How a Solar Module Works: Cells, Bypass Diodes and the I-V Curve

A module datasheet gives you a handful of numbers: Pmax, Voc, Isc, Vmp, Imp and three temperature coefficients. Behind them is a physical object holding 144 small silicon diodes, some copper and a sheet of glass. Why does a 550 Wp module deliver about 410 W on a clear afternoon? Why does one shadow take away a third of its output? What does a 30-year performance warranty mean in actual watts?

This guide builds the answers from first principles. It assumes you know the single-diode model of one cell, which our guide to the physics of a solar cell develops from photons to current. We restate it, connect cells with Kirchhoff’s laws, derive the half-cut benefit and the maximum power point, translate the datasheet to real weather, and then shade one cell.

Every module value below is an illustrative “typical datasheet” number for a 550 Wp module, not a specific product. Use the datasheet of your own module for real work. Where we mention Ray2Volt’s warranties, we state them as quoted on this site: a 12-year product warranty and a 30-year performance warranty on panels.

What is inside a module

A module is a laminated stack in which each layer has one engineering job. From the sunny side inwards:

• Front glass: tempered, low-iron glass a few millimetres thick, usually with an anti-reflective coating. It passes light and resists hail and wind.
• Front encapsulant: an EVA or POE polymer film that melts and cross-links in a vacuum laminator. It bonds glass to cells, excludes moisture and cushions the brittle silicon.
• Cells: thin silicon wafers, each a diode that converts light to current.
• Ribbons and busbars: tinned copper strips or wires soldered to the cell contacts. They carry current from cell to cell and to the junction box. Their resistance is a loss, which is why section 3 matters.
• Rear encapsulant and backsheet or rear glass: a polymer backsheet gives electrical insulation and a moisture barrier. Glass-glass modules use a second glass sheet, as many bifacial modules do.
• Frame and junction box: an anodised aluminium frame gives stiffness, mounting points and an earthing point. The junction box houses the bypass diodes and output cables.

Design qualification and safety are tested under IEC 61215 and IEC 61730, including thermal cycling, damp heat and hot-spot endurance. These tests set a minimum and do not rank modules. For how Indian brands compare, see our Tier-1 panel brands guide.

Series and parallel connection

First the notation for this section.

• $I$: current at the terminals, in amperes (A). $V$: voltage at the terminals, in volts (V).
• $I_L$: photocurrent generated by absorbed light, in A. $I_0$: diode saturation current, in A.
• $n$: diode ideality factor, dimensionless.
• $V_T = kT/q$: thermal voltage in V, with Boltzmann’s constant $k$, cell temperature $T$ in kelvin and elementary charge $q$. At 25 °C, $V_T = 25.69\,\mathrm{mV}$.
• $R_s$ and $R_{sh}$: series and shunt resistance of one cell, in ohms (Ω).
• $N_s$: number of identical cells in series, dimensionless.

Under uniform light and temperature, the single-diode model gives one cell this relation between current and voltage.

Now connect $N_s$ such cells in a chain.

Kirchhoff’s laws on a string of cells

Kirchhoff’s current law says charge is conserved at every node. The only path between neighbouring cells is the ribbon joining them, so the same current flows through every cell. Kirchhoff’s voltage law says the terminal voltage is the sum of the cell voltages.

Assume identical cells, so each carries $V/N_s$ at the common current $I$. Substitute that into the cell equation and write $V/N_s + I R_s$ as $(V + I N_s R_s)/N_s$. The module equation follows.

It is the same diode equation with an effective thermal voltage $N_s n V_T$, a module series resistance $R_{s,m} = N_s R_s$ and a module shunt resistance $R_{sh,m} = N_s R_{sh}$. The photocurrent and saturation current are unchanged, because the same current runs through every cell.

