Solar PV fundamentals: from photons to the maximum power point

Prerequisites: DC circuit fundamentals: voltage, current, power and the circuit laws · Buck converter fundamentals: from switch states to ripple and conduction modes

Module: re.pv · Also covers: re.mppt (introductory) · Level: undergraduate · Status: available Suggested time: 120 minutes · Labs: labs/renewable-energy/pv_model.py

A photovoltaic module has no moving parts, yet in the same sunlight it can deliver its full rated power or less than a quarter of it, depending on how it is loaded. This lesson explains where a solar cell’s current comes from, why its current–voltage curve has its shape, why one best operating voltage exists and moves with sunlight and temperature, and how a converter can follow it. Every number comes from a reproducible Python model of a synthetic 60-cell module.

Learning Objectives

After this lesson the learner can:

explain how photons above the band gap create a photocurrent in a p-n junction, and why a cell behaves as a current source in parallel with a diode;
define $I_{sc}$, $V_{oc}$, the maximum power point and the fill factor with SI units, and derive the single-diode equation with $R_s$ and $R_{sh}$ from Kirchhoff’s current law;
predict how $I_{sc}$, $V_{oc}$ and $P_{mp}$ change with irradiance and temperature, and reproduce the lesson’s numbers with the lab;
explain why a converter is needed, describe perturb-and-observe tracking and its oscillation versus speed trade-off, and state the model’s limits, separating analytical, numerical and (absent) hardware results.

Prerequisites

DC circuits: Kirchhoff’s laws, $P = VI$, series and parallel networks, the load line (module f.dc-circuits; lesson DC circuit fundamentals, slug dc-circuit-fundamentals; Chinese bridge in the prerequisite path).
The diode’s exponential law and thermal voltage (module f.electronics, introductory level); no further semiconductor physics is needed.
Running a Python script and changing one argument at a time (module f.computation).
Mathematics used: logarithm and exponential, one derivative for a maximum, one logarithmic derivative (signposted where it appears) and one fixed-point iteration; Newton’s method and golden-section search run inside the lab.
Helpful: the buck converter’s input resistance (module pe.buck; lesson Buck converter fundamentals, slug buck-converter-fundamentals; legacy buck lesson and open lab).

Physical Intuition

Photons and the band gap

Sunlight arrives as photons whose energy is set by colour: about 1.8 eV for red, 2.8 eV for blue. In silicon an electron must be lifted across the band gap of 1.12 eV before it can carry current. A photon below the gap passes through unused; a photon above it frees one electron and one hole, and its surplus becomes heat within picoseconds. These two losses are the main reason a silicon cell converts only about a fifth of the sunlight.

The junction separates the charges

Where n-type silicon meets p-type silicon, a thin region with a built-in electric field forms. Pairs generated nearby are pulled apart before they can recombine: electrons to the n-side, holes to the p-side. This flow is the photocurrent, proportional to the irradiance (light power per square metre) and almost independent of the terminal voltage.

A current source in parallel with a diode

In the dark the junction is an ordinary diode. Under light the photocurrent flows out of the p-terminal into the circuit, against the diode’s forward direction, so the terminal current is the photocurrent minus what the diode takes back:

   +----------+-------------+-----------+---/\/\/\---o  +
   |          |             |           |     R_s
  (I_ph)     ---  diode    | | R_sh    |
   |  light   \ /  (p-n     | | leakage |             V
   |  current  V   junction)| |         |
   +----------+-------------+-----------+--------------o  -

Short the terminals and the diode has almost no voltage, so the whole photocurrent flows out: the short-circuit current. Open them and the photocurrent is forced through the diode, whose voltage rises until it carries it all: the open-circuit voltage, which grows only logarithmically with light because the diode current grows exponentially. Two parasitic resistances complete the picture: the series resistance $R_s$ (Ω) of fingers, contacts and wiring costs voltage in proportion to current; the shunt resistance $R_{sh}$ (Ω) of leakage paths steals current in proportion to voltage and hurts most in weak light.

Predict before any formula: if irradiance halves, what happens to the short-circuit current and to the open-circuit voltage? If the cell heats up and the diode conducts more at the same voltage, what happens to the open-circuit voltage?

