Earthing and Surge Protection for Solar Plants: The Physics of Fault Currents and Lightning

A solar plant is a large metal structure bolted to your building, wired with live DC cables and left outdoors through every thunderstorm season. Two threats follow. Internally, insulation fails and someone touches a live frame. Externally, lightning or a grid switching event drives a fast current into the structure and the wiring.

Earthing answers the first threat and surge protection the second. Both rest on Ohm’s law, on how current spreads through soil and on the voltage a changing current produces across an inductance. This guide defines every symbol before using it and works clearly labelled illustrative numbers. It is an explanation, not a design method. For questions to ask before you buy, see our guide to choosing a solar installer.

Why earthing exists: touch voltage and the fault loop

When insulation fails, a live conductor can touch an exposed metal part such as a module frame or an inverter enclosure. Earthing gives the fault a low-resistance return path, so the metal stays near earth potential and the current is large enough to trip a protective device.

Touch voltage and body current

• $V_t$ is the touch voltage in volts (V) between the exposed metal and the ground under a person’s feet.
• $I_f$ is the fault current in amperes (A).
• $R_{\text{path}}$ is the resistance in ohms (Ω) of the protective path between the touched metal and the earth reference.
• $R_{\text{body}}$ is the resistance of the body path, including skin contact, in ohms.
• $I_{\text{body}}$ is the current through the person, in amperes.

Assume steady current and feet at reference-earth potential. Ohm’s law gives the touch voltage across the path, and the body current across the body.

Body impedance is not constant. IEC 60479 describes how it varies with voltage, current path, contact area and skin condition, and how the effect depends on current and duration. Without quoting its curves, tens of milliamperes through the chest can be dangerous, which is why a 30 mA residual current device (RCD) is the common choice for personal protection. We use a round $R_{\text{body}} = 1\,\mathrm{k\Omega}$ for illustration only.

The fault loop: will the breaker trip fast enough?

Let $U_0$ be the nominal line-to-earth voltage, 230 V on a single-phase Indian supply, and $Z_s$ the earth-fault loop impedance in ohms, covering the source winding, line conductor, fault and protective conductor. Ohm’s law gives the fault current.

Take an illustrative $Z_s = 0.8\,\Omega$, of which $R_{\text{path}} = 0.3\,\Omega$ is the protective conductor.

That body current is far above the tens-of-milliamperes region, so speed protects the person. A 16 A Type B breaker trips magnetically at typically three to five times its rating, or 48 A to 80 A, so 287.5 A opens it almost at once. Standards set maximum disconnection times, so check IS 732 and the current edition.

TN and TT earthing systems

In a TN system the source neutral is earthed at the transformer and exposed metal is bonded back to it, so $Z_s$ is small. In a TT system the installation has its own electrode $R_A$ and the fault returns through soil to the source electrode $R_B$, with conductor resistance $R_c$, all in ohms.

With illustrative values of 10, 2 and 0.5 Ω, $I_f = 230/12.5 = 18.4\,\mathrm{A}$. The frame sits at $18.4 \times 10 = 184\,\mathrm{V}$, yet the 16 A breaker needs 48 A to 80 A and may never trip. TT systems therefore rely on an RCD of rated residual current $I_{\Delta n}$. A common condition follows, where 50 V is a conventional AC touch-voltage limit to confirm in the current edition.

A 30 mA RCD allows $R_A$ up to about $1{,}667\,\Omega$, and a 300 mA RCD about $167\,\Omega$. Ask which system your site uses.

A PV string is not a 230 V grid

A module is close to a current source. For an illustrative module with $I_{sc} = 11.0\,\mathrm{A}$ and $I_{mp} = 10.4\,\mathrm{A}$, the short-circuit current is only 1.06 times the operating current. A DC earth fault cannot deliver hundreds of amperes, and fuses rated above $I_{sc}$ will not see it. DC-side safety therefore leans on bonding and insulation monitoring, covered below.

Earth electrode resistance from first principles

An electrode’s resistance is not in the metal. It is in the soil around it. Start with the shape that can be solved exactly.

The hemispherical electrode

• $\rho$ is the soil resistivity in ohm-metres (Ω·m), and $I$ is the current leaving the electrode in amperes.
• $r_0$ is the electrode radius and $r$ the distance from its centre, in metres.
• $J$ is current density in A/m², $E$ is electric field in V/m, and $V(r)$ is potential in volts relative to remote earth.

Assume uniform soil, a hemisphere with its flat face at the surface, current spreading evenly and radially, steady current and remote earth at 0 V. At distance $r$ the current crosses a hemispherical shell of area $2\pi r^2$, and Ohm’s law in local form gives the field.

