Re: quantized resistor doesn't save power
From: Zigoteau (zigoteau_at_yahoo.com)
Date: 01/10/05
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Date: 10 Jan 2005 14:09:18 -0800
Hi, Fred,
> The electrical resistance corresponding to operating at the
fundamental
> unit of conductance is 0.5*h/e^2, or 13000 ohms. Practical room
> temperature voltage shouldn't go below 0.25 V (or 10*kT) for a single
> switch. This means a single quantized conductor (like a nanowire)
> hooked up across this voltage would dissipate at least V^2/R=4.8
> micro-Watt. A billion of these conducting currents (in parallel)
would
> add up to 4.8 kW!
A nice calculation, and you're not alone in going down that alley, but
does it mean anything?
> Obviously by increasing the wire cross-sectional area, this
> resistance will go down. And also obviously, by putting many
> of these quantized conductors in series across the same voltage
> difference, the power consumption would go down (higher resistance,
less current).
The "quantum of conductance" that you are talking about certainly
exists, and is e^2/h. The value you give is its reciprocal. It is the
conductance per mode of a lossless metal wire that does not attenuate
any of the modes in the vicinity of the local Fermi level. The factor
of 2 you put in the denominator is valid for spin-degenerate systems.
Many, but not all, systems are spin degenerate, so that the modes
always occur in pairs.
However, not all modes have zero attenuation. Most active electronic
devices are constructed out of semiconductors, which attenuate electron
waves with energies in the vicinity of the Fermi level. A tunnel
junction with a single mode and transmission probability T has a
conductance of T*e^2/h. T varies exponentially with the width of the
tunneling gap, and can be made as small as you like.
Have you ever designed an electronic circuit, and/or do you understand
how they work? For good reasons, 99.9% of electronic circuitry has a
couple of connected pieces of metal, called the power supply rails,
between which the voltage is maintained as accurately as possible at a
certain value, the power supply voltage. The voltage of 0.25V you are
talking about is this value. It is not usually considered good practice
to connect a switch directly across the power supply rails. If you do
this, and then turn the switch "on", the resultant situation is called
a short circuit. Have you heard of the term? Yes, the power dissipation
of a circuit can be high in this condition, and it can do a lot of
damage, which is why electronic equipment is usually equipped with a
fuse or circuit breaker. Usually what you connect across power supply
rails is a series combination of two or more devices. There is a
voltage across the component in series with the switch, so that the
voltage across the swich does not remain constant. The voltage across
the switch is high only when the current through the switch is low, and
falls to a low value when the switch turns on.
Cheers,
Zigoteau.
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