Fundamental Principles

Quantities and units

Irradiance $G$ (W/m²), light power per unit area in the module plane; standard test conditions (STC): 1000 W/m², 25 °C cell temperature, standard spectrum.
Cell temperature $T$ (K) or $T_c$ (°C), $T = T_c + 273.15$.
Terminal voltage $V$ (V) and current $I$ (A) in the generator convention; delivered power $P = VI$ (W).
Short-circuit current $I_{sc}$ (A) at $V = 0$; open-circuit voltage $V_{oc}$ (V) at $I = 0$; maximum power point (MPP) $(V_{mp}, I_{mp})$ where $P$ is largest, $P_{mp} = V_{mp} I_{mp}$ (W).
Fill factor $FF = P_{mp} / (V_{oc} I_{sc})$ (dimensionless), the fraction of the $V_{oc} \times I_{sc}$ rectangle the curve fills.
Photocurrent $I_{ph}$ (A), diode current $I_D$ (A), saturation current $I_0$ (A), ideality factor $n$ (1 to 2 for silicon), $R_s$ and $R_{sh}$ (Ω), series cells $N_s$, band-gap energy $E_g$ (eV; 1 eV $= 1.602 \times 10^{-19}$ J).
Thermal voltage $V_t = k_B T / q$ (V), with $k_B = 1.380649 \times 10^{-23}$ J/K and $q = 1.602176634 \times 10^{-19}$ C: 25.69 mV at 25 °C; for $N_s$ cells of ideality $n$ the module thermal voltage is $a = n N_s V_t$ (V).
Relative temperature coefficients $\alpha$, $\beta$, $\gamma$ of $I_{sc}$, $V_{oc}$ and $P_{mp}$ (%/K on datasheets, 1/K in equations): $\alpha$ is a model input, $\beta$ and $\gamma$ are outputs.

Model assumptions

Superposition: terminal current equals photocurrent minus diode current minus shunt leakage; the photocurrent is independent of voltage.
$I_{ph} \propto G$; spectral changes are ignored.
Diode law $I_D = I_0 (e^{V_d / a} - 1)$ with constant $n$; $V_d$ (V) is the junction voltage, not the terminal voltage.
Temperature: $I_{ph}$ rises linearly through $\alpha$ (the datasheet’s +0.05 %/K); $I_0$ rises steeply through $E_g$; $a \propto T$.
Identical, equally lit cells at one temperature (one lumped diode with $N_s$ times the voltage); partial shading, bypass diodes and reverse breakdown are outside the model.
Constant $R_s$, $R_{sh}$, $n$ and $E_g$; cell temperature is an input (no thermal model from ambient temperature, irradiance and wind).
Steady state; in the MPPT demonstration a lossless converter imposes each commanded voltage exactly and the power is measured without noise.

The lab implements exactly these statements; its outputs are numerical simulations labelled validation_status: numerical-simulation, not measurements or validated hardware performance.

The synthetic module

An original 60-cell module, similar in scale to a residential crystalline-silicon module but not a copy of any product: $I_{sc} = 9.0$ A and $V_{oc} = 37.5$ V at STC, $n = 1.3$, $R_s = 0.3$ Ω, $R_{sh} = 300$ Ω, $\alpha = +0.05$ %/K, $E_g = 1.12$ eV, $N_s = 60$.

Mathematical Derivations

The single-diode equation

Kirchhoff’s current law at the node joining the current source, the diode and the shunt resistor gives the current flowing towards the terminal:

\[I = I_{ph} - I_0 \left( e^{V_d / a} - 1 \right) - \frac{V_d}{R_{sh}}\]

With $V_d = V + I R_s$ this becomes the module equation used throughout:

\[I = I_{ph} - I_0 \left[ \exp\!\left( \frac{V + I R_s}{a} \right) - 1 \right] - \frac{V + I R_s}{R_{sh}}, \qquad a = \frac{n N_s k_B T}{q}\]

$I$ appears on both sides, inside an exponential: the equation is implicit, but it has exactly one solution for each $V$ (the right-hand side falls as $I$ rises while the left-hand side rises), so a numerical root-finder is safe.

