Solar PV Cells and Modules: I-V Characteristics, Efficiency, Fill Factor, Series-Parallel Connection and MPPT

This chapter opens the Solar photovoltaic (PV) systems part of the Engineering Sciences (XE) paper. It belongs to Section XE8.1 (Energy Resources and Conversion Technologies) of XE8 Energy Science, one of the nine optional sections, any two of which a candidate chooses in the exam. It answers the first half of the heading “Solar photovoltaic (PV) systems: I-V characteristics, efficiency, fill factor, series and parallel connections, sizing of PV systems (load factor, days of autonomy, battery size, inverter size, PV array size), maximum power point tracking”: the physics of the cell, the I-V curve, efficiency and fill factor, series and parallel connection, and maximum power point tracking (sizing follows in the next chapter). The numericals are fill factor and efficiency from module data, open-circuit voltage from the diode equation, string voltage at low temperature, array current and power, cutoff wavelength, and converter duty cycle for maximum power transfer.

1. How a solar cell works: the physics that matters

A solar cell is a large-area p-n junction diode. A photon whose energy E = hc/λ is at least the band gap Eg lifts an electron from the valence band to the conduction band, creating an electron-hole pair. The built-in electric field of the junction sweeps electrons to the n side and holes to the p side, so a photocurrent IL flows through an external circuit with no applied bias. Using hc ≈ 1240 eV·nm, the longest wavelength that can be absorbed is λc = 1240/Eg (nm, with Eg in eV). Crystalline silicon has Eg ≈ 1.12 eV, giving λc ≈ 1107 nm in the near infrared; gallium arsenide has Eg ≈ 1.42 eV.

Two unavoidable losses set the ceiling for a single junction. Sub-band-gap photons (E < Eg) are not absorbed at all. Thermalisation: a photon with E > Eg gives the carrier only Eg of useful energy and the excess is lost as heat. A low Eg wastes the high-energy photons, a high Eg wastes the low-energy ones, so the best single-junction gap lies near 1.1 to 1.5 eV and the theoretical efficiency limit (the Shockley-Queisser limit) is roughly one-third. Practical losses add reflection, shading by contact fingers, and recombination of carriers before they are collected.

Cells are made mainly of crystalline silicon, either single-crystal or multi-crystalline, doped with boron on the p side and phosphorus on the n side to form the junction. Thin-film cells (amorphous silicon, CdTe, CIGS) use far less material but have lower efficiencies in commercial modules. A silicon cell is an indirect-gap absorber and needs a thickness of order a hundred micrometres to absorb most of the light; a direct-gap thin film absorbs the same light within a few micrometres. A multijunction cell stacks junctions of different Eg so that each absorbs its own part of the spectrum, which lowers thermalisation loss and passes the single-junction limit at higher cost.

⚠️ Photon energy, not intensity, decides absorption
Brighter light at a wavelength longer than λc still produces no electron-hole pairs in that cell. Brightness changes only the number of photons, hence the photocurrent, not whether a photon can excite a carrier.

2. I-V characteristics

The ideal single-diode model puts the photocurrent source IL in parallel with a diode: I = IL − I₀[exp(qV/nkT) − 1], where I₀ is the diode saturation current, n the ideality factor and kT/q ≈ 25.85 mV at 300 K. Real cells add a series resistance Rs (contacts, bulk) and a shunt resistance Rsh (leakage). The two end points of the curve are the short-circuit current Isc ≈ IL (V = 0) and the open-circuit voltage Voc = (nkT/q) ln(IL/I₀ + 1) (I = 0). Power P = VI is zero at both ends and peaks at the knee, the maximum power point (Vmp, Imp).

Effect of operating conditions on the I-V curve
ChangeIscVocPmax
Irradiance G risesrises in proportion to Grises only logarithmicallyrises nearly in proportion to G
Cell temperature risesrises slightlyfalls (about 2 mV per cell per °C for silicon)falls (crystalline Si: of order −0.4 % per °C)
Series resistance Rs risesnearly unchangednearly unchangedfalls, knee is rounded

Module ratings are quoted at Standard Test Conditions (STC): irradiance 1000 W/m², cell temperature 25 °C, air-mass 1.5 spectrum. Field cells run hotter than 25 °C, so the real output is below the nameplate even in full sun.

