3.3 Surge Protective Devices (SPDs) & Voltage Surge Arresters

Surge Protective Devices (SPDs) & Surge Arresters

Surge arresters and Surge Protective Devices (SPDs) are the primary defense against transient overvoltages caused by lightning strikes, utility switching operations, capacitor bank switching, and inductive load shedding. Understanding arrester selection criteria, ratings, traveling wave physics, and installation geometry is vital for the NCEES PE Electrical: Power exam.

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1. Traveling Wave Physics & Surge Impedance

When a lightning stroke hits an overhead line or a switching transient is initiated, the disturbance travels along the line as a distributed electromagnetic wave.

Wave Velocity ($v$) and Surge Impedance ($Z_0$)

$v = \frac{1}{\sqrt{L C}} \approx 3 \times 10^8\ \text{m/s}\quad \text{(in air for overhead lines)}$

$Z_0 = \sqrt{\frac{L}{C}}$

Where:

• $L$ = Distributed line inductance per unit length ($\text{H/m}$)
• $C$ = Distributed line capacitance per unit length ($\text{F/m}$)
• Typical Overhead Line: $Z_0 \approx 300\text{--}400\ \Omega$
• Typical Underground Cable: $Z_0 \approx 30\text{--}50\ \Omega$

Boundary Discontinuities: Reflection and Refraction

When an incident voltage wave ($E_{inc}$) traveling on a line of impedance $Z_1$ reaches a junction with impedance $Z_2$ (e.g., overhead line terminating at a power transformer), reflection and refraction occur:

$\text{Reflection Coefficient (}\Gamma_R\text{):}\quad \Gamma_R = \frac{Z_2 - Z_1}{Z_2 + Z_1}$

$\text{Refraction (Transmission) Coefficient (}\Gamma_T\text{):}\quad \Gamma_T = \frac{2 Z_2}{Z_2 + Z_1} = 1 + \Gamma_R$

   Incident Wave (E_inc) ------>       Junction
   ---------------------------------------+-----------------------------------
   Line 1: Surge Impedance Z1             |      Line 2 / Transformer: Z2
   <------ Reflected Wave (E_refl)        |      ------> Refracted Wave (E_trans)
                                          |

Critical Boundary Conditions for the PE Exam

1. Open-Circuit Termination ($Z_2 = \infty$):
   $\Gamma_R = +1.0,\quad \Gamma_T = 2.0$ The voltage doubles ($E_{trans} = 2 E_{inc}$) at an open terminal or high-impedance transformer winding! This makes surge arresters placed directly at transformer terminals essential.
2. Short-Circuit Termination ($Z_2 = 0$):
   $\Gamma_R = -1.0,\quad \Gamma_T = 0$
3. Matched Line ($Z_2 = Z_1$):
   $\Gamma_R = 0,\quad \Gamma_T = 1.0\quad \text{(no reflection)}$

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2. Surge Arrester Technologies: MOV vs. Silicon Carbide

Modern power system surge arresters utilize Metal Oxide Varistor (MOV) technology, completely superseding legacy Silicon Carbide (SiC) designs.

Comparison of Arrester Technologies

The Non-Linear V-I Power Law

$I = k \times V^\alpha$

Because $\alpha \approx 30\text{--}50$ in MOVs, a $10%$ increase in applied voltage causes a hundred- to thousand-fold increase in conducted current, clamping overvoltages with extreme precision.

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3. High-Voltage Surge Arrester Ratings (IEEE C62.11 & C62.22)

Selecting a surge arrester requires matching key electrical parameters to system operating characteristics:

1. Maximum Continuous Operating Voltage (MCOV):
   The maximum designated RMS power-frequency voltage that may be applied continuously across the arrester terminals.
   $\text{MCOV} \ge V_{L-G,max}$
2. Duty Cycle Voltage Rating ($V_{rated}$):
   The designated maximum permissible RMS power-frequency voltage at which the arrester is designed to correctly discharge surges and reseal against power-frequency follow current during standard duty-cycle testing.
3. Temporary Overvoltage (TOV) Capability:
   The capability of the arrester to withstand power-frequency overvoltages (e.g., unfaulted phase voltage rise during single line-to-ground faults or load rejection) for specified durations ($0.1\text{ s}$ to $100\text{ s}$) without thermal runaway.
4. Discharge Voltage (Clamping Voltage, $V_c$):
   The crest voltage appearing across the arrester terminals while conducting a specified discharge surge current (typically evaluated at $5\text{ kA}$, $10\text{ kA}$, or $20\text{ kA}$ with an $8/20\ \mu\text{s}$ waveshape).

