The timestamps for the different topics covered in the video: 0:27 Advantages of Self Bias Configuration over Fixed Bias Configuration 3:04 DC analysis of Self Bias Configuration 7:42 Example 1 12:03 Example 2 17:37 Example 3 (For practice)

your explanations are almost equivalent to class ones. i wonder how they have so less views considering your are one of the most genuine channel that I've discovered especially useful for the ones who missed their class lectures or want to clear up his/her basics. Anyways, really very helpful videos sir. Mera semester ap k wageh se hi clear ho paega so thanks a ton !!!!!!!!!!!!!!!!!!!

Answer of Example 3: Vgs=2.112V Id=4.2mA Vds=-4.5V It is in the saturation region.

Best online channel for electrical engineering . Keep it up

Great video simple and straight to the point!

What a dedicated teacher you are sir.... Great teacher..

Nice voice and teaching capacity.we should support these videos

Kudos. Thanks for making these videos.

In example 2, the validation for the p-channel Jfet operated in saturation should be |Vds|>= (Vp-Vgs) right??

V_gs = 2.106 V, I_d = 4.212 mA, V_ds = -4.47 V . This also works in the saturated/active region because | V_ds | >= V_p - V_gs is satisfied.

thanks..it's very useful

Very nicely explained 👍

Great video sir 👍👌💚

What we should do, if the jfat is not working in saturation?

For example 3 I got : Vgs = 2.106v Id= 4.212mA Vds= -4.45v And it is in saturation region Are these values correct? And thanks a million for your great explanation.

@chavanabhinay7743

3 жыл бұрын

yes buddy correct

@kimverlycasem3398

3 жыл бұрын

Isn't Vds = -4.47V?

@fadltac761

3 жыл бұрын

@@kimverlycasem3398 you are right .Vds must be - 4.47 but as you know very small differences

what is the answer for example 3?

sir please complete the biasing syllabus. and also mosfet moscap cmos lessons

How that Vds greater than or equal to Vgs - Vp comes???

Better than my prof fr fr

Im confused for the id formula for self biasing. If Vgs= Id×Rs. Original formula for Id is Id = Idss ( 1 - Vgs/Rs)^2. Then why in the formula after substitution, Id = Idss(1-Id/Rs)^2?

@lololo2270

4 ай бұрын

Yep bro is skipping tons of steps or maybe he is just yappin nonsense

I can't solve the quadratic equation part. Any help?

@surajitroy_roll-5023

3 жыл бұрын

Apply Sridhacharya's formula

How to solve VGS without ID?? I need helppp pls

what about ac analysis ?

Subtitles are blocking the diagram.

Will you be posting example 3 practice solution, thank you.

@jashwantsingh3907

4 жыл бұрын

I= 4.21mA Vgs= 2.10 Vds=-4.4

Why is vgs is ID not IDxRS? In solved question

Sir, we want a video also on voltage divider bias configuration of JFET.

@ALLABOUTELECTRONICS

4 жыл бұрын

will try to provide it soon.

i am confused in the polarity of gate terminal in the case of example 3 and example 2 in both cases polarity of gate is positive why it is so as the polarity in example 2 should be of negative as gate terminalfor n type j fet should be negative for reverse biasing if anyone can explain please tell me i ll be thankful to him / her

@ALLABOUTELECTRONICS

Жыл бұрын

In the example 2 and 3, there is a p-type JFET. For p-type JFET, Vgs is positive. I hope, it will clear your doubt.

Sir, can you make a tutorial on Phase Locked loop?

@ALLABOUTELECTRONICS

5 жыл бұрын

Soon, I will make a video on it.

Ans of the last question: ID= 4.212mA , VGS=2.106V, VDS= -4.47V, And this JFET is in the saturation region.

how is JFET used as a linear amplifier, where will the AC input be connected ?

How did you come up with 16.5 IB?

@ceyhunrehimli4685

Жыл бұрын

he multiplied the both side of equation with 36/8 and then rebuild the eq again

Sir, subtitles are blocking the circuit.

Vgs = -- Id x Rs not Vgs = Id x Rs, Vgs is positive with respect to ground, Id x Rs is negative with respect to ground.

why I got Id for first example 6mA?

@mayaisa6087

3 жыл бұрын

me too..where did he get the 16.5 Id...I think it should be 12Id

@muhammadsolehin8064

3 жыл бұрын

@@mayaisa6087 😆😆😆

I'd = 4.2mA, Vgs = 2.1V, Vds = 4.5V

While solving why don't u consider units of Id and Vgs I mean one is in mA ani other was Kv

@ALLABOUTELECTRONICS

3 жыл бұрын

Yes, but the multiplication of mA with kΩ will be in V.

13:53 its IdRs sir, u missed Rs

sir around 9:00 Vgs = -ID x Rs but in the equation u only provide ID for Vgs where does Rs go?

@tuchau5615

Ай бұрын

Vgs (V) = - ID(A) x Rs( Ohm) Rs =1k (Ohm ) Vgs (V) = - ID(mA) *10^-3 * Rs (Ohm)= - ID(mA )* 10^-3*1*10^3= -ID(mA)*1(Ohm) =-ID Because prof uses mA for ID.

9:01 , VP is -6 V, why become 6 V ?

@ALLABOUTELECTRONICS

3 жыл бұрын

Vgs is also negative. Since both Vgs and Vp are negative, the eventual sign in the equation after putting the values remains negative. I hope, it will clear your doubt.

Can someone tell me how did he got the 16.5 Id in the quadratic equation ex.1...I think it should be 12Id...