Worked example (Voc). n = 1.2, kT/q = 25.85 mV, IL = 8 A, I₀ = 1.0 × 10⁻⁹ A. Voc = 1.2 × 0.02585 × ln(8/10⁻⁹ + 1) = 0.03102 × ln(8 × 10⁹) = 0.03102 × 22.80 = 0.707 V. Halving IL lowers Voc by only 0.03102 × ln 2 = 21.5 mV: a large change in light, a small change in voltage.

3. Efficiency and fill factor

Fill factor FF = Vmp Imp / (Voc Isc) measures how square the I-V curve is; it is the ratio of the largest rectangle under the curve to the rectangle Voc × Isc, and is always below 1 (good silicon cells reach about 0.75 to 0.85). Therefore Pmax = FF · Voc · Isc, and the efficiency is η = Pmax/(G·A) = FF · Voc · Isc/(G · A), where A is the area receiving the irradiance G. Efficiency is quoted for the module's total area unless the cell area is stated.

Worked example. A module of area 1.6 m² has Voc = 38 V, Isc = 9.0 A, Vmp = 31 V, Imp = 8.3 A at STC. Pmax = 31 × 8.3 = 257.3 W. FF = 257.3/(38 × 9.0) = 257.3/342 = 0.752. η = 257.3/(1000 × 1.6) = 0.1608, i.e. 16.1 %.

Temperature in power terms. If the power temperature coefficient is γ per °C (negative for silicon), the power at cell temperature Tc is Pmax(Tc) = Pmax,STC [1 + γ(Tc − 25)]. With γ = −0.004 per °C, a 257.3 W module at 55 °C gives 257.3 × [1 − 0.004 × 30] = 226.4 W, a 12 % loss even at full STC irradiance. Cell temperature is higher than ambient by an amount that grows with irradiance, which is why a hot, sunny day is not the best day for efficiency.

⚠️ Do not use Voc × Isc as the power
Voc × Isc is a product of two end points that never occur together, so it is not power. Dividing the power by Voc × Isc gives FF, not efficiency; efficiency always needs the incident power G·A in the denominator.

4. Series and parallel connection of cells and modules

One cell gives about 0.5 to 0.6 V at maximum power, too little for any use, so cells are wired in series to raise voltage (a 36-cell module has Voc near 21 V) and strings in parallel to raise current. The rules are those of any sources: in series the currents are equal and voltages add; in parallel the voltages are equal and currents add. An array of Ns modules in series and Np strings in parallel has Voc = Ns·Voc,module, Isc = Np·Isc,module and Pmax = Ns·Np·Pmax,module for identical modules.

Series versus parallel wiring
ConnectionVoltageCurrentMismatch problem
Series stringsum of the cell voltagessame in all; limited by the weakest or shaded cella shaded cell throttles the string and can be reverse-biased and overheat (hot spot)
Parallel stringssame across all; a weak string drags voltage toward its ownsum of the string currentsa weak string can be driven backwards by the others

Two diodes protect the array. A bypass diode across a group of cells in series gives the string current a path around a shaded group, at the price of losing that group's voltage. A blocking diode in series with each string stops reverse current flowing into a weak string or, in a battery system, back into the array at night.

Worked example. Ten modules in series have Voc = 37.5 V each at 25 °C and Voc falls 0.12 V per module for each °C of rise in temperature (so it rises 0.12 V per °C on cooling). At a cell temperature of 5 °C each module gives 37.5 + 0.12 × 20 = 39.9 V and the string 399 V. The inverter or charge controller must tolerate this cold-morning peak, not the 375 V of the nameplate.

5. Maximum power point tracking (MPPT)

A module delivers its maximum power only at one operating point, where the load line meets the curve at the knee. A load of resistance R operates at V = IR; the module is matched when R = Vmp/Imp. Because the knee moves with irradiance and temperature, a fixed load is rarely matched. An MPPT controller places a DC-DC converter between module and load and adjusts its duty cycle D so that the module sees the matched resistance whatever the load really is.

