A question about Vce of an NPN BJT in saturation region

Below is an NPN transistor symbol and the voltages at its terminals are Vb, Vc and Ve with respect to the ground:

I read that: during the saturation the Vce = (Vc-Ve) settles to around 0.2V and the further increase in base current will not make Vce zero.

But why doesn't Vce become zero?

As far as I know: when the transistor is saturated, the base-collector junction turns on, like a diode, so the collector voltage will follow the base voltage increase, only it will be a diode-drop below. But the same thing happens between the base voltage and the emitter voltage. So at saturation and beyond one can write the following(?):

Let's call the diode drop as Vd between p and n junctions, so the collector and the emitter voltages can be re-written in terms of the base voltage as:

Vc = Vb- Vd

Ve = Vb -Vd

Vce = Vc -Ve = 0

Where am I wrong here?