Load Line Method with Transistors (19-Transistors)

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  • เผยแพร่เมื่อ 1 ม.ค. 2025

ความคิดเห็น • 12

  • @Hellhound604
    @Hellhound604 5 หลายเดือนก่อน +6

    Thank you, your lectures takes me back 50 years or so… thanks for the memories and also showing me I really need to relearn the basics. So often, as you grow older you forget all the basics…. Grrrrrr, getting old totally sucks. I mean, I did work as a design-engineer for 30+ years, but watching some elementary design videos really brings you back to how much you have forgotten in those years…

  • @davidreichert9392
    @davidreichert9392 2 หลายเดือนก่อน

    Best transistor tutorials on the web.

  • @harveyellis6758
    @harveyellis6758 5 หลายเดือนก่อน +2

    Videos in the past were more interesting when circuits were built and tested, compared to recent videos in which just theory and algebra are presented.

  • @ehsanbahrani8936
    @ehsanbahrani8936 5 หลายเดือนก่อน +2

    Thank you professor ❤

  • @samuelmatos9124
    @samuelmatos9124 5 หลายเดือนก่อน +1

    I love your explanation... Thanks!

  • @TheRevenant-pn2xi
    @TheRevenant-pn2xi 5 หลายเดือนก่อน +2

    Can't thank you enough

  • @dwk000
    @dwk000 5 หลายเดือนก่อน

    Useful Videos
    Thank you ...

  • @TOMTOM-nh3nl
    @TOMTOM-nh3nl 5 หลายเดือนก่อน +1

    Thank You

  • @stefano.a
    @stefano.a 5 หลายเดือนก่อน

    The last example raise a technical problem that I never solved in my life: for the "static" load line you obtained the line equation using Ohm's and Kirchhoff laws in the circuit: Ic = (Vcc-Vce)/Rc but how can the line equation be obtained for the "dynamic" load line (considering the effect of RL)? For the dynamic circuit the AC part of the supply is zero and in the axes of the graph we don't represent the AC values but only the instantaneous values. In other words I'm not so sure that the dynamic load line concept is correct. What do you think? Thank you for your very appreciated work.

  • @breedj1
    @breedj1 5 หลายเดือนก่อน

    Excellent videos. I am watching all of them on this channel.
    Question:
    In the first example Vbe = 0.9V was larger then Vce = 0.8V. So the transistor must be in saturation mode. Because Vcb = -0.1 V. Then why is the dot of the oparation point in the right graph in de active region? Or am I missing something.

  • @paulpaulzadeh6172
    @paulpaulzadeh6172 5 หลายเดือนก่อน

    Transistor is more voltage control device than current control, it is IB =IC/B not IB = IC x B .

  • @RasoulMojtahedzadeh
    @RasoulMojtahedzadeh 5 หลายเดือนก่อน

    When IB = 20 mA, IC is about 620 mA, so Beta shall be about 31 not 12! :)