part03

Chapter 3: Carrier Action

Source slice: Pierret, Semiconductor Device Fundamentals, Chapter 3 (book pp. 75-148). Equations keep the book’s numbering. OCR damage was repaired using physics; where a number in a problem statement was unreadable, the assumption is stated explicitly.

---

Prologue: from a sleeping crystal to a working device

In Chapter 2 we studied a semiconductor at rest (equilibrium). That picture is essential as a reference, but nothing happens in it: the net current is exactly zero. A device is useful only when we poke the crystal (apply a voltage, shine light, inject carriers) and the carriers respond.

Think of a crowded stadium at night after the match ends. In equilibrium everybody mills about randomly (thermal motion), and no net flow goes out of the gates. Now:

• someone tilts the whole stadium (an electric field), so everyone drifts downhill: drift;
• the north stand is packed and the south is empty (a concentration gradient), so people wander toward the empty side just by random bumping: diffusion;
• people vanish through hidden trapdoors and new people appear from the entrances: recombination-generation (R-G).

Those three actions are the whole story of Chapter 3. For each we will (1) picture it, (2) turn it into an equation, and (3) identify its “constant of the motion” (the one material parameter that sets its strength):

Carrier action	Gives rise to	Material parameter (“constant of the motion”)
Drift	drift current	mobility ${\mu }_n,{\mu }_p$
Diffusion	diffusion current	diffusion coefficient $D_N,D_P$
Recombination-generation	change of carrier concentration in time	minority-carrier lifetime ${\tau }_n,{\tau }_p$

Finally we glue the three together into the equations of state (continuity equations and the minority-carrier diffusion equation), solve a few canonical problems, and add two supplemental ideas: diffusion length and quasi-Fermi levels.

---

3.1 Drift

3.1.1 Definition and visualization

Drift is the motion of charged particles in response to an applied electric field $\mathcal{E}$.

Microscopic picture. Apply a field across a bar of semiconductor (Fig. 3.1a). The field pushes holes ($+q$) along $\mathcal{E}$ and electrons ($-q$) against $\mathcal{E}$. But a carrier is never allowed to accelerate for long: it collides with thermally vibrating lattice atoms and with ionized dopant atoms (it is scattered). The result (Fig. 3.1b) is a jerky, zig-zag path: accelerate, crash, restart, accelerate, crash. The overall trend is along (or against) the field.

Macroscopic picture. Tracking one carrier through every collision is hopeless, and nobody can measure it anyway. What we can measure is the average motion. Averaged over all carriers of one type, the motion is a steady drift velocity $v_d$ along (holes) or against (electrons) the field (Fig. 3.1c).

Analogy: a pinball machine. The ball is pulled down by gravity (the field) but keeps bouncing off pegs (scattering). It never speeds up indefinitely; it reaches an average downhill speed. That average speed is the drift velocity.

Common confusion: “isn’t the carrier already moving fast?” Yes! Even at equilibrium, thermal motion is ferocious: typical thermal speeds are about $1/1000$ of the speed of light at room temperature. But thermal motion is completely random (Fig. 3.2): for every carrier going left another goes right. Averaged over many carriers it is zero, so it carries no net current. Drift is a tiny, orderly bias superimposed on a huge random jitter, like a slow river current carrying a crowd of frantically swimming fish.

3.1.2 Drift current

Definition of current. Current $I$ is the charge per unit time crossing an arbitrarily chosen plane that is perpendicular to the flow.

Derivation of hole drift current. Take a p-type bar of cross-sectional area $A$ (Fig. 3.3), a plane normal to the flow, and holes with drift velocity $v_d$.

$v_{d}t\text{(distance)}$

All holes within a distance $v_{d}t$ upstream of the plane will cross it in time $t$.

$v_{d}tA\text{(volume)}$

All holes in this volume cross the plane in time $t$.

$p{v}_{d}tA\text{(number of holes)}$

$p$ is the hole concentration (holes per cm${}^3$) times the volume.

$qp{v}_{d}tA\text{(charge crossing in time }t)$

Multiply by the charge per hole.

