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?

Explanation:

Na = Acceptor Concentration (cm-3) = 0

Nd = Donor Concentration (cm-3) =

= Resistivity (Unit: ohm-cm)  

n = electron concentration (cm-3)

q =Charge on electron =

 are fit parameter

Arsenic is used

=52.2

=1417

=222.2

For n type =

(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

R=0.28/5*10-4=560

scientific laws are the result of testing theories and proving them to be facts.

dang that balls went far