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Drift velocity of electron

  1. Jun 6, 2013 #1
    I need help to understand the concept of drift velocity.
    Here is part from my book:
    Does this velocity refer to the velocity of each electron?
    And I don't understand the sentence:
    The acceleration due to the field and the collision with the crystal counteract, leading to a constant velocity for the carriers.

    Can you make it clear?

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  3. Jun 7, 2013 #2


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    It refers to an average velocity of the electrons. (The term 'charge carrier' might be so as to include electron holes.)
    In the absence of the collisions with atoms, the field would accelerate the electrons. But the collisions slow the electrons down in a manner related to their speed, so instead the field determines the average velocity. It's very like terminal speed of an object falling through a fluid. Stronger gravity would increase the terminal speed.
  4. Jun 7, 2013 #3


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    Both the free carriers (electrons of holes) are in motion in the crystal and the ions are vibrating about their equilibrium positions, according to the temperature of the crystal. Imagine that an electron moves with a certain velocity and then some electric field is established in the crystal. The electric field exerts -eE force: the electron accelerates and its velocity changes by Δv(t) in time. But it regularly collides with the vibrating ions in the crystal, at τ times in average. During the collision, the electron losses all extra velocity it gained from the electric field. The average change of momentum is - mΔv/τ is . The change of momentum divided by the time is the average force acting on the particle because of the collisions: md(Δv)/dt=-eE- mΔv/τ The terminal extra velocity is u=-eEτ/m, it is the average difference from the equilibrium velocity of the electron in the absence of the field, called the drift velocity.

  5. Jun 7, 2013 #4
    Thanks for helpful answers!
    The terminal extra velocity is u=-eEτ/m, it is the average difference from the equilibrium velocity of the electron in the absence of the field, called the drift velocity.
    Here is an example from the book:
    Here the mobility of electron μn= -eτ/m = 1350 cm^2/(V.s), right?
    The number 1350 cm^2/(V s) taken from the book is the mobility of electrons in silicon.
    And according to the result, the drift velocity is relatively large. I am a bit confused about the number because I have heard that electron move at very very low speed, approximately mm/s.
  6. Jun 7, 2013 #5


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    Omit the "-" sign. It only indicates that the electrons move against the electric field.

    Yes, that is a very large drift velocity, but the electric field is also very strong. That might happen around a pn-junction. If you just take a silicon sheet, used to fabricate diodes of transistors, its is about 0.5 mm thick. When connecting 1 V across the sheet, E is only 2000 V/m

  7. Jun 7, 2013 #6
    Why we call this is average velocity?
    Assume that the velocity of electron at equilibrium is v0 (m/s). Then the velocity of it at the time right before collision v = v0 + Δv(t)= v0 + eEτ/m.
    Δv(t) is called average velocity of the electrons. Therefore, I think that everage velocity of v0 is zero.
    Can you help me?
  8. Jun 7, 2013 #7


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    The electrons move in random, when no external field is present. They move with equal probability to the right and to the left, forward and backward... so the average velocity is zero.
    In the presence of electric field, the electrons get some extra velocity and the average is no more zero: it is the drift velocity.

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