In [4]:
tempK = linspace(1, 400)
semilogy(tempK, sqrt(ni2(tempK)))
title('Thermally-generated free carrier density, $n_i$')
ylabel('carrier density ($cm^{-3}$)')
xlabel('Temperature (kelvin)')
grid(True)
Get a sense of these formulas by seeing the huge effect of temperature on semiconductor properties.
The concentration of thermally-generated electron-hole pairs varies (lots!) with temperature, and is:
$$n_i = B \cdot T^{3/2} \cdot \exp\left(\dfrac{-E_g}{2 k_B T}\right) \;\; \mathrm{\#/cm^3}$$%matplotlib inline
import matplotlib as mpl
from pylab import *
# increase sizes for display
mpl.rcParams.update({'figure.figsize': (12.0, 8.0), 'font.size': 18,})
B_Si = (5.2e15)**2 # Razavi (2.1)
Eg_Si = 1.12 # eV (=1.602 x 10-19 J)
KB = 8.617e-5 # eV / K
def ni2(T, B=B_Si, Eg=Eg_Si):
return B * T**3 * exp(-Eg / (KB * T))
def ann_temp(T=25, text='Room temp'):
"""Helper to add an arrow to room temperature."""
ax = gca()
x,y = (T, sqrt(ni2(T+273.15)))
plot(x, y, 'o')
ax.annotate(text, xy=(x,y), xytext=(x, y/5), arrowprops=dict(arrowstyle='->',))
tempK = linspace(1, 400)
semilogy(tempK, sqrt(ni2(tempK)))
title('Thermally-generated free carrier density, $n_i$')
ylabel('carrier density ($cm^{-3}$)')
xlabel('Temperature (kelvin)')
grid(True)
For a non-bogus formula, check that we get $n_i = 0$ at $T=0\,K$.
commercial =(0, 85)
industrial = (-40, 100)
automotive = (-40, 125)
military = (-55, 125)
for tr,name in ((military, 'Military'),
(automotive, 'Automotive'),
(industrial, 'Industrial'),
(commercial, 'Commercial'),):
tempC = arange(tr[0], tr[1], 1)
tempK = tempC + 272.15
semilogy(tempC, sqrt(ni2(tempK)),
label='%s (%i to %i)' % (name, tr[0], tr[1]),
linewidth=4)
legend(loc='upper left')
ann_temp(300-272)
ylabel('Carrier density, $n_i$ ($cm^{-3}$)')
xlabel('Temperature ($^\circ C$)')
title('Intrinsic free carrer density over temperature')
grid(True)
THIS IS A HUGE RANGE !!!
... and plain huge numbers
How do they compare to the number of atoms in the crystal? ($5 \times 10^{22}$ atoms/cm$^2$)
Adding impurities to the crystal from Group III or IV elements (replacing an Si atom with another like B or P) also adds extra electrons or holes to the structure.
Impurities which add extra free electrons are called donors, with concentration $N_D \;\; \mathrm{cm^{-3}}$.
Impurities which add extra (free) holes are acceptors: $N_A \;\; \mathrm{cm^{-3}}$.
Even when doped, a semiconductor crystal has no net charge
$$\sum \text{charges} = q \left( +N_D + p - N_A - n \right) = 0$$This, combined with the fundamental relationship
$$n\cdot p = n_i^2$$Allows us to calculate the concentration of free electrons and holes for any doping condition (always choose $+$ version):
$$n = \dfrac{(N_D - N_A) \pm \sqrt{(N_D - N_A)^2 + 4 n_i^2}}{2} \; \text{ and } \; p = \dfrac{n_i^2}{n}$$or, solved the other way:
$$p = \dfrac{(N_A - N_D) + \sqrt{(N_A - N_D)^2 + 4 n_i^2}}{2} \; \text{ and } \; n = \dfrac{n_i^2}{p}$$Q: For the first formula, why do you always choose the $+$ version (or when would you choose the $-$ version) before the sqrt?
Q: Are the above two sets of solutions equivalent?
Q: Does it matter which set to use depending on which of $N_D$ or $N_A$ is larger?
When we dope in practice, the impurities $N_D$ and/or $N_A$ are in the range $10^{14} \text{ to } 10^{21} \; \mathrm{cm^{-3}}$, which is much greater than $n_i$ at normal temperatures. This results in us using the following approximations all the time:
$$\text{when $N_D$ is largest: } n \approx N_D \text{ and } p \approx n_i^2/N_D$$$$\text{when $N_A$ is largest: } p \approx N_A \text{ and } n \approx n_i^2/N_A $$$\rightarrow$ Notice how only the minority carrier concentrations are affected by temperature (holes for n-type doping, electrons for p-type doping).