Short circuit and open circuit

At $V = 0$ the junction voltage is $I_{sc} R_s = 2.70$ V, so the diode term $I_0 (e^{I_{sc} R_s / a} - 1) = 1.9 \times 10^{-7}$ A is about 0.2 µA and the shunt term a fraction $R_s / R_{sh}$ of the current:

\[I_{sc} \approx I_{ph} \left( 1 - \frac{R_s}{R_{sh}} \right) \approx I_{ph}\]

The model identifies $I_{ph}$ at STC with the nominal 9.0 A, so its own $I_{sc}$ is 0.1 % lower, 8.991 A.

At $I = 0$ no current flows in $R_s$, so the series resistance drops out:

\[0 = I_{ph} - I_0 \left( e^{V_{oc} / a} - 1 \right) - \frac{V_{oc}}{R_{sh}} \quad \Rightarrow \quad V_{oc} \approx a \ln\!\left( \frac{I_{ph}}{I_0} + 1 \right) \approx a \ln \frac{I_{ph}}{I_0}\]

Hence $V_{oc}$ rises by $a \ln 2 = 1.39$ V per doubling of irradiance while $I_{sc}$ doubles, and $V_{oc}$ is independent of $R_s$: poor series resistance bends the knee and shows only in the fill factor.

Saturation current from the datasheet

Solving the open-circuit condition at STC for the unquoted $I_0$,

\[I_0^{STC} = \frac{I_{ph}^{STC} - V_{oc}^{STC} / R_{sh}} {\exp\!\left( V_{oc}^{STC} / a_{STC} \right) - 1}\]

The only approximation is $I_{ph}^{STC} = I_{sc}^{STC}$; the model then reproduces $V_{oc}$ at STC exactly, by construction.

Temperature

In the usual single-diode array models,

\[I_0(T) = I_0^{STC} \left( \frac{T}{T_{STC}} \right)^{3} \exp\!\left[ \frac{E_g}{n k_B} \left( \frac{1}{T_{STC}} - \frac{1}{T} \right) \right]\]

with temperatures in kelvin, $T_{STC} = 298.15$ K and $E_g$ in joules (1.12 eV = $1.794 \times 10^{-19}$ J), so that $E_g / (n k_B) = 9.998 \times 10^{3}$ K for $n = 1.3$; the lab stores the gap in electronvolts and writes the same factor as $q E_g / (n k_B)$. The ideality factor sits in the exponent because the dominant recombination current scales as $\exp(-E_g / n k_B T)$. At fixed irradiance $I_{ph}(T) = I_{ph}^{STC} [1 + \alpha (T - T_{STC})]$.

One new piece of mathematics: a product of powers and exponentials is easiest to differentiate through its logarithm, since $\mathrm{d}(\ln y)/\mathrm{d}T = (\mathrm{d}y/\mathrm{d}T)/y$ turns products into sums. Applied to $V_{oc} \approx a \ln (I_{ph} / I_0)$ with $a \propto T$, $\mathrm{d}(\ln I_{ph})/\mathrm{d}T = \alpha$ and $\mathrm{d}(\ln I_0)/\mathrm{d}T = 3/T + E_g / (n k_B T^2)$:

\[\frac{\mathrm{d} V_{oc}}{\mathrm{d} T} = \frac{V_{oc} - 3a - N_s E_g / q}{T} + a \alpha\]

Every term is nearly constant, so $V_{oc}$ falls almost linearly, and since $N_s E_g / q = 67.2$ V exceeds $V_{oc} = 37.5$ V the slope is negative: $(37.5 - 6.01 - 67.2)/298.15 + 2.004 \times 0.0005 = -0.1188$ V/K, or $-0.317$ %/K, within the $-0.25$ to $-0.35$ %/K of crystalline-silicon datasheets. The lab’s numerical value, with the shunt term, is $-0.316$ %/K: an output of the model, not an input.

The maximum power point

$P(V) = V I(V)$ vanishes at both ends of the curve, so its maximum satisfies

\[\frac{\mathrm{d} P}{\mathrm{d} V} = I + V \frac{\mathrm{d} I}{\mathrm{d} V} = 0 \quad \Longleftrightarrow \quad \frac{\mathrm{d} I}{\mathrm{d} V} = - \frac{I}{V}\]

the incremental conductance equals minus the static conductance. For the idealised module ($R_s \to 0$, $R_{sh} \to \infty$) this rearranges to

\[V_{mp} = V_{oc} - a \ln\!\left( 1 + \frac{V_{mp}}{a} \right), \qquad I_{mp} = I_{ph} \, \frac{V_{mp} / a}{1 + V_{mp} / a}\]

solvable by a few substitutions; with the resistances there is no shortcut and the lab finds the maximum numerically.

