BJT: Voltage Divider Bias Configuration

In this video, the Voltage Divider Bias Configuration of BJT is explained with an example.
By watching this video, you will learn the following topics:
0:00 Introduction
0:23 Introduction to Voltage Divider Bias Configuration
1:43 Approximate DC Analysis of Voltage Divider Bias Configuration
4:30 Exact DC Analysis of Voltage Divider Bias Configuration
9:50 Solved Example on Voltage Divider Bias
Voltage Divider Bias Configuration:
It is one of the most frequently used DC biasing configurations of the BJT.
Unlike the other biasing configurations, it requires only a single power supply and the base voltage is derived using the voltage divider circuit.
In this configuration, if the values of the voltage dividing resistors are chosen properly then it is possible to make the operating point almost independent of the variation in β.
In this video, the approximate and exact DC analysis of the voltage divider Biasing configuration is provided and in the end, one example is also covered.
Check out my second channel for more Quiz on Electronics:
/ @allaboutelectronics-quiz
This video will be helpful to everyone in understanding the Voltage Divider Biasing Configuration of the BJT.
Follow my second channel:
/ @allaboutelectronics-quiz
Follow me on Facebook:
/ allaboutelecronics
Follow me on Instagram:
/ all_about.electronics
Music Credit:
www.bensound.com/

[@ALLABOUTELECTRONICS]

The timestamps for the different topics covered in the video: 0:23 Introduction to Voltage Divider Bias Configuration 1:43 Approximate DC Analysis of Voltage Divider Bias Configuration 4:30 Exact DC Analysis of Voltage Divider Bias Configuration 9:50 Solved Example on Voltage Divider Bias For more Solved Examples on the BJT, check out this playlist: kzbin.info/www/bejne/mKqzc42jeMummJo

[@jacquelineurena5860]

Great video! Your explanation was very clear, thank you.

[@aashishgautam258]

Rc coupled amplifier and others , sir you are an outstanding teacher pls keep it up

[@lautarodelucia6480]

Best video I found on this subject.

[@soumadipgarain]

As i can't join u so i see full ads coming in ur video.❤️❤️❤️.

[@deathstroke4424]

no one cares

[@HemanthKumar-yk8iy]

Sir your explanation is easily for understood

[@silverspin]

Best channel ever for electronics

[@mayurshah9131]

Excellent 👌

[@rajkamaldey6168]

Nice video, nice explain Last slide, 15:54 Rb ≤ .1 time...

[@abeyjoseph4727]

thank you so much

[@PatrickRibbsaeter]

great content / info as always..

[@edgbaston149]

Thank you so much Sir

[@quicktech7450]

thank you sir

[@priyajitmondal6661]

Thank you sir❤🙏👍

[@jstro-hobbytech]

Great video

[@AhsanKhan-sj8yr]

sir very informative video. can you please tell me which software you use for making these videos.

[@vinaykumarkv5830]

Superb👌👌🔥

[@dilibanb2226]

What factor leads to the stability in voltage divider bias?

[@soujanyaas9074]

Superb

[@technologyiskole9626]

best vedio

[@patrickmagombo4500]

How does current stabilisation achieved this configuration

[@taymoortariq6803]

thanx

[@EmCeyyyyyyy]

How do you calculate parallel? The one giving 1.66k ohms in the example at the end?

[@jbdlssnts7351]

Why is the equation for Vce, Vce=Vcc-Ic(Rc+Re) instead of Vce=Vcc-IcRc-IeRe? Is it because we assume that Ie=Ic? if so, is it still under exact method or approximation method? sorry for bad english

[@teeggindarshannagappa8858]

Hello, at 9:45 it has been said that if Rth

[@dasilvaleandro21]

Correct one is Rth

[@NakulMandwal-gt5cd]

Iske notes ke pdf download kar sakte hai kya for fast revision Is there any link or website of this channel from where i can download this presentation ppt

[@033_rkpragyajitsingh6]

Can we find the value of Ve using Vc-Vce instead of Vb-Vbe?

[@ALLABOUTELECTRONICS]

If Vc and Vce are known then yes, you can find Ve using that formula.

[@lawrencemayfield9669]

need a video having it in saturation and finding all things including beta

[@shuvbhowmickbestin]

What about the DC stabilisation?

[@aalabalu674]

ic =0.7mA,vce=9.8v huge variation in operating point under approximate analysis

[@aashishgautam258]

Sir pls make further videos in this series

[@ALLABOUTELECTRONICS]

Yes, very soon all the topics will be covered.

[@vijaysri4849]

Is self baising and voltage divider baising same

[@lawofselfcontrol]

which app are you using?

