Help - Amplifier with BJT and Mosfet

Discussion in 'Homework Help' started by Alfa_ET, Nov 26, 2011.

  1. Alfa_ET

    Thread Starter New Member

    Feb 26, 2011
    17
    0
    Hi everybody,

    I need help in this exercise of electronica. I'm not sure about the results I got. Please see my resolution and tell me where I am wrong.

    Exercise:
    http://img7.imageshack.us/img7/2826/digitalizar0001pi.jpg

    Resolution part 1:
    http://img259.imageshack.us/img259/2006/digitalizar0002d.jpg

    Resolution part 2:
    http://img832.imageshack.us/img832/4278/digitalizar0003v.jpg

    Resolution part 3:
    http://img12.imageshack.us/img12/3479/digitalizar0004wd.jpg

    Resolution part 4:
    http://img522.imageshack.us/img522/3710/digitalizar0005im.jpg

    Thank you all
    greetings
     
    Last edited: Nov 26, 2011
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Keep in mind for the PMOS case the source must be at a higher potential than the gate.

    You have calculated for the PMOS

    Vgs = 3.61V and Vg = 3.64V

    Thus Vs2 should be 3.61V above Vg2.

    Hence I would think Vs2=3.61 + 3.64=7.25V rather than 0.03V ....
     
  3. Alfa_ET

    Thread Starter New Member

    Feb 26, 2011
    17
    0
    Yes you are right. And about Vdrain 2?
     
  4. Alfa_ET

    Thread Starter New Member

    Feb 26, 2011
    17
    0
    the gain of first stage is -gm1*Rd. And about the gain of second stage?
    I have to consider the output resistance of the first stage??? PLEASE HELP
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I would think the 'logical' thing to do would be to set VD2 [Vo] static bias voltage to 0V or mid-poiint between the ±10 rails. This is an unusual amplifier configuration I've not met before so I would have to look at the circuit more carefully to be convinced in my own mind.

    In any case if VD2 is 0V and ID2 ≈ 7.27mA then RL would be 10V/7.27mA=1376Ω.

    I also think your estimate for ID1 & RE might be off a little.

    One can show for the NMOS

    ID≈(gm)^2/(4*Kn)≈(4E-3)^2/(4*2.4E-3)=1.67mA

    hence

    RE≈0.815/1.67mA=488Ω
     
  6. Alfa_ET

    Thread Starter New Member

    Feb 26, 2011
    17
    0
    Why set Vd2 to 0V? maximum excursion of the signal?
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    That was my reasoning - yes. But as I noted earlier I'd have to take a closer look at the large signal conditions to confirm that supposition.
     
    Last edited: Nov 27, 2011
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Did a simulation of the configuration out of interest.

    I note that the onset of large signal output cut-off is fairly insensitive to the DC bias operating point of the PMOS drain.

    The output THD is ~10% with an input of 400mV p-p.

    Peak voltage gain is about 6x with phase inversion.
     
    Alfa_ET likes this.
  9. Alfa_ET

    Thread Starter New Member

    Feb 26, 2011
    17
    0
    Good work.
    RL and RD had given the same value, except RE.
    What simulation software do you use?
    can you check the output resistance of the amplifier? I think the value is close to the value of RL.
    the gain is gm1*RL but i don't know explain why.
    thank you
     
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Yes I would think Rout≈RL. The PMOS effectively acts as a current source driver into the load, so looking back into the drain terminal one sees a high AC impedance. That leaves just RL in parallel with a high impedance.

    As to the overall voltage gain I think it would be given by

    A_v=-\frac{g_{m1}g_{m2}R_LR_D}{1+g_{m2}R_D}

    which can be approximated by

    A_v=-g_{m1}R_L

    provided

    g_{m2}R_D>>1

    The latter is true in my simulation setup as the apparent gm2 value is much greater than the value of 5mA/V stated in the original problem definition. That's dictated by the PMOS type and its parameters.

    Looking at circuit values in my simulation the gm1 value works out around 4.3mA/V and this equates well with the observed voltage gain of 6x. Since |Av|≈gm1*RL=4.3*1.4=6.0
     
Loading...