Consider a linearly graded junction $$N_D - N_A = ax$$ for some grading constant \$a\$. Using the standard simplifying assumptions one finds that the depletion region extent is $$x_{SCR} = \left[\frac{12 \epsilon (\phi_{bi}-V)}{qa} \right]^{1/3}$$ and a built-in voltage $$\phi_{bi} = 2\frac{kT}{q}\ln \frac{ax_{SCR}}{2n_i}.$$ These two equations determine \$\phi_{bi} \$ and \$ x_{SCR}\$. However, it seems that they also imply that \$\phi_{bi}\$ is bias-dependent. What should I make of this? It's hard for me to "see" but perhaps the dependence is sufficiently weak that we can still consider this "built-in"? How do I see how this dependence depends on \$a\$ in particular?
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\$\begingroup\$ Does the linear doping gradient extend indefinitely or do the acceptor and donor dopants have some maximum levels so that the net doping levels out at some x values on either side of the junction? (That's meant as a hint to get to the answer, not as an actual question for you, OP) \$\endgroup\$– The PhotonCommented Nov 17, 2023 at 16:07
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\$\begingroup\$ I suppose I haven't considered that and, in reality, there must of course be some end. But if we remain in some bias regime where \$N_D - N_A = ax \$ always holds then surely that doesn't matter? @ThePhoton \$\endgroup\$– EE18Commented Nov 17, 2023 at 16:11
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I think that the potential barrier is given by the following relation as demonstrated:
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\$\begingroup\$ Oh interesting. What is the source here? Is this Sze? \$\endgroup\$– EE18Commented Nov 17, 2023 at 18:11
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\$\begingroup\$ Touche, thank you :) I asked only because it was an image so I imagined it was from some external source. At any rate, if I can ask one last question...I don't know much about the Lambert W function, but should we conclude that the dependence here is superlinear in \$V\$ given that second term? That is, is the first term monotonically increasing in \$V\$? \$\endgroup\$– EE18Commented Nov 17, 2023 at 18:16
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\$\begingroup\$ @EE18 W is increasing or decreasing depending on which branch happens to be solved for. en.wikipedia.org/wiki/Lambert_W_function \$\endgroup\$ Commented Nov 20, 2023 at 3:29
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\$\begingroup\$ @TimWilliams Does the branch in question then need to be specified in the answer? Surely the built-in potential is single-valued, no? \$\endgroup\$– EE18Commented Nov 20, 2023 at 12:59