Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!usc!wuarchive!uunet!world!jon_sree From: jon_sree@world.std.com (Jon Sreekanth) Newsgroups: sci.electronics Subject: Transistor leakage current ? Message-ID: Date: 13 Feb 91 20:15:09 GMT Sender: jon_sree@world.std.com (Jon Sreekanth) Distribution: sci Organization: The World Lines: 48 Hopefully this is a simple question, but I forgot my transistor theory since it's been so long. This is the circuit : <--- Ic __________ Vcc | > 300 ohm < | | 1K ohm / C |/ NPN --- ^^^^^--- |\ MPSA42 \ input | E | ____ -- Vcc is about 100V; the NPN transistor MPSA42 is in a conventional common emitter circuit, collector resistance = 300 ohm, input resistor (base resistor) is 1K ohm, emitter is grounded. My question is : when I turn off the transistor, what is the leakage current drawn from the + 100V supply ? I have a Sprague data book, which says, for MPSA42, Icbo is 100nA at Vcb of 200V. Is Icbo the same thing as the leakage Ic in the above circuit ? Sprague conveniently ignored to show their measurement circuit, or define their terms. (To turn off the transistor, I have the choice of applying either 0V or -12V through the 1K resistor. -12V will break down the base-emitter junction, but the 1K prevents damage. I'll do whichever gets me lower leakage current from the 100V Supply) Thanks much, / Jon Sreekanth Assabet Valley Microsystems Fax and PC products 346 Lincoln St #722, Marlboro, MA 01752 508-562-0722 jon_sree@world.std.com