In many articles regarding transistor biasing I see a circuit like below and I don't really get what is the point of biasing like this WITHOUT emitter resistor Re?

When emitter branch is connected straight to the ground, wouldn't the VB voltage always be around 0.7V no matter the voltage divider output? "The rest" of biasing voltage would only result in greater base current. In other words - it's like we tried to bias a diode.

With emitter resistor it's different story. VB would be then VB=0.7V+VRe - we get a steady VB voltage with any value we want (e.g. 3V).

Could someone please clarify this?

[![r](https://i.stack.imgur.com/U6RI3.jpg)](https://i.stack.imgur.com/U6RI3.jpg)