solar cell is the current source $I_{L}$ in parallel with the diode. By Kirchhoff's current law (charge is conserved — what flows in must flow out), the current we actually get out at the terminals is:

I = light gives I_{L} - diode eats I_{0} (e^{V / (n V_{T})} - 1)

Reading it term by term:

$I_{L}$ pushes current out to the load — the good stuff.
The diode term steals current internally, and it grows exponentially with $V$ . At low $V$ it steals almost nothing; at high $V$ it devours everything.

WHY subtract. The light source and the diode share the same two terminals but point opposite ways. Charge conservation forces the output to be the difference. This is the ideal single-diode model — the parent note's central equation (with shunt/series resistance idealised away, $R_{s h} \to \infty$ , $R_{s} \to 0$ ).

PICTURE. Take the flat $I_{L}$ line and bend its right end downward by the exploding diode term — the familiar solar-cell curve emerges.

Step 5 — Read the two anchor points: $I_{sc}$ and $V_{oc}$

WHAT. Two special operating points pin down the curve's ends.

Short circuit — connect the tabs with a wire so $V = 0$ :

I_{sc} = I_{L} - I_{0} (e^{0} - 1) = I_{L} - 0 = I_{L}

$I_{sc}$ — the short-circuit current, the tallest current. With no push, the diode steals nothing, so all the photocurrent comes out.

Open circuit — leave the tabs disconnected so $I = 0$ . Set the whole expression to zero and solve for $V$ :

0 = I_{L} - I_{0} (e^{V_{oc} / (n V_{T})} - 1) ⟹ V_{oc} = n V_{T} ln (\frac{I _{L}}{I _{0}} + 1)

$V_{oc}$ — the open-circuit voltage, the widest voltage. The logarithm $ln$ is the exact inverse of the exponential from Step 2 — it answers "what voltage makes the diode eat all the photocurrent?"

WHY these two matter. $I_{sc}$ is the top-left corner, $V_{oc}$ is the bottom-right corner. Every real operating point lives on the curve between them. Notice: at $V = 0$ power is zero (no push), and at $I = 0$ power is zero (no flow). Both extremes give zero watts — so the best point must be somewhere in the middle. That sets up Step 6.

PICTURE. The full curve with $I_{sc}$ marked on the current axis, $V_{oc}$ marked on the voltage axis, and both endpoints labelled " $P = 0$ ".

Recall

At short circuit, output current equals? ::: $I_{sc} = I_{L}$ (the diode steals nothing at $V = 0$ ). $V_{oc}$ formula? ::: $V_{oc} = n V_{T} ln (I_{L} / I_{0} + 1)$ .

Step 6 — The power curve and its single peak (MPP)

WHAT. Multiply height by width at every point: $P (V) = V \cdot I (V)$ . Plot $P$ against $V$ . It starts at $0$ (at $V = 0$ ), rises to a single hump, then crashes back to $0$ (at $V = V_{oc}$ ).

The top of the hump is the maximum power point (MPP), at coordinates $(V_{m p}, I_{m p})$ .

WHY a derivative. To find the top of a hill exactly, we ask: where is the slope flat? The derivative $\frac{d P}{d V}$ is the slope of the power curve. At the peak the slope is zero:

\frac{d P}{d V} = 0

Why this tool and not guessing? Because "slope $= 0$ " is the precise mathematical statement of "can't go any higher." Expanding with the product rule ( $P = V \cdot I$ ):

\frac{d P}{d V} = I + V \frac{d I}{d V} = 0

$I$ — current we still have.
$V \frac{d I}{d V}$ — how fast we lose current as we push voltage up ( $\frac{d I}{d V} < 0$ on the falling shoulder).
Setting them equal and opposite means: the gain from more voltage exactly cancels the loss from less current. That balance point is the MPP.

PICTURE. The I-V curve and the P-V hump on the same voltage axis; a vertical dashed line drops from the power peak to mark $V_{m p}$ , and across to $I_{m p}$ . The largest inscribed rectangle under the I-V curve has area $P_{ma x}$ .

Step 7 — Fill factor: how "square" is the curve?

WHAT. Compare the real best rectangle $V_{m p} I_{m p}$ to the ideal (unreachable) rectangle $V_{oc} I_{sc}$ :

F F = \frac{V _{m p} I _{m p}}{V _{oc} I _{sc}} = \frac{P _{ma x}}{V _{oc} I _{sc}}

Numerator — the actual best power (real rectangle).
Denominator — the "dream" power if the curve were a perfect right angle.
$F F$ — a number between 0 and 1. Closer to 1 means the curve's shoulder is squarer and the MPP hugs the corner.