$qp{v}_{d}A\text{(charge per unit time)}$

Divide by $t$; this is precisely the definition of current. Hence

\boxed{I_{P|\text{drift}} = q\,p\,v_d\,A} \tag{3.1}

Current density. The area $A$ is baggage, and current as defined is a scalar although the flow has a direction. We therefore use the current density $\mathbf{J}$, a vector pointing along the flow with magnitude $J=I/A$:

\boxed{J_{P|\text{drift}} = q\,p\,v_d} \tag{3.2}

Relating $v_d$ to the field. Figure 3.4 (measured $v_d$ vs. $\mathcal{E}$ in ultrapure Si at 300 K) shows: at low fields $v_d\propto \mathcal{E}$ (a straight line of slope 1 on log-log axes), while at very high fields $v_d$ saturates at $v_{sat}\approx 10^7$ cm/s and no longer depends on $\mathcal{E}$. The book’s empirical form is

v_d = \frac{\mu_0\,\mathcal{E}}{\left[1 + \left(\mu_0\mathcal{E}/v_{sat}\right)^{\beta}\right]^{1/\beta}} \tag{3.3}

with $\beta \approx 1$ for holes and $\beta \approx 2$ for electrons in Si. Check the two limits:

$\mathcal{E}\to 0:v_d\to {\mu }_0\mathcal{E}(\text{linear; the proportionality constant is the mobility }\mu )$

$\mathcal{E}\to \infty :v_d\to \frac{{\mu }_0\mathcal{E}}{{\mu }_0\mathcal{E}/v_{sat}}=v_{sat}(\text{independent of }\mathcal{E})$

Why saturation? At extreme fields, carriers gain energy so fast that they dump it into the lattice by emitting vibrations (optical phonons) almost as soon as they acquire it. It is like a car whose engine power is entirely eaten by air drag at top speed.

Unless stated otherwise we use the low-field limit $v_d=\mu \mathcal{E}$, which is the case of most practical interest. For holes ($v_d={\mu }_p\mathcal{E}$) substitution into (3.2) gives

\boxed{J_{P|\text{drift}} = q\,\mu_p\,p\,\mathcal{E}} \tag{3.4a}

and similarly for electrons

\boxed{J_{N|\text{drift}} = q\,\mu_n\,n\,\mathcal{E}} \tag{3.4b}

${\mu }_n$ and ${\mu }_p$ are the electron and hole mobilities, always taken as positive numbers.

A sign puzzle worth thinking about. Electrons drift against the field: $v_d=-{\mu }_n\mathcal{E}$. But electrons carry negative charge, so the current they carry points opposite to their velocity, i.e., along $\mathcal{E}$. Two minus signs cancel: $J_{N|\text{drift}}=(-q)n(-{\mu }_n\mathcal{E})=+q{\mu }_{n}n\mathcal{E}$. Therefore both electrons and holes contribute current in the same direction as the electric field, and their contributions simply add. That is why (3.4a) and (3.4b) both carry a plus sign.

3.1.3 Mobility

Mobility is the star parameter of drift. Let us collect its facts in the book’s “information card” format.

Units. cm${}^2$/V$\cdot$s. (Check: $v_d=\mu \mathcal{E}$ gives $\text{cm/s}=\mu \cdot \text{V/cm}$, so $\mu$ is in ${\text{cm}}^2/\text{V s}$.)

Sample numerical values (300 K).

• Si: ${\mu }_n\approx 1360$, ${\mu }_p\approx 460$ cm${}^2$/V$\cdot$s (for $N_D=10^{14}$ and $N_A=10^{14}$ cm${}^{-3}$ doping respectively).
• Uncompensated high-purity GaAs: ${\mu }_n\approx 8000$, ${\mu }_p\lesssim 400$ cm${}^2$/V$\cdot$s.
• In the major semiconductors, ${\mu }_n>{\mu }_p$ for the same doping and temperature.

