Emitter-Bias Configuration (Solved Problem)

[@94D33M]

8:20 if understanding potential points seems difficult...just use intuition....take voltage difference between Rc and divide by current to get Rc...you'll get the same equation that you got using KVL

[@murtazabugti8305]

sir in C part why not write the IERE when u apply the input loop ..

[@dikukhanikar2837]

You can take it also...but it will just cost you a lengthier calculation process.

[@sizan2905]

6:54 he says start and end the loop with 2 potentials

[@maryjoymorzo]

At 4:40 is it not possible to use VBE instead of VB in finding RB?

[@tanmaymaloo3220]

I think you forgot the the condition of emitter bias in the example about Rb

[@rishabhgoyal908]

In input loop there is Re also where it is gone in the equation. @neso academy

[@Prakash-fh7ic]

sir to solve Rb we need to take kvl for whole loop including Ie*Re also but you didn't consider that...so please correct it sir ..remaining everything is greatttttt…..

[@soubhikdutta9642]

Bro we know the potential of point B i.e Vb so by general method u can do it ...it was unnecessary saying KVL...since u don't require a loop here

[@nirneetchanoria]

Sir Do u provide written notes of the same in pdf or any similar format? This may save us the time of noting down everything. Kindly let me know. Regards

[@arrabalimaz622]

Apn like kiya wha kya attitude hai bro bahut acchey maja aa gaya keep it up. .......

[@ameykhairnar5582]

Why is Vbe kept constant in F.B. Active region of transistor? Why Vcb isn't?

[@bashir7714]

what will be the changes in the equation if we give negative potential at the emitter instead of making it grounded?

[@ghost59066]

hello sir.i think at 5:28 Ie should have been Ic+Ib or not . maybe you neglected Ib ?

[@gurpritsingh5125]

sir g Ib is so small ,so that u can neglect it,so thats why we generally take Ie=Ic

[@drammeh1209]

Tamam... Am really get the stuff now... Thank you very much... May God bless you...

[@UECAshutoshKumar]

Thank you sir ❤️

[@xmartyplayer8492]

I got all my answers approximately right

[@prathumnaradhakrishnan837]

In subdiv(C) why we r using Vcc-Ib.Rb=Vb. why not Vcc +ibRb+Vbe+IeRe=0 eqn?

[@shravanbhilare8487]

In previous video you used the kvl for input circuit and the loop ended till ground point..... But in this problem the input loop ended at Vb it's how it is so.....its very very very confusing.... Please reply... 🚫🚫🚫🚫🚫🚫🚫⚠️⚠️⚠️⚠️⚠️⚠️⚠️⛔⛔⛔⛔⛔⛔⛔⛔⚠️⚠️⚠️⚠️⚠️⚠️🚫🚫⛔⛔⛔⛔⛔⛔⛔🚫🚫🚫🚫⚠️⚠️⚠️⚠️⚠️

[@bojanalasamrat5262]

hi

[@bojanalasamrat5262]

🚫🚫🚫🚫🚫🚫🚫⚠️⚠️⚠️⚠️⚠️⚠️⚠️⛔⛔⛔⛔⛔⛔⛔⛔⚠️⚠️⚠️⚠️⚠️⚠️🚫🚫⛔⛔⛔⛔⛔⛔⛔🚫🚫🚫🚫⚠️⚠️⚠️⚠️⚠️

[@adodiss-hz6tj]

Sir how do we know if we should consider IE nearly equal to IC or equal to IC+IB

[@Cybernetic1]

sir ur great teacher

[@divyaneesarwa6466]

Sir how could you not taking the voltage drop across emitter while applying kvl in input loop..

[@jyotiprakashray5976]

He is taking...V-be is that only...which is 0.7 for Si & 0.3 for Ge

[@aryankumar87771]

he is taking Vb as the point potential ....even though the circuit below could be used ....but we don't know the value of Re thats why we didn't proceed as you thought we should...if anyone still has doubt just comment.

