- #1
zeus_the_almighty
- 7
- 0
Hi everybody,
I study solid-state electronics, more precisely electron Fowler-Nordheim tunneling across the gate oxide of a Silicon-Oxide-Silicon capacitor.
I need help on the physics of electron tunneling across a triangular barrier.
It will be part of my PhD dissertation. I have been through many websites but couldn't find out any FULL, DETAILED, derivation of the so-called Fowler-Nordheim Current density. Looks like nobody knows exactly the math which lies behind!
Does anybody know how to derive the well-known formula of the Fowler-Nordheim current density resulting
from a triangular potential barrier, which is:
(E.1) J_{FN}=\alpha F^2 \exp{\frac{-\beta}{F}}
where \alpha and \beta are the so-called pre-exponential and exponential Fowler-Nordheim parameters, and F the electrical field across the tunnel oxide.
\alpha and \beta depend on the potential barrier height \Phi_{0} and the ratio of effectives masses (in the oxide conduction band and in the silicon conduction band) in the following way:
(E.2) \alpha=\frac{q^3}{8\pi qh\Phi_{0}}\frac{m_{Si}}{m_{ox}}
and
(E.3) \beta=\frac{8\pi}{3qh}\sqrt{2m_{ox}}(q\Phi_{0})^\frac{3}{2}
I know part of the derivation but there are some missing steps.
Here are a few hints for those who may help me:
the current density J_{FN} can be expressed as the product of:
i)
the number of electrons per unit area and time arriving at the Silicon/oxide interface,
and
ii)the tunneling probability T(E) for a triangular barrier.
From the above, and some straightforward homogeneity considerations, one finds out:
(E.4) J_{FN}=\frac{q}{m} \int_{0}^{E_{m}} n(E) f(E) T(E) dE
where n(E) is the density of states per unit energy,
f(E) the Fermi-Dirac function,
Em the highest energy of the electron gas.
Since we consider only cold emission, f(E)=1 .
Moreover, if the electrons are thought of as a free electron gas, the density of states per unit energy does not depend on the energy and is expressed (classicaly) as:
(E.5) n(E)=\frac{2\pi m^*}{h^3}
In addition to this, the tunneling probability, "seen" by an electron arriving at the Silicon/oxide interface with an energy "E", and resulting from a triangular barrier (whose height is q\phi_{0}-E and electric field F) can be derived easily in solving the steady-state one-dimensional Schrödinger equation, giving:
(E.6) T(E)=\exp{(-\frac{8\pi}{3qh}\sqrt{2m_{ox}(q\Phi_{0}-E) }\frac{1}{F})}.
From, this, the question is:
How does one go from:
(E.7)J_{FN}=\frac{q}{m} \int_{0}^{E_{m}} \frac{2\pi m^*}{h^3} T(E) dE
to (E.1)?
THANKS.
I study solid-state electronics, more precisely electron Fowler-Nordheim tunneling across the gate oxide of a Silicon-Oxide-Silicon capacitor.
I need help on the physics of electron tunneling across a triangular barrier.
It will be part of my PhD dissertation. I have been through many websites but couldn't find out any FULL, DETAILED, derivation of the so-called Fowler-Nordheim Current density. Looks like nobody knows exactly the math which lies behind!
Does anybody know how to derive the well-known formula of the Fowler-Nordheim current density resulting
from a triangular potential barrier, which is:
(E.1) J_{FN}=\alpha F^2 \exp{\frac{-\beta}{F}}
where \alpha and \beta are the so-called pre-exponential and exponential Fowler-Nordheim parameters, and F the electrical field across the tunnel oxide.
\alpha and \beta depend on the potential barrier height \Phi_{0} and the ratio of effectives masses (in the oxide conduction band and in the silicon conduction band) in the following way:
(E.2) \alpha=\frac{q^3}{8\pi qh\Phi_{0}}\frac{m_{Si}}{m_{ox}}
and
(E.3) \beta=\frac{8\pi}{3qh}\sqrt{2m_{ox}}(q\Phi_{0})^\frac{3}{2}
I know part of the derivation but there are some missing steps.
Here are a few hints for those who may help me:
the current density J_{FN} can be expressed as the product of:
i)
the number of electrons per unit area and time arriving at the Silicon/oxide interface,
and
ii)the tunneling probability T(E) for a triangular barrier.
From the above, and some straightforward homogeneity considerations, one finds out:
(E.4) J_{FN}=\frac{q}{m} \int_{0}^{E_{m}} n(E) f(E) T(E) dE
where n(E) is the density of states per unit energy,
f(E) the Fermi-Dirac function,
Em the highest energy of the electron gas.
Since we consider only cold emission, f(E)=1 .
Moreover, if the electrons are thought of as a free electron gas, the density of states per unit energy does not depend on the energy and is expressed (classicaly) as:
(E.5) n(E)=\frac{2\pi m^*}{h^3}
In addition to this, the tunneling probability, "seen" by an electron arriving at the Silicon/oxide interface with an energy "E", and resulting from a triangular barrier (whose height is q\phi_{0}-E and electric field F) can be derived easily in solving the steady-state one-dimensional Schrödinger equation, giving:
(E.6) T(E)=\exp{(-\frac{8\pi}{3qh}\sqrt{2m_{ox}(q\Phi_{0}-E) }\frac{1}{F})}.
From, this, the question is:
How does one go from:
(E.7)J_{FN}=\frac{q}{m} \int_{0}^{E_{m}} \frac{2\pi m^*}{h^3} T(E) dE
to (E.1)?
THANKS.