Question

In a uniform p-type Si sample, As is diffused to have a donor profile of 10^18/cm^3, Problems where acceptor concentration becomes negligible in comparison of donor concentration. Find the resulting value of the sheet resistance of the diffused layer up to the junction depth of 5 μm of the device. If the mobility for electrons is μn = 500 cm^2/ V-sec, then what will be the value for the conductivity?

Answer 1

`\sigma=80 (\Omega.cm)^(-1)`

Step-by-step explanation:

Na = Acceptor Concentration (cm-3) = 0

Nd = Donor Concentration (cm-3) =
`10^(18)`

`\rho`= Resistivity (Unit: ohm-cm)

n = electron concentration (cm-3)

q =Charge on electron =
`1.6* 10^(-19)`

`t=thickness=5 * 10^(-4) cm`

`\mu=\mu_(min) +(\mu_(max)-\mu_(min))/(1+(N)/(N_(r) )\alpha )`

`\mu_(max)`

`\mu_(min)`

`N_(r)` are fit parameter

Arsenic is used

`\mu_(max)` =52.2

`\mu_(min)`=1417

`N_(r)=9.68*10^(16) \\\alpha=0.68`

`\mu_(N)` =222.2

For n type =
`p=(1)/(qn\mu_(n) )`

`R=(\rho)/(t)`

(ND>>NA ---> n-type) // Here The p-type sample is converted to n-type material by adding more donors than acceptors

n=ND-NA =1018/cm3

`\rho=[1.6*10-19*1018*222.2]-1 = 0.28 (ohm-cm)`

R=0.28/5*10-4=560

`Given mobility, \mu=500cm2/V-sec`

`Conductivity= \sigma=\mu*N*q = (500*1018*1.6*10-19)\\Conductivity= \sigma=80 (\Omega.cm)^(-1)`