Relation to scattering. “Mobility” literally means ease of movement. More scattering means lower mobility. The two dominant scattering mechanisms in non-degenerately doped, device-quality material are:

1. Lattice scattering: collisions with thermally vibrating lattice atoms. (Note: it is the vibration, i.e., the displacement from perfect lattice positions, that scatters. The perfectly periodic crystal potential is already hidden in the effective mass.)
2. Ionized impurity scattering: Coulomb deflection by ionized donors or acceptors.

A quantitative relation is

$\mu =\frac{q⟨{\tau }_c⟩}{m_{\text{cond}}^{\ast }}$

where $⟨{\tau }_c⟩$ is the mean free time between collisions and $m_{\text{cond}}^{\ast }$ is the conductivity effective mass. (We write ${\tau }_c$ here to avoid mixing it up with the carrier lifetimes ${\tau }_n,{\tau }_p$ later in the chapter; the book writes $\mu \propto \tau /m^{\ast }$.) More collisions give a shorter free time and a smaller $\mu$. A small $m^{\ast }$ means a light, easily pushed carrier. This explains why electrons in GaAs ($m_n^{\ast }\approx 0.067m_0$) are far more mobile than in Si.

Intuition for $\mu =q{\tau }_c/m^{\ast }$. Newton: between collisions, a carrier accelerates at $a=q\mathcal{E}/m^{\ast }$ for an average time ${\tau }_c$, so its average extra velocity is about $a{\tau }_c=(q{\tau }_c/m^{\ast })\mathcal{E}$. The bracket is the mobility.

Doping dependence (Fig. 3.5). Mobilities are roughly constant at low doping (below about $10^{16}$ cm${}^{-3}$ in Si) and then fall monotonically with increasing $N_A$ or $N_D$. Table beside Fig. 3.5(a) for Si at 300 K:

$N_A$ or $N_D$ (cm${}^{-3}$)	${\mu }_n$	${\mu }_p$
$1\times 10^{14}$	1358	461
$2\times 10^{14}$	1357	460
$5\times 10^{14}$	1352	459
$1\times 10^{15}$	1345	458
$2\times 10^{15}$	1332	455
$5\times 10^{15}$	1298	448
$1\times 10^{16}$	1248	437
$2\times 10^{16}$	1165	419
$5\times 10^{16}$	986	378
$1\times 10^{17}$	801	331

Why? Use an electrical analogy (Fig. 3.6): each scattering mechanism is a “resistance to motion” and the resistances are in series:

$R_{TOTAL}=R_L+R_I$

($R_L$ = lattice, $R_I$ = ionized impurity.) $R_L$ has no doping dependence. At low doping $R_I\ll R_L$, so $R_{TOTAL}\approx R_L$ and $\mu$ is doping-independent. At higher doping $R_I$ grows (more ionized scatterers) and $\mu$ falls.

Temperature dependence (Fig. 3.7). For $N\lesssim 10^{14}$ cm${}^{-3}$ the curves merge into one near power law:

${\mu }_n\propto T^{-2.3\pm 0.1},{\mu }_p\propto T^{-2.2\pm 0.1}$

so mobility rises as T falls. For heavier doping, $\mu$ still rises with falling T, but more slowly; for $N\sim 10^{19}$ cm${}^{-3}$, $\mu$ even drops when T falls below about 200 K.

Why? Lattice scattering weakens as T falls (atoms vibrate less): $R_L$ decreases with decreasing T, so $\mu$ rises as a negative power of T. But ionized impurity scattering gets stronger at low T: slower carriers spend more time near each ionized impurity and are deflected more easily, so $R_I$ increases with decreasing T. At high doping $R_I$ is a larger fraction of the total, partly cancelling the lattice-scattering improvement, hence the flatter slope.

