Hello sir... We have studied a major part of our syllabus from your videos... Your explanation technique is amazing... After watching your videos we could understand every word of our book. Thank you so much sir... You are a great teacher!
due to the pandemic, not a single class was conducted by the college, moreover the govn declared no exams so when online classes started, they started teaching the next sem.. now suddenly they are conducting previous sem exam on a short term notice, do we pay fees only to give exams? where is the education? if i pass this exam its only because of this channel, may god bless you..please keep up the good work. topics which i never understood are so clear to me now..thank you.
nice video lecture sir Sir i am from IIIT and I attained your lecture .It helps me a lot in my examination. Thanks for the amazing teaching technique and excellent content.
6:11-depletion region of base collector junction which is reverse baised so you have shown that the depletion region penetrates more into base region but that is actually true for collector region as it is highly doped.
Ref 7:45 to 8:07: why would emitter current increase when effective base width reduces? less base width --> less recombination --> less base current and also the equation Ie = Ib+ Ic doesn't justify increase in emitter current. Awaiting for your response, thanks in advance..
Your answer lies in you question. ie= ib + ic. Here you see as you said less recombination >> more ie >> less ib, so as more ie and less ib the equation satisfies itself.
Ankit Garg Actually emmitter current is a combination of Diffusion of electrons from emitter to base ..and diffusion of holes from base to emitter .. but the diffusion from base to emitter do not contribute for the current from emitter to reach collector ... so that is the reason why doping is reduced in base so the max contribution of emitter current is through diffusion of electron from emitter to base . if the base width itself reduced due to RB of collector junction . emitter current is increased cause it is only achieved by diffusion frm emiitter to base increased..base to emitter diffusion current decreased
great work sir. With learning through online lectures i couldn't understand every bit our sir taught us. But you made it out. Thanks for this great teaching. Just one thing please can u improve your W its looking like K. nothing else thank you. i cant say it in words very very thank you
hello sir 8:01 you said that the chance of recombination is low and ie increases you're right. but my question is base is connected to ground and if there is recombination there is no potential to crete the floow of current of IB . increasing the the vcb even-though it increases the width the recombination will not have that much effect
At 2:38, you have said that an n-p-n transistor can be considered as two diodes connected back to back. This assumption is rarely true, for the arrangement, the p-region must be very thin and the doping concentration of p-region must be very less compared to the n-regions.
Sir I'm from nit I can't even understand which topic our faculty is teaching in class upcoming viva exams are there and your videos are at rescue 😅 thankyou sir
Sir ,in early effect the depletion layer penetrates more in collector than base due to high doping and electron being pulled away more from battery in conparison to wholes but due to less width of base the effective width still significantly decrease
at the end after drawing graph, you did say IE increases, but did not show that exact thing in graph, the corresponding height for IE also increases for increased VE.
Sir one correction please... Collector is lightly doped so more depletion region compared to base. Just to be digramatically correct, else concept is totally correct.
in my opinion, collector current should increase as a result of reduced recombination, whereas the increase in emitter current should be attributed to the fact that concentration gradient has increased. correct me if I am wrong.
The negative immobile ion concentration in base is increasing, so shouldn't the Ie (input current) decrease? Cause the electrons from emitter will be repelled by the negative ions of base.
The reason in concentration. Base is least doped part of the transistor when compared to Emitter or Collector. In REVERSE BIAS the LOWER the doping, Higher is the penetration of space charge region ie. higher the depletion width. Also higher doping means lesser pentration. The Weff region decreasing means lesser recombination, thus more electrons can pass to COLLECTOR region.