WHY it's useful. One number captures curve quality. Space-grade triple-junction GaAs cells reach $F F \approx 0.85$ – $0.88$ ; ordinary silicon sits at $0.7$ – $0.85$ .

PICTURE. Two nested rectangles — the small solid one ( $V_{m p} I_{m p}$ ) inside the large dashed one ( $V_{oc} I_{sc}$ ) — their area ratio is $F F$ .

Step 8 — Edge and degenerate cases (never let the reader fall off the curve)

Every corner case, with its own reasoning:

Total darkness / eclipse ( $I_{L} = 0$ ). The source line drops to the axis; the whole curve collapses onto the dark diode of Step 2. Output power is zero — this is exactly when batteries must take over. See Orbital Mechanics (Keplerian) for eclipse timing.
Hotter cell (temperature rises). $I_{0}$ climbs exponentially with $T$ (more electrons jump the gap thermally). Since $V_{oc} \propto ln (I_{L} / I_{0})$ , a bigger $I_{0}$ shrinks $V_{oc}$ (about $- 2.2 mV/^{\circ} C$ per cell). The whole curve slides left, and $P_{ma x}$ drops — a Spacecraft Thermal Control problem. Meanwhile $I_{sc}$ barely nudges upward.
Dimmer sun (off-angle, intensity $\times cos β$ ). $I_{sc} = I_{L}$ falls linearly with intensity, but $V_{oc}$ only falls logarithmically (a $ln$ of the intensity ratio) — so voltage is almost unchanged while current sags. The MPP moves down and slightly left.
Reverse bias ( $V < 0$ ). Push the wrong way and the diode term $e^{V / (n V_{T})} \to 0$ , so $I \to I_{L} + I_{0} \approx I_{L}$ : the cell still delivers $\approx I_{L}$ but now absorbs power ( $P = V I < 0$ ). This is the danger a shaded cell faces in a series string — it becomes a heat-dumping load (Radiation Effects on Materials and hotspot concerns).
Radiation ageing. Years of space radiation cut $I_{L}$ (fewer collectable carriers) and raise $I_{0}$ (more leakage). Both shrink the curve — the MPP wanders, which is precisely why the array needs active MPPT tracking rather than a fixed operating point.

PICTURE. One panel of overlaid curves: baseline (magenta), hot (orange, shifted left), dim (violet, shorter), and aged (navy, shrunk) — showing how the MPP migrates in each case.

The one-picture summary

Everything on a single figure: the flat photocurrent line ( $I_{L}$ ) minus the exploding diode term produces the solar-cell curve; its endpoints are $I_{sc}$ and $V_{oc}$ ; the largest rectangle under it touches the curve at the MPP $(V_{m p}, I_{m p})$ ; and the ratio of that rectangle to the $V_{oc} \times I_{sc}$ box is the fill factor.

Recall Feynman retelling (say it plainly)

A solar cell in the dark is a one-way valve — a diode — that only lets current through once you push hard enough, and it does so explosively (an exponential, because charges have to climb a hill and only the energetic few make it). Shine light on it and you add a steady pump, the photocurrent $I_{L}$ , that pushes a fixed amount of current no matter what. The current you actually get is the pump minus what the valve secretly drinks: $I = I_{L} - I_{0} (e^{V / n V_{T}} - 1)$ . Short the terminals and the valve drinks nothing, so you get all the pump: $I_{sc} = I_{L}$ . Open the terminals and voltage climbs until the valve drinks the whole pump, giving $V_{oc} = n V_{T} ln (I_{L} / I_{0})$ . Power is push times flow, and it's zero at both ends — so the best spot is a hump in the middle. Find the flat top of that hump ( $d P / d V = 0$ ) and you've found the maximum power point. How square the curve is — the fill factor — tells you how close that best rectangle gets to the dream rectangle. Heat, shade, wrong angle, and age all shove the hump around, which is why a spacecraft never nails the MPP once and forgets it: it hunts for it forever.

Prerequisites & connections

Exponential/diode physics: Semiconductor Physics
Boltzmann factor & thermal voltage: Thermodynamics & Heat Transfer
$V_{oc}$ temperature drop: Spacecraft Thermal Control
Eclipse timing (when $I_{L} \to 0$ ): Orbital Mechanics (Keplerian)
MPPT converters that chase the moving MPP: Power Electronics
Long-term curve shrink: Radiation Effects on Materials