The timestamps for the different topics covered in the video: 0:23 Introduction to Voltage Divider Bias Configuration 1:43 Approximate DC Analysis of Voltage Divider Bias Configuration 4:30 Exact DC Analysis of Voltage Divider Bias Configuration 9:50 Solved Example on Voltage Divider Bias For more Solved Examples on the BJT, check out this playlist: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-byPBZmCfsb8.html
Why is the equation for Vce, Vce=Vcc-Ic(Rc+Re) instead of Vce=Vcc-IcRc-IeRe? Is it because we assume that Ie=Ic? if so, is it still under exact method or approximation method? sorry for bad english
Iske notes ke pdf download kar sakte hai kya for fast revision Is there any link or website of this channel from where i can download this presentation ppt
Vce is the drop across the collector and emitter terminals of BJT, while VE is the drop across emitter terminal. To find Vce, first we need to find Vc and then Ve. The difference between them will give us the drop across those two terminals of BJT. Vc = Vcc - Ic Rc and Ve = Ie Re. And the difference between them will give us Vce. That’s what is shown at 3:50
The collector current is 1.266 mA. During the calculation, if you take the value of collector current up to 3rd decimal place, then Vce will come out as 6.174 V. If you consider the value of Ic up to 2 decimal places, then Vce = 6.204 V.
They are same. In the self biasing, we do not require another power supply for biasing the base of the transistor. Using external resistors, it is derived from the existing power supply. The same is done in the voltage divider bias.
As I mentioned in the video, you can check the condition for the approximation. That means if Rth is much less than (B + 1) RE, then you can do the approximate analysis. And as per the approximate analysis, you can find the Vb, just using the voltage divider rule. If condition mentioned above does not get satisfied, then you need to do the complete analysis by finding Vth and Rth.
I hope, you got the expression of Ib. Now, Ie = (beta + 1) Ib. Just put the value of Ib from the expression. And then divide the entire expression by (beta + 1). You will get that expression. If you still have any doubt then let me know here.
If you see, the base emitter junction is forward biased. So, it acts like a diode and the voltage drop across it can be assumed to be around 0.7V during the analysis. I hope it will clear your doubt.
Many such problems have been solved on BJT on the second channel. Here I am giving the link for the playlist on the solved problems on BJT. Ypu will find such example as well. Here is the link: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-8bePSIW9b9E.html Still if you find any difficulty, let me know here.
If there are few unknown, then must have been given other parameters. One such example is covered on the channel. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-rbcDdTOl-y0.html Please go through it.
They are used to block the DC biasing voltage. Typically these amplifiers are using in cascade configurations where more than one such amplifiers are cascaded. The DC biasing voltage of one amplifier should not affect the next amplifier or any other circuitry after the amplifier. The capacitor blocks the DC voltage and passes only AC signal.
Yes, that was considered for the approximate analysis. And it can be assumed when (B+1) Re >> Rth. At 7:21 the expression of the base current gives the exact value of the base current. I hope it will clear your doubt.
I have already covered it. Please go through the BJT playlist. If you go through the video, small signal analysis of CE configuration, you will get it.
Its emitter current. Just multiply the base current with ( beta + 1). Taht means, i nthe numerator you will have ( beta + 1) term. Now, multiply and divide the numerator and denominator by (beta + 1). So, finally, you will get {Rth/(beta+1)}+Re } in the denominator. I hope, it will clear your doubt.
