Voltage Divider Bias (Solved Problem)

[@ankitg2023]

The best part is the design intuition that you provide in the summary of each lecture. Thanks for these tutorials.

[@pashamshyamsundharreddy4438]

thanks sir .you are doing awesome job for the society even there are less views you are continuing to upload. thank you very much

[@shafinsarwar6448]

Best teacher on internet or maybe in the World. May Allah bless and help you all the time...

[@aryakrish4316]

really priceless content for someone like me as a electrical and electronic student .

[@fernandohood5542]

It be nice to see this applied to a practical circuit using known transistors.

[@abdullahsalik7078]

Impressive and genius way to teach! My professor in university is a Ph.D. but can't teach with this efficiency :')

[@shourayduggal9461]

sir can i say that potential drop across R2 = Vbe + pottential drop across 1.2 kohm, I can calculate the current through R2. Hence current through R1 will be 20uA + calculated current, hence apply kvl => Vcc - (I + 20uA)R1 = Vb.

[@akibjawed5076]

The difference between "with approximation" and "without approximation" is nearly 1.2k-ohm which is very large I think, isn't it?

[@josephguida5432]

The approximation method is an excellent way to reduce your final exam score.

[@riyasanthosh7124]

Thank You for your videos,they are really useful.Could you please add a presentation on MOSFETs too?

[@innocentlombanya7356]

I like the lecture... I'm understanding clearly.

[@arijit1204]

thank you Sir.. for this season..

[@RahulSingh-ey4wf]

U cleared the concept very well,Tysm sir

[@incredibleindia-withsid]

Thankyou sir for this playlist. I had a question, in the last part (f), when we use KVL instead of V(BE), what will be the equation because I always had this doubt, when using point potentials what if in the branch if we have more than 1 point potential how do we decide it's polarity?.

[@imminentfive5623]

Now I'm open thanks man kip it up nice and clear👍

[@AJ-et3vf]

Thank you very much sir! Lucid, insightful, and very understandable!

[@atharvak5748]

If you don't approximate Vth = Vb, then R1 = 26.857 KOhm

[@kv5539]

How did you find R1 .please help me

[@electro-magnetik528]

@@kv5539 Vth - Vb = Ib * Rth ---> (6.8/(R1+6.8))16 - 3.12 = 20micro(R1*6.8k)/R1 + 6.8k ---> solving this you will get finally 3256R1=87584 --> R1= 26.89kohm.

[@8xtreme]

thanks bro

[@vilayathshaik8733]

Sir for calculating Vce when we apply kvl for output loop Vcc-Vc-Vce-Ve=0 Vce=Vcc-Vc-Ve Vce=16-10.6-2.4 Vce=3v But the answer is completely different from urs pls clarify my doubt sir if I am wrong pls correct me And can we find as Vb=IbRb Sir ur reply is much needed pls clarify this

[@barathnatarajan8566]

You made my day sir

[@LuisMartinez-fj9vi]

you can also find the R1 using the value of IB

[@onlystudy3294]

as it is value of R1 =26857.91411 Ω, Vth= 3.233

[@rajkumarthakur7480]

Sir explain how you say that vth is equal to VB

[@sav7978]

why is VBE equal to 0.7V?

[@nccnitin]

Silicon

[@swethasri2276]

Thank U sir.. Please tell us which video you are uploading after transistor re model?

[@kshitishkumar5118]

Sir, can we obtain R1 with the help of ib formula ie :- ib=(Vth-Vbe)/(Rth+(beta+1)Re).......???

[@ruthwik8772]

Vcc is greater than Vb so current has to flow from Vcc to Vb .In that case Ic will be negative and also the R1 will be different.Can you please explain?

[@kamlakshimeena6042]

i think in the last part formula for voltage division is incorrect it should be VB= VTH= Vcc * R1/(R1+R2). Pls correct if i wrong??

[@aasthasinha4979]

Really helpful

[@akanshagupta1710]

why Vth = VB??

[@grandson_f_phixis9480]

Thank you very much sir🥳🥳🥳🥳

[@sayyad__md]

Helpful, thank you sir..!!

[@electronicsolutionindia3538]

Nicely explain

[@FaizanKhan-dp3sr]

You are great sir

[@babumalik7564]

Thanks for the given information....

[@kushalsiddeshwar7056]

thank you so much sir

[@themangoman5642]

how did you find R1 you didn't show the steps. please tell me.

