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User’s Manual 0 -1 I(L1-P) / A -2 -3 -4 -5 -6 0 Time/mSecs 0.2 0.4 0.6 0.8 1 200µSecs/div This shows that the operating current is less than 1.5A but peaks at over 6A. In practice you would want to use an inductor with a maximum current of around 1.5 to 2A in this application otherwise it would be over-designed and therefore over expensive! We will now replace the ideal component, with something closer to a real inductor. 46 1. Delete L1. 2. Select schematic menu Place|Magnetics|Saturable Transformer/Inductor.... A dialog box will be dispalyed. (See picture below). Select 0 secondaries then enter 34 in the turns edit box. Next check Select Core Type. Select EFD10-3F3-A25. This is part number for a Philips ferrite core. This is what you should have:
