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Transistor circuit: Determine source resistance

  1. Apr 9, 2015 #1
    1. The problem statement, all variables and given/known data

    Calculate Rs so that the draincurrent ID = 4 mA. Vad does the gain A become? uo/ui? For the transistor:

    k = 3 mA/V^2
    Vt = -2 V

    also:

    R1 = 50 k ohm
    Rd = 2.2 k ohm
    C = infinity for signalfrequencies

    image of circuit:

    http://sv.tinypic.com/view.php?pic=9pyq9z&s=8#.VSbHQfl_vxM


    2. Relevant equations
    [itex]i_{ds} = \frac{1}{2}k(v_{gs} - V_{t})^{2} [/itex] (1)

    3. The attempt at a solution

    [itex]I_{D} = 4 mA [/itex] (2)
    [itex]v_{d} = E - R_{D}\cdot I_{D} [/itex] (3)
    [itex]v_{s} = R_{S}\cdot I_{D} [/itex] (4)
    [itex]v_{ds} = ???? [/itex]

    From equation 1 we can get v_gs, that is the voltage between gate and source. But I still don't know how to get RS since I dont know u_in or v_ds.
     
  2. jcsd
  3. Apr 9, 2015 #2

    phyzguy

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    Science Advisor

    I think you are supposed to calculate Rs so that the quiescent current Id = 4mA. So you can assume that u_in = 0. Then you know Vg so you can write an equation for Ids (which you are given) in terms of Rs and solve for Rs.
     
  4. Apr 9, 2015 #3
    I don't understand why u_in = 0 just because it is the quiescent current?
     
  5. Apr 9, 2015 #4

    phyzguy

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    Well, I don't know for certain what the problem had in mind, but it looks like a small signal amplifier. So probably u_in is a small AC signal with an average value of zero, which best amplified and the amplified signal appears at u_out. The quiescent current is the current flowing when no input signal is applied, which means that u_in = 0. To be sure, maybe you should ask whoever assigned the problem whether you can assume that u_in = 0 when calculating Rs.
     
  6. Apr 9, 2015 #5
    From earlier examples I've seen in lectures that u_in is nonzero in quiesent operation. For example: U_gsq = 5 V, U_dsq = 5.5 V and I_dsq = 4.5 mA.

    I got the right answer so you are most probably right. But the quiescent current is a DC current so the quiescent input voltage must also to be a DC voltage. How do I know u_in = 0 produces a quiescent current of 4 mA? Since I am calculating on the quiescent current I am only taking the DC voltage part of u_in otherwise I would had to do my calculations on the small-signal model.
     
  7. Apr 9, 2015 #6

    phyzguy

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    The quiescent current is defined as the current that flows even when the input voltage and input current are zero. It is a current intentionally set up in the circuit so that the amplifier is biased at a point at which it will function properly.
     
  8. Apr 10, 2015 #7

    NascentOxygen

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    Staff: Mentor

    Gate-to-source voltage in your circuit will be non=zero too. It makes no difference to the transistor whether source is zero volts and the gate has a negative potential, or whether the gate is zero and the source is at some positive potential (as here). The important figure is that VGS is appropriate.
     
  9. Apr 10, 2015 #8

    Zondrina

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    Homework Helper

    For the DC analysis, you should be assuming ##u_{in} = 0##. You can superimpose the AC analysis separately afterwards.

    I'm assuming there is a common ground, so this would lead to ##V_G = 0## because ##I_G = 0##.

    The capacitor is an open circuit for DC, so all that is left is to find ##V_{GS}## using the saturation current ##I_D##.

    Then simply using the facts ##V_{GS} = V_G - V_S## and ##I_D = I_S##, you can solve for ##R_S = \frac{V_S}{I_S}##.

    As for the gain, simply look at the circuit with the AC source non-zero and all DC sources grounded. The capacitor will behave like a short circuit so its probably a good idea to re-draw the circuit to be careful.
     
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