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You may have noticed over the last few years that Moore's law is no longer applying to the real world. This observation states that over the history of computing hardware, the number of transistors on integrated circuits doubles approximately every two years.

However, as microprocessors in computers have continued to become smaller, the architecture size has become incredibly small:- the new Skylake architecture to be released by Intel will use 14 nm semiconductors.

We are beginning to reach the limit in size where the information being passed is still completely reliable.

Would it be reasonable to ask how much quantum tunneling would occur in a semiconductor 5 nm in length? We can assume that in this CPU

Voltage = 1.2 V

Current = 63 A

I enjoy physics, but I'm not that great at it; I'd love to hear the results! Even if massive approximations!

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  • $\begingroup$ I believe that "tunneling" in semiconductors, is another term for "leakage." When a transistor is reverse-biased, very little current flows (open circuit), when forward biased, a larger current flows (short circuit). If the leakage current is "large enough," the transistor will act as if it were permanently forward biased, and thus unable to change state. A CPU built with these transistors will not be a CPU! The leakage current (due to tunneling), depends on various parameters. One of them is the thickness of the p-n junction. The smallest thickness might be one atom of each type. $\endgroup$
    – Guill
    Apr 16, 2014 at 5:29
  • $\begingroup$ I think for tunneling to be observable, the device must be cooled to very low temperatures, like liquid Helium temperatures. At room temperature or at CPU heated temperature, one can't see tunneling in the loud noise of heat movement. Electrons are jumping physically as kicked but thermal movement. $\endgroup$
    – Dims
    Jun 19, 2015 at 20:44
  • $\begingroup$ it's a fact that quantum effects come at any temperature. They are more noticeable when the scale decreases and heat parasites fade. For semiconductors conceptors, that's all noise . They compete hard to remove them. Much of the circuit functions only serve to regulate and correct these errors , that is just ignored physics in the theoretical model that was used to develop them. While miniaturization increases, the designers take into account these effects more than before, sometimes hoping to gain some benefit. $\endgroup$
    – user46925
    Jun 25, 2015 at 17:53

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Dims is almost correct, in that you would only see resonant tunnelling effects at very low temperatures.

In other words, at very low temperatures the electrons will sit at very well defined energy levels within the transistors. Under certain biases (voltages), the energy levels on either side of the thin barriers between devices will line up, and electrons can tunnel between the devices without any resistance. This would give rise to sharp "spikes" in the current at very specific voltages, and the electrons would "spread" freely between adjacent devices.

At higher temperatures, tunnelling does still occur, but the electons are spread over a broad range of energies rather than in well-defined states. As such, you don't see the resonant "spikey" behaviour discussed above. Rather, you'll just see a general broad leakage of electrons between devices.

Now, in terms of quantifying this, take a look at any good textbook on heterostructure devices, for example, Chapter 2 in P. Harrison "Quantum Wells, Wires and Dots", 3rd Ed. Wiley (2009).

The probability (from 0 to 1) of an electron tunnelling through a single, thin barrier is given by: \begin{equation} T(E) = \frac{1}{1 + \left(\frac{k_b^2 + k_w^2}{2k_bk_w}\right)^2\sinh^2(k_bL)} \end{equation} where $L$ is the length of the barrier separating regions of the device, $E$ is the energy of the electron and $k_w$ and $k_b$ are the wave vector (proportional to momentum) of the electron within the device, and the decay constant of the wave function in the barrier respectively. These are given by: \begin{equation} k_w = \frac{\sqrt{2m_e^* E}}{\hbar} \end{equation} and \begin{equation} k_b = \frac{\sqrt{2m_e^* (V-E)}}{\hbar} \end{equation} respectively, where $m_e^*$ is the effective mass of the electron and $V$ is the barrier potential.

Let's consider electrons with energy half-way up the barrier (i.e., $E = V/2$) for a simple estimate of the "average" tunnelling probability. Therefore, $k_w = k_b$ and the tunnelling probability simplifies to: \begin{equation} T(E) = \frac{1}{1 + \sinh^2(k_bL)} \end{equation}

Now, to get a rough estimate (I don't have the parameters to hand), take an effective mass of $0.3m_0$, and a barrier potential of 1 eV. This gives $k_b = 2\,\text{nm}^{-1}$.

Now, calculating the tranmission probability for a range of barrier lengths gives the following:

Tunnelling probability vs. barrier length

It is, therefore (with this set of parameters!) unlikely that we'd see any tunnelling effects until the barrier thickness drops below a couple of nm. Note, though, that the effects would be much more significant for lower effective mass, lower barrier potential or higher electron energy. I'll happily update the plot if anyone feels like providing more realistic parameters.

Note also, that other classical breakdown effects like thermal leakage, and practical issues with quality of materials are likely to cause equally (or more) significant problems!

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  • $\begingroup$ So the tunneling probability is around T=1e-5, but we should still consider that many events happen. The tunneling probability is, if I understand correctly, the probability of transmission for an incident electron. So we should multiply by the rate of incoming electrons to get a rate of tunneling. Considering the number of transistors and their current error rate this could become significant even when T is very low. $\endgroup$ Jun 22, 2015 at 11:12
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    $\begingroup$ Yes, certainly, although this should also take into account the distribution of electrons with respect to energy i.e., density-of-states multiplied by distribution by tunnelling probability. This should then be integrated over the whole range of energies: $I\propto\int{}T(E)f(E)\rho(E)\text{d}E$. Most carriers lie at the lowest energies, where $T$ is very low, so actually the effect on current may not be very large. I suspect that other effects would have a more significant effect on leakage current. $\endgroup$ Jun 22, 2015 at 12:22

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