Electrical – Constant voltage-drop diode model

diodespn-junction

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:
enter image description here
Why is \$V_{bi}\$ then shown like this in the model:

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

Best Answer

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.