DC-DC converter relations used in MPPT
Converter (lossless, continuous conduction)Output voltageResistance seen by the source
Buck (step-down)Vo = D·ViRL/D²
Boost (step-up)Vo = Vi/(1 − D)RL(1 − D)²
  • Perturb and observe: nudge the voltage; if power rose keep going the same way, if it fell reverse. Simple, but it oscillates around the peak and can be misled by rapidly changing light.
  • Incremental conductance: at the peak dP/dV = 0, i.e. dI/dV = −I/V; compare the incremental and instantaneous conductance to decide the direction, and it can stop at the peak.
  • Constant-voltage method: hold the voltage at a fixed fraction of Voc; cheap but only approximate.

Worked example. A module has Vmp = 30 V, Imp = 8 A, so Rmpp = 3.75 Ω. It feeds a 0.6 Ω load through a lossless buck converter. The matching condition RL/D² = 3.75 gives D² = 0.6/3.75 = 0.16, D = 0.40. Check: the output is 0.4 × 30 = 12 V and 12/0.6 = 20 A, and 12 V × 20 A = 240 W = 30 V × 8 A. For a 30 V module charging a 48 V bus a boost converter is needed, with 48 = 30/(1 − D), D = 0.375.

🎯 MPPT gains energy, not Pmax
MPPT cannot raise the module's Pmax for the existing sunlight and temperature. It stops the operating point from sitting off the knee, which matters most with a battery whose voltage differs from Vmp and in cold weather when Vmp is high.

Key takeaways

  • A photon is absorbed only if hc/λ ≥ Eg; thermalisation and sub-band-gap loss limit a single junction to roughly one-third efficiency. Silicon: Eg ≈ 1.12 eV, λc ≈ 1107 nm.
  • I = IL − I₀[exp(qV/nkT) − 1]; Isc ≈ IL, Voc = (nkT/q) ln(IL/I₀ + 1). Isc ∝ irradiance while Voc grows only logarithmically and falls as temperature rises.
  • FF = Vmp Imp/(Voc Isc) and η = Pmax/(G·A) = FF·Voc·Isc/(G·A). STC is 1000 W/m², 25 °C cell temperature, AM1.5.
  • Series: same current, voltages add; parallel: same voltage, currents add. A shaded cell limits a series string; bypass diodes route around it, blocking diodes stop reverse current.
  • MPPT matches the load to Rmpp = Vmp/Imp through a DC-DC converter: buck Vo = D·Vi (source sees RL/D²), boost Vo = Vi/(1 − D) (source sees RL(1 − D)²). P&O oscillates; incremental conductance uses dI/dV = −I/V.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Light of a wavelength longer than the cutoff wavelength of a silicon cell falls on it, at very high intensity. What is the result?

    1. It produces a very large photocurrent, because intensity sets the photocurrent whatever the wavelength
    2. It produces no electron-hole pairs, because each photon carries less energy than the band gap
    3. It produces pairs, but only at the junction and so no external current flows
    4. It raises the open-circuit voltage by heating the cell above the band gap
    Show answer

    Answer: B — It produces no electron-hole pairs, because each photon carries less energy than the band gap

    Absorption needs E = hc/λ ≥ Eg; such photons fall below Eg, so no pairs are created however many arrive. Intensity only scales the count of photons that are able to excite carriers. The heating idea is wrong too: a higher temperature lowers Voc.

  2. Using hc = 1240 eV·nm, the longest wavelength absorbed by a cell material of band gap 1.12 eV is nearest to

    1. 554 nm
    2. 887 nm
    3. 1389 nm
    4. 1107 nm
    Show answer

    Answer: D — 1107 nm

    λc = 1240/1.12 = 1107 nm. The value 1389 nm comes from multiplying instead of dividing (1240 × 1.12), and 554 nm from dividing by 2.24 (a doubled gap).

  3. For a crystalline silicon module, the cell temperature rises from 25 °C to 60 °C at constant irradiance. Which set of changes is correct?

    1. Voc falls clearly, Isc rises slightly, Pmax falls
    2. Voc rises and Isc falls, so Pmax rises because the carriers are more energetic
    3. Voc is unchanged, Isc rises sharply with temperature, so Pmax rises as well
    4. Voc falls clearly, Isc falls, Pmax is unchanged
    Show answer

    Answer: A — Voc falls clearly, Isc rises slightly, Pmax falls

    Higher temperature raises the saturation current I₀ exponentially, so Voc = (nkT/q) ln(IL/I₀ + 1) falls; Isc rises slightly as the gap narrows. The voltage loss outweighs the small current gain, so Pmax falls. The idea that more heat means more power confuses the thermal effect with the irradiance effect.

