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

Answer:

$\sigma=80 (\Omega.cm)^{-1}$

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 = $1.6\times 10^{-19}$

$t=thickness=5 \times 10^{-4} cm$

$\mu=\mu_{min} +\frac{\mu_{max}-\mu_{min}}{1+\frac{N}{N_{r} }\alpha }$

 $\mu_{max}$

 

$\mu_{min}$

$N_{r}$  are fit parameter

Arsenic is used

 

$\mu_{max}$ =52.2

 

$\mu_{min}$=1417

$N_{r}=9.68\times10^{16} \\\alpha=0.68$

 

$\mu_{N}$ =222.2

For n type =$p=\frac{1}{qn\mu_{n} }$

$R=\frac{\rho}{t}$

(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

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

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