Physics of Electronics: Semiconductor Devices Problem sheet 2
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Physics of Electronics: Semiconductor Devices
Problem sheet 2
This problem sheet covers the material covered in Part 2 of the notes. The margins of the notes are also marked with relevant problems in Sedra and Smith (found at the end of each chapter), which you can also do for further practice.
Q.1 Holes
Explain the concept of a hole, and why it is useful to describe the properties of semiconductors.
Q.2 Intrinsic semiconductors
Germanium (Ge) is a semiconductor with a bandgap of 0.66 eV. The con- centration of electrons in the conduction band is given by:
n = NC exp (-∆gap /2kB T),
where the constant NC for Ge at room temperature is 1 . 1019 cm−3 . The electron and hole mobilities in Ge are 3900 and 1900 cm2 /Vs, respectively.
a) What is the cause of the exponential temperature dependence in this expression?
b) Deduce the concentration of free electrons in Ge at room temperature.
c) What is the hole concentration?
d) Calculate the conductivity of Germanium at room temperature.
e) Do you expect the conductivity of Ge to be higher or lower than Si at the same temperature? Explain your answer.
Q.3 Extrinsic semiconductors
Silicon has an intrinsic carrier concentration of 8 . 109 cm−3 at room temper- ature. The electron and hole mobilities in silicon are 1400 and 500 cm2 /Vs, respectively.
a) Suggest an element that could be used as a donor for silicon.
b) What concentration of donors is needed to make the resistivity of the silicon 0.2 Ωcm.
c) What is the hole concentration?
d) Is this material n-type or p-type?
e) Suggest an element that could be used as an acceptor for silicon.
f) What concentration of donors is needed to make the resistivity of the silicon 0.2 Ωcm?
g) What is the electron concentration?
h) Is this material n-type or p-type?
Q.4 PN junctions I
A pn junction is formed from two cubes of silicon, each of dimension 1 mm3 . One cube is doped with 5 . 1016 cm−3 of P, the other with 1017 cm−3 of B. For reference, the relative permittivity of Si is 12, and the permittivity of free space is 8.9 . 10 − 14 F/cm.
a) Explain the origin of the depletion region in a pn junction.
b) What is the built-in voltage across the junction?
c) If the total width of the depletion region is 180 nm, what is the extent of the depletion region in the n-type and p-type materials?
d) What is the total charge in the n-type part of the depletion region?
e) And in the p-type?
f) What is the net charge in the whole depletion region?
g) Estimate the capacitance of this junction, modelling it as a parallel plate capacitor.
h) What is the peak electric field in the junction, and at what point is this maximum?
Q.5 PN junctions II [Challenge]
A pn junction is formed from n-type and p-type silicon with equal dopant concentrations (NA = ND = N). Derive an expression for the peak electric field in the junction as a function of the dopant concentration.
Silicon breaks down at an electric field of about 3 . 105 V/cm. For a symmetrically doped junction, what is the maximum dopant concentration before this breakdown electric field is reached in the junction.
2022-03-11