What I am wrong here?

Discussion in 'Homework Help' started by screen1988, Apr 5, 2013.

  1. screen1988

    Thread Starter Member

    Mar 7, 2013
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    Please see my attached file. In the picture is a figure about the relation between Vin and Vout when Vin varies from 0 to VDD.
    As in the file, if Vin < Vth1, M1 is in cut off and M2 is in deep triode.
    Here is my reasoning:
    Call Ids1, Ids2 is the currents through M1 and M2.
    If Vin < Vth1, M1 is in cut off and Ids1= Ids2 =0.
    But when Ids2=0 then M2 must be in cut off not deep triode. This isn't consistent with the solution above?
    Could you tell me where I am wrong here?
     
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  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Even with M1 cut-off and the drain-to-source currents in both M1 & M2 being at or near zero, this doesn't imply automatically that M2 is also at cut-off. It's [M2's] simply starved of current from the source by virtue of M1's state. M2 may well be in a very low ohmic state but the fact that M1 is not conducting would prevent current flow from the source and down the series path through M1 & M2 to ground.

    To actually get M2 into cut-off one would have to set Vb [at M2 Gate] less than VTH2. The diagram in your post doesn't state what the value of Vb might be.
     
    Last edited: Apr 6, 2013
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  3. screen1988

    Thread Starter Member

    Mar 7, 2013
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    Thanks,
    Is there any an easy way to see that if Vin increase then Vout also increase?
    If I know this it will be more easier for me to figure out the relationship between Vin and Vout as in the picture.
    With this type of exercise I always have to assume all cases that can happen then solve and see if it is suitable.
    For example, in this case I have to consider all combinations between M1 and M2.
    M1 and M2 can be cut off, triode, saturation therefore I have to try 9 cases:(
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    As with a BJT in which you can make a rough claim, subject to some caveats, that Vbe is about equal to one diode drop, you can make a similar rough claim, again subject to some (stronger) caveats, that Vds is about one threshold voltage drop.

    Another way that is more general is to assume that the circuit is stable with some Vin and some Vout. Now make an incremental increase Vin and walk it over to what direction Vout will change in. In this case, an increase in Vin means that Vgs increases. That means that Ids increases. That means that Vds for the lower transistor increases (same Vgs, so more Ids means more Vds). That means that Vout goes up. Or, you can look at it as when Vgs on the top transistor goes up, Rds goes down. So you now have a voltage divider in which the upper resistor has gone down, hence Vout will go up.
     
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  5. screen1988

    Thread Starter Member

    Mar 7, 2013
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    I thought about this but it seems to me that MOSFET is unlike BJT. In BJT when
    BJT is in forwards bias, Vbe is about equal to one diode drop, but in MOS, Vgs will change accordingly to Ids (for example in saturation). Vgs can be any value???:(
    How can you see that when Vin increases, Vgs will increase? In this case Vgs= Vin - Vout. You said that if Vin increases, Vgs also increases but how can you know that because it may be Vout also increase?
     
  6. WBahn

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    Mar 31, 2012
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    Yes, which is why the caveats are stronger. But it is still often a useful initial assumption.

    Walk the changes through incrementally. When Vin increases a bit, it takes time for Vout to change. So, just as you make the increase in Vin, Vgs increases. It may well be that, later, Vout increases which will have the ultimage effect of reducing how much Vgs increases by, but you do NOT expect Vout to increase so much that Vgs will actually decrease, since that would result in LESS current flowing than was flowing previously which would require that Vout decrease from its original value.
     
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  7. screen1988

    Thread Starter Member

    Mar 7, 2013
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    Oh, thanks, I see it.
    Rds you said is the resistance of transistor in triode region. But in this case I don't know which region(cut-off, triode,, saturation) that M1 and M2 are operating?
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    Whether you are in the triode region or the saturation regions, if you increase Vgs you get higher Ids for the same Vds, correct? At any given moment you have an effective Rds which is Vds/Ids. Thus, if you increase Vgs the effective Rds tends goes down. That's the point. You can visualize the how things work several different largely equivalent ways. Pick whichever one floats your boat!
     
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  9. screen1988

    Thread Starter Member

    Mar 7, 2013
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    I think I need to go back here:
    I got this part.
    At the moment we increases Vin:
    With transistor M1:
    Vgs = Vin - Vout : increases (As you said Vout takes time to change)
    Vds= Vdd - Vout : remain unchanged
    => Ids(M1) = Ids (M2) increases
    With M2:
    Vgs = Vb = const
    Vds = Vout
    Ids (M2) increases
    => Vout has to increase
    => It seems that the assumption that Vout increases when Vin increases is appropriate.

    Now after some time, as you said above Vout will not increase so much that Vgs will actually decrease.
    But why after the time, Vout has to increase? Is it possible that Vout will decrease after the time?
    For now, let's assume that Vout will increase. For example, let's say that Vout increases so much that Vgs decrease.
    Then with M1:
    Vgs: decreases
    Vds: decreases
    => Ids also decreases
    => Ids through M2 also decreases and because the transistor has Vgs= const
    therefore its Vds= Vout has to decrease.
    Now I think I understand what you said here but why the case is impossible?

    How about the case that when Vin increase and Vout will decrease?
    Assume that Vin increases and Vout decreases.
    With M1:
    Vgs= Vin - Vout: increases because Vin increases and Vout decreases
    Vds= Vdd- Vout: increases
    => Ids increases
    With M2:
    Ids increases => Vout has to increases (which is not appropriate with initial assumption)
    => The case don't happen
    Is this right?
     
  10. screen1988

    Thread Starter Member

    Mar 7, 2013
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    Yes, I can see it from Id-Vds characteristics.
    As you have proved when Vin increases Vout also increases but it always increases in a maner so that Vgs = Vin - Vout also increases.
    With transistor M1:
    Vgs = Vin - Vout : increases
    Vds = Vdd- Vout: decreases
    => Ids can increase or decrease
    => Rds= Vds/Ids can also increase or decrease
    With transistor M2:
    Vgs = Vb = const
    Vds = Vout : increases
    => Ids increases
    => Rds = Vds/Ids increases
    Therefore, in this case if I model the transistors as Rds and they make a voltage divider. But there is a problem that Rds(M1) is unknown if it increases or decreases.
    Could you point out that? What I need to consider?
     
  11. screen1988

    Thread Starter Member

    Mar 7, 2013
    310
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    Could you confirm what I have done above? Do I understand it correctly?
     
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