Thread: SE transformer question View Single Post
 19th May 2022, 8:17 am #6 kalee20 Dekatron   Join Date: Feb 2007 Location: Lynton, N. Devon, UK. Posts: 6,300 Re: SE transformer question OK - so the core's cross-section area is 1,672mm˛ (generously sized!) and allowing for a lamination stacking factor of 95%, we get 1,588mm˛ of iron. It's fairly easy to derive a useful formula relating turns, inductance, current, area, and flux density: N = L x I / (B x A) where N= turns, L = inductance, I = current, B = flux density, and A = area , all in standard SI units of course. Morphing this around, we get L = B x A x N / I, and assuming 0.9T for flux density (thus leaving margin for flux swing); 1588mm˛ core area; 3600 turns; and 50mA current for the 2A3, we get: L = 103H. So you need to gap the core to bring down inductance to this figure. Too small a gap, the inductance will be higher, and from the above formula, you'll be closer to saturation (about 1.6T for most grades of iron) at 50mA. But - you've got 67H so that's more than enough gap! Which means that the inductance should be holding up to way more than 50mA! Something needs reconciling! Without access to your set-up, I'm going to postulate that it's down to the rather soggy nature of iron at small excitation levels, the incremental permeability is not the same as the large-scale permeability. I would be inclined to test the transformer on 50Hz AC and monitor the current waveform as the voltage is wound up. Downside is that it'll need quite a high voltage, but if it 'looks' nice and sinusoidal up to your 70mA peak, there's nothing to worry about. It might be easier to energise the transformer on the secondary, scaling voltages down by the turns ratio, and currents up by the turns ratio, just to keep voltages 'sensible'.