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Semiconductor basic

  1. Mar 24, 2014 #1
    I don't know if I am asking some stupid questions but I can't really move without knowing them.

    (1) In reverse bias situation electrons attracted towards positive terminal and hole attracted towards negative terminal widen the depletion region .My question is that why can't these hole combined and make a bond with the metal terminal because the negative terminal will give electrons to the p side [as battery always provides electrons].
    (2) True or false ?
    "The current in forward bias flows because the potential across the diode is more then the barrier potential but the current flows due the potential applied across it ,it is not the diffusion process in the diode which occurs after when the external voltage reduce barrier potential to zero."
    I am very confused with very basics .
  2. jcsd
  3. Mar 24, 2014 #2

    Simon Bridge

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    Welcome to PF;
    It is not stupid unless you don't ask it. But after a while you learn ways to check out the question's "stupid quotient" ... :)

    The terms "positive" and "negative" terminal is not helpful here.
    The holes are always attracted to the less positive end and the electrons are always attracted to the most positive end. Changing the bias just changes which way the voltage goes.

  4. Mar 24, 2014 #3
    (2) is not worded clearly. Current generally exists because a source which converts energy is connected to the circuit with the diode. The barrier potential is the energy per charge lost crossing it. The source emf is the energy per charge gained. Obviously the source must be able to supply a voltage greater than the potential barrier in order to sustain current.

    When a diode is connected to a source plus resistor, current begins. When charges reach the diode, a potential barrier is encountered, and charges lose some energy, on each side of the barrier charges recombine. This barrier tends to oppose incoming charges, so the source must provide sufficient energy to sustain the current, eventually equilibrium is attained.

    One thing to note is that the current is not because of the barrier potential, it is due to the source emf, minus barrier potential, divided by net resistance. The source (battery, car alternator, etc.) is what does the work moving charges via chemical means (redox), fuel burned to generate power (car alternator), etc. THe potential barrier exists because there is current. Current comes first, then potential barrier attains equilibrium per Shockley's equation Id = Is*exp((Vd/Vt)-1), or equivalently Vd = Vt*ln((Id/Is)+1).

    THe voltage across the diode is a barrier, it does not determine the current directly, the source emf, resistance, determine the current with Vd having a second order influence. Did this help?

  5. Mar 24, 2014 #4
    Sir, what I think is

    When there is no external battery applied we says that due to diffusion phenomenon electrons travel towards p-type semiconductor and holes travel towards n-type .After some time due to recombination in middle depletion layer created ok ! till here it is good.Now after applying the source (battery) we know that the force on an electron by the battery is higher then the opposite force given by the potential barrier (in a way, I am taking potential applied as force) that's why it will travel to the opposite side.

    What my friend told me that
    battery reduces the potential barrier height, due to that there is no opposing force for the electrons and holes ( I agree) and they travel opposite side due to "only diffusion" because there is no potential barrier ( here I get confused).
    >> electrons of hole should move due to electric field produced by the external source or only by diffusion (according to my friend) or both diffusion and electric field of battery ?
  6. Mar 24, 2014 #5
    With no external bias, the holes and electrons move since they are knocked loose from their parent atoms due to thermal energy. At a temperature above 0 K, lattice vibrations carry energy, quantized modeling is that of phonons in motion striking atomes, and knocking electrons out of the valence band leaving a hole behind. This interaction is known as "ehp" generation (electron-hole pair).

    At room temp w/o external bias, these holes/electrons cross the p-n junction barrier and recombine forming a depletion region. The local electric field is formed by this layer and it opposes incoming charges, a potential barrier. THe intrinsic voltage value of this barrier (integral of E field over line integral of path) is around 25.7 mV, known as "VT", the "thermal voltage".

    When an external source , say a battery, biases the diode (with a voltage source a resistor must be used in order to control current and prevent thermal runaway), the E field from the battery exerts forces on the free electrons in the resistor and wires, imparting motion. These electrons enter/exit the diode junction and cross the junction, recombining on other side, adding to the depletion layer charge density.

    This results in an increase in junction voltage, VD, since increased charge motion has taken place, and it takes a finite time for these additional charges to recombine, so that the depletion region has more stored minority charges and higher voltage.

    The equation describing this action and the relation between diode current ID and voltage VD, is per Shockley:

    ID = IS(e-VD/VT - 1)

    VD = VT(ln(ID/IS) + 1)

    So as forward current is increased, potential barrier must rise. So in order to sustain a specific value of forward current, the source must have enough voltage to meet the potential barrier value. The balance is dropped across the resistor.

    However, I don't want to argue semantics, but "lowering the barrier" is a phrase often used, that IMHO is misleading. The barrier does not get lowered, but rather the external source, battery, provides a potential greater than the barrier. If the battery is 2 1.5V cells in series for 3.0V, the junction has a thermal voltage of 25.7mV at room temp, no bias. Connecting the battery and resistor to the diode results in current through the diode junction. This injection of charges results in raising the barrier potential since additional charges reside in the depletion region. The barrier opposes charges coming in, but the potential from the battery is greater. So a 3.0V cell pair imparts charge motion when the barrier is 25.7mV. WHen the charges build up the depletion region and barrier increases to 0.65V, the difference between 3.0V and 0.65V (2.35V) is smaller than the start (3.000 - 0.0257V - 2.9743V), so the current will decrease.

    If the cell voltage is much larger than the junction drop, say the cell is 12.0V, then the changes in junction voltage have less influence. Anyway that is a brief overview of diode relations and junction action. I hope I helped, will clarify if you wish.

  7. Mar 25, 2014 #6
    Sorry for the typo. The first equation should not have the negative sign in the exponential, but should read as follows:

    ID = IS(eVD/VT - 1)

    The 2nd equation has the parentheses in the wrong place, should be as follows:

    VD = VT(ln((ID/IS) + 1))

    My apologies.

    Last edited: Mar 25, 2014
  8. Mar 25, 2014 #7
    Thank you very very much .these things were troubling me for 5 days .:) [But I will be having a lots of doubts in future :P because it was the starting ]
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