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An Oddball Electrical Engineering Question

Started by Rick Lyons 5 years ago9 replieslatest reply 5 years ago157 views

I've been trying to learn about satellite antennas. And in my studies, unexpected ideas have occurred to me. For example, I just realized that the flashlight ("torch" for our friends in the U.K.) in my garage is an antenna. And the flashlight bulb's filament, or the diode in an LED, is the "feed element" of the antenna.

But that topic is not my question here. I've also been thinking about electric field intensity, measured in volts/meter, and then it hit me. If a 5-volt integrated circuit contains transistors whose elements are separated by a distance of 90 nanometers (90 billionths of a meter), then the electric field intensities within an integrated circuit are:

Electric field intensity = 5/(90x10-9) = 55.5 million volts/meter.

That seems like an awfully large number. Am I correct or have I lost my mind?
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Reply by kjhSeptember 11, 2017

That sounds correct to a first approximation.

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Reply by Tim WescottSeptember 11, 2017

I'm not sure how big a transistor has to be to withstand 5V, but yea, that sounds like you're in the ballpark.

This document lists the dielectric strength of intrinsic silicon as \(5 \cdot 10^5\)V/cm, which is pretty close to your number.

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Reply by WB6GTBSeptember 11, 2017

In the ballpark in terms of order of magnitude. Even with 1.8 volt cores and long (100 mm) channels it is hard to argue with the approximation. 

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Reply by SteveSmithSeptember 11, 2017

Hey Rick,

Yes, that's correct.  Keeping this large electric field manageable is one of the key challenges as IC feature sizes have gone down, resulting in the migration from 5v to 3.3v to 1.2v logic.  Your mind is intact!

Regards,

Steve

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Reply by BEBSynthesizersSeptember 12, 2017

Hi Rick,

don't worry, you did not loose your mind : that's a correct approximation of the electric field met in integrated circuits.

And by the way, those electrical fields are one of the biggest challenge semiconductor companies are facing, because silicon (and other semiconductors) start to react strangely (or more precisely "differently") with such fields.

The value looks strange to us because such a field in our world would be quite hard to generate (and would be a little bit dangerous for people around), but they are in fact common to the IC industry

Benoit

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Reply by Rick LyonsSeptember 12, 2017

Hi Guys.  Thanks for all your replies!

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Reply by JohnEhlersSeptember 12, 2017

Sorry to disagree, but you are off by at least an order of magnitude when the junction is conducting.  Although the B+ voltage may be 5V, voltage across the junction is closer to 0.5 V.  Of course, the dielectric strength in the reverse direction must withstand nearly the entire B+ voltage because the impedance is large.  The voltage breakdown of integrated circuits is one of the major concerns of an EMP attack.

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Reply by Rick LyonsSeptember 12, 2017

Hi JohnEhlers.  You're not disagreeing, you're informing. Thanks for the information.

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Reply by Coop.aa1wwSeptember 12, 2017

If fact, the electric field strength value might be even higher.  Consider the gate oxide thickness within a modern (i.e. 2016, 22nm) process [1].  This is a particularly constraining factor in integration scaling [2].  According to two references I searched (listed below), the equivalent oxide thickness (i.e. "EOT") of even a 0.9V, 2016, 22nm process is 0.5nm or 5Angstroms.  This hints at E-field strengths beyond 1GV/m.  So breakdown, tunneling, dopant migration, stress induced trap states (and leakage paths), etc. all trouble the process engineer.  Slide #17 of 21 in reference [2] indicates a range of 10-14MV/cm which is 1-1.4GV/m.

I think you've quite rightly deduced that the electric field strength, while a benefit to channel drift velocity, is otherwise looming over the scaling of integration.


[1] http://inst.eecs.berkeley.edu/~ee130/sp06/chp7full...

[2] https://web.stanford.edu/class/ee311/NOTES/Gate_Di...


Coop, aa1ww