Why a converter is needed

A resistor $R$ draws $I = V / R$, a straight load line through the origin, and operates where that line crosses the module curve. Only $R_{mp} = V_{mp} / I_{mp}$ crosses at the MPP, which moves with irradiance and temperature while the resistor does not. A lossless buck converter in continuous conduction with duty ratio $D$ (dimensionless) feeding a load $R_L$ (Ω) presents an input resistance $R_{in} = R_L / D^2$ (Ω) to the module. Since $D \le 1$, $R_{in} \ge R_L$: a buck can only raise the apparent resistance and reaches the MPP when $R_L \le R_{mp}$ (3.575 Ω at STC); a boost or buck-boost covers the other case. Maximum power point tracking (MPPT) is the control loop that chooses $D$; it is not a converter efficiency.

Perturb and observe

Each control period the tracker moves the voltage by a fixed step $\Delta V$ (V), measures the power, keeps the direction if the power rose and reverses it if it fell. It inevitably steps over the summit and back, so in steady state it cycles through $V_{mp} - \Delta V$, about $V_{mp}$ and $V_{mp} + \Delta V$ in four steps, averaging slightly below $P_{mp}$: a small step gives a small oscillation loss but a slow climb, a large step the reverse. Incremental conductance uses the derivative test $\mathrm{d}I/\mathrm{d}V > -I/V$ (left of the MPP, so raise the voltage) instead; in principle it stops at the MPP, but with measurement noise and a finite step it behaves much like perturb and observe.

Worked Examples

Numbers come from the lab’s default run, rounded for display.

Example 1: from datasheet to the current at 30 V

Thermal voltage: $V_t = k_B \times 298.15 / q = 0.025693$ V, so $a = 1.3 \times 60 \times 0.025693 = 2.0040$ V.

Saturation current: $V_{oc}/a = 37.5/2.0040 = 18.712$, $e^{18.712} = 1.3387 \times 10^{8}$; the shunt takes $37.5/300 = 0.125$ A, leaving 8.875 A for the diode, so $I_0 = 8.875 / (1.3387 \times 10^{8}) = 6.630 \times 10^{-8}$ A (66.3 nA); $I_{ph}/I_0 \sim 10^{8}$ is typical of silicon.

Current at 30 V by fixed-point iteration from $I = 8.3$ A: $V_d = 30 + 8.3 \times 0.3 = 32.49$ V, $V_d/a = 16.212$, $e^{16.212} = 1.099 \times 10^{7}$, $I_D = 6.630 \times 10^{-8} \times 1.099 \times 10^{7} = 0.729$ A, shunt current $32.49/300 = 0.108$ A, so $I = 9.000 - 0.729 - 0.108 = 8.163$ A. The next passes give 8.178, 8.1765 and 8.1766 A, the lab’s value. It converges because an error of one ampere in $I$ changes $I_D$ by only $I_D R_s / a \approx 0.72 \times 0.3 / 2.0 = 0.11$ A; that slope, $I_D R_s / a + R_s / R_{sh}$, exceeds 1 above about 36.3 V ($I_D > a / R_s = 6.7$ A; 1.33 at $V_{oc}$), where fixed-point iteration diverges, so the lab uses Newton’s method with a bisection fallback. The power at 30 V is $30 \times 8.1766 = 245.30$ W.

Example 2: maximum power point and fill factor

The lab finds $V_{mp} = 29.632$ V, $I_{mp} = 8.288$ A and $P_{mp} = 245.58$ W, so

\[FF = \frac{245.58}{37.500 \times 8.991} = \frac{245.58}{337.16} = 0.7284\]

Per cell: $V_{oc} = 0.625$ V, $V_{mp} = 0.494$ V, $P_{mp} = 4.09$ W; the MPP sits at 79 % of $V_{oc}$ and 92 % of $I_{sc}$, and the matched load is $R_{mp} = 29.632/8.288 = 3.575$ Ω.