@ALLABOUTELECTRONICS

3 жыл бұрын

At 9:18, just multiply L.H.S and R.H.S by 36/8. And further, by moving (36/8) Id on the other side, you will get it 16.5 Id.

@agnesasopaj

Ай бұрын

@@ALLABOUTELECTRONICS but the value of Idss is 8mA not 8A

@ALLABOUTELECTRONICS

Ай бұрын

@@agnesasopaj Yes, that is why the final answer Id1 and Id2 are also in mA.

9:32 Sir, how did you get 16.5 Id in the first example? Can someone enlighten me?

@ALLABOUTELECTRONICS

3 жыл бұрын

At 9:18, just multiply L.H.S and R.H.S by 36/8. And further, by moving (36/8) Id on the other side, you will get it 16.5 Id.

@kazuto2151

3 жыл бұрын

​@@ALLABOUTELECTRONICS Thank you! Now I understand :)

@crackdoh

Жыл бұрын

@@ALLABOUTELECTRONICS How?

@user-vg9qc7dd9g

6 күн бұрын

@@kazuto2151 how?

Where is vedio for voltage divider biasing of JFET?

@deepikabirthare6622

4 жыл бұрын

I am unable to find it

@ALLABOUTELECTRONICS

4 жыл бұрын

It's not covered yet. But it will be covered in the coming months. ( The remaining topics of FET)

@deepikabirthare6622

4 жыл бұрын

Thanks a lot!!😀

how did he get up to ID1=13.91?

at around 9:07 Ugs shouldnt be -Id*Rs? Rs [Kohm] also is needed for Idss [mA]

@ALLABOUTELECTRONICS

5 жыл бұрын

Rs is 1 kilo-ohm. So, it is already taken into account during the calculation.

@noweare1

5 жыл бұрын

@@ALLABOUTELECTRONICS Please show us how it was done. Maybe you could explain in the community section. Thanks.

@ALLABOUTELECTRONICS

5 жыл бұрын

Hi, I think many of you finding it difficult. Sol, let me explain it here. Here, Id and Idss are in mA. Rs is in kilo-ohm. Now, the equation is Id = Idss ( 1 - Vgs/Vp)^2. Here. Vgs = - Id x Rs And Vp = -6V And Idss = 8 mA. So, it can be written as Id (in mA) = 8 mA x [ 1 - ( - Id ( in mA) x (1 kilo -ohm)/ (-6V))^2 Now, mA and kilo-ohm is V. And in the denominator there is volt. So, overall term will be unitless. So, overall it can be written as Id ( in mA) = 8 x ( 1 - Id /6)^2 I hope it will clear your doubt.

@chile_en_nogada2090

4 жыл бұрын

@@ALLABOUTELECTRONICS you are stupid for leaving this out of this video.

@jabirnalicho6218

4 жыл бұрын

@@ALLABOUTELECTRONICS i got it thanks a lot!

How do you simplify the quadratic equation i dont understand

@noweare1

5 жыл бұрын

Good Question. I have the same.

@ALLABOUTELECTRONICS

5 жыл бұрын

Hi, I think many of you finding it difficult. Sol, let me explain it here. Here, Id and Idss are in mA. Rs is in kilo-ohm. Now, the equation is Id = Idss ( 1 - Vgs/Vp)^2. Here. Vgs = - Id x Rs And Vp = -6V And Idss = 8 mA. So, it can be written as Id (in mA) = 8 mA x [ 1 - ( - Id ( in mA) x (1 kilo -ohm)/ (-6V))^2 Now, mA and kilo-ohm is V. And in the denominator there is volt. So, overall term will be unitless. So, overall it can be written as Id ( in mA) = 8 x ( 1 - Id /6)^2 I hope it will clear your doubt.

@noweare1

5 жыл бұрын

@@ALLABOUTELECTRONICS Thanks that makes sense.

@curlyterise8469

5 жыл бұрын

How did you get Id^2 -16.5+36 from 8/36(36-12Id+36Id^2)?

@matejgalesic5982

4 жыл бұрын

@@curlyterise8469 i have same qouestion here :/

Time 9.04 ma vp =-6 tha tuna +6 kyu liya , aur.fir.13.55 pa vp=+6 tha tuna +6 liya Matlab dono jagah +6 kyu kaisa???

@ALLABOUTELECTRONICS

29 күн бұрын

At 9:02, Vgs = -Id *Rs (Where Rs is 1 kΩ). And Vp is also - 6V, so the minus sign in the numerator and denominator was canceled. That's why +6 is written. I hope it will clear your doubt.

9.18 is quadratic equations' subtitution correct?

@dinithilakna9846

3 жыл бұрын

anyway final answer is correct if drain currents

9:06 You state that Vgs = -IdRs yet you only substitute Id in for Vgs. Where does the R go?

@abhilashasaini8758

4 жыл бұрын

R=1kohm so he didn't mention it.

Your work is sketchy...how do you get the quadratic expression ? Don't assume everybody in here knows ..plz do explain how

I apologize for what I am about to say but you skip over some fricken important steps. Thanks for wasting my time. Like maybe I in the wrong but why do u say Vgs = -Id*Rs and then you just out Id in to subistute for Vgs..... I dont understand.

@chile_en_nogada2090

4 жыл бұрын

@firstname lastname thats not the point.

@chaimaachocho9472

4 жыл бұрын

Cuz Rs = 1 k ohm

Hi anvil

I can't solve the quadratic equation part. Any help?