Load Line Analysis of Diode

[@CYTOTIMUS]

The fact that I can get these amazing lectures for free is mind-blowing. This is the best electronics tutorial in the while internet.

[@kirtikansal684]

best source of analog on the internet...even books cant clear such concept... hats off

[@theuntormented6560]

andha hai kya ..andar khali hai kya.. vo basic coordinate geometry dung se kr nahi paya

[@saranshgupta9119]

@@theuntormented6560 Sweg lewel 100🤣🤣

[@sahanuratul5049]

The intercept points of both axes are given wrong. In x-axis, intercept point is (v,0) and in y-axis, intercept point is (0,Vd/RL)

[@AkshitSharma0]

right

[@sanjanar3394]

Correct bro

[@sujaysb999]

Correct bro our rnsit college teacher has saw this video and thought the same in our class in doubt i came to see here but i was totally confused😂

[@Yashodhan1917]

Thank you so much man, you are saving my semester!

[@rajneesh8960]

Mine too

[@Yashodhan1917]

@Hanzla voice me?

[@anshdalal3123]

@@Yashodhan1917 yup

[@KongreHackers]

You messed up the coordinates at 3:30 :p

[@JH09SUMIT]

I was looking for this comment 😁

[@cinemafx1909]

They are not coordinates but value in the form of (I, V)

[@theuntormented6560]

@@cinemafx1909 they should be in the form of (v,i) as voltage is X coordinate and current is y coordinate :X

[@vasudhanagesh5445]

Yea..

[@nehasiriwardane9926]

@Andreas Lorraine Why would you hack into someone else's account? Don't you have respect for privacy? 😡😡

[@jyotasorout3100]

Bhut km लोग milte h इस दुनिया m इतने अच्छे thnku so much sir.

[@pavansharmajanapati9687]

Mathematical analysis is ok ... apt reasonable Explanation of how a concept is used ... can be told ..

[@harshvadhanas1603]

I recommend this to my all friends

[@harshvadhanas1603]

this is eventually the best academy for analog electronic

[@sunidhimonga7219]

The points on y-axis when VD= 0 should be (0,V/RL) not (V/RL,0) and similarly for x axis , it is (V,0) not (0,V). Plus Q has coordinates (VDQ ,IDQ). Why do you mention (y,x) ? Coordinates are written as (x,y)...

[@bonganikirankumar9688]

Exactly

[@vishvajeetyadav818]

Bs yhi check krne k liye maine comment open kiya Mujhe bhi lga wrong h

[@pranaygoud3850]

thank you for doing this ...neso academy ... +helping me a lot in my semester exams .

[@saket_saurav]

sir please tell something about the significance of operating point, how the change in Q point changes the different quantities ?

[@SameerKhan-xq3he]

Still looking for answer?

[@naveennataraj1578]

For the given voltage applied and the given resistance we can tell that I(DQ) flows through the diode and V(DQ) is the potential across it. The graphical analysis is important because we are not assuming any approximations

[@3xf250]

See... Q point tells us about the current scene of a diode... like what is the voltage across it and the current through it. So, if the Q point is in the 1st quadrant then we know that the diode is in forward bias... and similarly if the Q point is 3rd quadrant then the diode is in reverse bias. Additional note: Q point stands for quiescent which means non varying or stationary point.

[@smooooth_]

@@3xf250 So it's similar to the purpose of the Vx = Rx(V/(R1 + R2)) equation? In that you're finding the value of that component given the specific circuit it's in? But instead its for a non-linear component instead of a resistor/capacitor/inductor?

[@pavansharmajanapati9687]

See simple .... q point is just for having a point where the diode characteristics could be linear when a very small ac signal is given... why load line means to know our q point... if we just draw a diode characteristics we cannot know at which point q point has to be taken ... now when u draw load line it definitely is a linear line...in load line voltage across diode is just a drop... now merge diode graph and load line graph ... where they intersect is the q point... note : graph is seen only across vd and Id... this is very important.... when slope changes dc load line shifts hence intersection point also changes...

[@rehmankhan-ve9vo]

Great sir you r much much better than our phd professors

[@deeppatel6027]

i think in above graph...the quiescent point is (Vdq,Idq).

[@liviu201]

Best Indian on the internet!

[@harithahimal1966]

Best Indian teacher

[@Mulakulu]

I hate when people say things don't follow Ohms law. They do. Their impedance just changes

[@NazmulHasan-zx5hz]

what is the significance of load line?why it is important?

[@saket_saurav]

Nazmul Hasan i am also searching the same..for very long..plzz someone help.

[@NirajKumarGangale]

Me too.

