Design an ideal one-sided n^+p junction diode to meet a breakdown voltage specifi cation. Consider a silicon pn junction diode at T = 300 K. Assume that Nd = 3 × 10^18 cm^-3. Design the diode such that the breakdown voltage is VB = 100 V.

Question

Design an ideal one-sided [latex]\mathrm{n}^{+} \mathrm{p}[/latex] junction diode to meet a breakdown voltage specification.

Consider a silicon pn junction diode at [latex]T=300 \mathrm{~K}[/latex]. Assume that [latex]N_{d}=3 imes 10^{18} \mathrm{~cm}^{-3}[/latex]. Design the diode such that the breakdown voltage is [latex]V_{B}=100 \mathrm{~V}[/latex].

Accepted Answer

From Figure [latex]7.15[/latex], we find that the doping concentration in the low-doped side of a one-sided abrupt junction should be approximately [latex]4 imes 10^{15} \mathrm{~cm}^{-3}[/latex] for a breakdown voltage of [latex]100 \mathrm{~V}[/latex].

For a doping concentration of [latex]4 imes 10^{15} \mathrm{~cm}^{-3}[/latex], the critical electric field, from Figure [latex]7.14[/latex], is approximately [latex]3.7 imes 10^{5} \mathrm{~V} / \mathrm{cm}[/latex]. Then, using Equation (7.61), we find the breakdown voltage as

[latex] V_{B}=\frac{\epsilon_{s} \mathrm{E}_{ ext {crit }}^{2}}{2 e N_{B}} [/latex] (7.61)

[latex]V_{B}=\frac{\epsilon_{s} \mathrm{E}_{ ext {crit }}^{2}}{2 \mathrm{e} N_{B}}=\frac{(11.7)\left(8.85 imes 10^{-14} ight)\left(3.7 imes 10^{5} ight)^{2}}{2\left(1.6 imes 10^{-19} ight)\left(4 imes 10^{15} ight)}=110 \mathrm{~V}[/latex]

which correlates very well with the results from Figure [latex]7.15[/latex].

Conclusion

As Figure [latex]7.15[/latex] shows, the breakdown voltage increases as the doping concentration decreases in the low-doped region.

Question 7.7: Design an ideal one-sided n^+p junction diode to meet a brea...

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