Protective Relay Time Calculator

A protective relay time calculator determines the operating time of an inverse-time overcurrent relay from the available fault current and the relay curve data supplied for the protection setting. The primary result is Calculated operating time, expressed in seconds.

Relay operating time is used during protection review to compare the clearing response of upstream and downstream protective devices. It can help evaluate whether a feeder relay, transformer relay, motor feeder relay, or other overcurrent protection element has adequate time separation from another device. The result is a relay-curve arithmetic value, not a complete coordination study.

This worksheet applies only when the relay curve constant and exponent are already known and entered directly. It does not select a relay curve, retrieve manufacturer data, program a relay, determine interrupting rating, or establish NEC compliance.

Relay Inputs and Current Multiple

The relay curve operates from the relationship between actual fault current and the relay pickup setting.

Input	Electrical purpose
Fault current (A)	The current expected to flow through the protective element during the fault condition being reviewed
Pickup current (A)	The relay pickup threshold used for the protection screen
Time dial (x)	The time-multiplier setting applied to the entered relay curve
Entered curve constant (x)	The externally supplied constant that defines the entered inverse-time characteristic
Entered curve exponent (x)	The externally supplied exponent that defines how rapidly operating time changes as current rises above pickup
Entered operating-time limit (s)	The operating-time boundary used to calculate the displayed time margin

The calculator first determines the Current multiple:

\(\displaystyle \text{Current multiple}=\frac{\text{Fault current}}{\text{Pickup current}}\)

A current multiple of 1.0 means fault current equals relay pickup. Values above 1.0 indicate current above pickup and permit the inverse-time characteristic to calculate an operating time.

For the entered values:

\(\displaystyle \frac{5000\ \text{A}}{500\ \text{A}}=10\)

The result is:

Current multiple = 10 x

The same 5,000 A fault can produce a very different relay time when the pickup current changes. Pickup must therefore match the protection setting being reviewed, including the applicable CT ratio, relay element setting basis, and device configuration.

Entered Inverse-Time Curve Formula

With externally supplied curve constants, the calculator uses the entered current multiple to calculate relay operating time:

\(\displaystyle t=\frac{\text{Time dial}\times\text{Entered curve constant}} {\left(\text{Current multiple}^{\text{Entered curve exponent}}\right)-1}\)

Where:

• t = calculated relay operating time in seconds
• Time dial = entered time multiplier
• Entered curve constant = externally supplied curve constant
• Current multiple = Fault current divided by Pickup current
• Entered curve exponent = externally supplied curve exponent

The formula produces an inverse-time response: as fault current rises above pickup, the denominator increases and operating time generally decreases. The actual curve shape depends entirely on the entered curve constant and entered curve exponent.

The calculator does not validate whether the entered values represent a particular relay family, IEC curve, ANSI curve, manufacturer curve, or programmed device characteristic. Use curve data from the applicable relay documentation, settings file, protection study, or engineering schedule.

Calculation Example

Using the worksheet values:

Field	Entered value
Fault current (A)	5,000 A
Pickup current (A)	500 A
Time dial (x)	0.5
Entered curve constant (x)	0.14
Entered curve exponent (x)	0.02
Entered operating-time limit (s)	1.5 s

First, calculate the current multiple:

\(\displaystyle M=\frac{5000}{500}=10\)

Then apply the entered inverse-time equation:

\(\displaystyle t=\frac{0.5\times0.14}{10^{0.02}-1}\)

\(\displaystyle t=1.4853\ \text{s}\)

Calculated operating time = 1.4853 s

The worksheet also compares that result with the Entered operating-time limit:

\(\displaystyle \text{Time margin}= \text{Entered operating-time limit}- \text{Calculated operating time}\)

\(\displaystyle 1.5\ \text{s}-1.4853\ \text{s}=0.0147\ \text{s}\)

Time margin = 0.0147 s

A positive time margin means the calculated relay operating time is below the entered limit. It does not, by itself, prove selective coordination, equipment protection, arc-flash performance, or acceptable fault clearing.

Protection Coordination Boundaries

Relay operating time must be reviewed with the actual protective system, not in isolation. A usable coordination decision may require the relay’s complete time-current characteristic, tolerances, instantaneous elements, definite-time elements, CT performance, breaker clearing time, fuse curves, available fault current at the fault location, and upstream and downstream device settings.

Fault current is not a fixed value throughout a distribution system. Source impedance, transformer impedance, conductor length, conductor size, parallel paths, motor contribution, utility conditions, and system configuration can affect the available current. A feeder fault near the source can produce a different current multiple and relay time than a fault at the remote end of the feeder.

The calculator’s arithmetic also does not address conductor ampacity, AWG or kcmil selection, terminal rating, insulation temperature rating, adjustment factor, correction factor, current-carrying conductors, voltage drop, raceway fill, or conduit layout. Those are separate design and installation checks.

Use the displayed operating time and time margin as entered-curve arithmetic only. Final relay settings, selective coordination, equipment duty, worker-safety analysis, and code compliance require the applicable protection study, manufacturer information, project specifications, and AHJ requirements.