Electronics - Bipolar Junction Transistors (BJT) - Discussion

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.

One of the most common uses for transistors in an electronic circuit is as simple switches. In short, a transistor conducts current across the collector-emitter path only when a voltage is applied to the base. When no base voltage is present, the switch is off. When the base voltage is present, the switch is on.

The areas of operation for a transistor switch are known as the Saturation Region and the Cut-off Region. And use the transistor as a switch by driving it back and forth between its "fully-OFF" (cut-off) and "fully-ON" (saturation) regions.

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.

The transistor is always work in digital as a switch on/off or close/open and 0/1 is similar as 0V(low) and 5V(high)so, saturation and cutoff regions is the answer for digital function.

Although transistor is work in the active region because it has a biased voltage 5V/15V so ip is forward biased and op is reversed biased when ip is apply >0.8V op is 5V(high) and ip is <0.7V op is 0.4V(low).

As far I think that Digital circuit require switching element (ON&OFF) , that why transistor whenever used in digital circuit always operate in saturation (ON) & cut off (off) in region. Basically the digital circuits operate only binary digits (i.e. 0 and 1) , when ever this is operates in bjt saturation region (the switch is on) , &the cut off region (the switch is off).

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.

Generally we are using min and max term in digitial circuit like that the term will be 0 to 1 or others like that one is min and second one is max then the transister will be linear region and it also oprate in satutration region which help us the switch will be on and off.

Digital circuits always operates in two modes that is 1 & 0, and we can also say that though its a circuit so the transistor can be act as a switch, and switch always obtained in saturation and in cut-off modes. Active mode is used only for amplification factor.

In saturation region transistor acts as closed switch and in cut off region it acts as open switch. In digital circuits o/p can be 0 or 1. So when in saturation region means closed switch it gives output 1 and in cut off region means open switch it gives o/p 0.

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).