Case Study: Inrush Current In Transformers

Suppose we have a 2.0KVA transformer with the following specifications:

Peak Inrush Current occurs in one cycle = 564 Amps as measured on the scope.
Practical Max Inrush Current occurs in 5 cycles = 188.0 Amps as measured on the scope.
Input Voltage = 120VAC
Ambient temperature = 80?C.
Frequency = 60 Hz.
Efficiency of transformer = 70%.

Select a thermistor which would reduce the inrush current in the inductive circuit described above

SOLUTION:

• Calculate the Steady State Current
  (KVA of transformer) / (efficiency of transformer) X (Minimum Input voltage)
   2.0 KVA / (0.70) X (90) = 31.75 Amps.
   
• Calculate the maximum amount of Energy that the Thermistor should handle.
  Inductive Energy = ? X (L) (I)**2 I Peak
  564 Ohms Inductive Reactance = XL = 2 X (p) X (F) X (L)
  Line voltage / Max peak Current for One cycle.
  120 VAC / 564 Amp = 0.213 ohm.
   
• Calculate the Inductance.
  L = (XL) / (2) X (p) X (f)
  = (0.213W) / (2) X (3.142) X (60)
  = 0.000565 = 565 Milli Henry.
   
• Hence the Energy
  = ? X (L) X ( I)**2
  = ? X ( 0.000565) X (564)**2 @ 90.0 Joules.
   
• Since the Ambient is at 80?C and thermistors are rated up to 65?C for their operating current, then a derating factor must be taken in to account.
  Using the de-rating curve at 80?C use 90% of max rated Steady State Current.
  = 0.90 X 36 Amps = 32.40 Amps.
   
• You can use any of the devices rated up to 36.0 Amp to meet your SSI and Energy Requirements.
• See the charts for part numbers. (i.e. SL32 0R536 OR SL32 1R036).