really awesome explanation. i was trying to understand this topic for so long and u made it so easy , neat and brief , thank u so much . keep going on .
thanks alot! at the university I go to they use a lot of models making it hard to understand whats going on.. learnt in 30 minutes what they tried to explain to me over a couple of weeks.
This still confuses me - 4:50 What happens at the BE junction during a small AC signal? Does it just short - or how else would they become copies of each other? The way I understand it now (probably wrong) if we were ignoring the 0.65V and make an approximation by replacing VE and VB (since they differ by 0.65V)
Nope.If you put 3V battery with 8V the resulting voltage will be 11V.Similarly if you have ac input ex: -4 to 4 volt and you add it with vcc lets say 8 volt the resulting waveform will be 4V to 12V.this is how bjt is operated.
Unfortunately, I could not follow your hand writing. It was so confusing to distinguish between E, e, B, b and so on. Perhaps, you need to greatly improve your handwritings or just simply use a Power Point presentation.
Great explanation explaining all the aspects professor. 💯
really awesome explanation. i was trying to understand this topic for so long and u made it so easy , neat and brief , thank u so much . keep going on .
thanks alot! at the university I go to they use a lot of models making it hard to understand whats going on.. learnt in 30 minutes what they tried to explain to me over a couple of weeks.
This is great. Been searching for a good explanation. Thank you
This professor is a real Godsend, I tell you!
Lovely video, many thanks Professor!
The world's best teacher thanks sir
Just to reiterate "The emitter voltage is just copy of the base voltage" --> The ac voltage of emitter is copy of ac base voltage 🙂
This still confuses me - 4:50
What happens at the BE junction during a small AC signal? Does it just short - or how else would they become copies of each other?
The way I understand it now (probably wrong) if we were ignoring the 0.65V and make an approximation by replacing VE and VB (since they differ by 0.65V)
this was so helpful! thank you so much! :)
Thanks so much Professor it really helped me a lot 👍
someone could please indicate me the video in which there is the demonstration that vb is a copy of ve, i cant find it
Does VBE have to be 0.65 V? I thought VBE = 0.7 V in order for the BJT to be in active mode?
well done. thank you professor
Nyc explanation..👍👍
My is the complex collector current equal to the emitter current?
During the negative half of the input signal doesn't the base emitter junction become reverse bias?
Nope.If you put 3V battery with 8V the resulting voltage will be 11V.Similarly if you have ac input ex: -4 to 4 volt and you add it with vcc lets say 8 volt the resulting waveform will be 4V to 12V.this is how bjt is operated.
@@oogway5472 pls do suggest some textbook for electronics
@aspirants aspirants I think millman's books are the best
5 years on and only 346 likes ( including that of mine). I wonder are all of the rest so enlightened on electronics?
Thank you so much Walter White😊
How to find the capacitance values in ce amplifier circuit?how to find R equivalent in formula?
there is no voltage drop without a branch.
Thank you so much sir
common with what?
I E = (VE + Ve ) / R E = ?
Unfortunately, I could not follow your hand writing. It was so confusing to distinguish between E, e, B, b and so on. Perhaps, you need to greatly improve your handwritings or just simply use a Power Point presentation.
You could hear what he was writing as he was speaking
Just turn on the CC Bro.😀