by BJT can work in one of the following modes of operation.
- Active Mode
- Cut off Mode
- Saturation Mode
For Vi = – Ve, the NPN transistor will be OFF, therefore start from Vi=0
Out of these three modes of operation, the cutoff and saturation modes of operation are used to operate the BJT as a switch. When a transistor is in cut-off mode, it doesn’t conduct any current and this mode of operation is considered the “OFF” position of a switch. Now, when the emitter junction voltage will be beyond the cut-in voltage the transistor enters the saturation mode and the maximum output collector current flow in it. It is the “ON” position of the Switch.
Operation in Cutoff
When the base to emitter junction is less than 0.5v, it will be reverse biased and practically no current flow in the transistor.
Vi<VBE<0.5V, IB=IC=IE=0
Since all the currents in the transistor are equal to zero that is the transistor is in “OFF” condition.
Operation in Saturation
When VBE>0.7v the transistor enters into active mode. The VBE can be increased just by increasing the input Vi since the emitter is directly grounded. Transistors begin to conduct.
in active mode
IB = (Vi-VBE)/R
IB = β IB [Active Mode]
Also Vo=VCC-ICRC
The base current can be increased by increasing Vi. It is clear from the above expression that collector current increases with increasing base current. When Ic increases Vc decreases. In this process, if Vc will is less than VB(0.7) then the collector-base junction starts to become forward biased and the operation of the transistor enters into saturation mode. In saturation mode, if IB increases Ic also increases and Vc will decrease. This means that the collector-base junction of the transistor becomes more and more forward biased and the transistor enters into a deep saturation region.
Thus, in saturation mode
Ic ≠ β IB
The equivalent circuit diagram for the described conditions can be expressed as follow.
Truth Table
Vi = 0 then Vo = Vcc
Vi = Vcc then Vo = ON
