The discussion focuses on the relationship between base voltage (VB), emitter voltage (VE), and base-emitter voltage (VBE) in NPN transistors. It establishes that with a constant VB, an increase in VE leads to a decrease in VBE, which is a critical aspect of negative feedback in transistor operation. The exponential relationship between collector current (IC) and VBE is sensitive to temperature changes, necessitating the use of an emitter resistor (RE) to stabilize current levels. The equation VBE = VB - VE is central to understanding this feedback mechanism.
PREREQUISITESElectronics students, circuit designers, and engineers working with NPN transistors and those interested in understanding feedback effects in transistor circuits.
Oooh, I get it now. VBE is the potential difference between the emitter and the base. Decreasing it will reduce the base current. I was viewing that VBE as the 0,7V potential barrier but it was just the difference between VB and VE :D Thank you, kind sir!LvW said:Rewriting your equation we have
VBE=VB-VE.
As you wrote - in case of constant VB (resistive voltage divider) an increase of VE (larger emitter current IE) will reduce the difference (VB-VE).
This reduction in VBE is the desired negative feedback effect.
The problem is that the IC=f(VBE) exponential relation is very sensitive to temperature changes (-2mV/K). Hence, for a constant VBE the current IC would change its value - more than allowed. For this reason we are using such an emitter resistor RE which produces an increasing voltage drop VE=IE*RE for the unwanted current increase. As a result, the voltage VBE again will decrease and, thus, bring back the current IE (and withit: IC) to nearly the previous value.