Week 8 Quiz Answers ECE 606: Solid State Devices Mark Lundstrom Purdue University, Spring 2013 Quiz 1: Answer the four multiple choice questions below by choosing the one, best answer. Then ask a question about the lecture. 1) What does the term “depletion approximation” mean in the context of a PN junction? a) The term refers to an undoped region near the PN junction. b) The term refers to a region for which n, p << ni . c) The term refers to a condition for which n − p << N D − N A . d) The term refers to a condition for which n − N D << p − N A . e) The term refers to a region near the PN junction where n and p are much less than the doping density 2)
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In a PN junction, where does the peak electric field occur? a) At the edge of the depletion region on the N-­‐side. b) At the edge of the depletion region on the P-­‐side. c) At the center of the depletion region. d) At the center of the part of the depletion region that lies on the N-­‐side. e) At the junction between the N and P sides.
The formula for the built in potential is an important one that we should remember. What is it? a) qVbi = k BT ln ( N A ) ni2 b) qVbi = k BT ln ( N D ) ni2 c) qVbi = k BT ln ( N A N D ) ni2 ( )
= k T ln ( n N ) d) qVbi = k BT ln ni2 N A e) qVbi
B
2
i
D
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If we apply a reverse bias to a PN junction, what happens? a) The magnitude of the peak electric field decreases, and the width of the depletion region decreases. b) The magnitude of the peak electric field decreases, and the width of the depletion region increases. c) The magnitude of the peak electric field increases, and the width of the depletion region decreases. d) The magnitude of the peak electric field increases, and the width of the depletion region increases. e) The magnitude of the peak electric field is increases, and the width of the depletion region unchanged. Quiz 2: 1) For a moderately forward biased PN junction, we solve the minority carrier diffusion equation in the quasi-­‐neutral N and P regions. We need 2 boundary conditions. If x = 0 is the boundary of the quasi-­‐neutral P region adjacent to the depletion region, what is the boundary condition for the excess minority electron equation at x = 0 for an applied bias of VA? (This result is known as the “Law of the Junction.”). qV k T
a) Δn ( x = 0 ) = N Ae A B . (
c) Δn ( x = 0 ) = ( n
)
)e (
b) Δn ( x = 0 ) = ni2 N A ln ( qV A k BT ) 2
i
d) Δn ( x = 0 ) = ni2 e
(
NA
q Vbi −V A ) k BT
qV A k BT
)
e) Δn ( x = 0 ) = ni2 N A e
2)
. qV A k BT
For very large forward bias, the current “rolls off” (i.e. the current is not as large at a given voltage as the ideal diode equation would predict). The reason for this is: a) For high currents, there are voltage drops across the N and P regions, so only a portion of the applied voltage actually gets to the PN junction and forward biases it. b) The applied bias exceeds the bandgap of the semiconductor. c) The diode enters breakdown. d) The depletion approximation begins to fail. e) The diode begins to overheat.
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If you observe an n = 2 slope to the high forward bias IV current of a PN junction, what physical explanation would you give? a) Diode is in low-­‐level injection. b) Zener tunneling. c) Diode is in high level injection with ambipolar transport d) Recombination is controlled by radiative recombination e) Recombination is controlled by Auger recombination What region of the IV characteristic of a PN junction does tunneling affect? a) Low forward bias (if the junction is heavily doped on both sides) and moderate to high reverse bias. b) Low forward bias (if the junction is heavily doped on one side) and moderate to high reverse bias. c) High forward bias (if the junction is heavily doped on both sides) and moderate to high reverse bias. d) High forward bias (if the junction is heavily doped on one side0 and moderate to high reverse bias. e) Only high reverse bias. Quiz 3: 1) The Schottky barrier height is a key parameter for a metal-­‐semiconductor junction. For a metal, N-­‐type semiconductor, Φ bN is given by which of the following expressions? a) Φ bN = χ S − Φ M . b) Φ bN = χ S + Φ M c) Φ bN = Φ M − χ S . d) Φ bN = χ S − Φ M e) Φ bN = Φ M × χ S Continued on next page 3 2)
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The forward bias current in a typical Schottky barrier is due to what physical mechanism? a) Drift b) Diffusion c) Recombination d) Thermionic emission e) Zener tunneling
At a given forward bias, how does the current in a typical Schottky barrier compare to that in typical PN junction? a) The SB current is much larger than the PN junction current. b) The SB current is a little larger than the PN junction current. c) The SB current is much smaller than the PN junction current. d) The SB current is a little smaller than the PN junction current. e) The SB current is about the same as the PN junction current. What is the forward bias ideality factor of a Schottky barrier diode? a) n = 1 b) n = 2 c) 1 < n < 2 d) n > 2 e) n < 1 Quiz 4: 1) The small signal model for a Schottky barrier diode contact contains what parameters? a) The dynamic resistance and the junction capacitance. b) The dynamic resistance and the diffusion capacitance. b) The junction capacitance and diffusion capacitance. b) The dynamic resistance. b) The junction capacitance. Continued on next page 4 2)
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Majority carriers respond in what characteristic time? a) The carrier lifetime. b) The carrier diffusion time, c) The scattering time. d) The dielectric relaxation time. e) The drift time.
To make a good ohmic contact to a semiconductor, what should be done? a) Choose a metal with a high Schottky barrier height. b) Use a lightly doped semiconductor. c) Introduce defects into the semiconductor to lower its lifetime. d) Dope the semiconductor very heavily. e) Reduce the contact area. What are the consequences of Fermi level pinning? a) The Schottky barrier height will be insensitive to the type of metal. b) The SB will behave like a PN junction. c) The SB will be ohmic – not rectifying. d) The thermionic emission theory will have to be replaced by drift-­‐diffusion theory. e) The ideality factor of the SB will decrease. 5