### Bjt As A Switch

In the last post, we studied different biasing techniques of the BJT transistor. So in this post and in the next post, we will be dealing with the application of BJT
transistor and at the end, we will be dealing with different projects of bjt.
To operate bjt as a switch mostly fixed bias is used as there is only one resistor on the input side but I had directly applied Sinusoidal(ac signal) 5 Vp-p at the input side so that I can explain the concept in a detailed way. As I had explained in my earlier posts that when Vin is greater than 0(+ve half cycle) the output would be inverted by 180 degrees, similarly when Vin is less than 0(-ve half cycle) then at the output we get a positive half cycle. If you forgot that concept then here is the video which helps you to remember it.

So we will understand this concept in 2 cases
Case 1:-During Positive(+ve) Half-Cycle of Vin:-

During positive half-cycle (ideally) or when Vin is greater than 0.7v (practically) i.e. Vbe>0.7 then electrons flow from emitter side to collector side, which means current(Ic) flows from collector terminal to emitter terminal. So, in short, we can say that during a positive half-cycle our transistor acts as a short circuit or we can assume it as a closed switch. Since our transistor acts as a closed switch therefore we can say that it is working in saturation mode and if you remember in our earlier posts we had studied that in saturation mode the output current(Ic) is maximum while the output voltage(Vce) is less. Now if Vce is less that means the voltage drop across led is more. So due to this reason, the led will glow in this case. You can see this in the image given below.

So this was about a positive(+ve) half-cycle. Now starting with case 2 is during the negative(-ve) half-cycle.

Case 2 :-During Negative(-ve) Halfcycle of Vin:-

During negative half-cycle (ideally) or when vin is less than 0.7v (practically) i.e. Vbe<0.7 then electrons don't flow from emitter to collector since both the junctions of the transistor are reverse biased. As no or very less electrons(minority carriers) flow therefore the output current(Ic) generated is very less or negligible. So, in short, we can say that during the negative half-cycle the transistors acts as an open circuit or you can assume it as an open switch. Since our transistor acts as an open switch, therefore, we can say that it is working in the cutoff region and as we had studied that in the cutoff region the output current(Ic) is less and the output voltage(Vce) is maximum. Now if Vce is maximum that means voltage drop across led is very less. So the led remains off in this case. You can see this image given below.

CONCLUSION:-
Thus we can say that the transistor acts as a variable resistor. If  Vbe < 0.7v then the transistor provides high resistance, therefore, no current flows through the circuit and the transistor operates in the cutoff region and if Vbe > 0.7v then the transistor provides less resistance, and therefore the maximum current flows through the circuit and the transistor operates in the saturation region.
As usual, I conclude this part by living the video, this will clear your concept in a more detailed way for sure.

VIDEO:-