ni = 10^10 fracelektroncm^3; mun = 1320 fraccm^2V.s; mup = 460 fraccm^2Vs; q = 1.6x10^-19 C; VT

Name: ni = 10^10 fracelektroncm^3; mun = 1320 fraccm^2V.s; mup = 460 fraccm^2Vs; q = 1.6x10^-19 C; VT = 0.026 V
Uploaded: 2023-05-01T16:15:17-08:00
Duration: 1 min 54 s
Channel: Penny Riley
Description: ni = 10^10 fracelektroncm^3; mun = 1320 fraccm^2V.s; mup = 460 fraccm^2Vs; q = 1.6x10^-19 C; VT = 0.026 V

we'd like to calculate the built -in voltage. So now, beginning with this, we can first calculat

Ni = 10^10 fracelektroncm^3; mun = 1320 fraccm^2V.s; mup =...

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(a) A one-sided silicon $\mathrm{n}^{+} \mathrm{p}$ junction at $T=300 \mathrm{~K}$ is doped at $N_{d}=3 \times 10^{17} \mathrm{~cm}^{-3}$. Design the junction such that $C_{j}=0.45 \mathrm{pF}$ at $V_{R}=5 \mathrm{~V} .(b)$ Calculate the junction capacitance at $(i) V_{R}=2.5 \mathrm{~V}$ and $(i i) V_{R}=0 \mathrm{~V}$.

An abrupt silicon p-n junction consists of a p-type region containing 10^16 /cm^3 acceptors and an n-type region containing also 10^16 /cm^3 acceptors in addition to 10^17/ cm^3 donors. a. Calculate the thermal equilbrium density of electrons and holes in the p-type region as well as both densities in the n-type region. b. Calculate the built-in potential of the p-n junction. c. Calculate the built-in potential of the p-n junction at 100 C.

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An ideal one-sided silicon $\mathrm{p}^{+} \mathrm{n}$ junction at $T=300 \mathrm{~K}$ is uniformly doped on both sides of the metallurgical junction. It is found that the doping relation is $N_{a}=80 N_{d}$ and the built-in potential barrier is $V_{b i}=0.740 \mathrm{~V}$. A reverse-biased voltage of $V_{R}=10 \mathrm{~V}$ is applied. Determine ( a) $N_{a}, N_{d} ;$ (b) $x_{p}, x_{n} ;(c)\left|\mathrm{E}_{\max }\right|$; and $(d) C_{j}^{\prime} .$

Transcript

00:01 We'd like to calculate the built -in voltage.

00:04 So now, beginning with this, we can first calculate our conductivity.

00:10 Our conductivity is 1 over our resistivity, which is 1 over 2, and so we get 0 .51 ohms per centimeter, or 0 .50.

00:21 Now we could find, we have our conductivity given as nq mu n, and sigma p is pq mu p.

00:35 So sigma n is point, is equal to 0 .5.

00:38 So we have 0 .5 equal to n times 1 .6 times 10 to the negative 19 times 1320.

00:46 And so we can solve for nd...

Question

n_i = 10^{10} frac{elektron}{cm^3}; mu_n = 1320 frac{cm^2}{V.s}; mu_p = 460 frac{cm^2}{Vs}; q = 1.6x10^{-19} C; V_T = 0.026 V

Question

n_i = 10^{10} frac{elektron}{cm^3}; mu_n = 1320 frac{cm^2}{V.s}; mu_p = 460 frac{cm^2}{Vs}; q = 1.6x10^{-19} C; V_T = 0.026 V

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