Constant voltage-drop diode model?

Question

I'm in the process of learning about diodes and I'm currently learning about diode models. I came across the model called the constant voltage-drop diode model. So, let's say the diode built-in voltage is \$V_{bi}=0.7V\$, which means the diode won't conduct forward current unless the source voltage exceeds the built-in diode voltage, or am I wrong?
My main question is: If the built-in voltage has a higher potential at cathode side of the diode and lower potential at anode side, as shown in this picture: Why is \$V_{bi}\$ then shown like this in the model:

so that the voltage \$V_{bi}\$ is higher at anode side of the diode.
What am I missing?

Answer 1

The whole point of “forward-biasing” the diode is to apply some external voltage that will counteract the built-in voltage that prevents the charge carriers from flowing between the p and n regions. That's why external voltage must be applied “in reverse” relative to the built-in voltage.

Answer 2

Let me start by stating two things that will help us understand the situation:

1. We are not talking about constant potential difference source (e.g. a battery) here. We are talking about a constant potential barrier, which means that it acts as a variable resistance that drops a constant voltage across itself.
2. In the first image (with the depletion region), the notation Positive Terminal indicates that we need to apply a positive potential at this terminal with respect to the other Negative Terminal to overcome this potential barrier. This is not the same as the notation for the battery in the second diagram.

If we keep these two things in mind and write an expression for the potential across forward biased diode, we see that it acts like a battery with its terminals as shown in the second image (the circuit).