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I'm stuck with this doping problem and can't figure out where to go with it.

The built-in voltage of a GaAs pn junction diode is 1.25 V when the diode's temperature is T = 320K. The cathode region of the diode is doped with phosphorus at a concentration of 1e17 \$cm^{-3}\$. Determine the required doping concentration in the anode region.

Would I go about this by using the equation \$V_{bi} = V_T\cdot ln(N_a\cdot\frac{N_d}{n_i^2})\$?

\$V_T\$ being the thermal voltage
\$N_a\$ being the acceptor concentration on p side
\$N_d\$ being the donor concentration on the n side
\$n_i\$ being the intrinsic carrier concentration

I have completed the following work for the above question. Can I be checked for accuracy please!

enter image description here

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That is the correct formulae for an abrupt junction, just be sure to use the right temperature for the thermal voltage \$ V_T=\dfrac{KT}{q_e}\$ and the right intrinsic carrier concentration \$n_i\$ at that temperature.

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  • \$\begingroup\$ what is q in thermal voltage? \$\endgroup\$ – user081608 May 20 '13 at 20:07
  • \$\begingroup\$ answer edited ... electron charge. \$\endgroup\$ – placeholder May 20 '13 at 20:12
  • \$\begingroup\$ Okay is that always 1 or is it 1.60217657 × 10-19? \$\endgroup\$ – user081608 May 20 '13 at 21:03
  • \$\begingroup\$ always look at the units in a calculation, that will tell you. \$\endgroup\$ – placeholder May 20 '13 at 21:13

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