FE Electrical Electrical Materials Practice Problems

Answer: A) \(1.68 \times 10^{-8}\) Ω·m

The Handbook states on p. 118 that conductivity is the reciprocal of the resistivity:

$$\rho = \dfrac{1}{\sigma} = \dfrac{1}{5.96 \times 10^{7}\text{ S/m}} = 1.68 \times 10^{-8}\text{ Ω·m}$$

The unit follows because a siemens is an inverse ohm, so S/m inverts to Ω·m. This is copper's standard value at 20 °C, and the Handbook's table of metals on p. 120 lists \(1.55 \times 10^{-8}\) Ω·m at 0 °C, the difference being the 0.4% per °C rise of a metal's resistivity with temperature; only silver, at \(1.47 \times 10^{-8}\) Ω·m, conducts better. The number becomes a resistance through \(R = \rho L/A\): one metre of 1 mm\(^2\) copper wire has \(R = 1.68 \times 10^{-8} \times 1/10^{-6} = 0.017\) Ω, the figure behind wiring tables. In the mixed units of semiconductor work the same resistivity is \(1.68 \times 10^{-6}\) Ω·cm, since an ohm-metre is a hundred ohm-centimetres.

\(1.68 \times 10^{-6}\) Ω·m is the value in Ω·cm labelled as Ω·m.

\(5.96 \times 10^{7}\) Ω·m is the conductivity itself, not inverted.

\(1.68 \times 10^{8}\) Ω·m has the mantissa of the reciprocal but the exponent's sign lost.

Answer: B) \(0.409\text{ eV}\)

In an N-type sample nearly every donor is ionized at room temperature, so the electron concentration is essentially the doping:

$$n \approx N_D = 10^{17}\text{ cm}^{-3}$$

which is seven orders of magnitude above \(n_i\), and the Fermi level sits above the intrinsic level by

$$E_F - E_i = kT\ln\!\left(\dfrac{n}{n_i}\right) = V_T \ln\!\left(\dfrac{n}{n_i}\right)$$

$$\dfrac{n}{n_i} = \dfrac{10^{17}}{1.5 \times 10^{10}} = 6.67 \times 10^{6}$$

$$\ln(6.67 \times 10^{6}) = 15.71$$

$$E_F - E_i = (0.026)(15.71) = 0.409\text{ eV}$$

Sanity check against the figure: half the gap is \(0.56\text{ eV}\), so \(0.409\) puts \(E_F\) about three quarters of the way from mid-gap up to \(E_c\), which is where it is drawn, and still below the donor level. That check kills two of the wrong answers on sight.

\(0.177\text{ eV}\) uses \(\log_{10}\) where the relation calls for the natural logarithm: \(\log_{10}(6.67\times10^6) = 6.82\) instead of \(15.71\), and the ratio between the two answers is exactly \(\ln 10 = 2.303\).

\(0.817\text{ eV}\) is the built-in potential of a symmetric junction, \(V_0 = V_T\ln(N_aN_d/n_i^2)\) with \(N_a = N_d = 10^{17}\). That is the expression the Handbook actually prints, and it is exactly twice the quantity asked for, because it counts the Fermi shift on both sides of a junction rather than the one shift inside a single N-type sample.

\(1.018\text{ eV}\) drops \(n_i\) altogether and takes \(V_T\ln(N_D)\). The figure rejects it immediately: \(1.018\text{ eV}\) above mid-gap would put \(E_F\) well outside the \(1.12\text{ eV}\) gap, above the conduction band edge.

Answer: D) \(4.39 \times 10^{-6}\) S/cm

In intrinsic material \(n = p = n_i\), so the Handbook's \(\sigma = q(n\mu_n + p\mu_p)\) of p. 387 becomes

$$\sigma_i = qn_i(\mu_n + \mu_p) = (1.6 \times 10^{-19})(1.5 \times 10^{10})(1350 + 480)$$

$$\sigma_i = (2.4 \times 10^{-9})(1830) = 4.39 \times 10^{-6}\text{ S/cm}$$

The resistivity is \(1/\sigma_i = 2.28 \times 10^{5}\) Ω·cm, a million times that of the doped sample of a few tenths of an ohm-centimetre and \(10^{13}\) times copper's: pure silicon is nearly an insulator at room temperature, and it is the dopants that make it useful. Both carriers count here, unlike in extrinsic material, and the electrons carry 74% of the current because of their higher mobility. The conductivity is also strongly temperature dependent, roughly doubling every 11 °C through \(n_i\), which is the opposite of a metal and the reason intrinsic silicon serves as a thermistor-like sensor but not as a resistor.