Sanity check against an ideal module ($R_s \to 0$, $R_{sh} \to \infty$) with the same $I_{sc}$ and $V_{oc}$: iterating $V_{mp} = 37.5 - 2.004 \ln(1 + V_{mp}/2.004)$ from 30 V gives 31.95, 31.83, 31.84 V, then $I_{mp} = 9.0 \times 15.89/16.89 = 8.467$ A, $P_{mp} = 269.6$ W and $FF = 0.799$ (lab: --r-s 1e-9 --r-sh inf, 269.55 W; keeping the real module’s $I_0$ instead would give $V_{oc} = 37.53$ V). The resistances cost about 24 W: $R_s$ alone moves the MPP from 31.84 V to 29.62 V, about 2.2 V (first-order estimate $I_{mp} R_s = 2.5$ V), and the lab confirms the split (248.3 W without the shunt, 245.6 W with both).

Example 3: a hot, hazy afternoon

At $G = 600$ W/m² and $T_c = 50$ °C the photocurrent is $9.0 \times 0.6 \times (1 + 0.0005 \times 25) = 5.4675$ A and the model’s $I_{sc}$ is 5.462 A. The linear coefficient gives $V_{oc} = 37.5 (1 - 0.00316 \times 25) = 34.5$ V at full sun and 50 °C, and the irradiance term removes $a(T) \ln(1/0.6) = 2.172 \times 0.511 = 1.11$ V, about 33.4 V; the lab gives 33.388 V. The MPP moves to 26.396 V, 4.950 A, 130.67 W, $FF = 0.7165$. The 3.575 Ω resistor that was perfect at STC now sits at 19.2 V and extracts 103.5 W (79 %); holding 29.632 V extracts 111.6 W (85 %); a tracker that finds 26.4 V extracts all 130.67 W. At 200 W/m² and 25 °C the resistor manages 11.3 W of a possible 45.0 W.

Numerical or Simulation Lab

Script: labs/renewable-energy/pv_model.py, Python 3.10+, standard library only; matplotlib optional, behind --plot.
Inputs: irradiance in W/m² (0 to 2000), cell temperature in °C (−40 to 100), module parameter overrides (--cells, --i-sc, --v-oc, --ideality, --r-s, --r-sh, --alpha-isc, --band-gap), --points (default 201) and the P&O settings --step-v (0.5 V), --iterations (40) and --start-v (default 0.4 $V_{oc}$ = 15 V).
Model: the single-diode equation, solved by safeguarded Newton iteration to a current residual below $10^{-9}$ A; the MPP is bracketed on a 401-point grid and refined by golden-section search to $10^{-6}$ V. Precision of $P_{mp}$: about $10^{-12}$ W from the voltage bracket; the current tolerance could add up to $3 \times 10^{-8}$ W, but Newton’s quadratic convergence leaves residuals near $10^{-15}$ A (measured error at STC $10^{-13}$ W). A teaching model, not validated against any measured module.
Outputs: indented JSON (model, validation_status, assumptions, limitations, parameters, key points, eleven I-V samples, sweeps, temperature coefficients), every quantity a {"value", "unit"} pair.

python labs/renewable-energy/pv_model.py                       # default STC example
python labs/renewable-energy/pv_model.py --mppt                # add the P&O trajectory
python labs/renewable-energy/pv_model.py --irradiance 600 --temperature 50
python labs/renewable-energy/pv_model.py --json                # full 201-point I-V arrays
python labs/renewable-energy/pv_model.py --plot --out pv.png   # optional figure
python -m unittest tests.test_lab_pv_model                     # unit tests of the model

Expected default values, abbreviated from the JSON:

derived_parameters.thermal_voltage_module_a    2.0040 V
derived_parameters.saturation_current_stc      6.6296e-08 A
key_points.short_circuit_current               8.9910 A
key_points.open_circuit_voltage                37.500 V
key_points.v_mp / i_mp / p_mp                  29.632 V / 8.2878 A / 245.58 W
key_points.fill_factor                         0.7284
sensitivity.temperature_coefficients_at_stc    alpha +0.0500 %/K, beta -0.3165 %/K, gamma -0.4103 %/K

With --mppt (step 0.5 V, 40 iterations, start 15 V) the tracker reaches the band within one step of $V_{mp}$ at iteration 29, then cycles through 29.0, 29.5 and 30.0 V (244.85, 245.55 and 245.30 W); the mean over the last eight iterations is 245.31 W, a tracking efficiency of 0.9989.

