Virtual Lab | Numerical IDMT Overcurrent Relay

Radial Feeder Protection Schematic

Primary pickup

400 A

Fault current

2400 A

PSM

6.00

Trip time

1.002 s

Numerical IDMT Relay Calculations

CT Secondary Fault Current

I_{f (\sec)} = \frac{2400}{400} = 6.00 A

Plug Setting Multiplier

PSM = \frac{I_{f (\sec)}}{I_{s}} = \frac{6.00}{1.00} = 6.00

IDMT Characteristic Equation

t = TMS \frac{0.14}{{PSM}^{0.02} - 1}

IEC Standard Inverse characteristic: $k=0.14,\ \alpha=0.02$

Numerical Operating-Time Calculation

t = 0.10 \frac{0.14}{{6.00}^{0.02} - 1} = 1.002 s

IDMT Time–Current Characteristic

Relay characteristic

Operating point

Fault Current Waveform

Current waveform

Fault region

Operating Point & Results

Parameter	Value	Engineering interpretation
System voltage	11 kV	Nominal feeder voltage
CT ratio	400/1 A	Current transformation ratio
Primary pickup current	400 A	Pickup threshold referred to primary
Fault current	2400 A	Simulated primary fault current
Relay secondary current	6.00 A	Current measured by numerical relay
Plug Setting Multiplier	6.00	Fault current divided by pickup current
Time Multiplier Setting	0.100	Controls relay operating time
Calculated operating time	1.002 s	IDMT operating time for selected characteristic

Working Principle

An inverse definite minimum time overcurrent relay operates when the measured current exceeds the preset pickup current. The relay operating time decreases as the magnitude of fault current increases. For a radial feeder, this inverse characteristic assists in obtaining selective coordination between upstream and downstream protection.

PSM = \frac{I_{f (\sec)}}{I_{s}}

t = TMS \frac{k}{{PSM}^{α} - 1}

Characteristic	$k$	$α$	Equation
IEC Standard Inverse	$0.14$	$0.02$	$t = TMS \cdot \frac{0.14}{{PSM}^{0.02} - 1}$
IEC Very Inverse	$13.5$	$1$	$t = TMS \cdot \frac{13.5}{PSM - 1}$
IEC Extremely Inverse	$80$	$2$	$t = TMS \cdot \frac{80}{{PSM}^{2} - 1}$