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SOLUTIONS: ECE 305 Homework: Week 4 Mark Lundstrom Purdue University
- Answer the following questions about resistivity at T = 300K.
1a) Compute the resistivity of intrinsic Si, Ge, and GaAs. 1b) Compute the resistivity of n-‐type Si, Ge, and GaAs doped at ND = 1019 cm-^3. Assume complete ionization of dopants.
Solution: According to eqn. (3.7) on p. 85 of SDF ρ = (^) nq μ^1 n +^ pq^ μ p
Ω-cm What are the units of 1 nq μ n or 1 pq μ p? In MKS (also called SI) units: 1 nq μ →^
m-^3 ×^
C ×^
m^2 /(V-s) =^
V
( C/s) −^ m^ =^
V
A −^ m^ =^ Ω-m^ MKS or SI But we are usually given carrier densities per cm^3 and mobility in cm^2 /(V-‐s), which are not MKS units. 1 nq μ →^
cm-^3 ×^
C ×^
cm^2 /(V-s) =^
V
( C/s) −^ cm^ =^
V
A −^ cm^ =^ Ω-cm so then the answer comes out in Ω - cm. Resistivity is usually quoted in Ω - cm. 1a) We need the carrier densities and mobilities: n = p = ni From Fig. 3.5, p. 80 of SDF for silicon: μ n ≈ 1400 cm
2 V-s^ μ p^ ≈^460
cm^2 V-s ρ = (^) nq μ^1 n +^ pq^ μ p
= (^) n^1
iq (^ μ n +^ μ p )^
Ω-cm
ρ = 1010 × 1.6 × 10 − 191 × ( 1400 + 460 ) = 3.4 × 105 Ω-cm
ρ = 3.4 × 105 Ω-cm (silicon) This is a very large resistivity – not as large as an insulator, but very large for a semiconductor.
HW4 Solutions (continued):
For Ge : ρ = (^) n^1
iq (^ μ^ n +^ μ p )^
Ω-cm From Fig. 3.5, p. 80 of SDF for Ge: μ n ≈ 4000 cm
2 V-s^ μ p^ ≈^2000
cm^2 V-s From Fig. 2.20, p. 54 of SDF for Ge:
ni ( 300 K) = 2 × 1013 cm-^3
ρ = 2 × 1013 × 1.6 × 10 −^119 × ( 4000 + 2000 ) = 5.2 × 101 Ω-cm
ρ = 5.2 × 101 Ω-cm (Ge) Intrinsic Ge is fairly conductive! This happens because the mobilities are higher, but mostly because the bandgap is much lower, so the intrinsic carrier concentration is much larger. For GaAs : ρ = (^) n^1
iq (^ μ^ n +^ μ p )^
Ω-cm From Fig. 3.5, p. 80 of SDF for GaAs: μ n ≈ 8500 cm
2 V-s^ μ p^ ≈^430
cm^2 V-s From Fig. 2.20, p. 54 of SDF for Ge:
ni ( 300 K) = 2.25 × 106 cm-^3
ρ = 2.25 × 106 × 1.6 × 101 − 19 × ( 8500 + 430 ) = 3.1× 108 Ω-cm
ρ = 3.1× 108 Ω-cm (GaAs) Intrinsic GaAs has a resistivity that is orders of magnitude larger than Si! This happens even though the electron mobility is much larger than Si because the bandgap is much larger, so the intrinsic carrier concentration is orders of magnitude smaller.
HW4 Solutions (continued):
Comments: For Si, we found that the range of possible resistivities is 3.0 × 105 Ω-cm < ρ < 6 × 10 −^3 Ω-cm which is about eight orders of magnitude. In practice, we could never have pure, intrinsic Si, there will always be some unintentional dopants no matter how pure we try to make Si, so the resistivities would not be as high as indicted here, but it’s common to find resistivities in the 1000’s. Also, one can dope Si to 1020 cm-^3 , so resistivities 10 times lower than indicated above can be obtained. How do these limits compare to insulators and metals? A Google search shows Resistivity of copper: ρ Cu = 1.68 × 10 −^8 Ω-cm Resistivity of diamond: ρ diamond = 1013 − 1020 Ω-cm So semiconductors are not great metals and not great insulators – their usefulness comes from the ability to vary their resistivity controllably with doping.
- You are given a 10 Ohm-‐cm silicon wafer at 300 K.
2a) If it is n-‐type, What is the electron density? 2b) If it is p-‐type, what is the hole density?
Solution:
From Fig. 3.8 on p. 86 of SDF, we can read off the doping density for a given resistivity. 2a) 10 Ohm-‐cm n-‐type corresponds to ND ≈ 5 × 1014 cm-^3. Under these conditions:
n ≈ ND ≈ 5 × 1014 cm-^3
2b) 10 Ohm-‐cm p-‐type corresponds to N (^) A ≈ 1.3× 1015 cm-^3. Under these conditions:
p ≈ N (^) A ≈ 1.3× 1015 cm-^3
HW4 Solutions (continued):
- Determine the diffusion coefficient for electrons in Si at T = 300 K for the following two conditions. 3a) Intrinsic Si 3b) Si doped at ND = 1019 cm-^3
Solution:
3a) intrinsic Si From Fig. 3.5 of SDF for lightly doped material. Si: μ n = 1360 cm^2 V-s Dn μ n^ =^
kBT q →^ Dn^ =^
kBT q^ μ n^ Dn^ =^
kBT q^ μ n^ =^ 0.026^ ×^1360
Dn = 35 cm^2 s 3b) Si doped at ND = 1019 cm-^3 From Fig. 3.5 of SDF for lightly doped material. Si: μ n = 110 cm^2 V-s Dn = kB qT μ n = 0.026 × 110 Dn = 2.9 cm^2 s
- For the energy band sketched below, provide sketches of the following:
4a) The carrier densities, n (x), and p (x) vs. position,
4b) The electrostatic potential, ψ ( x ) , vs. position.
4c) The electric field E vs. position.
4d) The space charge density, ρ ( x ) vs. position.
HW4 Solutions (continued):
4b) The electrostatic potential, ψ (^) ( x ) , vs. position (just turn EC ( x ) , or EV ( x ), or Ei ( x ) upside down.
4c) The electric field E vs. position (is proportional to the slope of EC ( x ) , or EV ( x ), or Ei ( x ).
HW4 Solutions (continued):
4d) The space charge density, ρ ( x ) vs. position. (Take slope of electric field vs.
position or deduce from the carrier densities and doping densities.)
- For the energy band diagram sketched below, answer the following questions.
5a) Sketch the electrostatic potential, V (^) ( x ) , vs. position, x. 5b) Sketch the electric field, E (^) ( x ) , vs. position, x. 5c) Sketch the electron density, n (^) ( x ) vs. position, x. 5d) Sketch the hole density, p (^) ( x ) , vs. position, x.
HW4 Solutions (continued):
5c) Sketch the electron density, n (^) ( x ) vs. position, x.
5d) Sketch the hole density, p (^) ( x ) , vs. position, x.
