solve it please For a particular junction, the acceptor concentration (NA) is 1018/cm3 and the
donor concentration (ND) is 1014/cm3. Without doing any calculation, guess which side should
have more depletion width under open circuit condition. Now, If a 2V reverse bias is applied
across the junction find the depletion width (W) and charge stored in the depletion layer (Q3).
Assume n = 1.5 Times 1010/cm3, the Cross-Section A=10-3 cm2. Instead of reverse bias
voltage if you apply a forward bias voltage of 400 mv, calculate the new depletion width of the
junction. Explain briefly, what would happen if you slowly increase the forward bias voltage.
Especially, when the forward bias voltage, VF = Vo (depiction voltage) and VF > V0. In a
forward-biased pn junction. NA = 1018 cm3. ND = 1016/cm3. Lp = 5 mu m, Ln = 10 mu m, Dp
10cm2/s. De = 20cm2/s and the forward bias current I = 5mA. Find the current components Ip
and Ia. For a forward biased pn junction with forward voltage V=700mV, cross-section (A) =
200 mu m2, NA = 1018/cm3, ND = 1016/cm3, Lp = 5 mu m. La = 10 mu m, Dp = 10 cm2/s, and
Dn = 20 cm2/s, calculate the forward bias current I. A p+n junction is one in which the doping
concentration in the p region is much greater than that in the n region. In such a junction, the
forward current is mostly due to hole injection across the junction. As can be seen from the
following equations, when NA>>ND, the tree-electron injection current, Ia, becomes negligible
thus I becomes equal to Ip. For the above specific ease in which, ND = 1016/cm3, Dp =
10cm2/s, Lp = 5 mu m, and cross-section (A) = 200 mu m2, calculate the forward voltage, if the
forward bias current I = 0.5mA. Assume, temperature is 300K and n = 1.5 Times 1010/cm3.
Solution
1.
(a). In the n type material, there are many electrons i.ec many donors. Therefore, the donor
concentration of the PN junction (which is given) is almost equal to the donor concentration in
the n type material
(b). Even in the reverse bias it is same.
2. The formulae for the In and Ip are there in the 4th question just substitute the values there in
the formulae, we will get the ans.
3. In the question after solving the In and Ip and add both of the values, we will get the forward
bias current.
4. V = T/ni.
solve it please For a particular junction, the acceptor concentratio.pdf
1. solve it please For a particular junction, the acceptor concentration (NA) is 1018/cm3 and the
donor concentration (ND) is 1014/cm3. Without doing any calculation, guess which side should
have more depletion width under open circuit condition. Now, If a 2V reverse bias is applied
across the junction find the depletion width (W) and charge stored in the depletion layer (Q3).
Assume n = 1.5 Times 1010/cm3, the Cross-Section A=10-3 cm2. Instead of reverse bias
voltage if you apply a forward bias voltage of 400 mv, calculate the new depletion width of the
junction. Explain briefly, what would happen if you slowly increase the forward bias voltage.
Especially, when the forward bias voltage, VF = Vo (depiction voltage) and VF > V0. In a
forward-biased pn junction. NA = 1018 cm3. ND = 1016/cm3. Lp = 5 mu m, Ln = 10 mu m, Dp
10cm2/s. De = 20cm2/s and the forward bias current I = 5mA. Find the current components Ip
and Ia. For a forward biased pn junction with forward voltage V=700mV, cross-section (A) =
200 mu m2, NA = 1018/cm3, ND = 1016/cm3, Lp = 5 mu m. La = 10 mu m, Dp = 10 cm2/s, and
Dn = 20 cm2/s, calculate the forward bias current I. A p+n junction is one in which the doping
concentration in the p region is much greater than that in the n region. In such a junction, the
forward current is mostly due to hole injection across the junction. As can be seen from the
following equations, when NA>>ND, the tree-electron injection current, Ia, becomes negligible
thus I becomes equal to Ip. For the above specific ease in which, ND = 1016/cm3, Dp =
10cm2/s, Lp = 5 mu m, and cross-section (A) = 200 mu m2, calculate the forward voltage, if the
forward bias current I = 0.5mA. Assume, temperature is 300K and n = 1.5 Times 1010/cm3.
Solution
1.
(a). In the n type material, there are many electrons i.ec many donors. Therefore, the donor
concentration of the PN junction (which is given) is almost equal to the donor concentration in
the n type material
(b). Even in the reverse bias it is same.
2. The formulae for the In and Ip are there in the 4th question just substitute the values there in
the formulae, we will get the ans.
3. In the question after solving the In and Ip and add both of the values, we will get the forward
bias current.
4. V = T/ni
