How to identify the Saturation in BJT? What is Hard Saturation? Transistor as a Switch Explained

In this video, how the transistor (BJT) acts as a switch is explained with an example.
Along with that, it is also explained, how to identify the saturation in BJT and what is Hard Saturation.

By watching this video, you will learn the following topics:

0:00 Introduction

1:03 BJT in the Cut-off Region

2:21 How to identify the Saturation in BJT

4:31 Solved Example ( Identify the Region of Operation of Transistor )

7:10 What is Hard Saturation in BJT

8:07 Solved Example on Hard Saturation

9:40 Transistor (BJT) as a Switch

How to Identify that the BJT is operating in Saturation?

When the transistor operates in Saturation then the voltage between collector and emitter terminal (Vce) is very low (typically it is 0.1 to 0.2 V), and the collector current Ic is maximum.

Steps to Identify the Saturation in BJT :
1) Calculate the Ic(sat), the collector current in the saturation. (Assuming Vce = 0 )
2) Find the base current Ib.
3) Assuming the transistor is operating in the active region, find the collector current. (Ic = β Ib)
4) If the assumed collector current is more than Ic (sat) then the transistor is operating in saturation.

In this video, it is explained with the example.

Hard Saturation in BJT:

With the change in the external parameters like temperature, or with the replacement of the transistor the β may change. And with that change, the transistor may come out of saturation.

So, to ensure the operation of BJT in the saturation region in all the conditions, the β in the saturation should be very low. Typically around 10. I,e β sat should be around 10.
That means the values of base and the collector resistors should be selected such that the βsat is around 10.

By ensuring this condition, it can be ensured that the transistor will operate in saturation, even if there is a change in the external parameter.

Transistor as a switch :

By operating the BJT in the cut-off and the saturation region, the BJT can be used as a switch.

In cut-off region, the transistor offers high resistance and in saturation, it offers very low resistance.
That means in the cut-off region, the BJT will act as an open circuit and in saturation, it will act as a closed switch.

In this video, it is explained in detail, how the BJT can be used as a switch.

The link of the other useful videos related to BJT (Playlist)

This video will be helpful to everyone in understanding how the BJT can be used as a switch and how to identify the saturation in BJT.

#TransistorAsSwitch
#SaturationInBJT
#BJTasSwitch
#HardSaturation
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Music Credit:

[ALLABOUTELECTRONICS]

The timestamps for the different topics covered in the video:


1:03 BJT in the Cut-off Region

2:21 How to identify the Saturation in BJT

4:31 Solved Example ( Identify the Region of Operation of Transistor )

7:10 What is Hard Saturation in BJT

8:07 Solved Example on Hard Saturation

9:40 Transistor (BJT) as a Switch


For more Solved Examples on the BJT, check out this playlist:

[armyofteachers]

i havent seen the simplest explanation of electronics concepts work

[thelegendarychicken]

I have gone through so many stack exchange posts, videos, and readings to find the information that your video gave and clear up t he misunderstanding that I had. Thank you so much for this video!

[alpali]

Thank you, very thorough. :)
I had to do some uni assignments after missing a couple of lectures, haha.
I'd just add, that if the Vcesat is given, Icsat = (Vcc-Vcesat)/Rc.

[RexxSchneider]

Well done. You covered all of the practical cases that would be encountered in real applications of a switching transistor. However, for students, you are much better off teaching them to identify saturation by comparing the ratio of the saturation collector to the real base current with the transistor β, as you did at 6:20 and showing that when the transistor is in saturation, its current gain is significantly less than its current gain (β) in the active region. If you stick to this method, nobody gets confused by imaginary collector currents such as the Ic=4.65mA that you calculated at 5:33 or the Ic=9.3mA that you calculated at 12:00 (using the test of whether the actual current gain is significantly less than β only deals with real currents, so less confusion).
One other small point: your calculation of the transistor's output impedance in hard saturation is naïve. In reality, the change in Vce(sat) with current is very small, typically around 200mV for a change in collector current of around 100mA (viz. BC547 datasheet), thus giving a dynamic resistance of around 2 ohms, and of course, it's the dynamic resistance that's important in real applications.

[VersatileAnthem]

the way you explain in your videos concepts become crystal clear.love from Bangladesh..

[donnake]

Excellent video, very well explained - thank you!

[sjlee]

Very good lecture about bjt as a switch!! Thank you!!

[praveenbijaraniya]

Bravo....cleared every doubt.... marvellous...

[kadirtetik]

Thank you so much sir. Very clear description finally I understood it :)

[Yusuf-dxhw]

Thank you so much for your labor.It came to me complicated at the beginnig but after your examples i got it very clear.Greetings from Turkey

[mehmetsahin]

Adamın dibisin sen. You are realy great man bro

[ratnakumarikillada]

Excellent explantion, it is very usefull

[hemanthsaiborra]

Excellent explanation even better than lecturers

[michaelcostello]

Wonderful practical educational video. Thank you.

[martinchan]

Thanks for the crystal clear explanation.

[jitanshushaw]

Sir please please add a video about frequency response systems.Thank You.

[periohome]

great vid. at 9:03 in the video where/how did you get Vbe as 0.7v?

[rajendranluxmikanth]

If Ic < Ic(sat) what will happen?I mean the what is the mode of the transistor in that condition???

[prashantsoni]

For npn transistor, practically Vcb >(-0.4) for active region. Please explain why the characteristics shift backward beyond origin.