[@navinsabban6521]

It is basic tool to find the exact valir of q point to find the Active region for proper amplification. And to see the behavior of ic on vce to check saturation state and cut off state.

[@kawaiiaqua9072]

@@navinsabban6521 thank you

[@kalaiselvanrajasekaran8352]

The significance of load line is that,by load line you get to know about maximum values of the output voltage and output current, so you will know the maximum power dissipation by the device,so u know the maximum operating range and we can prevent damge of devices.

[@majiskani]

Simply well explained concept

[@fathimazia1995]

Amazing explanation 👌

[@Fighter_Believer_Achiever]

Thank you very much sir🥳🥳🥳🥳

[@நக்மாசெல்]

Matches well with textbook

[@நக்மாசெல்]

Follow this lectures. Pass msc electronics

[@yj2jeon]

can you explain how you could use y = mx + c (straight line) when Id shows non-linear pattern?

[@iam_shivshaktii]

Bhai neso academy ka do unit ka notes hoga kya tumhare paas

[@saurabhkunwar3692]

sir on 5:58 min why we did not consider mx=V/Rl instead of mx=-Vd/Rl??????

[@Subhrajit033]

Thank you sir❤❤

[@EditorGuru]

I have a doubt. If RL increases than the current decreases but the voltage across diode will be constant which is 0.7 V. Then in the Q point only Y coordinate should change but here both the coordinates are changing , how? Can anybody explain it to me?

[@poonammishra8544]

Coordinate of Q point should be(Vdq,Idq)

[@shalonijeevanantham1809]

Thankyou so much. Well understood

[@sangachidam3219]

Thanks excellent video

[@SS-pn7ss]

Really good video

[@dipayanchakraborty2417]

really helpful

[@RazorIance]

does it make a difference if the resistor is after the the diode? The equation from the KVL should not change, right?

[@sarsax123]

Doesnt make a differnce. Since both parts are in series the current will be the same

[@dhananjaymehta3154]

How can we change VD?

[@ebroleo94leo31]

Thank you so much master

[@sreekanth2339]

Thank you SIR

[@epiNoesis]

I do not understand. According to diode characteristics, when I_d =~ 0, then V_d is also around 0. But according to Load line, V_d = V. So, for one value, i get two other values. I do not get that at all. Same for V_d = 0. If V_d = 0, what is the I_d??? Is it O, or is it V/R_L ??

[@SameerKhan-xq3he]

Actually they consider diode as a battery of x volts that is reversely connect to circuit containig a resistor The equation would look like V=x+IR I.e. total voltage needed to produce a current in the circuit circuit is sum of voltage drop caused by resistor and reversely connected battery Now if you remove the reversely connected battery (i.e. x=0) you would need a voltage greater than zero to produce a current(V-x/R=I) similarly if you add the reversely connected battery you would need voltage greater than x to produce a current in the circuit Now if you could vary x from zero to any number you will get different values of current ranging from zero to any number.Now plotting those current values against those of reversely connected battery's voltage value will give us a straight line But we have a special type of reversely connected battery (in this case a diode) which can only have a value of 0.7/0.3 volts(forward biasing voltage) So in the equation V=x+IR X becomes constant since it is no longer varying we will have only one value of current from above equation for a specific R Now since the circuit is in series the same current must flow through both diode and resistor so if you plot diode characteristics curve and graph of current through resistor and our reversely connected battery you will find only one value of current that could satisfy both Now if you analyse the above equation V=x+IR This could be rearranged like I=1/R(V-x)+0 Which is same as Y=mx+c I.e. Equation of straight line Where m is slope and c is y intercept(distance at which line cuts y axis) We could easily calculate y intercept if we put x=0 and similarly if we rearrange this equation for x we could calculate X intercept(distance at which line cuts x axis) now we just draw a line through those point [And that's about it we plot values of current through resistor against that of revesely connected battery and compare that to didode characteristics curve inorder to find a common current value ]

[@053_eshan9]

@@SameerKhan-xq3he Thanks bro. Perfect explanation ❤️👍

[@smrutisagarika4742]

@@SameerKhan-xq3he thank you for saving me

[@MohdShahilKhan-cw5dl]

Awesome🎉

[@snehan5821]

Why we do not consider diode resistance?

[@barooseif4390]

Because it is very small so we neglect it

[@sss2393]

We consider ideal diode. With neglegible resistance

[@citizenfour69]

You messed up the co-ordinates bro @ 3:30 ...but thanks for the vid.

[@gopalkrishna9130]

Is RL the resistance of the diode ?

[@AjayKumar-ob3vm]

sir in load line analysis we are using an ideal diode????????

