We know that in a forward biased transistor we have βIb = Ic. However in the model it's clear that the above equality doesn't hold due to the addition of another resistor Ro .We now have Ic > β.Ib. What's wrong in my understanding of the model. Any help is appreciated Thank you.
when we include the early effect, the relation Ic=βIb is still the same, but the Ic you mention is not the real Ic(kinda confusing to say that), when we change Vce the value of Ic changes by Va/Ro, so the Ic you mentioned above is the original component alone, to get the real Ic in the relation above when Ic changes you have to consider both components, namely: Ic = original Ic + Va/Ro = βIb.
from the previous lectures sir had said that collector current will not change with Vce. but at 25:28 shows the with respect to change in Vce there is a change in collector current..HOW....?
Prefer previous videos to under stand that. For your heads up- v(pi)= v(in)=The fluctuation viltage input. The small signal odel is only for the v(in) voltage. While V(BE)=v(in)+v(o), v(o)=the biase voltage or battery which use large signal model (general model) to calculate Ic(o)
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At 44:00 Rpi=100*1.9=190ohm
I was wondering about it myself.
Really enlightening. Thank you for explaining these concepts so clearly and patiently
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Vec must be greater than Veb for a pnp to be in FORWARD active region
yes for npn VC>VB
and for pnp VC
True
Thank you, that was a really good explanation.
I am a little confused, shouldn't the derivatives be "partial" derivatives? We're keeping other variables constant while changing one variable.
Does saturation mean both junctions are forward biased for NPN transistor?
Best man
Thanks
We know that in a forward biased transistor we have βIb = Ic. However in the model it's clear that the above equality doesn't hold due to the addition of another resistor Ro .We now have Ic > β.Ib. What's wrong in my understanding of the model. Any help is appreciated Thank you.
when we include the early effect, the relation Ic=βIb is still the same, but the Ic you mention is not the real Ic(kinda confusing to say that), when we change Vce the value of Ic changes by Va/Ro, so the Ic you mentioned above is the original component alone, to get the real Ic in the relation above when Ic changes you have to consider both components, namely:
Ic = original Ic + Va/Ro = βIb.
ro should be 100 * 1.9=190ohms instead and not 1900ohms
Does Razavi have any recommended practice problems from his book for each lecture?
did you find any recommendation?
from the previous lectures sir had said that collector current will not change with Vce. but at 25:28 shows the with respect to change in Vce there is a change in collector current..HOW....?
Watch Lec 17 at the end...
At 26:33 why did nt include number 1 ?? ( 1+dV/Va)
In numerical at 45:30
gm*v(pi)=???
Is V(pi)=V(be)=0.8V
So is gm*v(pi)=0.424 A??
Prefer previous videos to under stand that. For your heads up- v(pi)= v(in)=The fluctuation viltage input. The small signal odel is only for the v(in) voltage. While V(BE)=v(in)+v(o), v(o)=the biase voltage or battery which use large signal model (general model) to calculate Ic(o)
✅
27:00 Why didn't we write ...(1+(deltaV/Va)) for the secound part of deltaIc ?
He has just expanded the term {1 + (Vce + delta V)/Va } into two terms
@@ytaccount9420 how? i dont understand that part
pnp start at 47:10
shouldn't we have base current change due to early effect i.e. change in Vce???
+Aditya Gupta Ib=Ic/ß, therefore as Ic changes due to the Early effect, the Ib changes as well.
We assume vbe is constant. So no. Ic is only dependent on Vce
What's the name of the track that plays at the beginning of these videos?
ruclips.net/video/6Z6Pul2NVkA/видео.html
@@ashikjayamon3963
Thanks. :D
at vce=0 volt the collector current should be maximum but why it is zero as per the output characteristics ?? can anyone explain this to me?
When Vce = 0, the transistor is not biased in the forward active region. We want Vce > Vbe. That could be the reason..