48. A Multistage JFET Amplifier

Поделиться
HTML-код
  • Опубликовано: 30 июл 2024
  • Someone had asked me to cover a multistage JFET amp and this offering is the result. This video acts a little like a review of the earlier JFET videos. The circuit consists of two common source amplifiers feeding a common drain amp. Some of the math is discussed and, as usual, a circuit is built and tested. Hope you find it useful!
  • НаукаНаука

Комментарии • 21

  • @audiofreq
    @audiofreq 3 года назад +2

    My favorite college professor! Took his electronics course a few years ago! Excellent teacher!

  • @hubercats
    @hubercats 2 года назад

    Great presentation! Thanks so much for making and sharing it. It ties together many aspects of circuit design.

  • @Dazzwidd
    @Dazzwidd 14 дней назад

    The only reason to use a jfet is on the input to get a high impedance, after that i'd use bipolar transistors. That way you can get more voltage gain. You could possibly get all you want with 2 stages.
    An interesting idea would be a cascode where the lower device is a jfet and the upper one is a bipolar transistor

  • @peterwu831
    @peterwu831 3 года назад +3

    For audio application my JFET biasing would start with a "Q point" at Id = .382 * Idss, which would also mean Vds = 0.382 * Vp (this can be derived from the governing transfer characteristic curve Id = Idss (1-Vgs/Vp)/2 ). So for self biasing design, the starting value of Rs is simply |Vp / Idss|, and can be micro adjusted later to maximize the gain. The Rd is selected to achieve Vd ~ 1/2 of Vdd, the mid point of DC rail, using Rd = (Vdd + Vgs)/2/Id. This would maximize the signal peak-to-peak potential.
    Form the transfer characteristic curve, this "double .382 point" is a "trade off" between the "half Vp" and "half Idss" methods, and can be argued to give a more optimized line. The simplistic calculation for the biasing resistors Rs and Rd is just a bonus.

  • @BalticLab
    @BalticLab 11 месяцев назад

    When finding an "average" transconductance from the min and max values in the datasheet, it is recommended to use the geometric mean and not the arithmetic mean. It's the square root of the product of min and max value. In your case, the geometric mean is SQRT(2000*7500)=3873 uS. Which is close to your 3500. The same principle is also applied when finding an average hfe for bipolar transistors.

  • @johnlavere8675
    @johnlavere8675 6 лет назад +1

    Great analysis video. I teach the analysis method involving the graphic I-V solution process. One question on actual use of a configuration like this: the JFET transfer curve is significantly non-linear for larger signal ranges. Does it make sense to use a JFET beyond the first stage (input stage) in this type of design?

    • @acmefixer1
      @acmefixer1 5 лет назад +1

      That's a good question. I wouldn't use a JFET for the second stage.

  • @acmefixer1
    @acmefixer1 5 лет назад

    I think that 1.6k for the drain load resistors is not a realistic value, the voltage gain would be higher if they were a higher value. But JFETs spread of parameters are so wide that I've seen stages where they won't work until the gate voltage is a few volts above ground. I think that's the reason why we seldom see JFETs used in the standard common source amplifiers. They're just too weird! Thanks for the informative video.

  • @sakii1997
    @sakii1997 4 года назад

    What is the type of transistor that you used?

  • @JurekPrzezdziecki
    @JurekPrzezdziecki 7 лет назад

    very useful, thanks a lot. will simulate it and experiment in Multisim with it.

    • @TheOffsetVolt
      @TheOffsetVolt  7 лет назад

      JurekPrzezdziecki Thank you. I may try doing the Multsim experiment myself. Someday I may add Multisim file for each of my videos.

    • @JurekPrzezdziecki
      @JurekPrzezdziecki 7 лет назад

      this would be really great, what fets you use?

    • @TheOffsetVolt
      @TheOffsetVolt  7 лет назад

      I used MPF102's for all three stages. Not the best choice for the low frequency but I used it principally to show the losses incurred in the common drain section. You can substitute 2N5484, 85, or 86.

  • @revenge620
    @revenge620 6 месяцев назад

    06:45 why did you chosen to cross at -5(about) Vgs? What's determining it? i mean: you accept the 11mA starting from the datasheet's average.
    Then you calculate the 5.17V by using the 470 ohm resistor by applying Ohm Law V = R*I. but I don't see (or it's not celar to me) how you got the 470 Ohm value out.
    Did you determine upfront somewhere to get the -5.1V having that current? and if yes: why did you chose for that?

  • @alexloktionoff6833
    @alexloktionoff6833 Год назад

    Only 5, why so low gain?
    I’d expect at least 50!
    Also I’m not sure about the linearity. How much THD for this amplifier?

  • @majordabalert
    @majordabalert Год назад

    Sick

  • @SpinStar1956
    @SpinStar1956 8 месяцев назад

    😊 👍

  • @chronobot2001
    @chronobot2001 4 года назад +1

    Too many ads... come on dude.

    • @TheOffsetVolt
      @TheOffsetVolt  4 года назад

      I agree, that's why I turned off all of the new mid-rolls youtube just added. It should be better now.