polus Posted October 20, 2006 Report Share Posted October 20, 2006 Hi All,I Quote Link to comment Share on other sites More sharing options...
audioguru Posted October 20, 2006 Report Share Posted October 20, 2006 To have a 35VDC regulated output, the circuit needs a 30VAC transformer. 30 x 1.414= 42.4V peak, minus 2V for the rectifier bridge and minus 1V of ripple in the filter capacitor = 39.4V unregulated.If you power a 5VDC/4A regulator from unregulated 39.4V then the heat in the series pass transistor is (39.4V - 5V) x 4A= 138W. Two power transistors on a huge heatsink is needed.If you have an adjustable output voltage supply going down to 0V (maybe when shorted) then its series pass transistor will heat with 39.4V x 3A= 118.2W and will need two power transistors on a huge heatsink.The negative adustable output voltage supply will need the same.If you have a 24VDC/5A supply then its series pass transistor will heat with (39.4V - 24V) x 5A= 77W and will need one or two power transistors on a huge heatsink.The above totals 15A. The transformer should be rated at the peak voltage of 42.4V x 15A= 636VA.If all the power is used at the outputs of this supply then you can heat your entire neighbourhood with it! It will be enormous! It will be expensive! Quote Link to comment Share on other sites More sharing options...
polus Posted October 20, 2006 Author Report Share Posted October 20, 2006 Haha! I dont have heating so maybe its a good thing :DI have a heatsink in mind and its 150mm X 130mm X 55mm 0.5 Quote Link to comment Share on other sites More sharing options...
Paul_J Posted October 20, 2006 Report Share Posted October 20, 2006 You can use a step-down converter for the 5V part. It's more efficient so it doesn't require a large heatsink. At 3A the IC will need an external transistor.The value of the inductor can be calculated with the formulas in the Application Note.MC34063_ST.pdfTL494_AppNote_-_slva001d.pdf Quote Link to comment Share on other sites More sharing options...
polus Posted October 20, 2006 Author Report Share Posted October 20, 2006 Thanks Paul! Thats a great solution! ;)Is there a general (or otherwise) way to calculate the size of the capcitors for use on the unregulated incoming DC?Ive read it should be around 1000 per Amp. Been searching and cant find much :-\ Quote Link to comment Share on other sites More sharing options...
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