Path: utzoo!attcan!uunet!cs.utexas.edu!news-server.csri.toronto.edu!utgpu!cunews!dfs From: dfs@doe.carleton.ca (David F. Skoll) Newsgroups: sci.electronics Subject: Re: How to make 600VDC from 800VDC Message-ID: Date: 18 Oct 90 15:57:03 GMT References: <92@decabo.enet.dec.com> <1112@tsdiag.ccur.com> Sender: news@ccs.carleton.ca (news) Organization: Carleton University, Ottawa, Canada Lines: 30 In <1112@tsdiag.ccur.com> davet@tsdiag.ccur.com (Dave Tiller N2KAU) writes: >How about: > / / >------800V------>|------>|--------/\/\/-----600v-------- > / / R | > 100V zener 100V (I limit) Ouch! First of all, the Zeners should be facing the other way - we're interested in reverse breakdown voltage here. The above circuit will give you a very nice 798V or so supply Secondly, with the zeners in series with the load, one-quarter of the total power dissipation takes place in the diodes which is rather nasty. (Same current as load through diodes; 200V across diodes.) Thirdly, having a current-limiting resistor as shown before the load makes the output voltage rather heavily dependent on the load current. Suppose you have a 5-ohm I limiting resistor. Draw 5A and your output voltage drops by 25V! Some other circuits posted here, such as the emitter-follower types, are more appropriate. >-- >David E. Tiller davet@tsdiag.ccur.com | Concurrent Computer Corp. -- David F. Skoll | Department of Electronics | Opinions expressed here are dfs@doe.carleton.ca | Carleton University | my own and not necessarily (613) 786-7515 | Ottawa, Ontario, Canada | those of my employer.