Validation: ValueError for non-finite, boolean or out-of-range inputs, voltages below 0 V (reverse bias is outside the model) or above twice $V_{oc}$, inconsistent parameter sets ($V_{oc}/R_{sh} \ge I_{sc}$, $R_s I_{sc} \ge V_{oc}$ or $V_{oc}/a > 100$) and invalid sample counts or P&O settings (checked even without --mppt); the command line prints error: and the message to standard error and exits with code 2. Zero irradiance gives an all-zero curve with the fill factor null, never NaN. Equal inputs give identical JSON; record the command, reproducibility.python_version and the key points.

Parameter Exploration

Change one input at a time and write the prediction first.

Irradiance 200 to 1000 W/m² at 25 °C (sensitivity.irradiance_sweep). Predict $I_{sc} \propto G$, $V_{oc}$ up about 1.4 V per doubling, $P_{mp}$ nearly proportional. Result: $I_{sc}$ 1.798, 3.596, 5.395, 7.193, 8.991 A; $V_{oc}$ 34.17, 35.62, 36.46, 37.05, 37.50 V (1.42 V per doubling against $a \ln 2 = 1.39$ V); $P_{mp}$ 45.0, 95.5, 146.1, 196.3, 245.6 W. The fill factor peaks near 400 to 600 W/m² (0.745), between series loss at high current and shunt loss in weak light.
Temperature 0 to 75 °C at 1000 W/m² (sensitivity.temperature_sweep). Predict each sign. Result: $I_{sc}$ 8.879 to 9.216 A; $V_{oc}$ 40.45, 37.50, 34.51, 31.49 V, almost exactly 3.0 V per 25 K; $P_{mp}$ 270.5, 245.6, 220.2, 194.5 W ($-0.41$ %/K); $V_{mp}$ 32.7 to 23.6 V. A module 30 K above STC loses over 12 % of its power and 3.6 V of MPP voltage: a tracker must follow temperature as well as light.
Resistances. --r-s 0.15 and --r-s 0.6: $V_{oc}$ stays 37.500 V, $I_{sc}$ moves about 0.1 %, $P_{mp}$ 255.9, 245.6, 225.2 W, $FF$ 0.759, 0.728, 0.669, a pure fill-factor loss. --r-sh 50 costs 5.5 % of $P_{mp}$ at full sun (232.2 W) but 28 % at 200 W/m² (32.4 W, $FF$ 0.54), because the leakage barely changes while the photocurrent falls with the light.
P&O step --mppt --iterations 200 with --step-v 0.1, 0.5, 1.0, 2.0. Predict climb time and power ripple. Result: iterations to the MPP band 146, 29, 14, 7; power band 0.03, 0.70, 4.15, 9.21 W; tracking efficiency 0.99995, 0.9989, 0.9942, 0.9839. The oscillation loss grows roughly as the square of the step (parabolic P-V top, modulated by where the step lattice falls around $V_{mp}$) while the climb time falls only linearly, hence small or adaptive steps in practice.
Series connection --cells 120 --v-oc 75 --r-s 0.6 --r-sh 600 (two identical modules). Predict $V_{mp}$, $I_{mp}$, $FF$. Result: 59.264 V (double), 8.288 A (unchanged), 491.16 W, $FF = 0.7284$ (unchanged): series connection adds voltages and leaves current alone, exactly so only for identical, equally lit cells.

Common Misconceptions

“A 245 W module delivers 245 W whenever the sun shines.” Only at STC and only at 29.6 V. At 600 W/m² and 50 °C the best is 131 W, and a resistor matched at STC extracts 104 W of that. Check: Example 3.
“A hotter module produces more power.” Photocurrent rises 0.05 %/K, $V_{oc}$ falls about 0.32 %/K, power falls about 0.41 %/K. Check: the temperature sweep and the $V_{oc}$ derivative.
”$V_{oc}$ scales with irradiance like the current.” It scales logarithmically: a tenfold drop in light removes about $a \ln 10 = 4.6$ V plus a small shunt contribution. Check: --irradiance 100 gives $V_{oc} = 32.66$ V, 4.8 V (13 %) below STC, while $I_{sc}$ falls by 90 %.
“MPPT efficiency is the converter’s efficiency.” Tracking efficiency compares the power drawn with the power available at the MPP; converter losses multiply it. Check: the lab’s 0.9989 assumes a lossless converter.
“A smaller P&O step is always better.” Below some size the oscillation loss is negligible, but the tracker then needs many periods to follow a passing cloud. Check: Exploration 4.

