Path: utzoo!attcan!uunet!tank!ux1.cso.uiuc.edu!uwm.edu!cs.utexas.edu!tut.cis.ohio-state.edu!pt.cs.cmu.edu!b.gp.cs.cmu.edu!ralf From: ralf@b.gp.cs.cmu.edu (Ralf Brown) Newsgroups: comp.sys.ibm.pc Subject: Re: 20 Mhz 386 SX (?) Keywords: Citrus and how do you do it? Message-ID: <7982@pt.cs.cmu.edu> Date: 14 Feb 90 18:57:28 GMT References: <1109@watserv1.waterloo.edu> Distribution: comp Organization: Carnegie-Mellon University, CS/RI Lines: 39 In article <1109@watserv1.waterloo.edu> ssingh@watserv1.waterloo.edu ($anjay "lock-on" $ingh - Indy Studies) writes: }3.) BTW, WHY does the componentry heat up? The following is true for CMOS, and partially true for TTL devices: A CMOS switching element has two transistors, one between the supply voltage and output, and one between the output and ground, like so: power supply | | _____|/ | |\ | | input ----+ +------- output | | |_____|/ |\ | | ground The two transistors of the switching element have opposite polarities, so that one is off when the other is on. Due to this architecture, the current drain at rest is very low (only enough to drive the gates of the switching elements connected to the output). However, when the input switches from one state to the other, that switch is not instantaneous, so for a brief period of time, both transistors are partially on. We now have what amounts to a momentary short between power and ground. The duration of this short depends on how cleanly the input switches states, which is relatively independent of clock speed. However, the clock speed does influence how often the short occurs, so a higher clock speed results in more momentary shorts (of the same duration), causing a higher current draw, which heats up the chip. -- {backbone}!cs.cmu.edu!ralf ARPA: RALF@CS.CMU.EDU FIDO: Ralf Brown 1:129/46 BITnet: RALF%CS.CMU.EDU@CMUCCVMA AT&Tnet: (412)268-3053 (school) FAX: ask DISCLAIMER? | _How_to_Prove_It_ by Dana Angluin 24. by appeal to intuition: What's that?| Cloud-shaped drawings frequently help here.