AC Load Line of BJT Amplifier Circuit

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  • Опубліковано 22 лип 2024
  • The load line for a BJT amplifier indicates the range of allowable values for collector current and collector-emitter voltage. It is useful for predicting at what point clipping on the output would occur in the circuit.
    Outro Song Credit: Enter the Tesla Machine by Tri-Tachyon (CC BY 4.0)
    David Williams
    www.elen.ca

КОМЕНТАРІ • 28

  • @DAHBLIFE
    @DAHBLIFE 7 років тому +5

    Thanks David! I learned so much from your videos! Great review material!

  • @golski4878
    @golski4878 4 роки тому +5

    Finally someone have more detailed explanation thanks buddy

  • @aerofart
    @aerofart Рік тому

    Excellent video that leads to a great understanding of how transistors do their thing when amplifying AC signals. Another way to bias without all the math is to hook up your oscilloscope and run a sinusoid through the circuit to check for clipping and min/max levels. Also, a spectrum analyzer clearly reveals harmonic distortions.

  • @manaoharsam4211
    @manaoharsam4211 Місяць тому

    Very Nice explanation.

  • @ivyloca3443
    @ivyloca3443 Рік тому

    Excellent explanation!! I finally understand it!

  • @backupemail3273
    @backupemail3273 2 роки тому +1

    Thanks man, it was helpful.

  • @SaeedMurtaza
    @SaeedMurtaza 4 роки тому +1

    nice explanation

  • @nhutnguyen2909
    @nhutnguyen2909 2 роки тому +1

    it's helpful for me, thank you so much.

  • @aditikumari3677
    @aditikumari3677 5 років тому +2

    What is that re in ac model ?😥

  • @ronnachshon
    @ronnachshon 4 роки тому +1

    god bless you!!!

  • @TheZiggyman01
    @TheZiggyman01 7 років тому +2

    Hi David, @7:43 Can you please explain it in a bit more detail why you can use the Icq in place of the AC collector current?

    • @jasonw.2232
      @jasonw.2232 6 років тому

      I have the same question. Thanks in advance!

    • @jasonw.2232
      @jasonw.2232 6 років тому +3

      Nevermind, it's actually explained in the example video at 5:00. Because ic can swing down from the DC Q point to 0mA (when there is no collector current), the magnitude of ic is equal to the DC current Q value found before. Beyond this, we would have clipping, and therefore distortion of the incoming AC signal. Hope this helps!

    • @nevis2769
      @nevis2769 10 місяців тому

      ​@@jasonw.2232legend

  • @HardikJain_YT
    @HardikJain_YT 6 місяців тому

    Doubling isnt good there can be saturation constraints also which may reduce the Vp-p if the emitter voltage is considerably high

  • @likhithanakka
    @likhithanakka 4 роки тому

    nice

  • @BenjaminK2
    @BenjaminK2 6 місяців тому

    At @2:53 the guy seemed to indicate that the resistor "re" was part of the actual transistor and that the transistor "E" point was on the other side, below said resistor. Is this an error right? (Those who are more familiar with this stuff may glide over such a trivial detail, but I'm hanging on to every detail to trying to follow😅). Thanks

    • @ElectronXLab
      @ElectronXLab  5 місяців тому +1

      re is part of the model of the transistor while RE is the external resistor that is part of the circuit

  • @klipk7296
    @klipk7296 4 роки тому +7

    finally a video where it isn't someone trying to explain this stuff with a thick indian accent

  • @e_wa.n5036
    @e_wa.n5036 2 роки тому

    a little confusing with the way you draw your ac Vce i keep mistaking it for the root symbol. But otherwise...amazing job with the expliantion!

    • @ElectronXLab
      @ElectronXLab  2 роки тому

      I'm glad you got something from my explanation, but sorry about my v's. I needed a way to distinguish upper case from lower case V's, but I think my hand just can't make a very convincing lower case v.

  • @jamilhneini1002
    @jamilhneini1002 6 років тому +4

    Why is Vce = ic Rload and not Vce = - Ic Rload?

    • @yashgupta3127
      @yashgupta3127 5 років тому +2

      It's negetive actually see in diagram the slope is negetive

    • @workspaceofduc
      @workspaceofduc 4 роки тому

      Why is Ic=Vcc/(Rc+Re) not Ic=Vcc/Rc? I thought Re is in another current?

    • @oneinabillion654
      @oneinabillion654 3 роки тому

      @glyn hodges honestly I don't get it 😂