Want to destroy it - try higher voltages, here is a "soft breakdown". Maybe you need to know mechanisms of current flow?

Sharp breakdown may be in MS structures (see gif). Insulator between them add as resistance as capacitance, and change characteristics.

How big the thickness of each layer and square of contact?

http://www.globalspec.com/ImageRepository/LearnMore/20131/diode58cd4e2f1b9114708851ac53e3d63e9c7.gif

It is not a diode, it is an MOS capacitor. There is much information in books and papers on the electrical characteristics; I suggest you do some reading.

Note that if you put too much voltage on, it is the OXIDE that breaks down. Good, thick SiO2 will break down at a field of >10 MV/cm. The breakdown mechanism is complex (it has been studied for decades). It also depends on the amount of charge that has flowed through it - hence time that the current has been flowing.  Note also that the ideal leakage current (prior to breakdown) is due to the Fowler-Nordheim mechanism.

 I would like that you give more information about your MIS structure such as the thickness of the aluminum oxide and the doping concentration of the silicon.

The conduction mechanisms of the MIS structures are explained in full details in the reference books of the electron devices, See for example the physics of semiconductors by S M Sze. Before the breakdown of the insulator there will be some leakage currents through: Thermionic emission above the potential barrier between the  the metal and insulator to the semiconductor, there is also the thermionic emission from charges trapped in the conduction band of oxide by the Frenkel-Poole mechanism. There is also tunneling or field emission across the a thin barrier of the oxide.There is also the space charge limited current.

When the electric field in the oxide reaches its critical value for breakdown avalanche multiplication of primary charges occurs and the current grows reapidly with the applied voltage.

As your range of currents is ampere range, it is not a leakage current but it is an avalanche current.  The curve is inclined because of the ohmic resistance of the diode.

Specifically the resistance of the semiconductor substrate.

In order to determine the dominant leakage current mechanisms you have log I versus V . This shape of such plot may point to the dominant mechanism. Measuring I-V curves at different temperatures may give some evince for locating the  leakage current mechanism.

Best wishes

Careful.. FN does not involve traps - it is purely a tunnelling mechanism

Thank you Steve it was Frenkel Poole mechanism not Fowler-Nordheim.

 @ Steven I understand that We cannot call this as diode because there is no metal semiconductor junction or P-N junction

 But I want to know, will this I-V characteristics of  MIS structure is correct  to measure the breakdown strength of oxide. Because the Forward voltage and reverse voltage values are same and There is no sharp  breakdown in reverse biased condition. THis is my first attempt to make such structure so I am not able to tell whether it is right or wrong. 

I had Metal semiconductor ohmic contact. Is it required to have schottkey contact?

Thanks Steven, I appreciate your response

 @ Abdehalim Thickness og Al2O3 film is 100nm and  Si doping concentration is n type Si wafer 10^19(measured with hall  at 1micro A), 

Thank you sir, It will helpful to me.

My apologies Abdelhalim!

Sheyra - you will see more detail if you plot the y-axis on a log scale. I don't really know what you are trying to do ('correct'?)

If you want to break down the oxide, put negative voltage on the top electrode so as to accumulate the substrate. All the voltage will then fall mostly across the oxide, so you can easily calculate the electric field in the oxide and find the breakdown field as well as the voltage. But note that breakdown is complex - depends on time as well as field. To see if the characteristic is 'correct', see if it fits the current mechanism equations for Fowler Nordheim (ideal) or Poole Frenkel as Abdelhalim suggests - Sze's book or search on-line.

correction - it's an n-type substrate so put positive voltage on the top gate to accumulate the substrate.  Electrons will then be injected from the substrate.

Dear Sherya, Dear Steve,

The word diode can be assigned also to the MIS two terminal structure. This is not the main issue. 

According to your data, the n-silicon substrate is heavily doped which kills more or less the field effect and the applied voltage is dropped on the insulator. Accordingly,

the electric field E=V/d= the break down voltage/ 100x 10^-7= 3,75x10^5 V/cm,

This electric field is much lower than the breakdown field of aluminum oxide. So, i would assume either the film thickness is not correct or the film is not homogeneous.In fact any film breaks at the thinnest point. Even one may have very small pin holes.

So you have to measure the capacitance of the structure to get the thickness of the insulator. You can characterize it by C-V and C-t.  You have to be sure also that your film is homogeneous and free of pinholes by using microscopes. 

Best wishes

Dear Steve,

Got it, Applying +ve voltage in top gate creates accumulations and Applying -ve voltage on top gate will create depletion so  in order to measure breakdown voltage I should apply -ve voltage on top gate contact. (One quick question, Is it required to have ohmic contact with metal and n type silicon wafer) I have used Ag as back as well as top contact.

Thank you very much for your time and response. I really understand it now.

@  Abdelhalim  Al2O3 film is not  uniform because it is deposited with pulsed laser deposition technique and I show particulates formation on top surface. I still need to optimized deposition condition to eliminate such issues. 

Thank you  again for sharing your  knowledge.

 @Abdelhalim,  Ag work function(4.3 to 5.1eV) is greater than n type Si (approximately 4.1)work functions. Still it is showing ohmic contact. How that will be possible?

Dear Shreya,

welcome!

The n-Si substrate concentration is high and is nearly equal the effective density of state. Therefore if we build such an M-S contact the width of the potential barrier region will be very small such that it is tunnelable forming an ohmic contact. 

From the conceptual point of view if you want to build an ohmic contact with any metal you just need to make silicon near the interface have high doping concentration of about the effective density of state.

You can refer to the chapter of the MS contacts in any book about electron devices or you can see the effect of doping concentration of ntype material on the type of the MS contact formed by referring to the paper in the Link; https://www.researchgate.net/publication/232710923_A_Zekry_and_G_El-Dllal_effect_of_MS_contact_on_the_electrical_behavior_of_solar_cells_Journal_of_solid-state_El

Best wishes

Article A. Zekry and G. El-Dllal, “ effect of MS contact on the elec...