[@akashtr4766]

Not surely

[@shoumikacharya8060]

no this is an actual diode

[@Aliahmed-fv4gp]

Can anyone explain how he used the kvl?

[@shaikakhila9738]

In the video of load line analysis I think so coordinates considered are wrong

[@sanjivgorain590]

sir plz add the video lecture of power amplifier as soon as possible.i need of that

[@vibhavkumarpandey2143]

In the graph the point (0,V), which u have taken at that point current IL should be zero but here it is not so, why?

[@alterguy4327]

Thanku☺

[@chennayyasuragimath2054]

How to get notes for these presentations??

[@j4Naga]

What does operating point for diode tells?

[@mkiratul]

Thanks a lot....u saved my ass

[@niravshah9469]

need pdf for this practical

[@doforget399]

Not (0,v) on x axis but it is (v, 0) And don't use (v/RL,0) on y axis but use (0,v/rl) So (v, 0) =(x1,y1) And (0,v/rl)=(x2,y2) So slop m=(y2-y1)/(X2-x1) So m=(v/RL-0) /(0-v)=(v/RL)/(-v) m = -1/RL=slop of the line

[@jinzjaaz6872]

Was searching for thus comment.dint know why ppl dint notice.

[@divyarao6280]

And (0, V/Rl)

[@harshvadhanas1603]

hatss off sir.....

[@youmnaamr2022]

thank you

[@swatisinha9503]

what is steady state current and steady state voltage?

[@Shoron09]

?

[@davidlandry3487]

Assume a constant DC supply for this problem.

[@divyarao6280]

It should be (V,0)

[@yatinrastogi5949]

sir you have marked points wrongly on axes....please do correct them....

[@sruthijagannathan3667]

what is the use of DC load line..fir what it is uses

[@pavansharmajanapati9687]

See simple .... q point is just for having a point where the diode characteristics could be linear when a very small ac signal is given... why load line means to know our q point... if we just draw a diode characteristics we cannot know at which point q point has to be taken ... now when u draw load line it definitely is a linear line...in load line voltage across diode is just a drop... now merge diode graph and load line graph ... where they intersect is the q point... note : graph is seen only across vd and Id... this is very important.... when slope changes dc load line shifts hence intersection point also changes...

[@ponnekantimahesh1907]

thank u so much...................... NESO ACADEMY

[@ABHISHEKBP-m3v]

Bro whare is output voltage

[@1_billionair_stocks]

pls mention lecture no. in all videos..

[@dasanikitha5941]

Points on x and y axis are written wrong... (0,v/R subscript L),(v,0)

[@yashikagulati6514]

Why did the video end abruptly?

[@asumibaloch2485]

sir why we put Id is equal to zero to find Vd and viveversa??????//

[@susmitasapkota4726]

To find out where will the line cut at X and Y axis. i.e at Vd and Id axis.

[@bickyou4696]

Because the load line is linear, so we need 2 points to draw it.

[@KHSY-]

this has a similarity to the Euler’s graph xD

[@pronobroy8389]

what is y = mx +c?

[@pronobroy8389]

I wish all my teacher could teach me like you

[@ZEdds414]

No.1 Why VD is -ve in kvl equation. No.2 when we take slope then we take 1/RL(m) and didnt take VD.Why?

[@ExaFadli]

coordinates?

[@khushipaliwal3883]

Today is my miss and I'm sorry to bother but it was 😂🎉😢😂😂😂😂

[@D3drock]

Coordinates bro

[@ArielLorusso]

SOLUTION TO I_s [e^(Vd/VT)-1] = (v-vd )/R ? I want to calculate the value Drawing the solution for a given curve its useless for me

[@alaminbijoy1027]

I wonder why do you did that 6:30 ? 😎

[@satyaki7866]

The coordinates are jumbled

[@bpallavi8615]

Sound problem

[@khushalsharma7370]

Co ordinate galat liye gaye hain....x axis per x component zero nahi ho sakta aur naa hi y axis per y component

[@nielnicholsonsara4621]

Test na namun bwas 😭

[@shalinikumari8255]

Sorry sir But I didn't understand this lec. When Id is zero, Vd is also zero but you pointed a value on graph, which was not relatable. Same was the case for Vd

[@phoenixop4673]

❤️❤️

[@DhananjaySahaniD1h2a31235]

You have confusion in coordinates

[@simplemathematics7792]

You 've taken coordinates Ina wrong way sir

[@abhibanerjee6734]

Sir your coordinates are wrong.. 😅

[@physicsbyvishvassingh]

Wow

[@jatindodiya9425]

thank you