[@adreq3.05]

If haves the transistor current amplification I CEO about fifty, this can to be amplicate als electronic key only.

[@user-ub5su9vj6d]

😇😇😇👍👍👍🙏🙏🙏Thanks a lot, great job!!!!!!!Please make a design bias circuit ( Design the value of resistors and capacitors).

[@ALLABOUTELECTRONICS]

It’s already covered through solved problems. Please check BJT solved problem playlist on the channel page.

[@user-ub5su9vj6d]

@@ALLABOUTELECTRONICS You always do a perfect job.👍👍👍😇😇😇🙏🙏🙏

[@nikolaynetsov6529]

I need a little help. I must find Re,R1 and R2. I have h213=80,Rc=2kΩ,Ic=1mA,Uce=2V,Vcc=4,5V,Id=10Ib and Ube=0.6V

[@ALLABOUTELECTRONICS]

Many such problems have been solved on BJT on the second channel. Here I am giving the link for the playlist on the solved problems on BJT. Ypu will find such example as well. Here is the link: kzbin.info/www/bejne/bpPIgYZ_jJ6Vb6c Still if you find any difficulty, let me know here.

[@nikolaynetsov6529]

@@ALLABOUTELECTRONICS Thank you a lot for the support

[@niladrimallik3172]

We have assumed base current almost equal to 0. So, at 7:21, won't we take base current=0?

[@ALLABOUTELECTRONICS]

Yes, that was considered for the approximate analysis. And it can be assumed when (B+1) Re >> Rth. At 7:21 the expression of the base current gives the exact value of the base current. I hope it will clear your doubt.

[@niladrimallik3172]

@@ALLABOUTELECTRONICS Yeah got it. Thank you for the answer.😊

[@jacquelineurena5860]

On min 15:15 did you mean Rth

[@ALLABOUTELECTRONICS]

Yes, that's right

[@pandurangaiahshetty4064]

Please do videos on tranistor amplifiers

[@ALLABOUTELECTRONICS]

Yes, that too will be covered soon.

[@MKTheBlackKid]

what if there are numbers on the capacitors🤔

[@niharikagoulikar2921]

Why is capacitor connected at input and output side?Could someone explain me?

[@ALLABOUTELECTRONICS]

They are used to block the DC biasing voltage. Typically these amplifiers are using in cascade configurations where more than one such amplifiers are cascaded. The DC biasing voltage of one amplifier should not affect the next amplifier or any other circuitry after the amplifier. The capacitor blocks the DC voltage and passes only AC signal.

[@jenadolfo9358]

what if the R1 and beta is unknown? what is the formula for it plss help

[@ALLABOUTELECTRONICS]

If there are few unknown, then must have been given other parameters. One such example is covered on the channel. kzbin.info/www/bejne/qJPGdZeKhNFgr5I Please go through it.

[@veenagovekar3653]

Sir how to get voltage gain of this circuit?

[@ALLABOUTELECTRONICS]

I have already covered it. Please go through the BJT playlist. If you go through the video, small signal analysis of CE configuration, you will get it.

[@im_cpk]

5:44 could you please explain,how vth comes?

[@im_cpk]

I mean to say, where's open circuit voltage...

[@ALLABOUTELECTRONICS]

In this case, the Thevenin's equivalent voltage is the voltage seen across the R2 and ground.

[@im_cpk]

8:23 how you got the value of ie??how rth/beta+1 came?? plez reply 🙏🙏🙏

[@im_cpk]

I know you got the value from i/p kvl equation...but how this rth/beta+1 came...

[@ALLABOUTELECTRONICS]

I hope, you got the expression of Ib. Now, Ie = (beta + 1) Ib. Just put the value of Ib from the expression. And then divide the entire expression by (beta + 1). You will get that expression. If you still have any doubt then let me know here.

[@creativearyaninnovative1846]

Actually imagine nahi ho pata h Apke lectures

[@deepanjalimishraa]

Want an answrr asap please

[@angelorison6455]

15:15 I think Vce is equal to 6.204 V right???? I got this value from my calculator if it is wrong check it and plz correct it.

[@ALLABOUTELECTRONICS]

The collector current is 1.266 mA. During the calculation, if you take the value of collector current up to 3rd decimal place, then Vce will come out as 6.174 V. If you consider the value of Ic up to 2 decimal places, then Vce = 6.204 V.

[@angelorison6455]

@@ALLABOUTELECTRONICS oh got it thank you

[@harshit9112]

16:15 0.1 will come instead of 10

[@ankitashirali9362]

Sir please make a video on ce amplifier

[@ALLABOUTELECTRONICS]

Yes, once I will cover the different biasing configuration, I will cover the CE amplifier and the AC analysis.