AI Tutor Exercises

Prompts written so that the tutor asks before it tells.

Diagnostic: “Before explaining anything, ask me to sketch a module’s I-V curve, mark $I_{sc}$, $V_{oc}$ and the MPP, and say what happens to each when irradiance halves and when the module heats by 25 K. Only then compare my predictions with labs/renewable-energy/pv_model.py.”
Explain-back: “Ask me to explain, without equations, why a solar cell is modelled as a current source in parallel with a diode, and probe with two follow-ups such as what the diode does at short and at open circuit. Do not give the explanation first.”
Change one condition: “Before I run the lab with --r-s 0.6, ask me which of $I_{sc}$, $V_{oc}$, $FF$ and $P_{mp}$ will change and by roughly how much; afterwards ask me to explain any wrong prediction with the open-circuit derivation.”
Numerical reasoning: “Give me $V_{oc}$, $a$ and $R_{sh}$ and ask me to compute $I_0$ by hand, checking each step instead of giving the answer; then ask what an ideality factor of 1.0 instead of 1.3 does to $I_0$ and $P_{mp}$.”
Debug my work: “I will paste my own perturb-and-observe code or my reading of the lab’s trajectory. Ask what steady-state pattern I expect, then help me find the discrepancy by asking questions, not by rewriting my code, and finish by asking which model assumption a shaded cell would break.”

Knowledge Assessment

Questions 2, 4 and 6 must be solved independently (no hints, no copied answers) before re.pv is marked demonstrated; assisted answers count as learning. An independent Q6 records re.mppt as introductory learning only, since demonstrated for re.mppt needs the converter-in-the-loop lab planned in Further Learning.

Conceptual. Why does a photon below the band gap produce no current, and where does the surplus energy of a photon well above the gap go?
Numerical. Another synthetic module has $N_s = 72$, $V_{oc} = 45.0$ V, $I_{sc} = 10.2$ A, $n = 1.2$ and $R_{sh} = 400$ Ω at STC. Compute $a$ and $I_0$ with steps and units, then check with --cells 72 --v-oc 45 --i-sc 10.2 --ideality 1.2 --r-sh 400.
Conceptual. A module’s $V_{oc}$ and $I_{sc}$ are normal but its fill factor has dropped from 0.73 to 0.65. Which model parameter has most likely changed, and what fault could cause it?
Numerical. Using $-0.316$ %/K for $V_{oc}$ and $-0.410$ %/K for $P_{mp}$, estimate both at 1000 W/m² and 60 °C and give one reason the exact model values differ slightly.
Conceptual. With the load-line picture, explain why a resistor matched at STC extracts only a quarter of the available power at 200 W/m², and what a converter changes.
Numerical and conceptual. At 600 W/m² and 50 °C (MPP 26.40 V, 130.67 W) a P&O tracker with a 1.0 V step starts at 20 V. Which voltages does it visit in steady state, what are its mean power and tracking efficiency, and what does halving the step change? Check with --mppt --step-v 1 --start-v 20 --irradiance 600 --temperature 50.
Model limits. Name two lab assumptions that fail for a module with one shaded cell and what the lab would then predict wrongly.