Impact of System Grounding on MCOV Selection

• Solidly Grounded Systems (Coefficient of Grounding $\le 0.8$): During a single line-to-ground fault, the healthy phase voltage rises to at most $\approx 1.4\times V_{L-G} \approx 0.8\times V_{L-L}$. Arrester MCOV is selected based on $V_{L-G,max} = V_{L-L,max} / \sqrt{3}$.
• Ungrounded or High-Impedance Grounded Systems: During a single line-to-ground fault, the unfaulted phase voltages rise to the full phase-to-phase voltage ($V_{L-G} = V_{L-L}$). The arrester MCOV must be rated for the full line-to-line voltage: $\text{MCOV} \ge V_{L-L,max}$.

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4. Insulation Coordination & Protective Margins

Surge arresters protect power equipment (transformers, switchgear, cables) whose insulation strength is defined by the Basic Lightning Impulse Insulation Level (BIL) and Basic Switching Impulse Insulation Level (BSL).

IEEE C62.22 Protective Margin Formulas

1. Lightning Impulse Protective Margin ($PM_1$): $PM_1 = \left( \frac{\text{BIL}}{V_{c,impulse}} - 1 \right) \times 100\% \ge 20\%$

2. Switching Surge Protective Margin ($PM_2$): $PM_2 = \left( \frac{\text{BSL}}{V_{c,switching}} - 1 \right) \times 100\% \ge 15\%$

Where $V_{c}$ is the arrester discharge voltage at the coordinating current magnitude.

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5. Low-Voltage SPDs: NEC Article 242 & UL 1449

In low-voltage ($< 1000\text{ V}$) premises wiring, NEC Article 242 and UL 1449 classify SPDs into four distinct categories based on their installation location:

   Utility Service Transformer
               |
   ============|======================================== Line Side of Service
               |   [Type 1 SPD] (Permanently connected on line side)
               v
       [Service Disconnect / Main Breaker]
               |
   ============|======================================== Load Side of Service
               |   [Type 2 SPD] (Branch panelboards)
               v
       [Distribution Panelboard]
               |
               | Conductor Length ≥ 10 m (30 ft)
               v
       [Type 3 SPD] (Point-of-use receptacles)
               |
               v
       [Sensitive Electronic Load]

Key UL 1449 Performance Parameters

• Voltage Protection Rating (VPR): The clamping voltage measured across the SPD terminals when subjected to a $6\text{ kV} / 3\text{ kA}$ combination wave impulse.
• Short-Circuit Current Rating (SCCR): The maximum symmetrical fault current the SPD can safely withstand at its terminals without rupture or fire hazard.

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6. Lead Length Inductance & Total Protective Voltage

The most critical installation rule for SPDs is minimizing connecting lead length. An SPD is connected in parallel with the protected equipment via line and ground lead conductors.

During a steep surge current wavefront ($di/dt$), the self-inductance of the lead wires ($L_{lead} \approx 0.35\ \mu\text{H/ft}$ or $1.2\ \mu\text{H/m}$) develops a substantial inductive voltage that adds directly to the SPD clamping voltage ($V_{SPD}$):

$V_{total} = V_{SPD} + L_{lead} \frac{di}{dt}$

$\text{Rule of Thumb:}\quad V_{lead} \approx 1.0\text{--}3.0\text{ kV per foot of lead wire}$

            Phase Conductor
   -----------------+---------------------> Protected Load
                    |       ^
                    |       |
                Line Lead   | V_lead,line = L_line * (di/dt)
                    |       |
                 +--+--+    v
                 | SPD |   <=== V_SPD (Clamping Voltage)
                 +--+--+    ^
                    |       |
               Ground Lead  | V_lead,gnd = L_gnd * (di/dt)
                    |       v
   -----------------+---------------------> Ground / Enclosure
                    Total Clamping Voltage V_total = V_SPD + V_lead,line + V_lead,gnd

If an SPD has a clamping rating of $V_{SPD} = 800\text{ V}$ but is installed with $3\text{ ft}$ of lead wire ($L = 1.05\ \mu\text{H}$) subjected to a modest $di/dt = 5\text{ kA/}\mu\text{s} = 5 \times 10^9\text{ A/s}$:

$V_{lead} = 1.05 \times 10^{-6}\text{ H} \times 5 \times 10^9\text{ A/s} = 5250\text{ V} = 5.25\text{ kV}$

$V_{total} = 800\text{ V} + 5250\text{ V} = \mathbf{6050\text{ V}}$

The equipment experiences $6050\text{ V}$ instead of $800\text{ V}$, guaranteeing insulation breakdown! Leads must be kept short, straight, and uncoiled.