[@ROHITKUMAR-lh3ps]

3.12=(16*6.8kohm)/(R¹+6.8kohm) 3.12(R¹+6.8kohm)=16*6.8kohm 3.12R¹+21.216kohm=108.8kohm R¹=(108.8kohm-21.216kohm)/3.12

[@jasonkimjabla1767]

R1 = (R2VCC-VBR2)/VB

[@themangoman5642]

@@jasonkimjabla1767 how soon you’ve replied to me

[@Followtheclass]

Without emitter resistance can solve thevenen voltage at base sir?

[@vinayaksood5677]

How was Vb=Vth ?

[@shubamsalaria9904]

sir , how did we get the value of vbe=0.7v???

[@lilianmaillot9449]

it's usual voltage for a "diode" to conduct, and a transistor could be seen as 2 diode back to back. So it 's not by calculation, it is only the conduction condition for a transistor to work.

[@lemuelcericos3395]

why do you consider Vth as voltage drop at vb ?

[@sandeepgouni2871]

sir if we solve with out considering thevenin circuit r1=(Vcc-Vb)/i b.which gives R1 as (16-3.1)/20micro amp=645 kilo ohm.why the answer is different?

[@Nilotpal_Chakravarty_]

It's wrong bcoz ib is not flowing through r1 it is ib+current flowing through 6.8kohm or r2 u need to use thevenin theorem

[@AmitKumar-es1gp]

Sir, how did u consider vcc =16 v and vbe=0.7v...Pls tell??????

[@antoinebadimame]

Hi I am not an expert but from what I have already seen in the video: Vcc can be easily gotten from calculating the pd drop across the R1 resistance as I1 * R1 = Vcc - Vb And lastly Vbe is a property of the B-E junction which is acts as a diode. We consider that it is a Silicon diode. And given that the Silicon diode has a barrier potential of 0.7V We conclude that Vbe = 0.7V. I hope this helps.

[@MCSGproject]

ur up electrical eng ineering shit creek amit

[@ViralShortsOfDuniya]

Amit Kumar if Vbe is not mentioned take it as 0.7V

[@saicharan9655]

Milan Samanta for silicon diode Vbe is 0.7v

[@sinchanghosal4342]

if you don't know this then go to hell. trivial stuff

[@newbie8051]

Difficult problem but cool thanks !

[@annamraghuram3713]

Is self bias and voltage divider bias are same?

[@lawrencemayfield9669]

How do you do a voltage Bias that is eing driven into saturation?

[@lawrencemayfield9669]

Voltage divider Bias in Saturation condition how do find all terminals and beta?

[@Ravi_Qant]

Sir, Ive been trying to make a single stage amplifier. By these calculations I'm getting confuse. I got whatever you taught. But when it comes to make an amplifier, I'm confuse in selection of base current. Even the law by which we free the collector current from beta is also getting miscalculate. Please help. And Thanks

[@fernandohood5542]

I have come to realize that many of these videos are not done for the real world. People forget there is a real world.

[@ekawakakonahi1174]

What’s the current thru R2 and isn’t the voltage across R2 =Vb?

[@manishramesh9344]

How did you Use Kvl in the particular you have used vcc-IcRc=10.64 how can we make the loop Like that I am confused ? Can you help me in figure it out?

[@kv5539]

Go through the before lecture that discus about the point potential.👍

[@Cybernetic1]

Sir Please make videos on OP AMP please..

[@MrBudgetneer]

good one thnk u

[@chandu1969]

sir can you make this problem video Design a Self bias circuit for following specificatins,V_CC=12V, V_CE=6V, I_C= 3mA, V_BE=0.6V, R_E=220OHMS, S=11. Find the resistance values of R_1 and R_2 and R_C.

[@naifalsefri1985]

Can u do an example where R1 and R2 are unknown ??

[@vivinraj8821]

What if base current is not given... How to find it?

[@Followtheclass]

Same procedure we follow?

[@nagabhooshanmesta9788]

Sir if vth=vb i think r1=5.16 ohms sir once it check otherwise say how r1=28ohms

[@charlesenricosalinas5368]

Is it always that Ic is nearly equal to Ie?

[@sunitparida6176]

How can I find out base current if asked

[@bibhutibhusansahoo6019]

Thank u sir

[@nickino211]

thank u, sir

[@KevinRodriguez-ip7wu]

i have a question is IC and ICQ similar? if you where to find ICQ....sorry i have lecturers that cannot explain at all if this question sound stupid.

[@ajithkumarthanakala3082]

wtf

[@sohalgate9668]

Icq is Ic at which our transistor is operating.

[@ViralShortsOfDuniya]

Thank you very Much sir🙏

[@ramnagaboina3339]

how can we consider ie=ic can u explain

[@ShubhamNahar26]

Would you plz tell how did you get V(BE) = 0.7 ?