[@abhilashasaini8758]

Vbe=0.7V how? Is it given? Pls reply fast

[@ALLABOUTELECTRONICS]

If you see, the base emitter junction is forward biased. So, it acts like a diode and the voltage drop across it can be assumed to be around 0.7V during the analysis. I hope it will clear your doubt.

[@abhilashasaini8758]

@@ALLABOUTELECTRONICS ok thanku thanks a lot

[@southcurseone1579]

I wanna ask, what if the value of beta is not given and only Ib is known, how can you solve value of beta?

[@ALLABOUTELECTRONICS]

If beta is not given in the example, then you must be given some other parameter instead of that.

[@southcurseone1579]

@@ALLABOUTELECTRONICS oh i see. So it is impossible to solve without beta. Thank u.

[@realmccoy813]

🙏🏾🙏🏾🙌🏾🙌🏾

[@bhaveshchaudhari3406]

Sir plz make video on IGBT..

[@ALLABOUTELECTRONICS]

Soon I will try to cover IGBT also.

[@bhaveshchaudhari3406]

@@ALLABOUTELECTRONICS thank you.....

[@abhijitkumarmanna2031]

How you have done {Rth/(beta+1)}+Re in place of {Rth+Re(beta+1)} ? 8:35

[@ALLABOUTELECTRONICS]

Its emitter current. Just multiply the base current with ( beta + 1). Taht means, i nthe numerator you will have ( beta + 1) term. Now, multiply and divide the numerator and denominator by (beta + 1). So, finally, you will get {Rth/(beta+1)}+Re } in the denominator. I hope, it will clear your doubt.

[@abhijitkumarmanna2031]

@@ALLABOUTELECTRONICS got it I have also worked it in my copy 😅 forgot to delete this query 😅

[@user-tv6gx1ct1s]

i have a small doubt is self bias and voltage divider bias same or different

[@ALLABOUTELECTRONICS]

They are same. In the self biasing, we do not require another power supply for biasing the base of the transistor. Using external resistors, it is derived from the existing power supply. The same is done in the voltage divider bias.

[@user-tv6gx1ct1s]

Thank you very much for your reply got clarified

[@blisstheprodigy]

How can one find VB in this type of configuration? Is VB = VTH?

[@ALLABOUTELECTRONICS]

As I mentioned in the video, you can check the condition for the approximation. That means if Rth is much less than (B + 1) RE, then you can do the approximate analysis. And as per the approximate analysis, you can find the Vb, just using the voltage divider rule. If condition mentioned above does not get satisfied, then you need to do the complete analysis by finding Vth and Rth.

[@blisstheprodigy]

@@ALLABOUTELECTRONICS Yes, I realised that after commenting. Thank you. Your videos are so helpful

[@deepanjalimishraa]

Is Ib is equal to Ie

[@ALLABOUTELECTRONICS]

No, Ie is emitter current while Ib is base current. Ie = ( β +1) Ib. The Ic (collector current) and Ie (emitter current) are almost comparable.

[@Shibomdas411]

3:45 then why you addd VE

[@ALLABOUTELECTRONICS]

Vce is the drop across the collector and emitter terminals of BJT, while VE is the drop across emitter terminal. To find Vce, first we need to find Vc and then Ve. The difference between them will give us the drop across those two terminals of BJT. Vc = Vcc - Ic Rc and Ve = Ie Re. And the difference between them will give us Vce. That’s what is shown at 3:50

[@kamleshchavan7451]

Please can it be possible for u to remove the subtitles of ur videos. Secondly can this video will be available in Hindi

[@ALLABOUTELECTRONICS]

You can turn off the subtitles in the KZbin video settings.

[@kamleshchavan7451]

@@ALLABOUTELECTRONICS thanks sir

[@malik-fl1yl]

4:15 1 8:35 2

[@yee-haa]

I am sorry. I still don''t understand how to choose R1 and R2 values in practice when designing a circuit. There must be a more practical way...

[@i.mandeepgoyal]

16:00 pe relation galat likh diya

[@a.keerthikiran7970]

ic=1.3mA,,,,,,,,,vce=6.02v

[@sankhadeepdatta9374]

Sir...i have a query.... could you please provide your mail id...

[@ALLABOUTELECTRONICS]

It has been already mentioned in the about section of the channel. You will find the details of all social handles and email id.

[@rakeshkundu]

Thode acche se explain krdo sirf here here na bolke

[@sutapasaha5615]

Informative but way of explaining should be improved... this could be more interesting if u would not speak like a robot .

[@PreetiYadav-ng1kw]

Thanks sir