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Drift velocity in semiconductor

  1. Apr 16, 2015 #1
    1. The problem statement, all variables and given/known data

    (a) Explain the terms intrinsic, extrinsic, mobility and effective mass in semiconductors
    (b) What is a hole and explain its mass and charge
    (c) Why is mobility of holes often less than mobility of electrons? Find the number density of holes and electrons
    (d) Find the mobility of the metal
    (e) Find the drift velocities of electrons in the metal and germanium

    Only major problem I have is part (e).

    2014_B6_Q3.png

    2. Relevant equations


    3. The attempt at a solution

    Part(a)

    Intrinsic: No impurities, density of holes = density of electrons
    Extrinsic: Impurities present, N-doping or P-doping
    Mobility: Ease of movement of electrons and holes through semiconductor ##\mu = \frac{v}{E}##.
    Effective mass: mass near the bottom of Conduction band or top of valence band

    Part(b)
    A hole is an absence of an electron. It has opposite charge, and opposite velocity to the electron, since overall charge must be conserved. It has the same effective mass of electrons, due to conservation of momentum.

    Part(c)
    Holes are surrounded by a sea of bounded electrons in the valence band whereas free electrons are not as exposed in the conduction band due to the absence of states. So it is easier for free electrons to move around inhibited compared to holes.

    Using ##J = nev = \sigma E##, we have ##\frac{v}{E} = \frac{\sigma}{ne} = \frac{1}{ne\rho}##.
    [tex]n_e = \frac{1}{e \rho \mu_e} = 6.94 \times 10^{19} m^{-3} [/tex]
    [tex]n_h = \frac{1}{e \rho \mu_h} = 3.47 \times 10^{19} m^{-3} [/tex]

    Part(d)
    The number density for this metal is ##n = \frac{N}{a^3} = \frac{4}{a^3} = 8.57 \times 10^{28} m^{-3}##.
    [tex]\mu_{metal} = 4.28 \times 10^{-3}[/tex]

    It seems that the mobility in this metal is about 100 times less than the mobility in germanium.

    Part(e)
    Since ##v_d = \mu E##, for a given mobility the drift velocity is dependent on electric field. But since the electric field is ##E = \frac{V}{l}##, if we do not know the length of the metal, how could we figure out the voltage or eletric field across it? Surely an infinitely long metal would dominate the voltage across it than the germanium.

    [tex]v = \mu E = \mu \left(\frac{V}{l}\right)\left( \frac{\rho l}{A} \right) = \frac{\mu V \rho}{A} [/tex]

    Can't find the drift velocity without knowing it's voltage across it, which I need to know its length to figure out its resistance to figure out its voltage across it by the potential divider principle.
     
  2. jcsd
  3. Apr 16, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    The mobility of electrons in the metal should be much better than the mobility in germanium. How did you calculate the value?

    (e) You can calculate the current flow through the germanium (assuming the resistance of the metal wires is very small in comparison). That allows to find the drift voltage in the metal.

    Free electrons are surrounded by empty states ("bound holes"), free holes are surrounded by filled states. Where is the difference?
     
  4. Apr 16, 2015 #3
    [tex]\mu = \frac{1}{n e\rho} = \frac{1}{(8.57 \times 10^{28})(1.6 \times 10^{-19})(1.7 \times 10^{-8})} = 4.28 \times 10^{-3}[/tex]


    So we assume that the potential difference across germanium is ##2V##. Using that, we find the current using ##I = \frac{V}{R}##. Assuming current in metal is same in germanium, using ##I = nevA## we can find the drift velocity in the metal.
     
    Last edited: Apr 16, 2015
  5. Apr 16, 2015 #4
    If the electron and hole have the same effective mass, why would electrons have higher mobility?
     
  6. Apr 17, 2015 #5

    mfb

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    Staff: Mentor

    Ah, forget my comment about electron mobility, sorry.

    Right.
    Why do you expect the same effective mass?
     
  7. Apr 17, 2015 #6
    Consider for an electron near bottom of conduction band:
    [tex]E' = E_0 + \alpha |k-k_{min}|^2 + \cdots [/tex]
    [tex]\alpha = \frac{1}{2} \frac{\partial^2 E}{\partial k^2} = \frac{\hbar^2}{2m^{*}}[/tex]

    Similarly for a hole, near the top of a valence band:
    [tex]E' = E_0 - \alpha |k_{max}-k|^2 + \cdots [/tex]
    [tex]\alpha = -\frac{1}{2} \frac{\partial^2 E}{\partial k^2} = \frac{\hbar^2}{2m^{*}}[/tex]

    My book says that a hole is at the highest possible energy while the electron is at the lowest possible energy configuration, and that driving a hole away from the maximum is like "pushing a balloon under water". I suppose that is why the mobility of electrons is higher?
     
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