Electronics - Bipolar Junction Transistors (BJT) - Discussion

Digital circuit means 0 or 1 therefore cutoff means 0 and saturation means 1, that's all.

As we know that when we talk about digital circuit then one thing is very clearly understand the operation of the circuit is totally depend on logic 1 (i.e high voltage) or 0 (i.e. Low voltage) and we can say that saturation is logic 1 and cut off region is logic 0.

So no doubt when transistor are used in digital circuit they operate in saturation and cut off region only. Because it full fill the digital circuit requirement.

In digital electronics world transistor is nothing but a switch, switch has two modes can be ON or OFF. To get the desired output you decide what you need it ON or OFF. Now we conventionally assume if ON is 1 then OFF is 0.

So saturation mode means your circuit is closed which also means your switch is ON which lead us to say our output is 1 and cutoff region is opposite. Common sense if you transistor in saturation mode output is max which is 1 however in cutoff it is opposite.

Key points : Saturation MEANS ON what is ON your switch also means circuit is closed. And cut OFF is OFF what is OFF your switch means circuit is open means 0 output.

When transistor is used as an amplifier then it has to be operated in active region and when the transistor is use in a digital circuits then it can be operated in Cut off (open s/w means 0) or in saturation region (closed s/w means 1).

Is transistor operates in saturation or active region itself (as the amplify or switching) or we do this?

The whole digital world considers only two states. That is on and off/ 1 and 0. So now the transistors saturation and cut-offs can be considered as on and off respectively.

Digital system as very clear having two states namely high or low, saturation implies high state and cut off implies off state.

I just want to know the voltages in both the modes i.e. cutoff and saturation.

What can we say about the voltages in cut off as well as saturation ? I mean when Ic is zero in cut off Vce should be maximum right?what about other voltages i.e. (base emitter and collector base) I know that base emitter junction will be reverse biased but can we come to know about the exact voltage? kindly guide me.

Cut off (no collector current), useful for switch operation.