Worked example. The example module has $V_{oc} = 49.6\,\mathrm{V}$ and $V_{mp} = 41.8\,\mathrm{V}$ with $N_s = 72$ (section 3 explains why a 144-half-cell module counts as 72). Per cell that is $49.6/72 = 0.689\,\mathrm{V}$ at open circuit and $41.8/72 = 0.581\,\mathrm{V}$ at maximum power. Sanity check: silicon cells give roughly 0.65 to 0.70 V at open circuit, so these are plausible.

Parallel connection

For parallel branches the roles swap. The voltage law forces the same voltage across every branch, and the current law makes the branch currents add at the shared node.

Here $M$ is the number of parallel branches. For $M$ identical branches the photocurrent and saturation current multiply by $M$, while the series and shunt resistances divide by $M$. The curve keeps its voltage range and stretches along the current axis. Half-cut modules use exactly this property.

Half-cut cells and ribbon loss

A half-cut module uses 144 cells, each half the size of the original full cell. Laser-cutting each wafer in two is the only change to the cell. The benefit appears in the wiring. The symbols are as follows.

• $I$: module current, in A. $P_{\text{loss}}$: power dissipated as heat in the ribbons, in watts (W).
• $R_c$: effective ribbon-path resistance of one full-size cell, in Ω. We take an illustrative $0.5\,\mathrm{m\Omega}$.
• $N_c$: number of full-size-cell equivalents in the module, here 72.

Ohm’s law gives the voltage across a resistor as $IR$, and the heat is that voltage times the current.

In a conventional module all $N_c$ cells sit in one string and every ribbon path carries the full module current.

Now arrange the 144 half-cells as two strings of 72, wired in parallel. By the parallel rule each string carries $I/2$. A half-cell has a ribbon path half as long, so its resistance is $R_c/2$. There are $2N_c$ half-cells in all.

Halving the current alone divides the loss by four, because current enters squared. Twice as many segments, each with half the resistance, cancel one another. The net loss is a quarter of the original. Each half-string still has 72 cells in series, so open-circuit voltage is unchanged, and the two strings in parallel restore the full current. Electrically, 144 half-cells equal 72 full cells.

Worked example. At the maximum power point $I = 13.16\,\mathrm{A}$, so $I^2 = 173.2\,\mathrm{A^2}$. Each full cell loses $173.2 \times 0.5\times10^{-3} = 0.0866\,\mathrm{W}$, and the module loses $72 \times 0.0866 = 6.23\,\mathrm{W}$. The half-cut layout loses a quarter, $1.56\,\mathrm{W}$. The saving is about 4.68 W, or 0.85% of 550 W. Sanity check: ribbon loss near a percent is the right order of magnitude.

The 0.5 mΩ figure is an assumption. It is about 41% of the fitted total series resistance of the example module (next section).

The I-V and P-V curves

The terminal behaviour of the module is its I-V curve: the current it delivers at each voltage under fixed light and temperature. The short-circuit current $I_{sc}$ is the current at $V = 0$. The open-circuit voltage $V_{oc}$ is the voltage at $I = 0$. The P-V curve plots power, the product of the two terminal quantities, against voltage.

The maximum power point

Power is zero at both ends, so there is a maximum between them. Treat $I$ as a function of $V$ along the curve and differentiate by the product rule. At the maximum, the slope of the P-V curve is zero.

A resistive load line from the origin to the operating point has slope $I/V$. The maximum power point is where the curve’s slope has equal size and opposite sign, so its dynamic resistance $-dV/dI$ equals the load’s $V/I$. A tracker seeks this balance, which our guide to MPPT and DC-DC converters develops. We call the point $(V_{mp}, I_{mp})$. The fill factor measures how square the curve is.

Worked example. The example datasheet gives $V_{oc} = 49.6\,\mathrm{V}$, $I_{sc} = 14.0\,\mathrm{A}$, $V_{mp} = 41.8\,\mathrm{V}$ and $I_{mp} = 13.16\,\mathrm{A}$. First check consistency: $41.8 \times 13.16 = 550.09\,\mathrm{W}$, matching the 550 Wp rating. Then $FF = 550.09/(49.6\times14.0) = 0.792$. Also $I_{mp}/I_{sc} = 0.94$ and $V_{mp}/V_{oc} = 0.843$, so the point sits near the knee, as it should. The required slope is $-I_{mp}/V_{mp} = -0.3148\,\mathrm{A/V}$, a dynamic resistance of 3.18 Ω.

