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voltage gain amplifier model

Discussion in 'Electronics Homework Help' started by kbcheong, Mar 26, 2012.

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  1. kbcheong

    kbcheong

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    Mar 24, 2012
    The figure shows a model of a real power amplifier. The dependent source is a voltage-controlled current source with characteristic

    iD = K/2 (vGS – VT)2 if vGS > VT
    iD = 0 if vGS < VT

    This device model has two parameters K = 2.0 A/V2 and VT = 2.0V

    Assume that the power supply voltage VDD = 145.0 V, and resistance RS = 22.0 ohm and RL = 120 ohm,

    Assume the input bias voltage VI = 6.5V

    What is the bias current ID, in Amperes, and output bias voltage Vo in volts?

    I tried to substitute to the iD formula but unable to get a correct answer, do we need to include the voltage of VDD?
     

    Attached Files:

  2. Laplace

    Laplace

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    Apr 4, 2010
    The voltage drop across the resistors is determined by the drain current, so the voltage drop across the current source is determined in part by VDD.
     
  3. kbcheong

    kbcheong

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    Mar 24, 2012
    What about the resistor Rs and RL . Can it be calculate by (VDD/ RL) + (VT/Rs) + id?
     
  4. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    The model describes a MOSFET. Have a look at the datasheet of a real MOSFET, e.g. http://www.fairchildsemi.com/ds/BS/BS170.pdf.
    Figure 1 on page 3 shows how Ids depends from Vgs and Vds. If Vds is high enough, Ids depends only from Vgs.
    In your problem:
    1) write down UGS in terms of Vi, VI and the voltage drop across Rs (hint: V=R*I, which I do you need?). You do not need Vds here.
    2) solve the above equation for Vgs.
    3) use Vgs to arrrive at Ids
    4) use Ids to calculate the voltage drops across all resistors. Consistency check: the sum should be less than the supply voltage.
    5) From the supply voltage and the voltage drops across the resistors you find Vds.
    You now have all the information to completely analyze the circuit. Finding Vo now is a no-brainer.

    Harald
     
  5. kbcheong

    kbcheong

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    Mar 24, 2012
    Is UGS = VI - vi and VGS = RS/RS+RL X VDD?
    The value of vi is not known, how to derive the equation to get the VGS?
     
  6. Laplace

    Laplace

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    Apr 4, 2010
    When you are asked to find the bias point, the input signal voltage vi is quiesced to zero. The bias point is also referred to as the quiescent point, hence vi=0.
     
  7. kbcheong

    kbcheong

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    Mar 24, 2012
    For the similar question, If we increment the input voltage vi = VI + delta vi. the output voltage will change v0 = VO + delta vo.
    If delta vi = 1V what is the delta vo, in Volts?
    May I enquire how can we obtain the delta vo from the equations?
     
  8. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    1) No, UGS <> VI
    2) VI sets the operating point whereas Vi is a small AC signal that wil be amplified by the transistor. You don't need to know a value for Vi if you want to calculated the amplification of the MOSFET. Using the next tip and my previous explanation you will arrive at an expression for Vo which includes Vi as a variable. From Vo/Vi you then get the amplification.
    3) Start at the lower left node (- of VI). work upwards and add voltages VI -> Vi -> Vgs -> V(Rs). Use Kirchhoffs laws to arrive at the sum of the voltages. Then continue as I describes above.

    Harald
     
  9. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    One more hint:
    VI+Vi-Vgs-Id*Rs = 0V
    By inserting the equation for Id into above formula you will arrive at a quadratic equation.
    Solve the equation for Vgs.
    With known Vgs you can calculate Id.
    With known Id you can calculate the voltage drop over Rs and Rl.
    Knowing these voltage drops and Vdd, you can now calculate the voltage drop acrosss the current source (i.e. drain-source voltage) and thus Vo. Now you have all the required bias values (assume Vi=0 for this purpose, as Laplace already wrote).

    Since VI is given, you have now Vo=f(Vi). Derive dVo/dVi=f ' (Vi) to arrive at the amplification of the stage.

    To make things easier you can substitute Vin= VI+Vi and use Vin in the calculation instead. Just remember to replace Vin by Vi+Vi in the final solution.


    Harald
     
  10. kbcheong

    kbcheong

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    Mar 24, 2012
    Is vGS = VI - vi?
     
  11. Laplace

    Laplace

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    Apr 4, 2010
    Is that a guess, or did you fail to read the hint provided by Harald?

    VI+Vi-Vgs-Id*Rs = 0V or Vgs = (VI+Vi)-Id*Rs

    Vgs is defined as the voltage from the gate to the source, Vgs=Vg-Vs, where in this case, Vg = (VI+Vi) and Vs is the voltage across the source resistance so Vs= Id*Rs.

    Vgs=Vg-Vs or Vgs=(VI+Vi)-Id*Rs
     
  12. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    No, that would be too easy, wouldn't it?
    Please rad my previous posts again. You have to take the voltage drop across Rs into account. All tha information you need is in my previous posts.

    Harald
     
  13. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Ome more hintg:
    out of pure curiosity I've done the calculation. It fits on a single sheet of paper. And no, I will not give you the full answer, but I will gve you the range where your result should be:
    2 V <Vgs < 3 V
    0.1 A < Id < 0.2 A
    4 < Vo/Vi < 7

    When you've done the math, you may post your results and I will check against mine.

    Harald
     
  14. jack Sparow

    jack Sparow

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    Apr 5, 2012
    ).

    Since VI is given, you have now Vo=f(Vi). Derive dVo/dVi=f ' (Vi) to arrive at the amplification of the stage.

    can some1 tell me what Vo=f(Vi)
     
  15. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    I could (at least in about 10 days since I currently do not have access to my notes).
    But I won't. This is the homeworks section and you're supposed to find the solution yourself - once you're a full grown engineer you will have to know how to do it anyway.

    My explanations have been not too long due to the nature that this is a forum. If you want to read up on Mosfet source followers, look e.g. here: http://whites.sdsmt.edu/classes/ee320/notes/320Lecture36.pdf or Google MOSFET SOURCE FOLLOER

    What I offer is that I can check your solution for correctness. But, as said above, in 10 days.

    Harald
     
    Last edited: Apr 5, 2012
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