dummer Posted March 28, 2006 Report Share Posted March 28, 2006 I need to build a power supply for a CNC machine. I've been told that the resultant DC voltage will be slightly higher than the non-rectified AC. Can someone tell me the precise factor used to compute the new voltage? Quote Link to comment Share on other sites More sharing options...
Staigen Posted March 28, 2006 Report Share Posted March 28, 2006 Hi dummer, velcome to the forum, nice to have you here ;DThe DC voltage is roughly the square root of 2 times higher, minus thevoltage drop across the rectifier diodes, when properly smoothed, andwithout a load.//Staigen Quote Link to comment Share on other sites More sharing options...
dummer Posted March 28, 2006 Author Report Share Posted March 28, 2006 The DC voltage is roughly the square root of 2 times higher, minus thevoltage drop across the rectifier diodes, when properly smoothed, andwithout a load.Could you show the math? Let's say the AC voltage came out of a transformer at 40VAC. Quote Link to comment Share on other sites More sharing options...
Staigen Posted March 28, 2006 Report Share Posted March 28, 2006 HiOops, that was a BIG question, there are written a whole book about it, but i willtry to show you. first we have to talk a little about the transformer, it have 40 Vout only at its rated current, at low or none current output it have larger outputvoltage, that you have to measure. But lets say that you have a ideal transformerthat have 40V output at all its current output up to its rated load, then therectified and smothed output DC voltage is 40 times 1.41 = 56.4 minus the dropover the rectifier diodes, about 1.2 V = 55.2 Volts at zero current. But, due to thetransformer is rising its output at low or zero output the voltage is going to be higher.Hope this solve some of your questions.//Staigen Quote Link to comment Share on other sites More sharing options...
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