The potential is the integral of the field from $r$ out to infinity.

At the surface $r = r_0$ this is the ground potential rise. Dividing by the current gives the electrode resistance.

For $\rho = 100\,\Omega\cdot\mathrm{m}$ and $r_0 = 0.5\,\mathrm{m}$, $R = 100/(2\pi \times 0.5) = 31.8\,\Omega$, and at $I = 10\,\mathrm{A}$ the electrode rises to 318 V. The soil between $r_0$ and $r$ holds the fraction $1 - r_0/r$ of the resistance, so at $r = 10 r_0$ already 90% is counted. Nearly all the resistance sits within a few electrode radii, the electrode’s “resistance area”.

Step and touch voltage from the same field

Let $s$ be a stride and $r_1$ the nearer foot’s distance from the electrode centre. The step voltage is the potential difference between two points on the surface.

With the same soil and current, $r_1 = 1\,\mathrm{m}$ and an illustrative stride $s = 0.8\,\mathrm{m}$, $V_{\text{step}} = 159.2 \times (1 - 1/1.8) = 70.7\,\mathrm{V}$. Someone touching a frame bonded to the electrode, with feet at 1 m, has $318 - 159 = 159\,\mathrm{V}$ across them. Such voltages from only 10 A are why electrodes belong to a bonded, equipotential system.

The driven rod

A real electrode is usually a vertical rod. Sum the point-source contributions along its length, include the mirror image above the surface and average the potential on the rod. The closed result, in the form usually credited to Dwight, uses rod length $L$ and diameter $d$, both in metres.

It assumes uniform soil, a vertical rod driven flush with the surface and $L \gg d$. IS 3043 and other handbooks give a closely related logarithmic form, often written with $\ln(4L/d)$, so use the form in the edition you work to. We computed both for a 3 m rod of 16 mm diameter with python3.

Sanity checks. Resistance scales exactly with $\rho$, as the hemisphere predicted. The rod acts like a hemisphere of radius $\rho/(2\pi R) = 0.475\,\mathrm{m}$. A hemisphere of the rod’s own 8 mm radius would give 1,989 Ω at 100 Ω·m, so length does the work: doubling it to 6 m cuts resistance by about 45%, while doubling the diameter trims only 11%. Soil varies with season, so these are illustrations. The resistance a plant must achieve comes from its design standards and DISCOM.

Measuring soil resistivity and using several rods

The Wenner four-pin method

Four small pins are driven in a line at equal spacing $a$ in metres. A tester passes current $I$ through the outer pair and reads the voltage $\Delta V$ across the inner pair, giving $R = \Delta V/I$. It works through the hemisphere result and superposition. The first outer pin is a hemispherical source and the last an equal sink. The first inner pin is $a$ from the source and $2a$ from the sink, and the second inner pin is its mirror image.

This holds for uniform soil and pin depth much smaller than $a$. In layered soil it gives an apparent resistivity, and a spacing $a$ senses soil to a depth of roughly the order of $a$. Separate current and voltage pins also keep the current-pin contact resistance out of the voltage reading. The first row below checks directly: $2\pi \times 1 \times 15.9 = 99.9\,\Omega\cdot\mathrm{m}$.

The falling trend suggests more conductive soil at depth, which would favour a longer rod. One season can mislead, and the test belongs to a competent person with a calibrated instrument.

Rods in parallel

Ideal parallel resistors give $R/n$ for $n$ rods, but rods in soil are not independent. Each raises the potential of the soil around it, and a neighbouring rod sits inside that raised region, so it collects less current. Their resistance areas overlap. For two identical rods each carrying $I/2$, let $R_1$ be the single-rod resistance and $R_{12}$ the mutual resistance, the extra voltage one rod causes at the other per ampere.

Far apart, $R_{12} \to 0$ and $R_2 \to R_1/2$. Coincident, $R_{12} = R_1$ and nothing is gained. A far-field estimate $\rho/(2\pi s)$ gives 5.3 Ω at $s = 3\,\mathrm{m}$ and is crude there. Our own numerical line-source model reproduces the 33.5 Ω single rod and gives $R_{12} = 4.4\,\Omega$, so $R_2 = (33.5 + 4.4)/2 = 18.9\,\Omega$.

More rods always help, with diminishing returns, and wider spacing recovers more of the ideal. A common rule of thumb is a spacing of at least the rod length. It is approximate, since connecting conductors and soil layers change the answer.

