The timestamps for the different topics covered in the video: 0:21 Fixed Bias with Emitter Resistor (Emitter Stabilized Bias) 4:19 Variation in operating point with the change in current gain (β) 8:10 Condition for the stable operating point in Emitter Stabilized Bias Configuration 10:48 Emitter Bias Configuration (DC Analysis) 13:33 Condition for the stable operating point in Emitter Bias Configuration
For those who are wondering why Ie = (b+1)ib Note that Ie = Ic + Ib and Ic = B x Ib so just add Ic to the Ie expression/equation then factor Ib to form : Ie = (B+1)Ib
I have one confusion here in this lecture,you have told as the emitter current increases the voltage drop across emitter resistance increases as a result VBE also increses.But how?as we know Vbe=Vb-Ve,if voltage drop across emitter resistance increases then Ve will increase and Vbe will decrease.
What I mean to say was Vb. Considering Vbe is approximately constant for analysis, with the increase in Ve, Vb will increase and that reduces the base current. I hope it will clear your doubt.
@@ramapatipatra5384 As I see it (see at 1:22 into the video) : As the emitter current increases, the voltage across the emitter resistor increase which (for a fixed base voltage) reduces the voltage between base and emitter which tends to reduce emitter current?
Personal Note: for the fixed bias the reason for deriving the (B+1) is to find the VCE which is done by doing the KVL at the output side. Not input side.
Sir, when beta reduces to 100 to 50,, lb increases to 18.56 to 20.6 also lc is reduced to 1.85mA to 1.03mA then why New Q point lie on same load line ... if new Q point lie on same load line then on Decreasing lc ,, lb must be decreased. Sir please clear the confusion 6;30
Here there is a base resistor. So, even during the AC analysis, there will be a base resistor between the base and ground terminal. So, it is not common base configuration. Moreover, here the circuit is only shown for the DC biasing perspective. But actually there will be bypass capacitor between emitter and ground. And because of that, for the AC analysis, the emitter terminal will act as a ground terminal. (Emitter resistor will act as a short circuit because of the emitter bypass capacitor ) Therefore, still it is common emitter configuration from AC analysis perspective. I hope, it will clear your doubt. Generally, when we say, common base, common emitter or common collector configuration, it is from AC analysis perspective. The DC biasing circuit might look different.
if Rb needs to be less than (beta+1)Re , why did you use Rb=400k Ohm, and Re=1k Ohm with a beta of 50 or 100, both would be much smaller than Rb. Can you please explain what I did wrong here?
At 4:32 you described how Q-point varies according to changes in beta. In this example you used a value for Rb bigger than the expression (beta +1)Re, which you will later explain in the video that Rb value should be lower or equal to that expression. However, I still can't understand why you would write Rb with a bigger value if we wanted the Q-point to not change and be stable with the change of beta or temperature.
@@straxescraft Because I just wanted to show that if Rb is larger than (β +1 )*Re, then with the change in the β, operating point changes. That's why intentionally the value of Rb was chosen bigger than (β +1 )*Re. I hope it will clear your doubt.
@ALL ABOUT ELECTRONICS thank you so much for your time and help! I just wanted to clear this confusion for myself. Side note : I have seen many examples of the emitter stabilized bias configuration on the internet, yet they do the same thing I was talking about in my question.
There is a seperate channel for the question. ( ALL ABOUT ELECTRONICS- QUIZ) And there is a entire playlist for the questions related to BJT. Here is the link : ru-vid.com/group/PLH9R5x7JVXCFyPB1oN2oYSqXZv0_2A_AS
The feed bCk resistor Re included to provide fixed Q point or stop the variation in the Q point u said..but u are saying still there is a variation.what the meanibg of that
It will reduce the variation or improve the stabilization of Q point. But it can not eliminate the variation completely. But for many applications, that much stabilization of Q-point will do the job.
@@ALLABOUTELECTRONICS thank you...one thing please can you explain op amp internal circuit by 741 IC ,how transistors on and off during inverting and non inverting inputs supply.if you can olease do it through skype i will pay you.non of the youtube videos can fount about explanation for that ic internal circuit please please
In case if you find it difficult, then just draw VCC and - Vee source with positive and negative terminals. And then apply KVL. It will get clear to you.
Great series of lectures but, as has been pointed out, there is an error at 13;23. As beta is decreased from 100 to 50 the collector curves move downwards at the same base current. So at 13:23 what is labeled as 10 microamps is actually the new curve for 30 microamps when beta has been reduced from 100 (original set of curves) to 50. Similarly at 14:00 the new operating point with a beta of 150 is still on the 30 micro amp collector curve which has now moved upwards and is now occupying the line that previously was labeled as 60 micro amps with a beta of 100.
Assuming VBE, the base-emitter voltage is almost constant, as the voltage across the emitter increases, base voltage also increases. Because Vb = Vbe + Ve. And Ib = (Vbb - Vb) / Rb. Therefore, as the base voltage increases, the base current reduces. I hope, it will clear your doubt.
@@ALLABOUTELECTRONICS Could you please make videos on Physics of semiconductor including introduction to quantum mechanics and Schodinger equation. That would be very nice of you Sir.
In this case, current Ib = (Vbb - Vb )/ RB. Therefore as Vb increases, the numerator will reduce and hence base current Ib will reduce. I hope it will clear your doubt.
