Load Line Analysis (Numerical Problem)

[@abhishekjain3344]

For all those wondering how Idq = 9.2, just put Vd = 0.8 in the equation of load line. The equation of load line is Id + Vd = 10 (Id in milli ampere).

[@SathvikReddySL]

Vb*

[@rishabsud7535]

Why are we taking the value of y axis first in Q point? We shd have (0.8,9.2) according to the conventional solution.

[@kamleshnehra407]

Ha bhai aise hi hoga , galti se mistake ho gaya bhaiya se 😝

[@swagatpalai415]

Same doubt.

[@facedekiadam2916]

I think he did not show the full solition here. You should use iteration method for the equation to match each other. I guess when Id =0.8 and etc. The Id equation is 0.8 = I0 x (e^(Vd/Vt)+1) this satisfies the output with Vd = 9.2

[@harshpratapraghav2342]

@@facedekiadam2916 how did u go Is as 10?

[@narasimha22]

Sir generally we right x axis value first then why we are writing ID value first for Q point

[@alterguy4327]

Commenting Watching the ad☺ Thats the only way i can help u

[@chathushkajayashan2603]

sir, barrier potential of this diode is 0.8v, without load line can't we just find the v(r) by 10- 0.8? can we use this if there is no constant voltage drop?

[@gamar1226]

I believe not. It worked here because the slope of characteristic curve is infinity (linear )so it wouldn`t matter where the load line would hit because the value of VD is the same.

[@ramubhandari5907]

thank you so much sir...from the bottom of my heart.

[@SohamChakraborty42069]

The explanation is so simple and good that I can understand all this as a 1st-year student of engineering.

[@rollercoaster9719]

Bro I have seen you in Mohit tyagi sir lectures comments , now I see u here lol? BTW which college ?

[@Vernacularpressact]

pehle saal me hi kaam aa gya kya

[@Josuke217]

@@rollercoaster9719 he's in iit kgp

[@TechnicalCEO]

Best of the best concept clearing video

[@mwansachikonde4193]

Really good explanation, I love this. Question; What exactly does the q-point mean? What is it used for?

[@chayandas8204]

dc operating point

[@bintezafarimam6739]

Ur way of teaching is very nice 🙂🙂

[@syedahamed7334]

In the previous lecture u have said that for solving numerical problems, we hav to use constant voltage drop model for diode equivalent circuit. But u used ideal equivalent circuit model...??? Can u please explain that am i asking the correct question?

[@yashjoon3889]

Did you got the answer.

[@micekey8401]

He is using constant voltage drop model not the ideal one.

[@i_m_hk]

Excellent ssir🔥

[@prithvirajsingh4150]

sir please clear one thing when Vd=.8,then how u write in the axis of ID=9.2

[@xtinee]

How did u get the ID of 9.2? Will u just assume it?

[@SamriddhaDhali]

just use the equation from KVL and put Vd value 0.8. you will find 9.2ma

[@UrlOn2Wheels]

Your Id is equal to 10 and your Vb is equal to 0.8, so 10-0.8 = 9.2

[@sumitsouravamohanty8935]

are graph se pata lagana he

[@xtinee]

Thank you for replying. Even though we've been past this lesson a few semesters ago, it's still nice to know how it's actually done.

[@alterguy4327]

Jem Quizon ☺

[@harshitgupta4169]

Sir here if we have written Qpoint(0.8,9.2) then is it correct or not.

[@20EUEE104-YOGESHWARANS]

crct

[@Skarabacak]

I have a queistion because really I didn't underistand.Well, how did you find IDQ=9.2ma? only kvl? 10-0.8=9.2V IDQ=V/RL ?

[@Cosmiclights6666]

Sir isme VR find krne time Id ka value kyu nh liye IdQ ka value kyu liye??

[@Josuke217]

Extremely helpful , thank you sir.

[@merajuddinansari7643]

Thanks man😎🤘

[@mhsrinath9069]

Id value is 10*(-2). how is it 10mA?

[@dhirajsingh10710]

1 Ampere= 1000 mA . So converting Ampere value into mA = 10*(-2) × 1000 = 10 mA

[@salmansyed9962]

Instead u use 1k ohm as it is without converting it to ohm... It'll give you answer directly in mA

[@syedahamed7334]

Id =10×10^-3 Id=10mA

[@kirtikansal6946]

How without graph..we can calculate the quiescent point values

[@smruti6684]

Is graph given to findout Q point from graph ... If don't what has to do ?

