GATE 2016 EC SET B QUESTION 37

GATE 2016 EC SET B QUESTION 37 :- Consider a region of silicon devoid of electrons and holes, with an ionized donor density of N+d=1017cm-3. The electric field at x = 0 is 0 V/cm and the electric field at x = L is 50 kV/cm in the positive x direction. Assume that the electric field is zero in the y and z directions at all points.

Given q=1.6×10-19 coulomb, ∈0=8.85×10-14F/cm,∈r=11.7 for silicon, the value of L in nm is______________.

Solution

We know that, dE/dx=qND/∈ . Hence from problem statement, we get
50×103/L =(1.6×10-19×1017)/(8.854×10-14×11.7)
=> L=(50×103×8.854×10-14×11.7)/(1.6×10-2 )
=> L=32.372×10-9m

Answer to GATE 2016 EC SET B QUESTION 37

The answer is obtained by finding the length of region of silicon devoid of electrons and holes, with given ionized donor density and for which electric field has been given to be  32.372 nm

GATE 2016 EC SET B QUESTION 36

Consider avalanche breakdown in a silicon p+n junction. The n-region is uniformly doped with a donor density ND. Assume that breakdown occurs when the magnitude of the electric field at any point in the device becomes equal to the critical field Ecrit.Assume Ecrit to be independent
of ND.If the built-in voltage of the p+n junction is much smaller than the breakdown voltage,
VBR, the relationship between VBR and ND is given by
(A)VBR × √ND= constant            (B) ND × √VBR =constant
(C) ND × VBR =constant             (D) ND/VBR =constant