Friday, 9 August 2019

voltage - Doping concentration in anode region


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




Answer



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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