The current emitted by the emitter is IE when IE (electrons in case of npn transistor) enter into base region recombines with holes in base region (suppose 5 holes are there in base region and 15 electrons in emitter region)so 5 electrons recombine with electrons and remaining electrons enter into collector region.so number of electrons is decreased from emitter to collector.so IE decreased. So if W effective is less.less recombination takes place .so IE increases
because the electrons coming from emitter will have less chance for recombination on base region so that all the electron will flow to the collector and finally to source which is ie hence ie increases
@@ArunKumar-qg9hm because the electrons that came from emitter region behave like minority charge carrier in base region and as reverse bias is favorable for minority carrier ,so the electrons can easily cross the J2 junction..
@@ArunKumar-qg9hm The collector juntion is more positve so electron will be attracted by the collector terminal and the barrier J2 will not have much impact because of low potential barrier
I am not from this background. But we generally look for output against changing values of Input. But here we are changing output and looking for its effect on input variables. Can someone help me in understanding it on broader level that if I have a exact physical replica of this circuit, how can I change output variable in layman terms? And second question is, when we decrease area the resistance should be more and hence less current but here less Weff is causing more current, this one I didn’t get either.
1. NPN Emitter base juntion Forward Biased, CBJ reverse biased: i. large concentration of electrons are injected into the base region from emitter. Some of these electrons recombine with the holes in the base and the rest of the electrons travel to the collector region. ii . there is a BASE current, which is due to the flow of holes into the base region to compensate for the holes lost during the recombination. 2. when CBJ revere bias is increased: i. Depletion region penetrates more into the base region, which also means that there is less mobile charge carrier(hole) in the base compared to previous situation. ii. therefore the electrons entering from the emitter wont recombine at the base as much as the previous case, which means the electrons can flow more freely to the collector, ie there is a reduction in resistance to the current flow.
Here the case is that the conventional direction of current is from positive to negative(i.e. p to n) but the actual flow of electrons are from emitter to base
in the ouput characteristics..Ie remains constant for one curve; it is just according to the equation Ie=Ic(approx)..but during breakdown Ic increases tremendously even when Ie remains constant...where is the rest of the current coming from when Ie is constant
Sir We know that in the region of base due to early effect ic current increase due to less recombination but the part that increase in ic is that which did not recombine so Ib is less and ie should not increase but your explanation say ie increase may you explain this??
Why does the I-V curve of the f.b. diode starts to exponentiially grow only after Vbe > 0.7(the voltage buil-in). We learned that it should ideally start to grow exponentially already when Vbe >0 , no?
This is the case where the diode is made of silicon.The Barrier voltage of silicon is 0.7 .Hence for the diode to conduct the Biasing voltage must be greater than or equal to 0.7
Now i understand if the area of the base decreases more and more electrons will flow through the collector and less recombination will take place in the base hence results in the increase of emitter current and decreased of base current
how, less no. of holes are available to recombine with electrons, therefore most electrons will move to collector region, increasing both Ic and Ie as Ic = alpha(Ie)
Hello sir... We have studied a major part of our syllabus from your videos... Your explanation technique is amazing... After watching your videos we could understand every word of our book. Thank you so much sir... You are a great teacher!
Yes .... exactly
due to the pandemic, not a single class was conducted by the college, moreover the govn declared no exams so when online classes started, they started teaching the next sem.. now suddenly they are conducting previous sem exam on a short term notice, do we pay fees only to give exams? where is the education?
if i pass this exam its only because of this channel, may god bless you..please keep up the good work. topics which i never understood are so clear to me now..thank you.
The Early effect is often mentioned in textbooks but rarely explained...thank you sir !
your lectures are really helping me for my electronics course,thanks dude
No one explained i/p character so easily sir . Hats off
The only teacher who explained early effect so brilliantly. Thank you so much sir
only video over youtube which explains early effect
Only one you found i think! there're more and this one explains it wrong
@@imalkavindawickramasingha4331 hahahaha
@@imalkavindawickramasingha4331 are you from Sri Lanka ???
nice video lecture sir
Sir i am from IIIT and I attained your lecture .It helps me a lot in my examination. Thanks for the amazing teaching technique and excellent content.
