Path: utzoo!utgpu!news-server.csri.toronto.edu!rpi!zaphod.mps.ohio-state.edu!wuarchive!uwm.edu!ogicse!zephyr.ens.tek.com!gvgpsa!gold.gvg.tek.com!grege From: grege@gold.gvg.tek.com (Greg Ebert) Newsgroups: sci.electronics Subject: Re: MOSFET Puzzle Message-ID: <2178@gold.gvg.tek.com> Date: 11 Apr 91 18:28:29 GMT References: <1285@tusun2.mcs.utulsa.edu> Distribution: na Organization: Grass Valley Group, Grass Valley, CA Lines: 53 In article <1285@tusun2.mcs.utulsa.edu> lansford@tusun2.UUCP (Wendell Wayne Lansford) writes: >The following problem was recently posed by my professor in Advanced >Networks: > >TRANSISTOR IN A BOX PUZZLE > > ------------- > + ----| |---- + > | Black | > Vin | Box | Vout > | | > - ----| |---- - > ------------- > > [...] Awright, how 'bout this : A depletion-mode MOSFET is actually a 4-terminal device. It looks like this: o Drain | ----||---+---K|--- | Cgd | | | \ *---o substrate gate o--* | | | | | ----||---*--K|---- Cgs | o Source Those -K|- devices are parasitic diodes. The switch inside is controlled by Vgs: OFF if Vgs < Vp; on otherwise. Connect an AC source across the gate and substrate. Now connect the source and drain terminals together. Finally, connect an imagionary load across the source-drain terminal and the substrate. Look familiar ? It's a capacitive voltage doubler. Imagine the gate is negative with respect to the substrate. Cgs and Cdg are charged 'negatively', thus the MOSFET is pinched-off. In fact, you can now forget that it's even a MOSFET. Now let the gate go positive with respect to the substrate. There is no path to discharge Cgs & Cgd (thanks to the diodes), so the device remains pinched-off. Looking at the drain-source node, the voltage with respect to the substrate is the voltage on Cgs/Cgd PLUS the gate-substrate voltage. Thinking about it some more, I think the issue of whether or not the MOSFET is pinched-off is irrelevant.