written 8.3 years ago by | modified 2.8 years ago by |
Mumbai University > FE > Sem 1 > Basic Electrical and Electronics Engineering
Marks: 4 M
Year: May 2015
written 8.3 years ago by | modified 2.8 years ago by |
Mumbai University > FE > Sem 1 > Basic Electrical and Electronics Engineering
Marks: 4 M
Year: May 2015
written 8.3 years ago by |
Cut-off region:- Both the emitter to base and collector to base junctions are reverse biased. The base current $I_B=0$ and the collector current is equal to the reverse leakage current $I_{CEO}$.
Active region:- The emitter to base junction is forward biased and the collector to base junction is reverse biased. The collector current $I_C$ increases slightly with increase in voltage $V_{CE}$ However the collector current is largely depend on the base current $I_B$.
Saturation region:- The emitter to base junction and the collector to base junction are forward biased.
$$\boxed{\text{output resistance},r_0=\dfrac{\Delta V_{CE}}{I_C}\bigg|I_B=Constant\bigg|}$$