Where series and shunt resistance act

From here $R_s$ and $R_{sh}$ mean module-level resistances. Write $a = N_s n V_T$ for the effective thermal voltage and $u = V + I R_s$ for the voltage across the diode and shunt. Let $g$, in siemens (S), be the combined conductance of the diode and the shunt.

Differentiate the module equation with respect to $V$, using $du/dV = 1 + R_s\,dI/dV$. This gives $dI/dV = -g\,(1 + R_s\,dI/dV)$, and solving for the slope:

Near short circuit the exponential term is tiny, so $g \approx 1/R_{sh}$. Provided $R_s \ll R_{sh}$, the slope there is about $-1/R_{sh}$: a low shunt resistance makes the flat part of the curve droop. Near open circuit the diode conducts strongly and $g$ is large. If $R_s\,g \gg 1$, the slope approaches $-1/R_s$. In general the diode’s own dynamic resistance $r_d = 1/g$ adds in series with $R_s$, so the exact end slope is the following.

Worked example. We fit the module equation to the four datasheet points, assuming $R_{sh} = 350\,\Omega$ (illustrative), and solve for the other four parameters. The fit gives $I_L = 14.004\,\mathrm{A}$, $I_0 = 3.54\times10^{-9}\,\mathrm{A}$, $R_s = 0.0882\,\Omega$ and $n = 1.214$, so $a = 2.246\,\mathrm{V}$ at 25 °C. The curve then reads as follows.

The slope at short circuit is $-2.856\times10^{-3}\,\mathrm{A/V}$ against $-1/R_{sh} = -2.857\times10^{-3}\,\mathrm{A/V}$, an excellent match. At open circuit $r_d = 0.162\,\Omega$, so the exact slope is $-1/(0.0882 + 0.162) = -4.00\,\mathrm{A/V}$. The bare $-1/R_s$ would give $-11.3\,\mathrm{A/V}$, because $R_s\,g = 0.54$ is not much greater than one. Reading $R_s$ from the slope at $V_{oc}$ therefore needs $r_d$ subtracted. At the maximum power point the model gives $-0.3148\,\mathrm{A/V}$, exactly $-I_{mp}/V_{mp}$. The loss in the series resistance there is $13.16^2\times0.0882 = 15.3\,\mathrm{W}$, or 2.8% of $P_{mp}$. A larger $R_s$ pulls the knee inward and lowers the fill factor.

STC, cell temperature and irradiance

Datasheet values are measured at Standard Test Conditions (STC): irradiance of 1000 W/m² in the module plane, a cell temperature of 25 °C and the AM1.5G reference spectrum. STC fixes the cell temperature, not the air temperature. A module on a Tirupati roof in April never sees a 25 °C cell in full sun. STC says nothing about low-light behaviour, soiling, wind cooling or ageing. It compares modules; it does not forecast output.

Cell temperature from NOCT

Datasheets also state the Nominal Operating Cell Temperature (NOCT): the cell temperature on an open rack at 800 W/m², 20 °C ambient and about 1 m/s wind. Let $T_c$ be cell temperature and $T_a$ ambient temperature, both in °C, and $G$ irradiance in W/m². In steady state, absorbed heat equals heat lost to the air. If the loss coefficient is constant, the temperature rise above ambient is proportional to $G$, and the NOCT point fixes the constant.

For a NOCT of 45 °C the constant is $25/800 = 0.03125\,\mathrm{K\,m^2/W}$, so 1000 W/m² adds 31 K to ambient. The model ignores wind and the electrical power extracted, so expect an error of several kelvin.

