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Single stage amplifier with current mirror biasing

  1. Mar 24, 2015 #1
    1. The problem statement, all variables and given/known data
    123.jpg

    2. Relevant equations

    ID= kn (Vgs - Vth)^2
    3. The attempt at a solution

    I asked to find the current I D , M2.


    I think in order to solve this question I need to find Vgs but how do I do that?

    RC1 = 2k ohm
    R = VDD - Vgs / I D ,M2

    VDD = 5V , and R = 2k ohm and I have 2 unknown variable. Vgs and I D ,M2

    I think i need to make use of this ID= kn (Vgs - Vth)^2 to solve this question
     
  2. jcsd
  3. Mar 24, 2015 #2

    rude man

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    Good idea! But you must first find ID for M1. Combine your last equation with another expression for ID summing currents at the drain.
    Then, you assume M1 and M2 have identical I-V characteristics so knowing IDM1 gives you IDM2, and you're done.
     
  4. Mar 24, 2015 #3
    Could I have more information why the characteristics are identical for M1 and M2? They only have the same value for Vth while their Kn is different by a factor of 2?

    I forgot to mention that I D,M1 is also given as 2mA for this question so I D, M2 is also 2mA?
     
  5. Mar 24, 2015 #4

    rude man

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    I didn't see that, sorry.
    No. For any MOSFET in the 'saturation' region with λ = 0, which this is since Vgs > Vth & Vds > Vgs - Vth),
    I = k(Vgs - Vth)2.
    So, since k2 = 2k1, what is I,M2 if I,M1 = 2 mA?
     
  6. Mar 24, 2015 #5
    Hi

    So I find the value for Vgs by subbing in Vth and I D,M1. Once I gotten the value of Vgs , i sub this value into the equation to find i D,M2 with a k2? Am I correct?
     
  7. Mar 24, 2015 #6

    rude man

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    No point in solving for Vgs. From my last post it should be obvious that I_M2 = 2I_M1. Solve for I_M1 as I discussed earlier.
     
  8. Mar 24, 2015 #7
    Sinc k2 = 2k1 , I2 = 2I1. I2 = 4mA?
     
    Last edited: Mar 24, 2015
  9. Mar 24, 2015 #8
    I have another question for this. I am asked to find the small signal parameter of the 2 MOSFET .
    The formula are
    Output resistance in the small signal equivalent circuit of a MOSFET operating in the saturation region is given as R0 = 1 /(channel length modulation * Current ID)

    I have ID value for both 1D M1 and 1D M2 but in this question , channel length modulation for both are equal to 0
    when you sub in these value , you will get an error.
     
  10. Mar 24, 2015 #9

    rude man

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    Good guess!
     
  11. Mar 24, 2015 #10

    rude man

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    Suppose the channel modulation was a very small number but not zero - what would be the output resistance then?
     
  12. Mar 24, 2015 #11
    hmm good guess meaning is ID , M2 = 4mA correct or wrong?

    For channel length modulation
    In the picture it says that the channel length modulation is equal to 0 for channel length modulation M1 and channel length modulation M2.
     
  13. Mar 24, 2015 #12
    I would also like to check if this is a configuration of a Common source with source degeneration single stage amlifer?
     
  14. Mar 24, 2015 #13

    rude man

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  15. Mar 24, 2015 #14

    rude man

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    What terminal is common to the input and output?
    What is the definition of "source degeneration"? You want answers on a platter and we're not allowed to give you that - which is to your benefit.
     
  16. Mar 24, 2015 #15
    The source is connected to the gate and Output is connected to the drain
     
  17. Mar 24, 2015 #16

    rude man

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  18. Mar 24, 2015 #17

    rude man

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    Yes, but what terminal is common to the input and the output? What is the input referred to, and what is the output referred to?
     
  19. Mar 24, 2015 #18
    Input is connected to the gate , Output is connected to the drain. Source is connected to neither input or output
     
  20. Mar 24, 2015 #19
    I am looking at the amplifier at M2
     
  21. Mar 24, 2015 #20

    rude man

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    Good.
    An input is a voltage. A voltage is always between two points.
    An output is also a voltage. Same thing.
    So, again - what terminal is common to input and output?
     
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