\(3.24 \times 10^{-6}\) S/cm counts the electrons only, \(qn_i\mu_n\).

\(1.15 \times 10^{-6}\) S/cm counts the holes only, \(qn_i\mu_p\).

\(2.4 \times 10^{-9}\) S/cm is \(qn_i\), the charge density in C/cm\(^3\), with both mobilities forgotten.

Answer: C) \(5000\ \Omega\)

The Handbook (p. 388) gives exactly this geometry:

$$R = R_s \dfrac{L}{W}, \qquad R_s = \dfrac{\rho}{d}$$

where \(R_s\) is the sheet resistance in ohms per square, \(d\) is the layer thickness (called \(t\) on the figure), and \(L\) and \(W\) are the length and width of the diffusion. Two steps, and each has one trap.

Sheet resistance. Keep \(\rho\) and the thickness in the same length unit. With \(\rho\) in \(\Omega\cdot\text{cm}\), put the thickness in centimetres:

$$t = 0.1\,\mu\text{m} = 0.1 \times 10^{-4}\text{ cm} = 10^{-5}\text{ cm}$$

$$R_s = \dfrac{0.01}{10^{-5}} = 1000\ \Omega/\text{sq}$$

Number of squares. Sheet resistance is the resistance of one square of the layer, so the slab's shape enters only through how many squares fit between the contacts, along the current:

$$\dfrac{L}{W} = \dfrac{50}{10} = 5\text{ squares}$$

$$R = (1000)(5) = 5000\ \Omega$$

Note the ratio is length over width, not the other way round: a long narrow strip is a high resistance, exactly as \(R = \rho L / A\) would say. That is what \(200\ \Omega\) gets wrong — it uses \(W/L\) and so divides by five where it should multiply. \(50000\ \Omega\) reads the thickness as \(0.01\,\mu\text{m}\), giving \(R_s = 10000\ \Omega/\text{sq}\). \(2000\ \Omega\) makes both mistakes together.

A check on the whole thing without sheet resistance at all: the cross-section the current crosses is \(W t = (10)(0.1) = 1\ \mu\text{m}^2 = 10^{-8}\text{ cm}^2\), so \(R = \rho L / A = (0.01)(50 \times 10^{-4}) / 10^{-8} = 5000\ \Omega\), the same number.

Answer: D) \(0.308\text{ °C/W}\)

All the heat crosses all three layers one after the other, so the three conduction resistances are in series and add. The Handbook gives both parts on p. 210: the plane-wall conduction resistance is \(R = t/(kA)\), and resistances in series are added.

The area is common to all three, \(A = 1\text{ cm}^2 = 10^{-4}\text{ m}^2\), and the thicknesses go to metres:

$$R_1 = \dfrac{0.5\times10^{-3}}{(150)(10^{-4})} = 0.033\text{ °C/W}$$

$$R_2 = \dfrac{0.1\times10^{-3}}{(5)(10^{-4})} = 0.200\text{ °C/W}$$

$$R_3 = \dfrac{3\times10^{-3}}{(400)(10^{-4})} = 0.075\text{ °C/W}$$

$$R_{total} = 0.033 + 0.200 + 0.075 = 0.308\text{ °C/W}$$

The result worth carrying away is which layer dominates. The paste is the thinnest layer by a factor of six, yet it contributes about two thirds of the total, because its conductivity is thirty times worse than the silicon's and eighty times worse than the copper's. Thickness and conductivity trade off directly in \(t/k\), and a thin bad conductor beats a thick good one. This is why thermal interface material is chosen and applied so carefully in practice.

\(0.108\text{ °C/W}\) is the stack with the paste left out, \(0.033 + 0.075\): it is what you get by counting only the layers that look substantial. \(0.033\text{ °C/W}\) is the die alone. \(0.021\text{ °C/W}\) combines the three in parallel instead of in series, which would mean the heat had three separate routes out rather than one route through all three.