Download Part 5. Semiconductors

PHYS 353 Solid State Physics
Supplemental Problems
Part 5. Semiconductors
5-1. We know that the intrinsic carrier concentrations are equal for semiconductors, ni = pi.
(a) Using the general expressions for the carrier concentrations in Eq. (5.16) and (5.20), show that
1
3
the chemical potential is given by Eq. (5.24). i.e.   EG  k B T ln( mh / me ) .
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4
(b) Evaluate the chemical potential for intrinsic Si near room temperature (T = 300 K) and show
that it is approximately in the middle of the band gap. Relevant quantities for Si follow. (A
handy thing to remember is that kBT = 25.9 meV at 300 K.)
EG = 1.12 eV at 300 K
me ~ 0.9 m
mh ~ 0.5 m
5-2. Consider a sample of n-type Si that has been doped with phosphorous at a level of 1022/m3 and
boron at a level of 1021/m3. Note that there are 10 times more donors than acceptors. You are
going to show that this sample is indeed n-type where the electron majority carriers vastly
outnumber the hole minority carriers. In part (c), you will verify that its Fermi level is between the
middle of the band gap and the donor energy level.
(a) At room temperature (300 K), you can assume that all of the donors and acceptors are ionized
so that the electron carrier concentration is given by Eq. (5.31), i.e. n = ND – NA. Find the
intrinsic carrier density ni using Eqs. (5.23), (5.17), and (5.21). Then use Eq. (5.25) to find the
hole carrier concentration.
(b) Find the ratio of majority carriers to minority carriers. Isn’t it amazing that just ten times more
donors than acceptors results in this huge difference in carrier concentrations?
(c) Calculate the Fermi level at 300 K using Eq. (5.32) and show that is indeed below the donor
energy level but above the middle of the band gap. (You can find information about the donor
and acceptor energy levels in Table 5.1.)
5-3. GaAs and In Sb are both III-V direct gap semiconductors. The band gap energies at 300 K are
1.424 eV for GaAs and 0.18 eV for InSb. Suppose you use them to make two optical detectors.
(a) Find the longest wavelength of light that each material can detect.
(b) Suppose the light that is being detected by the InSb detector is thermal radiation from an object
that can be approximated as a blackbody. If the longest detectable wavelength is the
wavelength at which the object emits the most radiation, what is the temperature of the object?
(Recall Wien’s Displacement Law!)
(c) If you cooled these detectors, would the longest detectable wavelengths increase or decrease?
5-4. You are going to show that semiconductors make much more sensitive Hall probes than metals.
Let’s compare the metal Al to the Si sample of Problems 5-1 and 5-2.
(a) Find the Hall coefficients for Al and this n-type Si. For Al, you can use Eq. (3.39). You have
already figured out the conduction electron concentration in Problem 3-5. For the Si, you can
use Eq. (5.43) where n is the majority carrier concentration you found in Problem 5-2.
(b) Suppose two identical samples are made of the Al and Si. Find the Hall voltages measured
across the height of each sample if the samples are 5 mm wide and 5 mm high, and each sample
is placed in a 0.1 T magnetic field with 100 mA of current flowing along the length of each
sample, perpendicular to the field. (See the diagram below.)
(c) Is the Si probe more sensitive than the Al probe?
V
5 mm B = 0.1 T
i = 0.1 A 5 mm
5-5. In a cyclotron resonance experiment, a sample of GaAs is placed in a variable dc magnetic field
and radio waves of frequency 25 GHz hit the sample. The absorption of the radio energy is
measured as the size of the magnetic field is varied. Three strong absorption peaks occur at 0.060
T, 0.072 T, and 0.402 T and are respectively identified as the electron, light hole, and heavy hole
peaks. Find the effective masses of these three charge carriers. Express the effective masses as a
factor times m, the bare electron mass.
