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My goal is to understand how to correct for the field drop and the work function difference when performing electrical measurements on a certain energy level of a sandwiched system.

The situation is the following: one has 2 metallic electrodes, their difference in work function being $\phi_{2}-\phi_{1}$. In the interface betwen one of the electrodes there is a dielectric which touches one electrode and then vacuum, which reaches to the other electrode (electrode-dielectric-vacuum-electrode). In the interface between the dielectric and the vacuum I have an energy level $E_{real}$ which one desires to find.

The measurement proceeds by ramping the voltage and measuring the respective voltage for which there is an increase in current - therefore you end up measuring a peak in current for $E_{meas}$. Another know value is the voltage drop percentage between the vacuum region (compared to the total), which is $r$.

Arguing from the stand point of levelling both fermi levels, the relationship between $E_{real}$ and $E_{meas}$ is

$$E_{real} = r\cdot[E_{meas} -(\phi_{2}-\phi_{1})]$$

Assuming, for example, that $\phi_{2}-\phi_{1}$ is $1 eV$, $r$ is $0.5$ and that we have measured $E_{meas}$ to be $3 V$, this gives a $E_{real}$ of $1 V$.

The main problem I have is that I cannot test it to another value and just judging from the perpective, I would say that the value is smaller than the measured one, which makes sense. But the main question is: is there a way how to test my reasoning?

Another hypothesis which I have had is that the real correction of the voltage must be performed to minimize the electric field and then one should add back the work function difference, yielding $$E_{real} = r\cdot[E_{meas} -(\phi_{2}-\phi_{1})] + (\phi_{2}-\phi_{1})$$, and in using the numbers given in here, $E_{real}$ would be then $2 V$. Which is still smaller than the measured voltage and would make sense.

Anyone would like to shed some light into helping me to find the conundrum in one of these mathematical descriptions? Any commentary or criticism is welcomed.

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1 Answer 1

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The quantity of electrons traveling across this interesting capacitor is necessarily hindered by the formula you used. Although there would be this consideration for electron emission, the potential through the dielectric would be chiefly concerned with the polarization and I would therefore consider that. Polarization is the net shift in dipole moments of the domains within a material, and this appears to me to be the most likely phenomenon to be effected by the work functions. However, this difference would occur over a small time frame, because the potential different of the circuit is larger than the P.D of the capacitor until the potential is dispersed equally and is resolved overall.

So therefore the factor to consider is the Electric Displacement in place of the electric field.

which is

$$ D = \epsilon_0 E + P $$

where

D = electric displacement

E = Electric field

P = Polarization

$$ E = \frac{q}{4 \pi \epsilon_0 \epsilon_r r^{2}} $$

&

$$ P = Np $$

n = number of molecules per unit volume p = average dipole moment per unit volume

As you said you need to account for the voltage drop of the work function differential, my suggestion for a resolve of this electrical engineering problem is to compute: (forgive to poor notation for contour integrals)

$$ \int_{C} E = e.m.f $$

and

$$ V = V_{0} e^{\frac{t}{\tau}} $$

with its ampere analogue to find some graphs to analyze whether and if this voltage drop is significant.

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