NLDD dose influence on threshold voltage

Discussion in 'Cadence' started by cheanglong, Sep 23, 2006.

  1. cheanglong

    cheanglong Guest

    Dear All

    Anyone can help to give me some idea what will happen to Vtn if the
    NLDD dose is increase or decrease?
    ANy BSIM equation I can relate Vtn and NLDD dose?
    I appreciate if you have any paper or book, you can refer me to.
    Kindly help me to understand the physics behind for the relations.

    Thank you all

    best rgds
    Jason
     
    cheanglong, Sep 23, 2006
    #1
  2. cheanglong

    Stephen Guest

    Jason,

    I will give you my best answer to your question, with quotations, but
    hopefully someone else can verify that my answer is correct.

    According to "Introduction to VLSI Circuits and Systems" by Uyemura, p.
    137:

    "A lightly doped drain MOSFET is designed to reduce the electric fields
    in the channel region by providing n- (lightly doped) drain and source
    regions instead of the usual n+ regions. Theoretically, this reduces
    the maximum electric field intensity, which in turn increases the
    reliability of the transistors."

    Since you did not ask about why reliability is increased, I will not go
    into that here.

    However, since you are asking about the effect on Vtn, my answer is
    that I'm not sure that there is an effect. If you look at the standard
    equation for the threshold:

    Vtn = Vto + lambda[ sq(-2*fermi_potential + Vsb) -
    sq(2*fermi_potential) ]

    Now if we assume that Vsb = 0, then we just have:

    Vtn = Vto

    Vto is a function of the work function difference between the gate
    material and the silicon substrate. It seems to me that Vtn is simply
    trying to create the channel between the source and drain that will
    allow electrons (or holes) to flow through, and that the doping
    concentration of the source/drain regions determines the magnitude of
    the electric field. (In small channel devices, this electric field
    causes velocity saturation - or the current to saturate very quickly,
    and other frusturating design effects such as reliability problems,
    which LDD helps solve by decreasing the doping concentration, and
    thereby decreasing the electric field). I cannot find any explainable
    links between Vtn and LDD, at least not directly.

    Do others agree with this?

    - zielstep
     
    Stephen, Sep 24, 2006
    #2
  3. cheanglong

    cheanglong Guest

    Dear Stephen

    Firstly I would like to thank you for posting to my message. It is my
    honour to have people around here who is willing to discuss with me. It
    is not easy to find people who willing to share ideas.

    After going through your message, I got to know a bit more of the short
    channel effects and the reason for using LDD.

    Currently I am sorting out my thoughts on this subject. I will post
    back within this week after I sort out my mind.

    As of now, I totally agree with you.
    But since the LDD will decrease the electric field at drain side, would
    not the lower electric field will help in improving the threshold
    voltage( not too low) ?
    Any comments?

    Thank you Stephen!

    Anyone has idea, kindly share with us
    Cheers


    best rgds
     
    cheanglong, Sep 25, 2006
    #3
  4. cheanglong

    Stephen Guest

    Jason,

    I've throught it through some more, and here's my new conclusion:

    At very low channel lengths, as the length continues to decrease we are
    seeing that Vth is actually decreasing! This is because the
    source/drain regions actually have a depletion region of their own, and
    this subtracts from the channel area where the Vgs has to overcome Vth.
    Therefore, at larger values of L, the source/drain depletion regions
    are very small and have no real effect, but at small values of L, they
    cause Vth to lower.

    In the same way, if we lightly dope part of the drain region, then this
    will cause the (rather small) depletion region to shrink just a little.
    Therefore, Vth will rise for very low L (it will stay the same
    otherwise, because the change will not be noticeable comparibly
    speaking). However, the trend of Vth decreasing will continue with
    smaller values of L, but by having a lightly doped drain it just won't
    decrease by as much.

    I hope this helps.

    - zielstep

     
    Stephen, Sep 30, 2006
    #4
  5. cheanglong

    cheanglong Guest

    Dear Stephen

    I totally agree with what you said. I have tried to link short channel
    effect to the Nldd and have the similar thought as you have shared.
    So whenever channel is very short, Vt will decrease. But when the Nldd
    is lighter, it will compensate the Vt roll off up to some degree which
    is unknown to me.
    Therefore, the Vt will stay the same or increase very slightly.
    I will write back if I read about some new stuff ...
    Thank you so much Stephen

    best rgds
    Cheang Long
     
    cheanglong, Oct 5, 2006
    #5
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