43 BJT Circuits at DC

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  • Опубликовано: 7 фев 2025

Комментарии • 47

  • @ahyungrocks5509
    @ahyungrocks5509 11 месяцев назад +9

    This is super helpful. 30 years ago when I took microelectronics while learning English, I was not able to comprehend despite I passed the course. Because of that I stayed away from the dark side (analog). Didn't realize this is so much more fun than 1s and 0s.

  • @mugiwara-no-luffy
    @mugiwara-no-luffy 2 года назад +14

    I hope you know how valuable these videos are! Cheers from Canada.

  • @mohamedqasim7787
    @mohamedqasim7787 3 года назад +6

    Thank you Prof. for this precious explanation 🙏

  • @raccoonlord9646
    @raccoonlord9646 Год назад +3

    Big W, U saved my ahh for this exam bro u the best

  • @jxcjxuan
    @jxcjxuan 9 месяцев назад +1

    clear and smooth explanation, suggesting to insert the special formula for reference

  • @tonykimani9933
    @tonykimani9933 8 месяцев назад +1

    perfect for what I have been looking for

  • @ramirosuarez9537
    @ramirosuarez9537 8 месяцев назад +1

    I am so grateful for your explanation. Thanks

  • @gilbertgonzalez7890
    @gilbertgonzalez7890 3 года назад +2

    Thank you! Superb explanation

  • @cucuteanualexandru-ciprian4244
    @cucuteanualexandru-ciprian4244 Месяц назад

    thank you for your video

  • @angelabryan7529
    @angelabryan7529 10 месяцев назад

    I hope you will consider adding videos for the second half of the textbook: chapters 9 -14

  • @thunderd7904
    @thunderd7904 Месяц назад

    19:45 I know I'm 4 years late and my test is in 2.5 hours, but couldn't you have just voltage divided it to get the voltage at that node and then solved normally from there?

  • @niccettajames6072
    @niccettajames6072 2 года назад +1

    Good job explaining

  • @ToolsAandLogic
    @ToolsAandLogic 4 месяца назад

    How did you derived 0.3 v value?
    How can electron pass base collector junction, when it is in reverse bias? Generally, in PN junction, in reverse bias, election can not cross junction in reverse bias.

  • @jonelberdejo9446
    @jonelberdejo9446 Год назад

    In your final analyis I think the transistor in saturation mode. Because the base current is approximately 129 micro amps and then the current gain is 100 so the collector current should be(100*129microA), which is 12.9 mA. So to find the collector voltage we have to subtract 15-(12.9mA*5000) and we get an impossible answer that is -49.5 which indicates that the transistor is operating in the saturation mode. A hard saturation

    • @gwav1a
      @gwav1a Год назад

      Yes it turns out like that but in actual fact you apply 0.7 across the base emmitter diode and the large part of the majority carriers from the emmitter cross the base collector junction to constitute the collector current. The smaller portion comes out through the base. The proportionality of collector current to base is beta. Viewing it this way shows causation - what results in what. To me base current is the result, not the cause of collector current. I would say the base current is a function of collector current not the other way

  • @SAhellenLily
    @SAhellenLily 9 месяцев назад

    Thank you 👍

  • @LukiePookieStudy
    @LukiePookieStudy 2 месяца назад +1

    19:10 isn't it 10V for Rth because 15V*(100/150) = 10 not 5

    • @El650Jefe
      @El650Jefe 2 месяца назад +1

      No because Vb is the voltage drop across the 50k resistor not the 100k resistor

  • @babatundetaiwo2817
    @babatundetaiwo2817 3 года назад +3

    So let me get this straight: If I apply a 0.7 Voltage bias from Base with respect to the emitter while also supplying current in the base of the NPN, I will get an output current from Collector to emitter that is a multilocation beta of the Base current. This the emitter is what I'm taking everything in respect to : Voltage form Base to emitter with a current applied to the base outputs a BETA times the input current flowing from collector to emitter.

  • @MrSomethingdark
    @MrSomethingdark 16 дней назад

    I have serious problems understanding how we arrive at these considering the real current direction and engineering norm agreed upon. Also, coming from a physics perspective, I would need to see way, WAY more graphical geometric explanations. The reasoning on some of these is triple backwards. Shouldn't it always be that we determine the input characteristics and downstream from that arrive at other values according to the input???

  • @arashyusefi1889
    @arashyusefi1889 Год назад +1

    Thanks Microelectronics 💖✨💫🙏🌹💖

  • @labanidharnaik1103
    @labanidharnaik1103 3 года назад

    Very clear.

  • @GOODBYE.SHAHZAIBNAJAF
    @GOODBYE.SHAHZAIBNAJAF 3 года назад

    Great man

  • @Lexyvil
    @Lexyvil 6 месяцев назад

    For the circuit at 9:30, why is there a current I_c when the diode at the collector point is against I_c? (Reversed)? Since it's reversed bias, shouldn't I_c be 0?

    • @dvx-ze1qz
      @dvx-ze1qz 6 дней назад

      This is how transistor works, and the CB junction shouldn't be compared against a separate diode. In fact, the base area of the transistor is much thinner than the collector or the emitter, so if you had two diodes connected back to back, it will not work as a transistor does

  • @sipoVvids
    @sipoVvids 9 месяцев назад

    Great!!

  • @sibilakshman9575
    @sibilakshman9575 Год назад

    In saturation why is Vce = 0.2 V ?

  • @makokakao6782
    @makokakao6782 Год назад

    which books is you took the problem examples ?

    • @nelsonkh001
      @nelsonkh001 Год назад +2

      Microelectronic Circuits by Sedra & Smith

  • @tinaquayum8910
    @tinaquayum8910 2 года назад

    How did you got .2v

  • @thokozaninhlakahh9255
    @thokozaninhlakahh9255 Год назад

    how much is alpha ,please show steps

  • @Epigonee
    @Epigonee 4 месяца назад

    OMG AN ACTAL ACCENT

  • @LegoBenjago
    @LegoBenjago 3 месяца назад

    big up da mandem

  • @VndNvwYvvSvv
    @VndNvwYvvSvv 2 года назад

    Thanks, but in the final analysis around 22:00 - 23:00 you dropped the k on the resistor. I think it's 33.3k(IB) plus (beta+1)IB. That's 33,300 +303, so then Ib = 4.7/(33,603) ≈ 139.87 uA.

    • @ohbaewan3147
      @ohbaewan3147 2 года назад +2

      not quite, he factors out the k at first because it is part of the units, so its (beta+1)IB*3 which is why it simplifies to 303 but is still in units of k, then in the end he adds them up, 303 + 33.3 = 336.3 and adds the k back during unit evaluation.

  • @StuartSegers-g6w
    @StuartSegers-g6w 4 месяца назад

    Caleb Prairie

  • @shantnukumar3982
    @shantnukumar3982 9 месяцев назад

    What is this book name

  • @worldgalaxyuniversemilkiwa1321
    @worldgalaxyuniversemilkiwa1321 9 месяцев назад

    Negative and positive 😊

  • @BraySetlla-h2b
    @BraySetlla-h2b 4 месяца назад

    Kallie Court

  • @malekhaboosh476
    @malekhaboosh476 3 года назад

    Use BJT transistor to design an amplifier circuit that operates in the frequency range 10kHz100kHz. Verify your design by simulating the design using the circuitmaker software. Use
    standard resistor values (10% tolerance). The design should meet the following criteria:

  • @ikmalhakimi2957
    @ikmalhakimi2957 2 года назад

    TQ OIIIIII !!!🤣