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2. You are given an unbiased Schottky diode at 300K for an ideal Schottky barrier formed between platinum\\ ($\phi$ = 5.3 eV) and silicon doped with $N_d$ = 1 × 10$^{16}$/cm$^3$. The area of the diode is 10$^{-5}$ cm$^2$,\\ $\epsilon_s$ = 11.9$\epsilon_0$, $\chi_s$ = 4 eV and $E_g$ = 1.1 eV. Note that $k_bT$ = 0.0259 eV.\\ Calculate the capacitance of the Schottky diode (assume charge on metal is infinite) in F.\\ Do not enter units, just the number (2 sig figs).\\ Make sure your answer is in the format xey, where x is numerical answer, y is power and e represents 10$^\wedge$ (ten\\ raised to the power of).

          2. You are given an unbiased Schottky diode at 300K for an ideal Schottky barrier formed between platinum\\
($\phi$ = 5.3 eV) and silicon doped with $N_d$ = 1 × 10$^{16}$/cm$^3$. The area of the diode is 10$^{-5}$ cm$^2$,\\
$\epsilon_s$ = 11.9$\epsilon_0$, $\chi_s$ = 4 eV and $E_g$ = 1.1 eV. Note that $k_bT$ = 0.0259 eV.\\
Calculate the capacitance of the Schottky diode (assume charge on metal is infinite) in F.\\
Do not enter units, just the number (2 sig figs).\\
Make sure your answer is in the format xey, where x is numerical answer, y is power and e represents 10$^\wedge$ (ten\\
raised to the power of).
        
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2. You are given an unbiased Schottky diode at 300K for an ideal Schottky barrier formed between platinum

(ϕ = 5.3 eV) and silicon doped with Nd = 1 × 10^16/cm^3. The area of the diode is 10^-5 cm^2,

 = 11.9ϵ0,  = 4 eV and Eg = 1.1 eV. Note that kbT = 0.0259 eV.

Calculate the capacitance of the Schottky diode (assume charge on metal is infinite) in F.

Do not enter units, just the number (2 sig figs).

Make sure your answer is in the format xey, where x is numerical answer, y is power and e represents 10^∧ (ten

raised to the power of).

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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You are given an unbiased Schottky diode at 300K for an ideal Schottky barrier formed between platinum (Φ = 5.3 eV) and silicon doped with Na = 1 x 10^16 /cm^3. The area of the diode is 10^-5 cm^2, εs=11.9εoεs=4eV and Eg=1.1eV. Note that kT=0.0259eV. Calculate the capacitance of the Schottky diode (assume charge on metal is infinite) in F. Do not enter units, just the number (2 sig figs) Make sure your answer is in the format xey, where x is the numerical answer, y is the power, and e represents 10 (ten raised to the power of).
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Transcript

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00:04 In this question we have to find the total number of excess electrons in p region and the total number of excess holes in n region when the bias is 0 .3, 0 .4 and 0 .5 voltage.
00:19 So first, the excess electron concentration is dlnp, which is np minus np0.
00:31 Or it is np 0 time e to the power of v a or v t minus 1 e to the power minus x over lm and then the total number of the excess electron or np it is a time the square time integral from 0 to infinity of del np d x and so we can evaluate the integral and have that that will be a ln n p 0 time e to the power of v a or v t minus 1 and so we find that d n is 25 cm square per s and ln is 50 micron and np 0 is n i square or n a or 2 .8 1 times 10 to the power 4 per kibb meter then np is 0 .106 time e to the power of v a or vt minus 1 so for the va of 0 .3 volt 0 .4 volt and 0 .5 volt have that np is 1 .51 time 10 to the power 4 7 .17 times 10 to the power of 5 and 3 .4 times 10 to the power of 7.
02:15 And similarly, for pn, the total number of excess holds, it is alppn0 time e to the power of v8 over vt minus 1...
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