Re: quantized resistor doesn't save power
From: Fred Chen (flipsu5_at_comcast.net)
Date: 01/10/05
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Date: 10 Jan 2005 15:15:15 -0800
Zigoteau wrote:
> 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.
Zigoteau, thanks for the extended discussion. What you say is entirely
correct, and I knew all that. You also brought up some additional
angles which are also useful (distance-dependent T, spin-dependence).
The point of bringing it up is precisely to indicate what happens when
you scale down a metallic wire to smaller and smaller dimensions (area
and length). Eventually you hit the lowest unit of conductance (if not
zero) and this corresponds to around 13 kilo-ohms. The conductor is
already a resistor. Between power and ground, you would need enough
resistance across all paths to reduce power. You won't get this by
scaling things down.
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