Transistor Biasing: Emitter Stabilized Bias and Emitter Bias Configuration Explained

Рет қаралды 134,032

Күн бұрын

Пікірлер: 75

@ALLABOUTELECTRONICS 5 жыл бұрын

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

@Psycho_Engineer786 2 жыл бұрын

Sir i m first year Engineering student . Can I see this lecture for College exam

@akashtandel9633 5 жыл бұрын

I wonder why do I pay the college fees. These videos are superb.

@vinaykumarkv5830 4 жыл бұрын

Same lol

@shriganeshayenamah3422 Жыл бұрын

for degree

@vienlacrose Жыл бұрын

for ABET certification, so that you can be held legally liable in the event something you manufacture goes wrong.

@louisanthonybernante6706 4 жыл бұрын

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

@eggxecution Жыл бұрын

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.

@lhaellor Жыл бұрын

6:57 mistake in the calculation of IB I am getting it as 20.6208 uA

@bhageshmaheshwari2807 3 жыл бұрын

Just wanted to say great job and thanks :)

@techtheguy5180 2 жыл бұрын

There are times when your faith in humanity is restored by some kind stranger who taught you something without even knowing you.

@ramapatipatra5384 3 жыл бұрын

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.

@ALLABOUTELECTRONICS 3 жыл бұрын

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 3 жыл бұрын

So here Vb=Vbe(constant value)+Ve? Am I right?

@ALLABOUTELECTRONICS 3 жыл бұрын

@@ramapatipatra5384 Yes.

@ramapatipatra5384 3 жыл бұрын

Ok..

@neilcullimore5798 Жыл бұрын

@@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?

@segniguta6569 4 жыл бұрын

These are very nice videos sir! Thank you! Can you provide examples also please?

@ALLABOUTELECTRONICS 4 жыл бұрын

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 : kzbin.info/aero/PLH9R5x7JVXCFyPB1oN2oYSqXZv0_2A_AS

@kalaeswareswar4030 Жыл бұрын

@@ALLABOUTELECTRONICS 😍tqs Sir

@saumyasingh1844 Жыл бұрын

At 2:30 ... Didn't get IB equation...from where (beta +1) RE came in denomination

@farhanupaul 4 жыл бұрын

Is two supply emitter bias configuration better for both stability and amplification?

@Jnglfvr Жыл бұрын

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.

@Fotosepken Жыл бұрын

why we say 10:49 emitter bias configuration. Isn't it common base configuration because base is grounded instead of emitter or collector

@ALLABOUTELECTRONICS Жыл бұрын

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.

@kalaeswareswar4030 Жыл бұрын

Tqs Sir ❤

@vinaykumarkv5830 4 жыл бұрын

Superrrrb🔥👌

@panneerselvam6241 4 жыл бұрын

Nice bro

@imucetaspirants6272 4 жыл бұрын

Please try to solve while you teach we really can't get most of the formulas please sir it's a request

@ALLABOUTELECTRONICS 4 жыл бұрын

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.

@straxescraft 4 жыл бұрын

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?

@ALLABOUTELECTRONICS 4 жыл бұрын

Would you please mention the timestamp where you are referring to in the video.

@straxescraft 4 жыл бұрын

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.

@ALLABOUTELECTRONICS 4 жыл бұрын

@@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.

@straxescraft 4 жыл бұрын

@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.

@narendrababukrishnan6541 Жыл бұрын

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

@mayurshah9131 5 жыл бұрын

Very good 👌👍👌

@gurpreetsingh206 5 жыл бұрын

@rukshanfdoable 3 жыл бұрын

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

@ALLABOUTELECTRONICS 3 жыл бұрын

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.

@rukshanfdoable 3 жыл бұрын

@@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

@zwehtetnainai6904 Жыл бұрын

how we we determine beta for different temperature ? pls let me know.

@ALLABOUTELECTRONICS Жыл бұрын

You will find the curve in the data sheet.

@sayalikathore559 2 жыл бұрын

In the output side while applying kvl how Vcc can be positive 😭😭 please reply to my questions

@ALLABOUTELECTRONICS 2 жыл бұрын

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.

@sowmyagolla9815 3 жыл бұрын

sir can you please tell why does the base current decreases if voltage across emitter increases

@ALLABOUTELECTRONICS 3 жыл бұрын

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.

@sowmyagolla9815 3 жыл бұрын

@@ALLABOUTELECTRONICS ok thank you sir

@polaroid2674 Жыл бұрын

@@ALLABOUTELECTRONICS thank you I had the same doubt

@kanishkasingh1439 10 ай бұрын

In 11:31 why is vb 0 if rb is very small . It might be a lame question sorry!

@ALLABOUTELECTRONICS 10 ай бұрын

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.

@teeggindarshannagappa8858 Жыл бұрын

Why increase in Vb, decreases Ib....? Ref:- 1:25

@ALLABOUTELECTRONICS Жыл бұрын

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.

@darshanteggin6505 Жыл бұрын

@@ALLABOUTELECTRONICS Got it, Thank you.

@saquibalam7871 3 жыл бұрын

@4:41 sir from where you take Vbe=0.7 Volt??

@ALLABOUTELECTRONICS 3 жыл бұрын

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.

@saquibalam7871 3 жыл бұрын

@@ALLABOUTELECTRONICS ohk, thank you sir!

@saquibalam7871 3 жыл бұрын

@@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.