Omg thank you so much. I seriously can't understand concepts that aren't well reasoned and this video has perfect reasoning 😍😍😍😍😍
How can depletion layer penetrate?
Thank you, Sir! Your explanation is very clear. May God Bless You!
6:11-depletion region of base collector junction which is reverse baised so you have shown that the depletion region penetrates more into base region but that is actually true for collector region as it is highly doped.
Wahh matlb kuch bhi 😂
@@nishantgautam7627 not kuch bhi. Go and read properly before commenting bullshit.
very much informative sessions...expecting a lot from u guys..
the best online lecture on trnsistr charactrstcs..
Sir the videos are very helpful in completion of our syllabus with clear understanding...thank u sir
Ref 7:45 to 8:07: why would emitter current increase when effective base width reduces?
less base width --> less recombination --> less base current and also the equation Ie = Ib+ Ic doesn't justify increase in emitter current. Awaiting for your response, thanks in advance..
I wanted to ask the same question also>>@neso_academy
Your answer lies in you question. ie= ib + ic. Here you see as you said less recombination >> more ie >> less ib, so as more ie and less ib the equation satisfies itself.
Ankit Garg Actually emmitter current is a combination of Diffusion of electrons from emitter to base ..and diffusion of holes from base to emitter .. but the diffusion from base to emitter do not contribute for the current from emitter to reach collector ... so that is the reason why doping is reduced in base so the max contribution of emitter current is through diffusion of electron from emitter to base .
if the base width itself reduced due to RB of collector junction . emitter current is increased cause it is only achieved by diffusion frm emiitter to base increased..base to emitter diffusion current decreased
when Ib decreases than according to the equation Ic should increase.why Ie is increasing?great confusion.
even I thought increase in base width would result in more recombination, thus decrease in ie
Awesome video! Best explanation on UA-cam
thank you sir,,,your explanation is very clear and easy to understand
ThanQ so much sir, now we can understand many things in electronics as we saw your videos. 👌🏻😍
Only a good thing teachers can do is that they refer your channel to students
Early Effect very well explained !
great work sir. With learning through online lectures i couldn't understand every bit our sir taught us. But you made it out. Thanks for this great teaching. Just one thing please can u improve your W its looking like K. nothing else thank you. i cant say it in words very very thank you
thank u sir...you explain concepts so well!
God gifted teacher to us❤❤❤
It gives details concept on early effect. thanks for this one
hello sir 8:01 you said that the chance of recombination is low and ie increases you're right. but my question is base is connected to ground and if there is recombination there is no potential to crete the floow of current of IB . increasing the the vcb even-though it increases the width the recombination will not have that much effect
I hope, you also add loud sound with this sweet explanation.
At 2:38, you have said that an n-p-n transistor can be considered as two diodes connected back to back.
This assumption is rarely true, for the arrangement, the p-region must be very thin and the doping concentration of p-region must be very less compared to the n-regions.
Hello sir
Thank you for explaining everything clearly ❤❤
Thank u sir... I fall short of words how, helpful this was to me.
Ur explanation are so clear pls do on difficult subjects like electromagnetics please
Sir I'm from nit I can't even understand which topic our faculty is teaching in class upcoming viva exams are there and your videos are at rescue 😅 thankyou sir
You are our God sir.Words are less for you, praisings are less for you, you are simply superb sir.Thank you sir.
no words,, ,really ur great man
Sir thank u for being there for us.
Sir ,in early effect the depletion layer penetrates more in collector than base due to high doping and electron being pulled away more from battery in conparison to wholes but due to less width of base the effective width still significantly decrease
if recombination due to early effect decreases ,then ic must increase rt because more electrons will move to collector how does ie increase?
Pllz add video for Operational amplifiers! 🙏.......ur explanation is much preferred than any other videos
Very complicated concept of Early Effect. No wonder I didn't follow it in college lecture.
Aditya Chopra Me too
this video is better than lecture in campus :')))))))))).
at the end after drawing graph, you did say IE increases, but did not show that exact thing in graph, the corresponding height for IE also increases for increased VE.
