Hello students now we have already seen different types of biasing circuits using BJT they are fixed bias and collector to base bias but there is another very powerful circuit of biasing which is called as voltage divider bias now first of all we will come to know why it is called as voltage divider bias voltage divider means there are arrangement of the resistor which will divide certain voltage okay so circuit having arrangement of registers which is dividing voltage is called as voltage divider bias let’s see the circuit now you can see that this is voltage divider bias circuit in which resistors r1 and r2 are called as biasing resistors which are used to divide voltage of VCC across base or getting these point students yes resistors r1 and r2 are called as biasing resistors which will divide voltage of VCC and base of the transistor and hence this circuit is called as voltage divider bias circuit now again whenever we want to find out Q points what we do is we will apply three steps step one is apply cable two input step two is find out value of IC and step three is apply KVL to the output but in this circuit there are two resistors at base R 1 and R 2 and hence first of all we need to find out only one resistor at the base and how to achieve this this is question mark so I know there are div there are different DC analysis theorems or you can say that DC theorems which we have studied in b-double one of the most important theorem was convenience theorem it used to say that we can resolve any part that is number of resistors and voltage source at any side of the circuit with the help of only one resistor and mono voltage source in series I’ll repeat if there are combinations of the resistor and the voltage then that can be resolved with the help of arrangement of one resistor in series with voltage source which is called as V th so what we try to do is we will try to resolve this spot with the help of Thevenin theorem so I’ll write that apply Dominions theorem so in order to apply the minions theorem or to find out Vth I’ll write V th is equal to now the V th can be easily calculated as VCC multiplied by r2 divided by r1 plus r2 or getting this point yes B th that is convenient equivalent voltage is equal to VCC into r2 divided by r1 plus r2 hence I can write V th is equal to VCC multiplied by r2 divided by r1 plus r2 similarly convenience equivalent resistance is equal to r1 parallel r2 so by applying Dominions theorem I have got Vth and rth now the next task is to replace base side with the help of Thevenin equivalent circuit so I’ll do that now so convenience equivalent circuit can be drawn as follows the remaining part of the circuit will remain same only change will be at input side where we are replacing entire portion by a single resistor in series with voltage source this is how we can replace bass side with the help of convenience equivalent circuit in which a register is in series with V th now this voltage can be VB whereas this voltage is nothing but we see current from this will be current from the base will be IB from this it will be IC and from here it will be IE now we can apply different steps step one to find IB for which apply KVL to input that is base to emitter so equation will become first of all I’m getting VT s so it is V th minus IB into R th minus VB minus ie into re is equal to zero but we can write I is equal to beta plus one times of IB and hence the equation will become IB into R th plus beta plus one times of IB into re is equal to V T H minus V B and hence IB is equal to V T H minus VB divided by RB plus beta plus 1 times of re where this RB is the thing but convenience equivalent circuit which we will mention in the bracket where RB is equal to R th some of the books will follow notation as RB and some other books will follow notation as rth now next is step 2 in which we are supposed to find out collector current so step 2 is to find I see where IC is nothing but beta times of IB and now step 3 is to find we see for which apply KVL to output loop hence equation will become VCC minus IC into RC minus VC minus IE into re is equal to 0 where IC is approximately equal to ie and hence equation we can rewrite as VCC minus ICRC minus VC minus IC into re is equal to 0 and hence VCE is equal to VCC minus IC into bracket RC plus re so by applying three steps step one step two and step three we can easily find out Q point in voltage divider bias circuit Thank You students

thq Sir. I like ur teaching, i have understand a lot. thq so much. 😊😊😊

useful and simplified concept 👍

use full is every students

Really nice video

THANK YOU

nicely explained sir . but what is the use of capacitor C(E)?

Nice explanations……..

sir please explain about rc coupled amplifier

Very nice explanation thanks sir

this YouTube channel is best

Very useful Sir. However, you made a slight mistake in the evaluation of IB at min 6:23 as you equated it to

(vth-vbe)/(RB + (B+1)RE) instead of (vth-vbe)/(Rth + (B+1)RE). Basically RB instead of Rth. Thank you.

6:38 we know that if sum number will changes its place to come another side of equal to then the sign will change but here it is not.

thankyou so much….unexplainable in words the feeling…..i got it totally(the concept)……thank q so much….god bless you ekeeda

Best teacher on Youtube presently

Super teaching

SuperB

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thank you Sir <3