[@Ravi_Qant]

Shubham Nahar that's fixed for silicone transistor.

[@ShubhamNahar26]

Ravi Kant i figured it out. btw thanks :-)

[@lifeandu3286]

what is the procedure if vth is not equal to vb?

[@shihabmahmood9135]

consider current through R1 is I1 and through R2 is I2. Applying KCL at Vb in the first circuit we have, I1=I2+Ib. Now, I2=Vb/R2. As we now know I2 and Ib we can calculate I1. Finally R1=(Vcc-Vb)/I1.

[@shihabmahmood9135]

This particular math can be solved without Thevenin equivalent circuit

[@94D33M]

@@shihabmahmood9135 Absolutely correct

[@mrcastor9093]

Sir, when calculating R1, we have to multiply Ohms with Volts. But that's against Ohm's Law, where U = R * I. There's no U * R ! EDIT : I've figured out that by continuing the calculations for R1, if you keep the units of measurements multiplied with each other, KOhm * Volts will be on the denominator and Volts on the numerator. By making the common factor the Volt unit, it will cancel out between the denominator and the numerator, in which only KOhms remain for R1.

[@hamzafarid6153]

Mr Castor here he used voltage divider rule not ohms law... V*R/Req

[@awaiskhan-yn8so]

Sir in our book capacitor is included Please help sir examanition is comming on head please sir

[@shriganeshayenamah3422]

Ha bhai HBTU mein bhi h

[@saikumarhalyal8337]

R1 calculate in the which formula

[@santhoshthota4778]

Excellent but how Vth =Vb at 8:42

[@thinker6655]

What if no beta value given?

[@surayakshsharma7883]

upload the amplifier part also please

[@creative_ajit_]

How to come 0.7 = Vbe

[@chetanbhatt5153]

Why not you take Vbe as 0.7 and get solution like every question.. Here too Vbe doesnot came 0.7 is there any mistake you had done🙄

[@kv5539]

How can i find R1 without approximation of Vth please anyone help me

[@elamparithiparithi3344]

Why Vbe is 0.7 volt?

[@janesalinamaris4321]

Can anyone pls explain how do we arrive at vth expression

[@manickamshanmugaraja7935]

Upload stability factor

[@asadarshad8628]

How Vcc is 16 can anyone tell me?

[@shaikshamsheer7559]

How u ggot R1=28pls tell me

[@ajithkumarthanakala3082]

can u tell me a good reference for solved problems for analog circuits..

[@kiranilyas079]

why do we take r1 always greater than r2?

[@Thisious]

Based on the equation Ve = (R2/(R1+R2)) * (Vcc), you will see that if R1 is not greater than R2 Ve will be very large and most of the circuits voltage will drop over Re. At extremes this causes Ic to lower to minimum and Vce to increase to maximum - not a particularly efficient circuit and nowhere near an ideal q point

[@JustinVarghese]

I got r1=5.117

[@All_in_one5678]

❤

[@bsbijoy]

joss.....

[@d96yt4]

NIT Agartala

[@asishswain1860]

Wow

[@lemuelcericos3395]

you doing it wrong, if you did it in kvl you will come up with a different answer

[@sajidmalik2596]

according to me it should beta +1 =ie/ib .because there is RE. so ie should be 2020

[@ehsanahmedsiyam4584]

You guyz are teaching great but why don't you guyz improve your english.

[@SimpleNishi_info]

Speak loud man

[@UECAshutoshKumar]

Thank you sir

[@chandu1969]

sir can you make this problem video Design a Self bias circuit for following specificatins,V_CC=12V, V_CE=6V, I_C= 3mA, V_BE=0.6V, R_E=220OHMS, S=11. Find the resistance values of R_1 and R_2 and R_C.

[@khalidsami5556]

thank you sir

[@chandu1969]

sir can you make this problem video Design a Self bias circuit for following specificatins,V_CC=12V, V_CE=6V, I_C= 3mA, V_BE=0.6V, R_E=220OHMS, S=11. Find the resistance values of R_1 and R_2 and R_C.

[@chandu1969]

sir can you make this problem video Design a Self bias circuit for following specificatins,V_CC=12V, V_CE=6V, I_C= 3mA, V_BE=0.6V, R_E=220OHMS, S=11. Find the resistance values of R_1 and R_2 and R_C.

[@chandu1969]

sir can you make this problem video Design a Self bias circuit for following specificatins,V_CC=12V, V_CE=6V, I_C= 3mA, V_BE=0.6V, R_E=220OHMS, S=11. Find the resistance values of R_1 and R_2 and R_C.