Suppose we have a Aluminium on p-Si wafer contact.

Calculations have shown that it will form a Schottky junction since the p-doping will move the fermi level towards the valence band, and hence the fermi level of the p-Si before contact will be lower than the fermi level of Al before contact (condition for a Schottky contact for metal-semiconductor).

 

In a non-ideal contact, a thin insulating layer of Si oxide is between such contacts.

If the applied voltage with respect to the Si-wafer is 10V, estimate the amount of field-assisted emission current.

Also estimate the amount of the tunnleling current.

Given that the thickness of the SiO2 layer is

(i) 2nm.

(ii) 10nm.

 

My answer:

Since holes are the majority carriers mainly responsible for the forward and reverse conduction currents for the system, we shall neglect the electrons as the carrier in the calculations.

The schottky barrier formed is such that if a positive voltage is applied on the n-Si, it will be forward bias because the Schottky barrier is lowered by the external applied field, resulting in more holes moving from p-Si to Al, as compared to the holes from Al to p-Si.

 

V = applied voltage = 10V

Wonder if because of this, we can apply the Schotkky diode equation:

I = I0*exp[-eV/kT] neglecting the -1 term

where

I0 = reverse saturation current

= A*Richardson constant*T^2*exp[-effective barrier/kT]

 

How do we calculate the effective barrier in the equation?

 

 

I am fumbled on how to calculate the tunnelling current too.

Suppose we have a Aluminium on p-Si wafer contact.

Calculations have shown that it will form a Schottky junction since the p-doping will move the fermi level towards the valence band, and hence the fermi level of the p-Si before contact will be lower than the fermi level of Al before contact (condition for a Schottky contact for metal-semiconductor).

 

In a non-ideal contact, a thin insulating layer of Si oxide is between such contacts.

If the applied voltage with respect to the Si-wafer is 10V, estimate the amount of field-assisted emission current.

Also estimate the amount of the tunnleling current.

Given that the thickness of the SiO2 layer is

(i) 2nm.

(ii) 10nm.

 

My answer:

Since holes are the majority carriers mainly responsible for the forward and reverse conduction currents for the system, we shall neglect the electrons as the carrier in the calculations.

The schottky barrier formed is such that if a positive voltage is applied on the n-Si, it will be forward bias because the Schottky barrier is lowered by the external applied field, resulting in more holes moving from p-Si to Al, as compared to the holes from Al to p-Si.

 

V = applied voltage = 10V

Wonder if because of this, we can apply the Schotkky diode equation:

I = I0*exp[-eV/kT] neglecting the -1 term

where

I0 = reverse saturation current

= A*Richardson constant*T^2*exp[-effective barrier/kT]

 

How do we calculate the effective barrier in the equation?

 

 

I am fumbled on how to calculate the tunnelling current too.