Friday tute week 2 solution PDF

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UNSW SOLA3507 TUTORIAL WEEK2

Term 1 2019

PC1D Simulations of p-n Junction Silicon Solar Cells Question 1: Built in potential and electric field A silicon p-n junction is formed with both p-type and n-type doping concentration of 2.5x1016 cm-3. (a) Calculate the built-in potential,

Plug in numbers….. Vbi = 0.760 V

(b) Calculate the depletion width at zero bias, and

W = wp + wn =

2V j q

 1 1    N + N  D   A

Vj is simply Vbi. W = 2.81 x 10-5 cm, which is 281 nm

(c) Calculate the maximum field at zero bias using parameters at 300K.

Since dopings are equal so are the depletion widths on both sides of junction. So take wn = 140.5 nm. Max electric field is at junction (x = 0). So take F = qNDwn/ε = 54,275 V/cm

Question 2: Built in potential and electric field An abrupt p-n junction is formed in a silicon wafer. The wafer has an area of 100 cm2, and is uniformly doped on the p-type side with 5x1018 boron atoms/cm3 and is uniformly doped on the n-type side with 2x1015 phosphorus atoms/cm3. (a) Calculate the built-in potential and depletion width at 300K under equilibrium conditions.

Vbi = 0.832 V (b) Calculate the total width of the depletion layer when the junction is biased with an applied external voltage of VD = 0.5 V (forward bias). What distance does the depletion layer extend on the p-type side and n-side?

Vj = 0.332 V

wp = wn =

SOLA3507 – Exercise #1

ND W NA +N D NA W NA +N D

W = wp + wn =

2 V j q

 1 1    +  NA ND 

Page 1

Term 1 2019

UNSW SOLA3507 TUTORIAL WEEK2

W = 463 nm then follows wp = 0.19 nm, wn = 462.81 nm. The depletion region is almost completely on the lightly doped side.

(c) If the diode is biased at VD= -10 V, what would be the width of the depletion layer, W, wn and wp?

In this case junction voltage is Vbi + 10 V = 10.832 V. So find W again. W = 2.648 microns Wn = 2.647 microns wp = 1 nm Still have the depletion region all on the n side, but the depletion region is much bigger.

Question 3: PN junction Consider a p-n junction at zero bias with an electric field distribution as sketched below. The metallurgical junction is placed at x=0.

(a) Calculate the built-in potential.

Vbi is simply the negative integral of the electric field. So count up the area under the profile. We get -0.75 V

(b) Estimate the doping type and doping level of the region between 50 < x...