Thank you sir ❤️
Perfect way of explanation..clear 👏👏👏
best teaching sir best
You explains so good...thnx brother
Sir one correction please... Collector is lightly doped so more depletion region compared to base. Just to be digramatically correct, else concept is totally correct.
W.r.t to base it is more doped that is why
thank u so much for such a good explanation :)
@2:06, shouldn't input current be -Ie and output be -Ic in cb npn transistor?
I Appreciate very much your lecture videos!
Just superb....keep it up!!!
Why do we take Vcb as constant for input characteristics? Why do we take current IE as constant for output characteristics?
Is cutoff voltage also decreases on increasing output voltage ( as there is decrease in effective width of base)?
Awesome ..... Keep making such videos Bro
your lecture very helpfull
your video was quite helpful.
Sir why we are taking graphs between input votage and input current and output voltage why not output current?
If any one know please answer
Because Ic depends on VCB.
Why is it shifting towards the left ? And shouldn't that first and middle line become longer ?
Nice Lecture
Sir why emitter current is input current ..? Please reply me .
Output voltage should constant in input chrc
Thank uuuuu so much 😊 sir 🎉
in my opinion, collector current should increase as a result of reduced recombination, whereas the increase in emitter current should be attributed to the fact that concentration gradient has increased. correct me if I am wrong.
You are right
@@Pablo-ho2rg thank you for letting me know
The negative immobile ion concentration in base is increasing, so shouldn't the Ie (input current) decrease? Cause the electrons from emitter will be repelled by the negative ions of base.
Sir can u plz explain why the depletion region is more concentrated in base region and how does emitter current increase. Thanks in advance
The reason in concentration. Base is least doped part of the transistor when compared to Emitter or Collector. In REVERSE BIAS the LOWER the doping, Higher is the penetration of space charge region ie. higher the depletion width. Also higher doping means lesser pentration.
The Weff region decreasing means lesser recombination, thus more electrons can pass to COLLECTOR region.
thank u sir
How IE increases with decrease in w effective plz explain me sir
The current emitted by the emitter is IE
when IE (electrons in case of npn transistor) enter into base region recombines with holes in base region (suppose 5 holes are there in base region and 15 electrons in emitter region)so 5 electrons recombine with electrons and remaining electrons enter into collector region.so number of electrons is decreased from emitter to collector.so IE decreased.
So if W effective is less.less recombination takes place .so IE increases
I didnt get it when recombination decreases how ie will increase
because the electrons coming from emitter will have less chance for recombination on base region so that all the electron will flow to the collector and finally to source which is ie hence ie increases
@@samishang.c4791 i have one doubt...there is depletion region(barrier) in base..then how the electron can flow from emitter to the collector?
@@ArunKumar-qg9hm the electron cmng from emitter is highly energetic and bcz j1 is forward biased which make the way easier to get into colletor
@@ArunKumar-qg9hm because the electrons that came from emitter region behave like minority charge carrier in base region and as reverse bias is favorable for minority carrier ,so the electrons can easily cross the J2 junction..
@@ArunKumar-qg9hm The collector juntion is more positve so electron will be attracted by the collector terminal and the barrier J2 will not have much impact because of low potential barrier
Please refer a best book for basic electronics....
thankyou very much for clearing concepts
I am not from this background. But we generally look for output against changing values of Input. But here we are changing output and looking for its effect on input variables.
Can someone help me in understanding it on broader level that if I have a exact physical replica of this circuit, how can I change output variable in layman terms? And second question is, when we decrease area the resistance should be more and hence less current but here less Weff is causing more current, this one I didn’t get either.
1. NPN Emitter base juntion Forward Biased, CBJ reverse biased:
i. large concentration of electrons are injected into the base region from emitter. Some of these electrons recombine with the holes in the base and the rest of the electrons travel to the collector region.
ii . there is a BASE current, which is due to the flow of holes into the base region to compensate for the holes lost during the recombination.