Answers

1. A sub-gap photon cannot free an electron and is not absorbed; the surplus of an above-gap photon becomes lattice heat within picoseconds. 2. $$a = 1.2 \times 72 \times 0.025693 = 2.2198$$ V; $$V_{oc}/a = 20.272$$, $$e^{20.272} = 6.37 \times 10^{8}$$; the shunt takes $$45/400 = 0.1125$$ A, leaving 10.0875 A for the diode, so $$I_0 = 10.0875/(6.37 \times 10^{8} - 1) = 1.58 \times 10^{-8}$$ A (15.8 nA); the lab gives $$1.5844 \times 10^{-8}$$ A. 3. $$R_s$$: it leaves $$V_{oc}$$ unchanged and barely changes $$I_{sc}$$ but rounds the knee; causes include corroded ribbons, degraded solder joints or a poor connector. 4. $$V_{oc} \approx 37.5 (1 - 0.00316 \times 35) = 33.4$$ V, $$P_{mp} \approx 245.6 (1 - 0.00410 \times 35) = 210$$ W; the lab gives 33.311 V and 209.92 W because the dependence is only approximately linear. 5. The load line $$I = V/R$$ is fixed; at 200 W/m² it crosses the shrunken curve near 6.4 V, giving 11 W instead of the 45 W available near 28.2 V ($$R_{mp} = 17.7$$ Ω). A converter changes the effective resistance: a buck with $$D = 0.45$$ presents $$3.575/0.45^2 = 17.7$$ Ω and moves the crossing to the MPP. 6. It reaches the band at iteration 6 and cycles through 25, 26 and 27 V (128.74, 130.49 and 130.20 W); the mean over the last eight iterations is 129.98 W, a tracking efficiency of 0.9947. A 0.5 V step cycles through 26.0, 26.5 and 27.0 V with 130.50 W and 0.9987 but needs 12 iterations to reach the band: smaller ripple, slower climb. 7. Identical, equally lit cells and one lumped diode: a shaded cell has a much smaller photocurrent and is driven into reverse bias or bypassed, so the real curve has several local maxima (which can trap a P&O tracker) and less power than the lab's smooth single-peak prediction.

Academic References

MIT OpenCourseWare

Primary academic reference:

MIT 2.627 — Fundamentals of Photovoltaics (Fall 2013) https://ocw.mit.edu/courses/2-627-fundamentals-of-photovoltaics-fall-2013/ Syllabus: https://ocw.mit.edu/courses/2-627-fundamentals-of-photovoltaics-fall-2013/pages/syllabus/

Supporting academic references:

MIT EC.S07 — Photovoltaic Solar Energy Systems (Fall 2004) https://ocw.mit.edu/courses/ec-s07-photovoltaic-solar-energy-systems-fall-2004/ Lecture notes: https://ocw.mit.edu/courses/ec-s07-photovoltaic-solar-energy-systems-fall-2004/pages/lecture-notes/
MIT 6.012 — Microelectronic Devices and Circuits (Fall 2005) https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-fall-2005/ Lecture notes: https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-fall-2005/pages/lecture-notes/

Relevant subjects:

Photon absorption, band gap and charge separation at a p-n junction (2.627, 6.012)
Solar cell current–voltage characteristics, equivalent circuit and losses (2.627)
Module behaviour under varying irradiance and temperature, load matching and power conditioning (EC.S07)
The p-n junction diode equation and its temperature dependence (6.012)

This lesson has been independently developed by AIPE Academy with reference to the academic scope of the above courses. MIT OpenCourseWare is an external educational resource released under its own licence (generally CC BY-NC-SA 4.0). AIPE Academy is not affiliated with or endorsed by MIT.

Other references

M. G. Villalva, J. R. Gazoli and E. Ruppert Filho, “Comprehensive Approach to Modeling and Simulation of Photovoltaic Arrays”, IEEE Transactions on Power Electronics, vol. 24, no. 5, 2009. Publisher record: https://doi.org/10.1109/TPEL.2009.2013862 (single-diode parameter approach and $I_0(T)$ law; no content reproduced).
W. De Soto, S. A. Klein and W. A. Beckman, “Improvement and validation of a model for photovoltaic array performance”, Solar Energy, vol. 80, no. 1,
1. Publisher record: https://doi.org/10.1016/j.solener.2005.06.010.
R. W. Erickson and D. Maksimović, Fundamentals of Power Electronics, 3rd edition, Springer, 2020. Publisher record: https://link.springer.com/book/10.1007/978-3-030-43881-4 (converter input-resistance relations).

Further Learning

Next in this track: a planned lesson on PV systems, inverters and grid interfaces (module re.pv-systems, with pe.inverters and ps.grid-converters).
Deeper MPPT (re.mppt): drive the buck model of the open buck lab from this module model, so that duty ratio replaces voltage as the control variable and the input capacitor and sampling period enter the loop (pe.buck, pe.control).
Related reading: the diode law and thermal voltage behind $I_0$ and $n$ in f.electronics (power diodes as switches in pe.devices); within the same domain, energy and efficiency bookkeeping in re.conversion.
Validation: no lesson yet compares this model with a switching simulation or a measurement; a planned follow-on lab would check it against an ngspice model of a buck-based tracker and, where equipment exists, a measured I-V trace. Until then every number here is a model output.

Academy learning map