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
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.
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.
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?
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.
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 : ruclips.net/p/PLH9R5x7JVXCFyPB1oN2oYSqXZv0_2A_AS
@@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.
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.
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.
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
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
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.
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.
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
Sir i m first year Engineering student . Can I see this lecture for College exam
I wonder why do I pay the college fees. These videos are superb.
Same lol
for degree
for ABET certification, so that you can be held legally liable in the event something you manufacture goes wrong.
6:57 mistake in the calculation of IB I am getting it as 20.6208 uA
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
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.
At 2:30 ... Didn't get IB equation...from where (beta +1) RE came in denomination
why we say 10:49 emitter bias configuration. Isn't it common base configuration because base is grounded instead of emitter or collector
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.
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.
So here Vb=Vbe(constant value)+Ve? Am I right?
@@ramapatipatra5384 Yes.
Ok..
@@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?
There are times when your faith in humanity is restored by some kind stranger who taught you something without even knowing you.
Just wanted to say great job and thanks :)
Is two supply emitter bias configuration better for both stability and amplification?
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.
These are very nice videos sir! Thank you! Can you provide examples also please?
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 : ruclips.net/p/PLH9R5x7JVXCFyPB1oN2oYSqXZv0_2A_AS
@@ALLABOUTELECTRONICS 😍tqs Sir
@4:41 sir from where you take Vbe=0.7 Volt??
It is the typical value of the base-emmiter voltage for the BJT. If it is not specified in the question, you can take it as 0.7V.
@@ALLABOUTELECTRONICS ohk, thank you sir!
@@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.
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?
Would you please mention the timestamp where you are referring to in the video.
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.
Why increase in Vb, decreases Ib....? Ref:- 1:25
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.
@@ALLABOUTELECTRONICS Got it, Thank you.
In 11:31 why is vb 0 if rb is very small . It might be a lame question sorry!
The base voltage Vb is -Ib*Rb. If Rb is small then Vb is also very small and it will be very close to 0. I hope, it will clear your doubt.
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
Tqs Sir ❤
Superrrrb🔥👌
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
how we we determine beta for different temperature ? pls let me know.
You will find the curve in the data sheet.
In the output side while applying kvl how Vcc can be positive 😭😭 please reply to my questions
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.
sir can you please tell why does the base current decreases if voltage across emitter increases
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 ok thank you sir
@@ALLABOUTELECTRONICS thank you I had the same doubt
Please try to solve while you teach we really can't get most of the formulas please sir it's a request
I would recommend you to follow the video in a series. Please check the BJT playlist. If you follow the videos in that order, you will get it.
not getting Ib after calcuclation,5.06
ok got it
Nice bro
How that 0.7 volts appear?
Will please mention the time where you are referring to in the video ??
@@ALLABOUTELECTRONICS at timestamp 4:44
In the previous video sir explained that the typical value of Vbe=0.7 volt
@@sarthakmutreja1559 It's the typical value of voltage Vbe.
Very good 👌👍👌
J
why ic =i E. plese explain it
collector current decreases when i increase base current.WHY?
Would you please mention the timestamp where you are referring to in the video?
опять индусы.... они во всем разбираются...
Hindi ma banao sir...
Amplifier banwane ke liya contacts kre 9878375665 WhatsApp kre amplifier ki accessories bi mil jaige apko
Timestamp @2.45
👍😁
Is two supply emitter bias configuration better for both stability and amplification?
YES