Current through stacked transistors

Discussion in 'Homework Help' started by lll, Feb 22, 2013.

  1. lll

    Thread Starter New Member

    Mar 7, 2012
    The current through nodes 1, 2, and 3 is 1/3 mA, but why?

    At first I thought the current would be 1mA since it has the same Vgs as the other transistors with 1mA currents, but after I saw that it wasn't I thought maybe it's in triode mode. But I am stuck as to what to do next.

    Assuming the kn's are all equal for all the transistors, how do I find the current through the transistors on the right?
    Last edited by a moderator: Feb 23, 2013
  2. WBahn


    Mar 31, 2012
    I would expect it to be less than 1mA, but would not expect it to be 1/3 of the programmed current.

    Only the bottom transistor has the same Vgs as the bias transistor. The Vgs for the two transistors above it are less because their source nodes are not at ground. Thus the top transistor will want to limit the current to something less than 1mA but the bottom transistor wants to make it 1mA, so it's output will go pretty hard into saturation resulting in a low voltage drop across it (but not zero and perhaps not even particularly close to zero, though your sim results on the left indicate that it is ~4mV, so pretty close in this case). Similar reasoning applies to the second transistor and so it will also probably have a low voltage across it, but not as low as the bottom one. The sim results on the left are consistent with that. But the top transistor has a Vgs that is reduced by 8.7mV from the programmed Vgs. To determine the effect of that, you can use the small signal model of the transistor and consider that you have the nominal Vgs as your DC bias and then you have a -8.7mV signal as your input.
  3. lll

    Thread Starter New Member

    Mar 7, 2012
    My book claims you can get 1/3 mA as the current through those transistors through hand calculations, but doesn't provide the reasoning. Is there maybe a way you can approximate the current as being 1/3 mA?
    Last edited: Feb 23, 2013
  4. tshuck

    Well-Known Member

    Oct 18, 2012
    As a side note, I would refrain from using that name as your sign in name ( or at least crop it out of your screenshot...).

    You may offend people, otherwise, willing to help you....
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    I imagine the "ideal" model would consider M6 & M7 replaced by their equivalent resistance ≈1/gm in each case. The resulting series combination [≈2/gm] acts as a degenerative feedback element which by virtue of the inter-related gm values effectively reduces the series stack current to approx 1/3 of the value obtained without degeneration of the reference Vgs [as generated by the diode configured mosfet M1]. The concept can be extended to any number of degenerative mosfets in a series stack being replaced by an "equivalent" 1/gm resistive element.

    The idea probably works most "ideally" with a very low current drive some orders of magnitude below the device current rating. I doubt it would work in a practical situation with the likely spread of gm values from device to device expected [albeit with the same device type number].