Question

= 8.05 - 1022 - 1.21 . 1023 VD (Linear Fit) 1/C? (1/F?) 0.0 0.5 100.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 Voltage (V) Figure 1: CV data for a Silicon p-n junction diode. 2. A colleague hands you CV data (capacitance vs voltage) of a silicon p-n junction diode shown in Figure 1. For this particular diode, the donor concentration on the n-side is much larger than the acceptor concentration on the p-side (i.e. Np > NA) and the cross-sectional area of the diode is 1 x 10-cm-2 (20 points) (a) Use this data to estimate the built-in potential. (10 points) (b) Use this data to estimate the effective doping concentration. Does the effective doping concentration more accurately represent the donor or the acceptor doping concentration? Explain. (10 points)

Answer 1

We can find the answers for these questions by comparing the given equation by the standard equation for a CV data. The standard equation is given in the attached image.

Capacitance of a on junction disde, CA/ Esa (Wat ta) ( ) ,i-Built in voltage potential I = 2 (Natwd . co Aqes Na Na e - 1 = 2 (Not Nd) di - 2 (Nat Nd) v A²q Es Na Nd ARGES Nand. Comparing it with the equath in the questa Śzz 8.05 X 1022 - 1.21x10?v, We get 2 Natna a $.2101023 ♡ A2gts Na Nd 2 Nathd diz8 .05 x 1022 @ AZQES Na Nd @ Using eqn 0 & we get, I pi= 0.665 v mana, te se na ($i-y : 2 = 1. 21 x 1023 Axq Es Na & Naz tz 2 fazl6x10-19 C A²QEX 1.20 x 1023 Es 211.7€. z 2 1.6410-19 x 11.788.85x10-12 x 1.21x 1023 x 10-4x10-4 Татья (м3 - Naz 9.97x 10 It more accurately refreports the acceptar concentratr. as Nd is vanishes from the ear after considering the fact that Na Na.