Earthing and bonding practice on a PV plant

Equipotential bonding

Every exposed metal part a person could touch together, such as frames, rails, trays and the inverter enclosure, should stay at nearly the same potential during a fault. Two bonded parts joined by resistance $R_b$ carrying fault current $I$ differ by $I R_b$ volts, so every joint matters. Conductors and clamps must follow the standards, notably IEC 60364-7-712, and the module maker’s manual. Copper resistance follows the $R = \rho\,l/A$ relation in our guide to solar wiring from first principles, but fault-current sizing comes from the standards.

Separate or interconnected electrodes

A plant may have an AC earth, an equipment earth and a lightning earth. Suppose 10 kA enters a lightning electrode of an illustrative $10\,\Omega$. By Ohm’s law it rises to $10{,}000 \times 10 = 100\,\mathrm{kV}$ above remote earth. If the inverter’s earth is a separate electrode near 0 V, all 100 kV appears across any wire joining the two systems, and across the equipment between them.

That is the physical case for bonding electrodes together, so the system rises and falls as one. Practice varies. Some plants use one interconnected grid, others keep separate lightning pits bonded directly or through an SPD. Follow IS 3043, IEC 62305 and the project design.

Insulation resistance monitoring

Most arrays are floating, with neither pole earthed. A first insulation fault then draws almost no current, which is safe, yet it leaves the array one fault from a short circuit through earth. Transformerless inverters have no galvanic isolation, so they measure the array’s insulation resistance $R_{iso}$ to earth before connecting, and often during operation. With $V$ the test voltage in volts and $I_{\text{leak}}$ the leakage current in amperes:

Leakage paths of $N$ strings are in parallel, so their conductances add and the total is below any single path.

Take 20 strings, each leaking through an illustrative 10 MΩ. The total is $10/20 = 0.5\,\mathrm{M\Omega}$. Now let one flooded connector drop one string to 100 kΩ.

At 600 V that leaks $600/84{,}000 \approx 7.1\,\mathrm{mA}$. If the datasheet threshold is, for illustration, 100 kΩ, the inverter will not start although most of the array is healthy. Typical causes are damp connectors, dew, a nicked cable on a metal tray or a damaged backsheet. Capacitive leakage also matters on transformerless designs, one reason the switching in how a solar inverter makes AC is chosen with care.

A recurring insulation alarm is a safety signal. Do not reset it repeatedly or probe the array. A qualified electrician should find the string, and checks in our solar panel maintenance guide can catch damp connectors early.

Lightning and the voltage induced in wiring loops

Two impulse waveforms

Standard current shapes are defined in IEC 62305 and the SPD standards. A label such as 10/350 µs gives the front time, roughly how fast the current reaches its peak, then the time to fall to half of peak. The 10/350 µs impulse represents the partial current of a direct strike. Integrating a double-exponential shape with those times, we find a charge near 49 C for a 100 kA peak. The 8/20 µs impulse represents induced surges and carries of the order of twenty times less charge.

Induced voltage from the magnetic field

A nearby strike need not touch your array. Its changing field through any wiring loop induces a voltage. The notation follows.

• $\mu_0 = 4\pi \times 10^{-7}\,\mathrm{H/m}$ is the permeability of free space, $I$ the strike current in amperes and $B$ the flux density in tesla.
• $h$ is the loop height parallel to the channel in metres. The loop’s near and far edges are at $d_1$ and $d_2$ from it, and $A$ is its area in m².
• $\Phi$ is flux in webers, $M$ is mutual inductance in henries and $V$ is the induced voltage in volts.

Treat the channel as a long straight conductor and the loop as a rectangle in a plane containing it, the strongest coupling. Ampere’s law gives the field at distance $d$.

Flux through a strip of height $h$ and width $dr$ is $B\,h\,dr$, so integrating across the loop gives the total.

Flux per ampere is $M$, and Faraday’s law gives the open-circuit voltage. The narrow-loop form follows from $\ln(1+x) \approx x$.

Take an illustrative 30 kA strike rising in 1 µs, so $dI/dt = 3\times10^{10}\,\mathrm{A/s}$, at $d_1 = 50\,\mathrm{m}$ from a loop 2 m high and 5 m wide, so $d_2 = 55\,\mathrm{m}$ and $A = 10\,\mathrm{m^2}$.

The peak field at 50 m is 120 µT, and the narrow-loop formula gives 1,200 V, about 5% high. Now a loop of 0.5 m by 0.2 m, which is 0.1 m², at the same distance.

