frequency response push pull CMOS as amplifier.

Discussion in 'General Electronics Chat' started by luma, Jan 28, 2016.

  1. luma

    Thread Starter Member

    Nov 5, 2015
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    Hello, I would like to ask you a question about amplifiers in particular a buffer. I was simulating a buffer made by a pmos and nmos, (push pull) and I don't understand why the frequency response of the system doesn't change with or without the transistor. Does it means that the mosfet will not amplify and the signal pass just through the parasitic capacitors? If that is the reason then why the mosfet don't work as amplifier? (p.s the capacitances placed around the transistor should model simply the parasitic capacitors)

    question_all_about.PNG

    Thank you for your help.
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Maybe because your VDD and GND is wrongly connected (swapped). And your MOS spice model do not include any frequency dependent components.
     
  3. luma

    Thread Starter Member

    Nov 5, 2015
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    Hi Jony, thank you for you answer, however the power supply is connected correctly if we want to build a push pull amplifier, if i swap it it will become an inverter (which i don't want). Even if the mosfet has no parasitic capacitance inside its model, the gain in the pass band should change when the mosfet is in the circuit and when there is not.
     
  4. Papabravo

    Expert

    Feb 24, 2006
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    Maybe because the input is capacitatively coupled to the output. In AC terms the input and output are effectively at the same potential. Why did you do this?
     
  5. luma

    Thread Starter Member

    Nov 5, 2015
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    thanks for your answer, papabravo, yes that was also my guess . the reason why i put those capacitance is because i wanted to model the parasitic cpacitance Cgs and Cgd for each mosfet externally. So even if i don't put them but i specify the parasitic capacitance in the model of mosfet probably it will happen the same. practially what i did is just explicit the parasitic capacitance of the mosfet.
     
  6. luma

    Thread Starter Member

    Nov 5, 2015
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    if you have a more detailed answer please let me know :)
     
  7. crutschow

    Expert

    Mar 14, 2008
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    Your transistors are not biased correctly.
     
  8. luma

    Thread Starter Member

    Nov 5, 2015
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    can you be a little more specific. Anyway thank you for your suggestion
     
  9. Bordodynov

    Active Member

    May 20, 2015
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    See:

    AmplCMOS.png
     
  10. luma

    Thread Starter Member

    Nov 5, 2015
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    Thank you, Bordodynov to try to answer the question, anyway I would like to remember you and crutschow that i m dealing with a push pull and not with an inverter . In particular your example are IC which contains inverter and not push pull amplifier. Indeed you cannot expect a gain higher than 1 in a push pull while in your examples tha gain is much higher. So let me know , if someone have any other suggestions
     
  11. tophericks

    New Member

    Jan 29, 2016
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    Maybe because the input is capacitatively coupled to the output.
    [​IMG][​IMG]
    [​IMG]
     
  12. crutschow

    Expert

    Mar 14, 2008
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    The output stage appears to be a source-follower push-pull.
    For that to work each gate-source voltage needs to be biased at or slightly above their threshold voltage (1.4V) for the transistor to amplify.
    In your circuit each transistor appears to be biased at essentially a Vgs of 0V (look at the voltage at the gates and V2 ).

    Note that a Spice AC small signal analysis requires some current through the transistors. A true class B stage with no bias current will have no AC gain.

    Post your .asc file.
     
    Last edited: Jan 29, 2016
    absf likes this.
  13. Bordodynov

    Active Member

    May 20, 2015
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    Your system is not correct. Floating gates. Signal transmission occurs through the output capacitance. set the current through the transistor. From this current depends on the gain.
    it is class AB. Class B Gain<<1.

    See:

    AmplCMOS2.png
     
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  14. luma

    Thread Starter Member

    Nov 5, 2015
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    This is the most precise explanation I ever had , thank you very much. Then how to polarize the circuit in order to make it work as amplifier?
     
  15. Bordodynov

    Active Member

    May 20, 2015
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    In my circuit, I did so. With the potentiometer I asked about a current of 50 mkA (actually sets the offset for the operating point of the transistors).
     
  16. crutschow

    Expert

    Mar 14, 2008
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    Use a voltage divider of three resistors for the gate biases.
    Use a pot for the middle resistor connected as a rheostat so you can vary its resistance and adjust the bias on the transistors to slightly turn them on.
    Connect the respective gates to each side of the pot.

    Note that this configuration does not provide a stable enough bias point to use in a real circuit.
    It's only for test purposes.
     
  17. dannyf

    Well-Known Member

    Sep 13, 2015
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    "d I don't understandwhy the frequency response of the system doesn't change with or without the transistor."

    Maybe you are just doing it incorrectly?
     
  18. luma

    Thread Starter Member

    Nov 5, 2015
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    Do you have more hints?
     
  19. luma

    Thread Starter Member

    Nov 5, 2015
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    Would you be able to give me a sketch of the characteristic curve Vin vs. Vo in order i can understand where to place the bias point?
     
  20. luma

    Thread Starter Member

    Nov 5, 2015
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    Thank you guys for your answer I guess this thread can be considered closed. Anyway I built my original circuit and I obtained some signals I cannot explain , however I post it in another thread thank you so much.
     
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