  4. The series resistance of a cell increases because of corroded contacts, with irradiance and temperature unchanged. Which is the most accurate effect on its I-V curve?

    1. Isc and Voc both fall by the same fraction, so the curve shrinks uniformly towards the origin
    2. Voc rises because of the extra voltage drop in the contacts, while Isc stays the same
    3. The fill factor and the power at the knee fall, while Voc and Isc are almost unchanged
    4. The curve becomes more square and the fill factor rises, so more power is available at the knee
    Show answer

    Answer: C — The fill factor and the power at the knee fall, while Voc and Isc are almost unchanged

    At open circuit no current flows through Rs, so Voc is unchanged; at short circuit the voltage drop across Rs is small, so Isc is nearly unchanged. Between them Rs rounds the knee and lowers FF and Pmax. A squarer curve would need less, not more, series resistance.

  5. Two identical modules, each with Voc = 22 V and Isc = 5 A, are connected in parallel. The combination has

    1. Voc = 44 V and Isc = 5 A
    2. Voc = 44 V and Isc = 10 A
    3. Voc = 22 V and Isc = 5 A
    4. Voc = 22 V and Isc = 10 A
    Show answer

    Answer: D — Voc = 22 V and Isc = 10 A

    In parallel the voltage is common and the currents add: 22 V and 10 A. The 44 V and 5 A pair describes a series connection.

  6. In a series string, one cell is fully shaded and the string has no bypass diode. What happens to the current the string can carry?

    1. It is held down to what the shaded cell can supply, and that cell may be driven into reverse bias and heat up
    2. It is the average of the cell photocurrents, since the cells share the load equally
    3. It is almost unaffected, because the shaded cell only loses its own voltage and the others carry on as before
    4. It is set by the best-lit cell, since the others adjust their voltage
    Show answer

    Answer: A — It is held down to what the shaded cell can supply, and that cell may be driven into reverse bias and heat up

    Series cells carry the same current, so the weakest cell sets it; the others force their current through the shaded cell, reverse-biasing it, which dissipates power as a hot spot. A bypass diode removes the problem by giving the current another path.

  7. A 36-cell module is rated 1.2 m² with 16 % efficiency at STC. Which statement about its output in the field at 800 W/m² irradiance and a 45 °C cell temperature is most reasonable?

    1. The output exceeds 0.16 × 800 × 1.2 W because the current rises with temperature
    2. The output is exactly 0.16 × 800 × 1.2 W, because efficiency is a constant of the module
    3. The output is below 0.16 × 800 × 1.2 W because the higher temperature lowers the efficiency
    4. The output is 0.16 × 1000 × 1.2 W because only the STC irradiance matters
    Show answer

    Answer: C — The output is below 0.16 × 800 × 1.2 W because the higher temperature lowers the efficiency

    The 16 % applies at 25 °C. At 45 °C Voc and Pmax are lower, so the output is less than the 153.6 W obtained at the STC efficiency. Power does scale roughly with irradiance (800 not 1000), which rules out the 1000 W/m² idea.

  8. A lossless boost converter connects a module at its maximum power point (Vmp = 30 V) to a battery bus held at 48 V. The duty cycle is

    1. 0.625
    2. 0.375
    3. 0.600
    4. 0.250
    Show answer

    Answer: B — 0.375

    Boost: Vo = Vi/(1 − D), so 1 − D = 30/48 = 0.625 and D = 0.375. The value 0.625 is 1 − D, and 0.600 = 30/50 is not the relation. A buck-type rule (D = Vo/Vi) cannot give a step-up at all.

  9. An incremental-conductance MPPT algorithm measures V and I. At the maximum power point the condition it detects is

    1. dI/dV = −I/V
    2. dI/dV = +I/V
    3. dI/dV = 0
    4. dV/dI = −I/V
    Show answer

    Answer: A — dI/dV = −I/V

    P = VI, so dP/dV = I + V dI/dV, which vanishes when dI/dV = −I/V. The condition dI/dV = 0 holds at short circuit (a flat part of the curve), not at the knee, and the sign must be negative because I falls as V rises.