[@31hitesh38]

@@aryankumar87771 sir in c part equation should be 12-IbRb-Vbe- IeRe= 0 agar ham derivation ke according kare. But sir ne 12- IbRb=3.1(Vbe) tak hi likha. Why??

[@abusinalnil8001]

Thanks dear; But is it Emitter- feedback bias or emitter bias?

[@anupamkumar-bb2yi]

In the previous video it is said that Re value should be high compared to Rb. However, in the problem, the obtained values of Re is very less compared to Rb. Can somebody advise on this please?

[@alterguy4327]

Thank You

[@shravanthipabba8122]

In C part where is Vbe?

[@belongstozorax4640]

watching this video in 2020 feels like TENET.

[@moabeloprince6854]

HOW DO WE THEN CALCULATE THE CURRENT ACROSS EACH RESITOR?

[@sltiktok6952]

Ohms law

[@dsp6742]

Sir, which book is this sum from? By the way, excellent approach. Loved how you started from the end.

[@vincentblaze8424]

It was just common sense so if you find such an obvious things surprising then you should probably get your IQ checked

[@sizan2905]

Rc = 1.41 kilo ohm or only ohm??

[@tinotendastasha1210]

On the (e) part why did you add to get VB

[@louisanthonybernante6706]

because to get Vbe we Vbe = Vb - Ve since we already have Vbe transpose -vbe to the other side so it became +vb , now we have Vbe + VE = Vb

[@josegiboyeaux5942]

to get d) why did you use VB instead of VBE in the equation?

[@bhavyanandwani3282]

If you use Vbe then also use Ve ...

[@santhoshthota4778]

Just awesome

[@jithin2107]

thanks a lot sir...!!!

[@ArmanKhan-dl8wb]

How can Ie =Ic.??

[@sunitapoptani1372]

We know Ie x alpha = Ic , but value of alpha is universally very small & neglecting it we get Ie = Ic

[@TheZohan933]

Good evening....how can you just ignore the I subscript e and R subscript e in the input loop when applied kvl???

[@shubhamsingh6884]

suraj singh yes please reply +neso academy

[@AkshayKumar-dz5ts]

look at his other lessons on how to perform proper calculations across two potential points you can have a kvl equation between any two potentials ,it doesn't have to be a full loop

[@shubhamsingh6884]

Akshay Kumar Dude kvl and kcl can only be applied in loops (I.e if there are points a->b->c->d then you can't go to d from a without going through b & c)

[@AkshayKumar-dz5ts]

shubham singh did you watch his other video where he explains how to solve these kinda things?

[@swesh6657]

Actually What the Tutor did was he just applied KVL between two Potential Points +12v and Vb....

[@rakeshnaidu7115]

Sir you violated the rule that BRe>>Rb

[@abelk1451]

From my understanding, VCC - (RB * IB) [12 - (273.3k * 0.0000375] should give you VB. But it gives 1.75V, which does not equal (VBE + VE) [0.7 + 2.4 = 3.1V] which likewise should give you VB. Furthermore, if you take KVL of the entire input loop, (VCC-IB*RB-VBE-VE) (12-10.25-0.7-2.4) = -1.35, but does not equal zero, hence KVL does not hold. So, if my calculations are not incorrect, then perhaps there is a problem with the question.

[@shriganeshayenamah3422]

Bhai sir ne solve shi kiya h Bas explain galat Kiya...... Waha pr Ohm's Law use Kiya h..... Kyuki KVL sirf full circuit pr hee laga skte h..... Wo Vcc-Vb=Ib×Rb Hai....... Resistance key across voltage

[@sandeepreddyvutakanti1189]

Sir how you can say that ic=iE

[@naduriba7032]

i think question gave too many known quantities

[@amoghaarrajashekhark2112]

c part answer is false, you have used fixed bias configuration formula instead of using emitter bias configuration

[@mahmmadhusen6794]

Exillient

[@lifelines9547]

sir notes upload kr do

[@thewaveclasses7740]

please make video on z transform

[@kundurupavan1429]

#neso_academy

[@kundurupavan1429]

#nesoacademy

[@demetriusdemarcusbartholom8063]

Thank you