Return-Path: Received: from smtp8.gateway.net ([208.230.117.252]) by ns1.olsusa.com (Post.Office MTA v3.5.3 release 223 ID# 0-64832U3500L350S0V35) with ESMTP id com for ; Wed, 19 Jul 2000 19:31:45 -0400 Received: from oemcomputer (2Cust70.tnt26.chi5.da.uu.net [63.28.14.198]) by smtp8.gateway.net (8.9.3/8.9.3) with SMTP id TAA06194 for ; Wed, 19 Jul 2000 19:38:03 -0400 (EDT) Message-ID: <000201bff1da$6b6e0480$c60e1c3f@oemcomputer> From: "bobjude" To: "Lancair Mail List" Subject: Diode Voltage Drop Date: Wed, 19 Jul 2000 18:36:57 -0500 X-Mailing-List: lancair.list@olsusa.com Reply-To: lancair.list@olsusa.com Mime-Version: 1.0 <<<<<<<<<<<<<<<<--->>>>>>>>>>>>>>>> << Lancair Builders' Mail List >> <<<<<<<<<<<<<<<<--->>>>>>>>>>>>>>>> >> After reading all this stuff about diodes connected in series with the charging circuit of a battery, I can not restrain myself any longer. I suggest that you take a look at the E/I characteristic curve of your diode. You will find that the voltage across the diode is not constant, but is a function of current. The voltage is relatively constant over a range of current values especially near the high end of its rated range. In the case of charging an auxiliary battery which has not been discharged, the current will be very low and the voltage drop across the diode will below the knee of the curve. You certainly can not expect that the voltage drop will always be .6 volts. Bob Jude >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> LML website: http://www.olsusa.com/Users/Mkaye/maillist.html LML Builders' Bookstore: http://www.buildersbooks.com/lancair Please send your photos and drawings to marvkaye@olsusa.com. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>