  10. A cell has Voc = 0.62 V and Isc = 35 mA/cm² at 1000 W/m² (which is 100 mW/cm²). Its efficiency is 17 %. Its fill factor is nearest to

    1. 0.55
    2. 0.93
    3. 0.86
    4. 0.78
    Show answer

    Answer: D — 0.78

    Pmax = 0.17 × 100 = 17 mW/cm². Voc × Isc = 0.62 × 35 = 21.7 mW/cm². FF = 17/21.7 = 0.783. A value of 0.55 would be a very poor cell, and a fill factor above 0.9 would be near-ideal and is not reached by silicon.

  11. Which of the following statements about the I-V characteristics of an illuminated solar cell are correct?

    1. Isc is nearly proportional to the irradiance
    2. Voc increases roughly logarithmically with irradiance
    3. The fill factor can exceed 1 for a very good cell
    4. The power delivered is highest at the open-circuit point
    Show answer

    Answer: A — Isc is nearly proportional to the irradiance; B — Voc increases roughly logarithmically with irradiance

    Isc ≈ IL follows the photon flux, while Voc = (nkT/q) ln(IL/I₀ + 1) grows only logarithmically. FF is a ratio of the knee rectangle to the Voc × Isc rectangle and is below 1 by construction. At open circuit I = 0, so the power is zero.

  12. Which of the following are correct for identical modules wired as Ns in series per string and Np strings in parallel?

    1. The array Voc is Ns times the module Voc
    2. The array Isc is Np times the module Isc
    3. The array Voc is Np times the module Voc
    4. The array maximum power is the module maximum power divided by Ns·Np
    Show answer

    Answer: A — The array Voc is Ns times the module Voc; B — The array Isc is Np times the module Isc

    Series count multiplies voltage and parallel count multiplies current, so Pmax is multiplied by Ns·Np, not divided. Using Np for the voltage swaps the two rules.

  13. Which of the following statements about maximum power point tracking are correct?

    1. The maximum power point shifts with irradiance and with cell temperature
    2. By changing the duty cycle, the converter changes the resistance the module sees
    3. MPPT raises the module Pmax above its value for the existing irradiance and temperature
    4. Perturb and observe settles at the peak with no oscillation at all
    Show answer

    Answer: A — The maximum power point shifts with irradiance and with cell temperature; B — By changing the duty cycle, the converter changes the resistance the module sees

    The knee moves with conditions, and the converter changes the apparent source-side resistance (RL/D² for a buck). MPPT only keeps the module at its existing Pmax; it cannot exceed it. Perturb and observe keeps perturbing, so it oscillates about the peak.

  14. A module of area 1.6 m² gives, at STC (1000 W/m²), Vmp = 31 V and Imp = 8.3 A. Its efficiency in percent, based on the module area and to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 16.1

    Pmax = 31 × 8.3 = 257.3 W; incident power = 1000 × 1.6 = 1600 W; η = 257.3/1600 = 0.1608 = 16.1 %. Verified twice by hand and in Python.

  15. Ten modules are connected in series. Each has Voc = 37.5 V at a cell temperature of 25 °C, and Voc of each module changes by 0.12 V for each °C (it falls as temperature rises). On a cold morning at a cell temperature of 5 °C, the Voc of the string in volts is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 399

    Cooling by 20 °C raises each module by 0.12 × 20 = 2.4 V, to 39.9 V; ten in series give 399 V. Using 25 °C data gives the trap value 375 V.

  16. A cell has ideality factor n = 1.2, kT/q = 25.85 mV, photocurrent IL = 8 A and saturation current I₀ = 1.0 × 10⁻⁹ A. Using the ideal single-diode equation, the open-circuit voltage in volts, to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 0.71

    Voc = nkT/q × ln(IL/I₀ + 1) = 1.2 × 0.02585 × ln(8 × 10⁹) = 0.03102 × 22.80 = 0.7074 V, which is 0.71 V. Checked in Python (0.7074).

  17. A module at its maximum power point has Vmp = 30 V and Imp = 8 A. It feeds a purely resistive load of 0.6 Ω through a lossless buck converter. For maximum power transfer the duty cycle, to two decimal places, must be ____.

    Numerical answer — type the value.

    Show answer

    Answer: 0.40

    Rmpp = 30/8 = 3.75 Ω. A buck converter presents RL/D² to the source, so D² = 0.6/3.75 = 0.16 and D = 0.40. Check: Vo = 12 V, Io = 20 A, 240 W in and out.