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From: poffen@sj.ate.slb.com (Russ Poffenberger)
Newsgroups: comp.sys.ibm.pc
Subject: Re: 20 Mhz 386 SX (?)
Keywords: Citrus and how do you do it?
Message-ID: <1990Feb21.191336.24704@sj.ate.slb.com>
Date: 21 Feb 90 19:13:36 GMT
References: <1109@watserv1.waterloo.edu> <7982@pt.cs.cmu.edu>
Reply-To: poffen@sj.ate.slb.com (Russ Poffenberger)
Distribution: comp
Organization: Schlumberger Technologies, San Jose, CA.
Lines: 49

In article <7982@pt.cs.cmu.edu> ralf@b.gp.cs.cmu.edu (Ralf Brown) writes:
>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.

This is incorrect.

Actually the reason why more current is drawn is because the output stage is
driving a capacitive load at the input of the following stage. This causes the
capacitance to constantly charge to an opposite state when it switches, meaning
the output stage must drive current to discharge the capacitance, then more
to charge it to the opposite polarity. The faster (more often) this is done,
results in more current draw from the supply and resulting heat generation.

Russ Poffenberger               DOMAIN: poffen@sj.ate.slb.com
Schlumberger Technologies       UUCP:   {uunet,decwrl,amdahl}!sjsca4!poffen
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