A few circuits are presented to understand the feedback of the triode (static analysis) Fig. 1 Schematic circuit, which is basically not applied in practice.
Fig. 1 Fixed bias common-injection amplifier circuit with the following static operating point.
Ic=βIbVce=VCC-IcRc=VCC-βIb*Rc, β value discrete and no feedback, the actual circuit is difficult to apply.
1, deformation circuit
1 Current negative feedback amplifier circuit.
Fig. 2 Deformation circuit 1 Fig. 2 has an additional shot-level resistor compared to Fig. 1, which is a negative feedback resistor. The negative feedback regulation process is as follows: first assume that the increase in temperature leads to an increase in the β value of the transistor -> leading to an increase in the collector current Ic, which leads to an increase in the emitter current Ie (Ib is very small, Ic ≈ Ie), Ie = Ue/Re, Ie increases, which inevitably leads to an increase in Ue, Ue increases will lead to an increase in the base potential, and because the voltage across Rb is equal to the supply voltage VCC and the base The voltage drop across Rb decreases, leading to a decrease in Ib, which in turn leads to a decrease in Ic. The end result is a process where the temperature rises, Ic increases, and the circuit feedback regulates the final Ic decrease.
- Deformation circuits
2 Stabilising the static operating point works better.
Fig. 3 Deformation circuit 2Vce=VCC-IcRc-IeRe≈VCC-Ic(Rc+Re)Ic≈Ie={[Rb2/(Rb1+Rb2)]VCC-0.7}/ReYou can see that there is no β in the above relation for Ic, so the static operating point is not affected by the triode β.
3, deformation circuit
Fig. 4 Deformation circuit 3 How does this circuit achieve a stable static operating point? Suppose the temperature rises and Ic rises –> Uc decreases –> voltage across Rb decreases –> Ib decreases — ->Ic(βIb) decreases.
Finally, did you learn to waste this little knowledge point?
