voltage biasing for BJT

I don't know of any such thing as "voltage biasing" of a bjt. It would be thermally unstable. Nobody ever controls Vbe directly. We emitter current bias the device. We do not base current bias the device because of beta dependency.

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I found this in an old thread here: https://www.electro-tech-online.com/threads/transistor-biasing-question.113993/
Does this mean that a battery is connected beween B and E?
Can you make clear about "thermally unstable"?
I think I know what does "voltage biasing" mean now but I curious about the case as Claude has thought. Why isn't it thermally stable?

Hi anhnha,
I agree with you - the answers in the referenced link are not satisfying - either they do not answer your question in a sufficient (detailed) manner or they are simply false.

The classical and most common way to bias a BJT in common emitter configuration is as follows (voltage biasing):
* Use a grounded voltage divider at the base, which is driven by the supply voltage.
The resistive niveau normally is chosen such that the current through the grounded resistor is approximately k*Ib=k*Ic/beta (k~6...10)
* Use, in addition, an emitter resistor Re to stabilize the selected Ic value against tolerances and temperature changes.
The value of Re normally is chosen to be app. 10...20% of the collector resistor Rc.
* Be aware that the signal gain drastically is reduced (but stabilized) due to Re - unless Re is bypassed with a capacitor Ce (gain increased again for frequencies above the corner frequency).

Winterstone

Supplement to the first point: The divider should create a voltage at the base node of Vb=Vbe+Ie*Re (with Ie~Ic and Vbe=0.65...0.7V)