For silicon at room temperature verify that Nc=1.8x1019 cm-3. Show both the numerical and units compution.
For silicon at room temperature verify that ni~1x1010 cm-3.
The probability that an electronic state with energy E is
occupied by an electron is given by the Fermi-Dirac distribution
function:
f(E) = E E kT Fe ( )/1 1 −+ (Equation 1.2)
where EF is the Fermi level, the energy at which the probability of
occupation by an electron is exactly one-half. At room temperature,
the intrinsic Fermi level lies very close to the middle of the
bandgap.
The effective density of states in the conduction band NC is equal
to 2[2πmnkT/h2]3/2. Similarly, the effective density of states in
the valence band NV is 2[2πmpkT/h2]3/2. At room temperature, NC for
silicon is 2.8 x 1019 atoms/cm3. For an intrinsic semiconductor,
the number of electrons per unit volume in the conduction band is
equal to the number of holes per unit volume in the valence band.
That is, n = p = ni where ni is the intrinsic carrier density.
Since n = NCexp{-(EC-EF)/kT} and p = NVexp{-(EF-EV)/kT}, where n is
the electron density and p is the hole density, np = ni2 =
NCNVexp{(EV-EC)/kT} = NCNVexp{-Eg/kT}. Therefore,
n N N
E kTi C V g= − ( ) exp{ / 12 2 }
For a doped, or extrinsic, semiconductor, the increase of one type
of carriers reduces the number of the other type. Thus, the product
of the two types of carriers remains constant at a given
temperature. For Si, ni = 1.45 x 1010 cm-3 . .
For silicon at room temperature verify that Nc=1.8x1019 cm-3. Show both the numerical and units compution....
For n-type silicon at room temperature, with a donor doping concentration of 10^(17) cm^(-3), approximately how much larger will be the electron concentration, compared to the hole concentration? Assuming ni=10^(10) cm^(-3).
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