Collector Feedback Biasing (Solved Problem)

[@dhn9987]

In collector feedback bias, if you increase the Rc value,more voltage drop takes place, so you need some voltage to turn on Transistor ,so more VCC is required.., so cost is high

[@sankhamitra3862]

Answer to the H.W. problem is Q point= (11.53 mA,8.39 volts)
Very conceptual lecture sir with very simple words 👌🙏 you are the best..

[@riteshrk18]

✌️✌️

[@travisbickle3841]

Neso Academy,you are the channel who brought udemy to youtube for free.Thank you very much.I am ce student and I utilize yours lessons too much.I am grateful for you each day Neso Academy.

[@muaazulhassan1721]

Really appreciate your work, you always helped me a lot to cover my course in a really less time.

[@shreyasawant4604]

Thankyou very much Neso sir Neso Academy, I am very thnkful each and everyday for ur free lectures in all electronics subjects, bcoz of u only im able to get good grades in collg exams, THANKYOU SM

[@sliderulelover]

There is a formula to calculate IC from the circuit parameters (components). Here it is: IC = Beta(Vcc - VB) / (RB + Beta(RC)). Now filling in the component values: IC = 150(20 - 0.7) / (100K + 150(1,000)) = 11.58 mA. Once you have IC, you can easily calculate IB and all the circuit voltages.

[@DataWiseDiscoveries]

Ic=11.58mA
Vce=8.34
SO the operating point is (8.34V,11.58mA).

[@chinmoy5936]

Q-pt is (11.53mA,8.39V)

[@merdenyesil559]

For B=100 ;
I calculated: Ic=9.6mA Ib= 96uA and Vce=9.44 V

[@prasannabiswas681]

Q(11.533,8.39)
thank u very nice explanation :)

[@makindeolalekanmonsuru5349]

The beta* Rc is not far greater than Rb so as a result change in beta shift the operating point to the left, hence no stability. Ic~11.53mA, VC~ 8.4V

[@UECAshutoshKumar]

Thank you sir ❤️

[@nandakumarg2320]

sir vce should be in the x cordinate and ic should be in the y cordinate not the reverse.is that right?

[@victorwainaina2584]

Unfortunately, you have not.

[@mohammadlafi4331]

I think he meant here :
th-cam.com/video/o5qiPytBEBQ/w-d-xo.html

[@RAGESOUL-js4rc]

Base current=77.2 microA
Collector current=11.58mA
V ce =8.3428V

[@sumanthchevula1601]

for collector bias is'nt the condition should be rc>rb

[@tannkyesha1573]

The condition is β*Rc > Rb

[@pranjalsutradhar_IITian]

HW ans : Q-pt = (Ic, Vce) = (11.533 mA, 8.390108 V)

[@waferlayout]

Much appreciated Sir 😊

[@yashpatil2536]

Thank you so much for all your videos it helps a lot :)

[@ghost59066]

VCE=8.47 V
Ic=11,53 mA

[@ABHISHEKKUMAR-.]

Vce is y coordinate of operating point???

[@steevenkenny9791]

thanks a million sir

[@sindhubhargaviveldhi4116]

Q-pt : (11.53mA 8.39V)
thank you sir ..,

[@demetriusdemarcusbartholom8063]

Thank you

[@murphysun4087]

Thank you sir

[@nkyampz]

Ic= 11.5mA
Vce = 8.43v

[@tulakchowdhury4906]

Homework Problem
Ic= 11.533mA
Vce=8.4661V

[@Bremontval]

Hello
Thank you for the video
How to calculate the voltage amplication and power amplification of this circuit ?
Thank you !

[@imskr2683]

(11.5338ma,8.39v) am i right plz tell me.

[@VaswataRanjan]

Perfect, I got the same values bro

[@vivekkeviv5573]

Q(8.46V,11.53mA)

[@Biswajit_Sahoo1]

answer to the H.W problem- Q point=(11.5 mA, 8.41 V)

[@naimafatima8163]

Q-pt : (11.53mA 8.39V)

[@xiaoshan9722]

Q-pt = (11.53mA,8.51V)

[@ankitvishwakarma4070]

11.4ma,8.524v

[@greyshadow37]

For Beta = 150
Q point = (11.4 mA, 8.524 V)

[@venkateshmamidi8407]

IcQ=11.53 mA
VcQ=8.393 volts

[@swapanghosh7550]

Ic = 11.4 mA
Vce=8.54 v

[@agaelema]

VCE = 8.46V
Ic = 11,53mA

[@sivajigadidala]

(11.58 , 8.34)

[@krmfahith7975]

Betta = 150
Ic = 11.534mA
Vce = 8.389V

[@victorwainaina2584]

Why the feck are people posting different answers and not explaining why?
Edit: It's because he asks a question at the end of the lecture, and asks people to post answers in the comments....goddammit.

[@AshishGusain17]

Victor Wainaina u are living in your own world

[@30_sandeepsaurav73]

F***k you

[@jayvarshney2171]

Vce=8.32v
Ic=11.6mA

[@arjunfitness3836]

Ic=11.53ma
Vce=8.39v

[@sushmareddy348]

Ic=11.52mA
Vce=8.4035v

[@thecodehubchannel1661]

Ic=11.533mA
Vce=8.391V

[@victorwainaina2584]

So did he do something wrong? What's up with what he's done? I think you're all talking about the loading line coordinates, right? I hope so.