Constant voltage-drop diode model?

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?

• n side to minus is conducting, and the bar in the schematic is the n side. There has to be a voltage with the shown polarity and greater than .7V for the Diode to conduct. Dec 10 '17 at 21:49