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7. Step-by-Step Worked Calculation Example

Problem Statement

A $138\text{ kV}$ (nominal, solidly grounded wye) substation transformer has a Basic Lightning Impulse Insulation Level (BIL) of $650\text{ kV}$. The maximum continuous operating line-to-line voltage is $V_{L-L,max} = 145\text{ kV}$.

A Metal Oxide Varistor (MOV) surge arrester is installed to protect the transformer primary windings. The arrester manufacturer provides the following ratings:

• MCOV rating: $84.0\text{ kV}$
• Maximum discharge voltage ($V_c$) at $10\text{ kA}$ ($8/20\ \mu\text{s}$ wave): $280\text{ kV}$

The total lead length between the phase bus, arrester, and ground grid is $6.0\text{ ft}$ ($L_{lead} = 0.35\ \mu\text{H/ft}$). The incoming lightning surge produces a rate of current rise $di/dt = 20\text{ kA/}\mu\text{s} = 2.0 \times 10^{10}\text{ A/s}$.

Calculate:

1. Verify if the arrester MCOV is adequate for the $138\text{ kV}$ system.
2. The lightning impulse protective margin ($PM_1$) neglecting lead length.
3. The inductive voltage drop ($V_{lead}$) developed across the $6\text{ ft}$ lead wire.
4. The total voltage ($V_{total}$) impressed across the transformer insulation and the effective protective margin ($PM_{1,eff}$).

Solution Walkthrough

Step 1: Verify MCOV Adequacy

For a solidly grounded system, the maximum continuous line-to-ground operating voltage is:

$V_{L-G,max} = \frac{V_{L-L,max}}{\sqrt{3}} = \frac{145.0\text{ kV}}{\sqrt{3}} = 83.72\text{ kV}$

Since Arrester $\text{MCOV} = 84.0\text{ kV} > 83.72\text{ kV}$, the MCOV rating is adequate and correctly applied.

Step 2: Calculate Impulse Protective Margin ($PM_1$) Neglecting Leads

$PM_1 = \left( \frac{\text{BIL}}{V_c} - 1 \right) \times 100\% = \left( \frac{650\text{ kV}}{280\text{ kV}} - 1 \right) \times 100\%$

$PM_1 = (2.3214 - 1) \times 100\% = \mathbf{132.14\%}$

Since $132.14% \ge 20%$, the margin is highly acceptable under idealized zero-lead conditions.

Step 3: Calculate Lead Inductive Voltage Drop ($V_{lead}$)

Total lead inductance for $6.0\text{ ft}$:

$L_{total} = 6.0\text{ ft} \times 0.35\ \mu\text{H/ft} = 2.10\ \mu\text{H} = 2.10 \times 10^{-6}\text{ H}$

$V_{lead} = L_{total} \times \frac{di}{dt} = 2.10 \times 10^{-6}\text{ H} \times 2.0 \times 10^{10}\text{ A/s} = 42{,}000\text{ V} = \mathbf{42.0\text{ kV}}$

Step 4: Calculate Total Impressed Surge Voltage & Effective Margin

$V_{total} = V_c + V_{lead} = 280.0\text{ kV} + 42.0\text{ kV} = \mathbf{322.0\text{ kV}}$

$PM_{1,eff} = \left( \frac{\text{BIL}}{V_{total}} - 1 \right) \times 100\% = \left( \frac{650\text{ kV}}{322.0\text{ kV}} - 1 \right) \times 100\%$

$PM_{1,eff} = (2.0186 - 1) \times 100\% = \mathbf{101.86\%}$

Assessment: The effective protective margin drops from $132.14%$ to $101.86%$. Because $101.86% \ge 20%$, the installation remains safe, but the calculation demonstrates how quickly excessive lead length degrades equipment protection margins.

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8. Common NCEES Exam Pitfalls

Pitfall 1: Confusing MCOV on Solidly Grounded vs. Ungrounded Systems
On an ungrounded system, a ground fault on one phase shifts the neutral and elevates the remaining two healthy phases to the full line-to-line voltage ($V_{L-L}$). Applying an MCOV rated for $V_{L-G}$ will result in thermal runaway and violent arrester failure during an uncleared ground fault!

Pitfall 2: Neglecting Open-Circuit Wave Doubling
At open switches, unloaded transformer terminals, or cable junction transitions from overhead line ($Z_1 = 400\ \Omega$) to open breaker ($Z_2 = \infty$), the reflection coefficient $\Gamma_R = +1$, causing the voltage wave to double ($2 E_{inc}$). Arresters must be located as close to the terminal as physically possible.

Pitfall 3: Overlooking Lead Length Inductance ($L \cdot di/dt$)
When computing the total surge voltage across protected equipment, always add $V_{lead} = L (di/dt)$ to the catalog discharge voltage $V_c$.