Translating the datasheet

The datasheet coefficients are relative changes per kelvin: $\alpha$ for $I_{sc}$, $\beta$ for $V_{oc}$ and $\gamma$ for $P_{max}$, in %/°C. Write $\Delta T = T_c - 25\,^\circ\mathrm{C}$ and $G_0 = 1000\,\mathrm{W/m^2}$. Photocurrent is proportional to photon flux, so $I_{sc}$ scales with $G$. Setting $I = 0$ in the diode equation gives $V_{oc} \approx a\ln(I_L/I_0)$, so a change in light shifts $V_{oc}$ by $a\ln(G/G_0)$: logarithmic, not linear. To first order:

In the thermal voltage $T_c$ is in kelvin. These are linear approximations, adequate over the normal operating range, and the power expression ignores the small efficiency loss at low irradiance. Because $P_{max} = V_{oc}\,I_{sc}\,FF$, the coefficients are linked by $\gamma \approx \beta + \alpha + \gamma_{FF}$, where $\gamma_{FF}$ is the fill-factor coefficient. With the example values, $\gamma_{FF} = -0.34 + 0.27 - 0.05 = -0.12\,\%/^\circ\mathrm{C}$. Heat costs power twice: voltage falls and the knee rounds.

Worked example: a 550 Wp module on a hot afternoon

Take the illustrative module, $P_{max} = 550\,\mathrm{W}$, $V_{oc} = 49.6\,\mathrm{V}$, $I_{sc} = 14.0\,\mathrm{A}$, $\gamma = -0.34\,\%/^\circ\mathrm{C}$, $\beta = -0.27\,\%/^\circ\mathrm{C}$, $\alpha = +0.05\,\%/^\circ\mathrm{C}$ and $\mathrm{NOCT} = 45\,^\circ\mathrm{C}$. Find its state at $G = 850\,\mathrm{W/m^2}$ and $T_a = 35\,^\circ\mathrm{C}$. First the cell temperature.

So $\Delta T = 36.56\,\mathrm{K}$ above STC. Next the maximum power, with $G/G_0 = 0.85$.

The short-circuit current follows from irradiance and the small positive $\alpha$.

For the open-circuit voltage, the temperature term gives $49.6 \times (1 - 0.0027 \times 36.56) = 44.70\,\mathrm{V}$. The irradiance term needs $T_c = 334.71\,\mathrm{K}$, so $N_s n k T_c/q = 72 \times 1.214 \times 0.02884\,\mathrm{V} = 2.521\,\mathrm{V}$, and $\ln 0.85 = -0.1625$.

Does it make sense? Irradiance alone would give $550 \times 0.85 = 467.5\,\mathrm{W}$, so heat removes 58.1 W, or 12.4%, which is $0.34\% \times 36.56$. The operating fill factor is $409.4/(44.29\times12.12) = 0.763$, against 0.792 at STC. The ratio is 0.963, close to the first-order estimate $1 - 0.0012\times36.56 = 0.956$. And the datasheet values are mutually consistent: $V_{mp} \times I_{mp} = 550.09\,\mathrm{W}$. This is module output only; wiring, soiling and inverter losses come on top. The cold-morning end of the voltage range matters for string design, covered in our guide to string sizing and temperature.

Mismatch, shading and bypass diodes

So far every cell was identical. In a series chain that fails once one cell receives less light. All cells carry the same current, yet a cell can only supply current up to about its own photocurrent. The weakest cell therefore limits the chain, and if the chain forces more current through it, the shaded cell is driven into reverse bias, where it absorbs power instead of producing it.

Reverse-bias heating

Let $V_r$ be the reverse voltage across the shaded cell in volts, $I$ the current through it in amperes and $P_h$ the heat it dissipates in watts. A reverse-biased cell is a passive load, so the power is the product of the two.