[@rishukumar8151]

He represented q point as (I dq,V dq)..if I am not wrong then it should be like (V dq,I dq)....I mean points represented like (X,Y) so how he can represent I dq first in coordinate. Anyone can help??

[@akashsuklabaidya8196]

He did a mistake there....

[@raginikhandelwal4633]

Sir, Why quiescent current is equal to diode current??? which concept you have used for this??

[@juliuscaesarengracia6760]

if u notice the circuit, it's in series. current in a series circuit are all equal

[@brandonfd789]

Could also take your 10v-0.7 and determine the voltage across the resistor since you know a voltage drop of 0.7v is across diode.

[@supreemh5026]

What is the significance of operating point

[@sneharajpoot8377]

Q point coordinates must be (0.8,9.2) according to my best knowledge.🤔 Is it mistakenly written or is there any other logic????

[@PANKAJKUMAR-ob4hr]

I also confused at there..!!

[@muhammedidid4823]

Hello sir, why do we need operating point

[@azzam9x]

or we can say Vr = V - Vd -> 10 - 0.8 = 9.2V BTW way won't you just use silicon * 0.7?

[@dhineshkola8691]

please upload lectures on IGBT S

[@SamriddhaDhali]

excellent sir.... thank you so much..

[@parchitmalhotra3738]

sir plzz upload videos of topic " SMALL SIGNAL ANALYSIS OF MOSFET"

[@dheerajkumarraghuvanshi532]

In the Circuit diagram you are representing diode voltage as V sub D... isn't its same to barrier potential that is V sub b

[@SamriddhaDhali]

no its not same.. Vd consists of Vb+id*Rd and after kvl the actual KVL (taking Vb under consideration) V=iLRL+Vb+idRd..

[@avinayakumar626]

Thanks sir

[@lhlhlhlhlhlhlllililIIIll]

Brilliant

[@sravaniishika422]

Sir please share the diffrentioal amplifiers and current mirrors chapter vedios

[@souradipkumarsaha9267]

sir the coordinate points are reversed i.e (x,y) is mistakenly reversed as(y,x)

[@neerajgupta-kt7uj]

pl illustrate q point, by giving practical example

[@mdbarjahan6695]

awesome & thanks a lot!

[@sriramdr6939]

thanks

[@Shoron09]

Why you are putting the value of y axis first sir(in the co ordinates) ?

[@davidcheung4987]

I am studying Engineering too and here is my interpretation. The graph represents the behaviors of the diode. The video makes use of the KVL to find the relations between VD and ID and plot the line onto the diode graph.SO you can imagine that once you know the VD, You will know the ID then. and the intersection point is exactly the value of VD in this circuit. then you can find the ID by the equation ( or by observation in graph) , make use of this information can find the voltage across the RL(VR)

[@hibak8196]

Thanks a lot! Really helped :)

[@rashmikapur2801]

sir the coordinate of Q(Vdq,Idq) not Q(Idq,Vdq) a/c to P(x,y).... am I right or wrong?

[@venugopalreddy3915]

yes it should be q(Vdq,Idq)....but the coordinates of Q are Q(Idq,Vdq)...

[@saragupta1074]

why is this circuit forward biased?

[@muriirakevintalemwa2358]

why did you take 1e-3 as a negative not a positive.

[@unknownv8castle266]

thanks a lot. :)

[@bateinhamareapnoki8726]

Sir please power system and power electronics and analog complete kar do plz plz

[@brandonfd789]

Just isolate the resistor, find the current and then the draw the load line, since the voltage is known across the diode you can find your Q point by graphing both. KVL seems excessive

[@dineshkoppula359]

but in the exam they wont give vi charecteristic graph than how to find vdq and idq

[@bateinhamareapnoki8726]

Gate aur ies ke liye

[@bateinhamareapnoki8726]

Reply karna jaroor sir

[@keshavrathore2191]

Bhai ye neeche wali line hata diya karo taki hum aap jo likho wo bhi dekh sake wo faltu line chalti rahti h

[@vivekkasar4415]

who in the right mind dislikes this video, Dislike my comment cat

[@danishansari-vd8yv]

thanx sir