Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!samsung!dali.cs.montana.edu!uakari.primate.wisc.edu!zaphod.mps.ohio-state.edu!sdd.hp.com!hplabs!hpda!hpcuhc!dclaar From: dclaar@hpcuhc.cup.hp.com (Doug Claar) Newsgroups: sci.electronics Subject: Re: Need info on a diode rectifier Message-ID: <107710002@hpcuhc.cup.hp.com> Date: 10 Jan 91 20:29:32 GMT References: <42276@ut-emx.uucp> Organization: Hewlett Packard, Cupertino Lines: 29 / hpcuhc:sci.electronics / ifar355@ccwf.cc.utexas.edu (David H. Huang) / 12:43 pm Jan 8, 1991 / a) 1____|\|____2 | |/| | --- --- ^ ^ /_\ /_\ | | 3____|\|____4 |/| ---------- Now would you like to know where the wires go? 1 & 4 = AC 2 = + 3 = - Theory: A diode passes postive voltage thru the arrow, negative into the arrow. (Ok, it's not real precise, but...) So when 1 is positive, it goes to 2, but can't get to 3. When it is negative, it goes to 3, but can't get to 2. The same is true for 4: Positve goes to 2, negative to 3. I'm sure that someone will say it much more eloquently, but that's the terms in which my EE-wannabe CS mind remembers it. Doug Claar HP Computer Systems Division UUCP: mcvax!decvax!hplabs!hpda!dclaar -or- ucbvax!hpda!dclaar ARPA: dclaar%hpda@hplabs.HP.COM