Congratulations on your videos, your teaching is amazing. Just one question, the correct assumption for "zi" is not: "zi = (Bcc + 1)*r'e" if "Rb >= 10*(Bcc + 1)*r'e"? (as in the last video)?
At 4:47 Current divider rule is applied wrongly, So final expression is also Wrong. Please try to mention your faults in the comment section @neso academy.
@@bomberman0023 Nah. He is right but I get why you think that. It's because of the direction of the currents that's why you're confused. BIb is not a resistance, remember? So Ic will remain the same value because it is determined by Ib as it is a current controlled current source. Therefore, the only resistances there are ro||Rc and Rl. and if you don't let the directions confuse you it's by all means a current divider between those two resistances.
thank you, I have learn so much with yor videos. there is something i dont understand. when there are two resistences in parallel and one is 10 times less than the other one, the equivalent resistance is the less of them, then is Rb | | (β+1)re and Rb≥10(β+1)re then Zi=(β+1)re. I think the error is in the condition and actually is Rb≤10(β+1)re
@@svaditya2896 the approximation was for Rb >= 10(B+1)Re. So if you remove (B+1)Re from the denominator, Rb cancels Rb, leaving (B+1)Re on top. It's totally wrong.
If RL is not connected shouldn't the Output Current be = βI_b = I_c? My professor in class was using current divider with ro for some reason to find I_o. It makes no sense. I_o should be the same as I_c when RL is not connected. Am I correct?
imagine you have two resistors in parallel. one is much larger than the other. the large one will act like an open circuit. current will flow through the smaller resistor. So you can effectively ignore the large resistor and only consider the smaller resistor. So When ro>>Rc, ro||RC = RC (approximately)
oyou gotta watch the previous lecture. he did take ro. Its just that it was a part of a fraction and canceled out. [ Zo =(roRc)/(ro + Rc)]...Rc is removed from denom...ro canceled with the numerator
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Zi=(β+1)re will the correct one
Kesav Kumar beta is large so
(B +1 )~= B
yep..exactly
yeeeees
Congratulations on your videos, your teaching is amazing.
Just one question, the correct assumption for "zi" is not: "zi = (Bcc + 1)*r'e" if "Rb >= 10*(Bcc + 1)*r'e"? (as in the last video)?
You are simply awesome. Studying for my EE exams is so much more productive thanks to your videos.
Zo=Rc and Zi=(B+1)re
Thanks a lot for these videos sir! Godbless You!
Neso is gold... thanks a million sir...
sir alhamdulillah ur lectures r awesome.
i havve not understood why Zi is Rb?
At 4:47 Current divider rule is applied wrongly, So final expression is also Wrong. Please try to mention your faults in the comment section @neso academy.
how that is wrong?
Yes io and ic will be replaced
Yes ic should be some fraction of io depending on the resistances in the two branches, not the other way around.
@@bomberman0023 Nah. He is right but I get why you think that. It's because of the direction of the currents that's why you're confused.
BIb is not a resistance, remember? So Ic will remain the same value because it is determined by Ib as it is a current controlled current source.
Therefore, the only resistances there are ro||Rc and Rl. and if you don't let the directions confuse you it's by all means a current divider between those two resistances.
@@therealgoat3367 thx for explaining
thank you, I have learn so much with yor videos.
there is something i dont understand.
when there are two resistences in parallel and one is 10 times less than the other one, the equivalent resistance is the less of them, then is Rb | | (β+1)re and Rb≥10(β+1)re then Zi=(β+1)re. I think the error is in the condition and actually is Rb≤10(β+1)re
it's not Rb≤10(β+1)re
it's 10Rb≤(β+1)re
Wonderful lectures. Sir, please complete the analog electronics series. It is very helpful. Thanks.
THANKS SIR
Zi=(beta+1)re is Right one
Source current(Is) and input currenr(Ii) are same if i am correct
In previous video u said zi is equals to (B+1)re bt in this lecture u r saying that zi is now equals to rb how is it possible
approximation
@@svaditya2896 the approximation was for Rb >= 10(B+1)Re. So if you remove (B+1)Re from the denominator, Rb cancels Rb, leaving (B+1)Re on top. It's totally wrong.
bhai itna achha padhate ho silly mistakes karke q confuse kar dete ho
?>
Please reply asap🙏...what happen to current gain if the load is replacde by open circuit or RL=∞?
Just one correction ... Zi = (beta+1) Re. That's it
If RL is not connected shouldn't the Output Current be = βI_b = I_c? My professor in class was using current divider with ro for some reason to find I_o. It makes no sense. I_o should be the same as I_c when RL is not connected. Am I correct?
zi is equal to (beta+1)re
sir how could
be the infinity/infinity will be the finite value?
Why we are not considering negtive output current while calculating current gain.....?
Is is equal to i1, both currents are In series?
What if Rb=Rc=0(no bias)
Then Ai=0?
But Ai =Beta is the right answer
zi=RB !!!!! how.
Zi will be beta*Ib
would u mind answering why Zi is Rb and not beta re??????
I think (beta +1)re is the right one
if r0>=10Rc then r0||RC shouldbeequal to r0 not RC
imagine you have two resistors in parallel. one is much larger than the other. the large one will act like an open circuit. current will flow through the smaller resistor. So you can effectively ignore the large resistor and only consider the smaller resistor. So When ro>>Rc, ro||RC = RC (approximately)
Sir I have a doubt. When ro>>10rc.we should take ro value. Why u select rc. Pls rly anyone
oyou gotta watch the previous lecture. he did take ro. Its just that it was a part of a fraction and canceled out. [ Zo =(roRc)/(ro + Rc)]...Rc is removed from denom...ro canceled with the numerator
How you people study Electronics..most boring subject of any group.....