
60 Hz Inductor Design Requirements: 500 μH minimum from 1 to 5 amps of 60 Hz Current. 
Consider minimum current level. 
For this example, the inductor must maintain 500mH minimum from 1 to 5 amperes, or down to 20% of fullrated current (I max). The importance of this consideration is illustrated by the graph above. This shows that, in the case of 40 Material, if the inductor is designed to operate at 10 kG at I max, that the inductance will be greater than or equal to L at I max down to 2.5 / 42 = 6.0% of I max. Likewise, if the inductor operates at 8 kG at I max, the inductor can only be operated down to 4.6 / 19.5 = 25% of I max before lower inductance will result. 
Calculate Energy Storage Required (1/2 LI^{2} 
1/2 LI^{2} = (1/2) (500) (5) = 6250 μJ

Select appropriate core size. 
In this example 40 Material will be used. In order to maintain a minimum inductance down to 1 amp (20% of I max), the inductor must be designed to operate at greater than 8 kG at I max. This requires a core no larger than the E137 core or T131 toroidal core. To keep temperature rise down, the T131 will be selected. 
Calculate the required Energy Storage (1/2 LI2) 
1/2 LI^{2} = (1/2) (45) (7.5)^{2} = 1266 μJ 
Determine number of turns. 
At 6250 μJ, the T13140 indicates 235 ampereturns.
NI = 235 N = 235 / 5 = 47 turns

Select wire size. 
Since the “simple” winding results are a rough approximation of typical singlelayer windings, the Single Layer Winding Table can be used as a guide in selecting the wire size. #19 will fit in a single layer and yield about 20C° temperature rise due to the winding losses.
Solution: Part no. T13140
47 turns #19 