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According to my textbook:

The drift current is relatively insensitive to the height of the potential barrier. The reason for this anomaly is the fact that the drift current s limited not by how fast the carriers are swept down the barrier but rather by how often.

From what I know, the drift current(let's say for holes) is given by, $$\vec J_p=pq\mu_p\vec E$$ where p is hole concentration, q is a charge, \$\vec E\$ is the net electric field and \$\mu_p\$ is hole mobility. The net electric field here is the vector sum of the applied field(due to bias) and the built-in field due to the donor and acceptor ions.

Even when there is no bias and the junction is at equilibrium, the above equation still holds. In fact, at equilibrium, it can be said that \$\vec J_{diffusion}+\vec J_{drift}=0\$ and the expression for \$\vec J_{drift}\$ is derived from the above equation.

So I don't understand how the electric field does not affect the drift current.

PS: It may be that I'm wrong about \$\vec E\$ being the net electric field(while it should be the built-in electric field). But even then, I don't get how the applied electric field(due to the bias) does not affect the drift current.

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  • \$\begingroup\$ It isn't exactly independent it is just drift current is so small and below the breakdown voltage it changes very slowly. \$\endgroup\$ – Se1fie Oct 21 '20 at 9:16
  • \$\begingroup\$ @Se1fie If the drift current is small then how does it cancel out the diffusion current at equilibrium? \$\endgroup\$ – ElonTusk Oct 21 '20 at 9:37
  • \$\begingroup\$ The diffusion current in equilibrium (zero bias) is also very small. \$\endgroup\$ – Se1fie Oct 21 '20 at 9:39
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The drift current is limited by the available charge carriers arriving at the reverse-biased junction. As a higher field draws more carriers, carrier concentration decreases. The resulting dependence of the current on the height of the potential barrier is weakened by the limited availability of charge carriers.

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