5-6. When analyzing the motion of majority carriers in non-equilibrium conditions, we concluded that
they will “rush” where they need to go to keep electrical neutrality ( n'  p ' ). We are going to
investigate this in more detail in this problem. Consider an n-type sample. The continuity equation
for the electron majority carriers is Eq. (5.66),
D
n'
 2 n'
 ( p'n' )  2D
t
x 2
where the dielectric relaxation time  D ~1 ps and the Debye length  D ~ 400 Å for a typical n-type
semiconductor. Let’s assume that our sample occupies the region x>0 and that we introduce a
disturbance in neutrality such that n' p'  no at x = 0.
(a) Let’s first analyze the time-independent solution. In other words, let’s see how the disturbance
of the electron majority carrier concentration looks after a steady state has been reached.
1. Rewrite Eq. (5.66) if we want to examine the time-independent solution.
2. Let’s make a key assumption that the disturbance in the hole minority carrier
concentration does not vary much over the distance that we will examine, i.e. let’s
approximate p' as a constant. If that’s the case show that the solution to the previous
differential equation is n' ( x)  p ' no e  x /  D .
3. Sketch a plot of n' p ' vs. x.
4. Over what distance does it take electrical neutrality to be restored? To answer this,
approximate that n' p ' ~0 when n' p'  0.001no . (This is like asking how long it takes
to discharge a capacitor…same sort of exponential dependence.)
(b) Let’s now analyze the change in the carrier disturbance with time at one position. In other
words, let’s focus on one position so we ignore the spatial dependence of n' .
1. Rewrite Eq. (5.66) if we want to examine the position-independent solution.
2. Let’s make a key assumption that the disturbance in the hole minority carrier
concentration does not vary much over the time scale that we will examine, i.e. let’s
approximate p' as a constant. If that’s the case show that the solution to the previous
differential equation is n' (t )  p' no e t /  D .
3. Sketch a plot of n' p ' vs. t.
4. Over what time scale does it take electrical neutrality to be restored? To answer this,
approximate that n' p ' ~0 when n' p'  0.001no .
(c) What you have shown is that any disturbance in electrical neutrality only will occur over a
distance of roughly 2000 Å and for only about 7 ps. The minority carrier concentration doesn’t
change much over this distance and time scale (recall we assumed p' was constant). Since this
distance and time scale are so small compared to characteristic distances and times for most
semiconductor behavior, can we then assume electrical neutrality for most semiconductor
behavior?
5-7. Use the data in Table 5.1 and Eq. (6.2) to do the following.
(a) Find the value for the diffusion potential  o for a Si p-n junction at 300 K if both doping
levels are at typical values of 1022/m3. Verify that it is close to 0.7 V.
(b) Show that the diffusion potential is a rather weak function of doping levels by evaluating it at
relatively low doping levels of 1019/m3, a thousand times less than dopant concentrations in (a).
By what factor did the potential decrease?
5-8. When using Si diodes, circuit designers assume around a 0.7 V drop across the diode in the circuit
when it is “on”. That is, when current is flowing through it in forward bias, it is assumed that 0.7 V
is dropped across it. Explain why this is a valid assumption based upon the I-V curve of a diode
and the band diagram of the p-n junction. Specifically, answer the following.
(a) It is assumed that no significant current flows until what special voltage is reached? It is also
assumed that any increase in current is accompanied by a very small increase in voltage.
Sketch what this assumed I-V curve looks like.
(b) For a forward bias at this special voltage, where is the top of the valence band on the p-side
relative to the top of the valence band on the n-side? What about the bottom of the conduction
band on the p-side relative to the bottom of the conduction band of the n-side?
5-9. Use the information and results in Problem 5-7 to do the following.
(a) Calculate the width of the depletion region for a Si p-n junction at 300 K if the doping levels
are both 1022/m3.
(b) Plot the electric field versus position from one side of the depletion region to the other side.
What is the size of the peak field in the depletion region?
(c) Recall that the voltage difference across a region of length d is V = Ed for a uniform field. The
field across the depletion region is not uniform but we can assume that it is with a value that is
the average of the field in the region. Calculate this average field and then the voltage
difference across the region using V = Ed. What is significant about the result?