By Kirchhoff’s voltage law, the reverse voltage equals the voltage generated by the other cells in the protected group, plus any diode drop. Take an unprotected 72-cell string at 13.16 A with one dark cell. The other 71 cells would present $71 \times 0.581 = 41.2\,\mathrm{V}$. Reverse breakdown clamps the dark cell well below that, at an illustrative 15 V. The heat is then $15 \times 13.16 = 197\,\mathrm{W}$ in a cell of about 0.033 m², a clear route to a burnt spot. The hot-spot endurance test in IEC 61215 exists for this reason.

What a bypass diode does

A typical module has three bypass diodes in the junction box, each wired antiparallel across one third of the cells: 24 full-cell equivalents, or 48 half-cells. A diode with forward drop $V_f$ (0.7 V, illustrative) stays off while its substring generates positive voltage. When a shaded substring would go negative, the diode conducts and clamps that substring at $-V_f$. Module voltage at current $I$ is then the sum over substrings $j$ of the greater of the substring voltage and the clamp.

With the diode, the shaded cell’s reverse voltage is at most $V_f$ plus the voltages of the other 23 cells in its substring: $0.7 + 23 \times 0.689 = 16.5\,\mathrm{V}$. The table models a fully dark, leaky cell as a reverse shunt resistance $R_{sh,c}$ (illustrative).

The diode cuts the heating by roughly an order of magnitude but does not remove it, which is why thermal imaging is useful during inspection.

The stepped curve and its voltage cost

Suppose one substring receives a fraction $s$ of the light the other two get, so its photocurrent is about $s\,I_{sc}$. Below that current all three substrings contribute and the curve is normal. Above it, the shaded substring is bypassed and the voltage drops by about one third. The I-V curve gains a step and the P-V curve gains two local maxima.

For the voltage cost, take one diode conducting with the module still at 13.16 A. Then $V = \tfrac{2}{3}\times41.8 - 0.7 = 27.17\,\mathrm{V}$, a loss of 14.63 V, and the power is $27.17\times13.16 = 357.5\,\mathrm{W}$, or 65% of 550 W. An ideal two-thirds would be 66.7%, so the diode drop costs a further 1.7 percentage points. The table comes from the fitted model with three 24-cell substrings.

Near $s = 0.6$ the best operating point switches peaks, a jump of about 17 V. A tracker that only perturbs voltage gently may stay on the lower peak. Modern inverters therefore scan the whole voltage window periodically for the global peak, a behaviour covered in MPPT explained. Layout matters too: a horizontal shadow across a portrait module can reach all three substrings.

Degradation and warranty arithmetic

Warranties usually express the slow loss of output as a straight line. Let $P_0$ be initial power in W, $d_0$ the first-year drop as a fraction, $d$ the annual loss as a fraction per year and $t$ the elapsed years.

The average over a life of $T$ years is the mean of a straight line, and a stated end-of-life power $P_T$ gives the rate directly.

Worked example. The rates are illustrative, not a promise: $d_0 = 1\%$ and $d = 0.4\%$ per year for the 550 W module.

Over 30 years the average is $P_0\,(1 - 0.01 - 0.004\times15) = 0.93\,P_0$. Working backwards, a floor of 87% at year 30 with a 1% first-year drop gives $d = (1 - 0.01 - 0.87)/30 = 0.4\%$ per year, matching the table.

Two promises sit behind Ray2Volt’s panel terms: a 12-year product warranty on the product and a 30-year performance warranty on the output curve. Read the exact guaranteed power at each year, and the claim procedure, on your module’s own warranty sheet. Our guide to solar panel warranties explains what to look for.

To apply this to a real roof you need the module datasheet, the shade profile through the day and the inverter’s voltage window. Ray2Volt can do this analysis for your own site, including cell temperature, shading and string layout, as part of a free power audit.

Frequently asked questions

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We design, install, and service rooftop solar for homes and businesses across Tirupati district and Andhra Pradesh, from PM Surya Ghar residential systems to commercial and industrial plants. Every enquiry starts with a free power audit.