feedback amplifier

Discussion in 'Homework Help' started by amangupta1219, May 17, 2013.

  1. amangupta1219

    Thread Starter New Member

    Oct 18, 2012
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    I have to find the feedback polarity. I proceeded as follows :

    1. As Iin = iF , increasing Iin will increase iF
    2. Now drain current for MF is increasing so Vgs of the transistor should increase
    3. That means increasing Iin increase Vout
    4. Now as Vout increases the current through RD2 decreases
    and all current through RD2 will flow through M2 , this means drain current through M2 is decreasing
    5. This will cause gate voltage of M2 to decrease, this will cause current through RD1 to increase which means drain current through M1 increasing which further implies that gate of M1 or drain of MF increasing

    Now in short i found that increasing Iin will cause drain voltage of MF to increase

    Now my question is how to proceed further ???
     
  2. Jony130

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    Feb 17, 2009
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    I think that you should start from Vout and Iin = const. It is much easier way.
    At first we assume that Vout voltage increases. If so, Vgs gate (MF) also must increases. This will cause gate voltage of M1 to decrease and the voltage at the M1 drain and Vgs M2 will increases. And since Vgs_M2 increases, the drain voltage (Vout) will decrease. And so the output voltage returns to its previous value. So we have a negative feedback in the circuit.
     
  3. amangupta1219

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    Oct 18, 2012
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    thanks i got your solution

    Can you tell me that if Iin is increased then why gate of M1 rises ????
     
  4. Jony130

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    M1 Vgs voltage rise because there is no other way to Vout to increase.
    Also notice that MF cannot respond until Vout voltage start to change.
     
  5. screen1988

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    Mar 7, 2013
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    Hi Jony,
    Can you explain it more detail?
    Consider transistor MF:
    Vgs increase, iF = iIN = const
    How the voltage at drain of MF increase?
    And the condition "Vgs increase and iF = iIN = const" seems impossible to me.
     
  6. Jony130

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    In the second post I describes the situation when the input current is held constant (does not change). But output voltage has change under some external disturbance.
    So if Volut ---> rise. The Vgs_MF also rise. Increase in Vgs voltage increase MF drain current. But as we know increase in drain current cause decrease in Vds voltage in CS amplifier (in this circuit all three transistors work as a CS (common source) amplifier).
     
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  7. screen1988

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    Mar 7, 2013
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    But in the circuit IF = Iin = constant despite Vgs_MF rising. And this make me confused.
     
  8. Jony130

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    OK I see your problem. When Vgs rise how can Id change if we assume Iin constant?
    Simply ID will not change, the only think that will change will be Vds voltage.
    If Vgs rising, the MOS "resistance" will decrease. So ther will be less voltage drop across rds resistance. So Vds will drop. In order to avoid confusion in this type of a theoretical circuit analysed as if you was dealing with the ordinary CS amplifier.
     
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  9. screen1988

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    Mar 7, 2013
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    I still can't understand it.:confused:
    [​IMG]

    As in the MOS characteristic when Vgs increase Id has to increase and Vds has to decrease (I see it along the load line).

    Ah, now I think when you said about its resistance, you imply is operating in triode region and therefore it acts as a resistor?
     
  10. amangupta1219

    Thread Starter New Member

    Oct 18, 2012
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    Can we say that if Iin increases then iF will have to increase because gate current is 0 for M1 (always).
    Now if above statement is right then how can we relate it to Vds of MF ???
     
  11. Jony130

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    Feb 17, 2009
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    Find a transfer function for CS amplifier with active load.
    http://www.southalabama.edu/enginee...urses/EE334-Fall07/Chap16-1-NMOS-Inverter.pdf
     
  12. Jony130

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    Add the biasing resistor. And now the situation looks more obvious.
     
  13. screen1988

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    Mar 7, 2013
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    Hi Jony,
    I want to ask about large signal analyse for the circuit:
    [​IMG]

    If the transistor is in saturation:
    Id = 1/2*Kn* (Vgs - Vth)^2
    But in your analysis, Id = const and Vgs increase => the equation is impossible!
    If the transistor is in triode:
    Id = 1/2* Kn * ((Vgs - Vth)*Vds - Vds^2)
    If Id = const and Vgs increase, then the only way to make the equation hold is Vds change.
    => In CS amplifier with active load the transistor always operate in triode region, right?
     
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