One hundred times less area gives about one hundred times less voltage (a 1 m² loop gives about 120 V). The practical lesson is to run the positive and negative string conductors together along the same route, and to keep any bonding conductor beside them. A string whose two cables run on opposite sides of a roof encloses a large area, and the surge finds it.

Treat these as orders of magnitude. A 1 µs rise is deliberately steep: the 10/350 µs shape peaks near 7.7 kA/µs for 30 kA by our fit, inducing about 290 V in the 10 m² loop, though subsequent strokes can be faster. The model ignores screening by metal structures, and the 0.17 µs propagation delay over 50 m is not negligible. The robust result is proportionality to area.

Surge protective devices and why lead length matters

A surge protective device (SPD) is nearly an open circuit in normal use and conducts when voltage rises. Type 1 devices are tested with the 10/350 µs current, for places that may carry part of a direct strike, and Type 2 with 8/20 µs, for induced and switching surges. IEC 61643-31 covers SPDs on the DC side of PV systems and IEC 61643-32 their selection and application. Use only SPDs rated for PV DC, since a DC arc has no natural zero crossing. Their rated voltage must also exceed the array’s cold-morning open-circuit voltage, as in our string sizing guide.

• $U_p$ is the voltage protection level in volts, the highest voltage across the SPD while it conducts its rated current. It is on the datasheet.
• $U_w$ is the impulse withstand voltage of the protected equipment, in volts, also on its datasheet.
• $L$ is the lead inductance in henries and $dI/dt$ is the surge current’s rate of change in A/s.

The equipment sees the SPD’s drop plus the inductive drop in the leads. Adding the peaks is conservative. Before any safety margin, the sum must stay below $U_w$.

A straight wire has about 1 µH per metre, nearly independent of diameter because the dependence is logarithmic, so a thicker wire does not rescue you. For 1 m of lead and 10 kA/µs, which is $10^{10}\,\mathrm{A/s}$:

That is the “keep leads short” rule. A 10 kA, 8/20 µs surge rises more gently, at about 1.25 kA/µs along its front, so the same metre adds 1.25 kV. The table uses an illustrative $U_p = 1.5\,\mathrm{kV}$ and $U_w = 4\,\mathrm{kV}$. Both the line lead and the earth lead count, so use their total length.

A protector with $U_p = 1.5\,\mathrm{kV}$ can leave the equipment facing 11.5 kV after one metre of lead against a steep front, almost three times this example’s 4 kV withstand. Keep leads short and straight, mount the SPD close to the equipment and follow the maker’s wiring instructions.

Air terminals and the rolling sphere

IEC 62305 places air terminals with the rolling-sphere method. Imagine a sphere of radius $R$ rolled over the structure. Wherever it can touch, a strike could attach, so terminals go there. Wherever it cannot touch, the volume is protected. The radius represents the striking distance and depends on the protection class. IEC 62305 uses classes I to IV with radii of 20, 30, 45 and 60 m.

Let $h$ be the mast height above the reference plane (ground or roof), with $h \le R$, and $r$ the horizontal distance at which a sphere resting on the plane just touches the mast tip. The sphere’s centre is at height $R$ and horizontal distance $r$, and its distance to the tip at height $h$ must equal $R$. Pythagoras gives the result directly.

Check it geometrically. If $h = R$, then $r = R$: the centre sits level with the tip, one radius away. As $h \to 0$, $r \to 0$. For $h \gt R$ the sphere touches the mast’s side and $r = R$. Every point closer than $r$ is out of the sphere’s reach, so it is protected. The protected region is a curved surface, not a 45° cone.

As a worked check, a 3 m mast with $R = 45\,\mathrm{m}$ gives $r = \sqrt{2\times45\times3 - 9} = \sqrt{261} = 16.2\,\mathrm{m}$. The shortcut $\sqrt{2Rh} = 16.4\,\mathrm{m}$ is only 1.7% higher. A smaller sphere demands more terminals.

For a PV array, treat the module plane as the reference plane. Real roofs have edges and parapets that the sphere also touches, so a full design checks them all. The standards also require a separation distance between the lightning system and PV conductors, and a terminal must not shade modules. Where separation cannot be kept, parts are bonded and Type 1 SPDs may be needed. Whether a plant needs lightning protection at all is a risk-assessment question under IEC 62305.

Ask for the evidence on your own plant: the soil survey, measured electrode resistances, bonding records, SPD datasheets with lead lengths, and the commissioning insulation record. Ray2Volt’s engineers can review the earthing, bonding and surge-protection arrangement of your site against those records, and ongoing inspection is part of what RayCare service plans are for.

Frequently asked questions

Ray2Volt Solar Private Limited

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.