2. when CBJ revere bias is increased:
i. Depletion region penetrates more into the base region, which also means that there is less mobile charge carrier(hole) in the base compared to previous situation.
ii. therefore the electrons entering from the emitter wont recombine at the base as much as the previous case, which means the electrons can flow more freely to the collector, ie there is a reduction in resistance to the current flow.
Sir I'm confused. How IE is input when it is leaving the transistor and IC is output when it is entering the transistor?
Here the case is that the conventional direction of current is from positive to negative(i.e. p to n) but the actual flow of electrons are from emitter to base
Respected Sir, How did depletion region move further into the base region at 5:47?
nishant gautam haha, never got into electronics :) but thanks man! Baki ke kids ko help karega tumhara revert
ie =ib+ic incresing vcb decrease ib increse ic ,somehow ie is constant?
Yes bro same doubt bro 😔
Why Emitter current increases with increase in collector to base voltage in common base configuration of a BJT?
because now more electrons can diffuse through base region as recombination decreases
Graph of Vbe vs Ie should be start line.
I don't understand why it's curve.
Please explain sir
very helpful video sir
sir why Vcb move towards left side in graph?
Sir what about depletion region of n-p which is in forward bias as you wrote W (eff) is region without depletion region
in the ouput characteristics..Ie remains constant for one curve; it is just according to the equation Ie=Ic(approx)..but during breakdown Ic increases tremendously even when Ie remains constant...where is the rest of the current coming from when Ie is constant
Sir what is the effect on input voltage actually when Vcb increases input voltage is decreasing
thank you srila prabhupad , krishna , and sir
The Early effect will have no impact on breakdown voltage across EB?@Neso
tnxx...it really helped me a lot!!
Excellent ❤️
Sir
We know that in the region of base due to early effect ic current increase due to less recombination but the part that increase in ic is that which did not recombine so Ib is less and ie should not increase but your explanation say ie increase may you explain this??
Yes bro ur right
Which is the Ebers Moll Model of BJT?
Sir why the depletion region is decreased during the active mode?? In revrse bias the depletion region is increased
in taking input characteristics why cant we draw graph for different values of Ic instead of Vcb
Sharmila Arem Ic depends on Vcb
In input characteristics Ie and Vbe is constant instead of taking Vce in different what will happen when we take different output current Ic
Why does the I-V curve of the f.b. diode starts to exponentiially grow only after Vbe > 0.7(the voltage buil-in). We learned that it should ideally start to grow exponentially already when Vbe >0 , no?
This is the case where the diode is made of silicon.The Barrier voltage of silicon is 0.7 .Hence for the diode to conduct the Biasing voltage must be greater than or equal to 0.7
Depletion width of collector will be thicker than in the base region due to higher charge density in base region(smaller size). Isn't it?
sir will you make videos on small signal and large signal model of bjts and fets?
At 8:40 , how and which side the concentration is increasing with decreasing effective area of the base,
can u explain
As the area is decreased the concentration of what increased in the base?
Now i understand if the area of the base decreases more and more electrons will flow through the collector and less recombination will take place in the base hence results in the increase of emitter current and decreased of base current
Sir
If increasing VCB would increase IE
Then according to the graph their is no change in the current ?
Please explain
Slope is increasing that is why
How is Ie input current if it is exiting from the transistor?
Nice video
how ic and vcb are output current and voltage, as we are only supplying the voltage to the collector junction?
Why we are getting the output characteristics in first quadrant. Are u taking the magnitude only( neglecting the direction of Ic and sign Vcb)
why decrease in recombination increases the emitter current
God bless Nesco Academy
I thought increase in base width would result in more recombination, thus decrease in ie
how, less no. of holes are available to recombine with electrons, therefore most electrons will move to collector region, increasing both Ic and Ie as Ic = alpha(Ie)