A flash memory stick (also known as a travel-drive) contains billions of transistors for storing charge....

 

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A flash memory stick (also known as a travel-drive) contains billions of transistors for storing charge. Each charge storing transistor gives rise to a 'bit', with a 1 or 0 depending on whether or not the charge is stored. The way charge is stored is by having electrons tunnel through an oxide barrier to charge a capacitor. When there is no voltage applied to the transistor, the barrier is too high for significant tunneling and thus charge storage does not take place. When a voltage is applied to the transistor, the tunneling barrier is lowered enough so that tunneling and charging and thus storage of 'bits' takes place. a. If the rectangular barrier is 1nm thick and 4.2eV high, write down the time independent SWE for the three regions of interest. Take the potential energy on either side of the barrier to be OeV. b. Solve the SWE to obtain the wave-function in each of the three regions. Take the electron energy to be 1.1eV, which is obviously lower than the barrier energy. You do not have to obtain values for the constant coefficients in front of the exponents. c. Now lower the barrier to 1.2eV by applying a bias of 3.0V to the device. How has the solution to the SWE changed? (The electron energy is still 1.1eV). Describe quantitatively, by referring to your solutions above, the difference in the barrier penetration and thus charging between cases in parts 4b and 4c. A flash memory stick (also known as a travel-drive) contains billions of transistors for storing charge. Each charge storing transistor gives rise to a 'bit', with a 1 or 0 depending on whether or not the charge is stored. The way charge is stored is by having electrons tunnel through an oxide barrier to charge a capacitor. When there is no voltage applied to the transistor, the barrier is too high for significant tunneling and thus charge storage does not take place. When a voltage is applied to the transistor, the tunneling barrier is lowered enough so that tunneling and charging and thus storage of 'bits' takes place. a. If the rectangular barrier is 1nm thick and 4.2eV high, write down the time independent SWE for the three regions of interest. Take the potential energy on either side of the barrier to be OeV. b. Solve the SWE to obtain the wave-function in each of the three regions. Take the electron energy to be 1.1eV, which is obviously lower than the barrier energy. You do not have to obtain values for the constant coefficients in front of the exponents. c. Now lower the barrier to 1.2eV by applying a bias of 3.0V to the device. How has the solution to the SWE changed? (The electron energy is still 1.1eV). Describe quantitatively, by referring to your solutions above, the difference in the barrier penetration and thus charging between cases in parts 4b and 4c.

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a The potential energy in the three regions of interest is Region 1 left of barrier 0 eV Region 2 ba... View the full answer