Scattering	Doping dependence	Temperature dependence
Lattice ($R_L$)	none	decreases with decreasing T
Ionized impurity ($R_I$)	increases with $N_A$ or $N_D$	increases with decreasing T

Exercise 3.1: Empirical mobility fit (with MATLAB)

Problem. The majority-carrier mobility in Si at 300 K is computed from the empirical fit

$\mu ={\mu }_{min}+\frac{{\mu }_0}{1+(N/N_{ref}{)}^{\alpha }}$

where $N$ is the doping concentration ($N_A$ or $N_D$) and the others are fit parameters. Temperature dependence enters through

$A=A_{300}{\left(\frac{T}{300}\right)}^{\eta }$

where $A$ stands for ${\mu }_{min}$, ${\mu }_0$, $N_{ref}$, or $\alpha$, $A_{300}$ is its 300 K value, $T$ is in kelvin, and $\eta$ is the temperature exponent. Parameters for Si:

Parameter	Electrons	Holes	Temp. exponent $\eta$
$N_{ref}$ (cm${}^{-3}$)	$1.3\times 10^{17}$	$2.35\times 10^{17}$	2.4
${\mu }_{min}$ (cm${}^2$/V s)	92	54.3	$-0.57$
${\mu }_0$ (cm${}^2$/V s)	1268	406.9	$-2.33$ (electrons), $-2.23$ (holes)
$\alpha$	0.91	0.88	$-0.146$

(a) Construct a log-log plot of ${\mu }_n$ and ${\mu }_p$ vs. $N_A$ or $N_D$ for $10^{14}\le N\le 10^{19}$ cm${}^{-3}$ and compare with Fig. 3.5(a). (b) Plot ${\mu }_n$ vs. $T$ and ${\mu }_p$ vs. $T$ for $200\le T\le 500$ K with $N$ stepped in decades from $10^{14}$ to $10^{18}$ cm${}^{-3}$ and compare with Fig. 3.7.

Logic of the code. The formula is a “floor plus a falling step”: ${\mu }_{min}$ is the floor at very high doping; ${\mu }_{min}+{\mu }_0$ is the ceiling at low doping (lattice-limited); $N_{ref}$ is the doping at which the step is half-way down; $\alpha$ controls how steeply it falls. We use logspace to get points equally spaced in $\log N$ and element-wise operators (./, .^) so MATLAB evaluates the formula for the whole vector at once.

Solution (a). (The book’s listing, cleaned of OCR noise.)

% Mobility versus Dopant Concentration (Si, 300 K)
% Fit parameters
NDref = 1.3e17;  NAref = 2.35e17;
mun_min = 92;    mup_min = 54.3;
mun_0 = 1268;    mup_0 = 406.9;
an = 0.91;       ap = 0.88;
 
% Mobility calculation
N = logspace(14,19);                        % 50 log-spaced points
mun = mun_min + mun_0./(1 + (N/NDref).^an);
mup = mup_min + mup_0./(1 + (N/NAref).^ap);
 
% Plotting results
close
loglog(N, mun, N, mup); grid;
axis([1.0e14 1.0e19 1.0e1 1.0e4]);
xlabel('N_A or N_D (cm^{-3})');
ylabel('Mobility (cm^2/V-sec)');
text(1.0e15, 1500, 'Electrons');
text(1.0e15, 500,  'Holes');
text(1.0e18, 2000, 'Si, 300 K');

Quick hand-check of one point ($N=10^{16}$): $(10^{16}/1.3\times 10^{17}{)}^{0.91}=(0.0769{)}^{0.91}=0.0969$, so ${\mu }_n=92+1268/1.0969=92+1156=1248$ cm${}^2$/V s, which matches the table. The plot is numerically identical to Fig. 3.5(a).

Part (b) (the book leaves it to the reader; Problem 3.3 asks for it, and it is completed here).

% Mobility versus Temperature (Si) with doping as parameter
T = linspace(200,500,100);
Nlist = 10.^(14:18);
 
% 300 K parameters and temperature exponents (k=1 electrons, k=2 holes)
Nref0 = [1.3e17 2.35e17];  etaNref = 2.4;
mmin0 = [92 54.3];         etamin  = -0.57;
m00   = [1268 406.9];      eta0    = [-2.33 -2.23];
al0   = [0.91 0.88];       etaal   = -0.146;
 
for k = 1:2
    subplot(1,2,k)
    for N = Nlist
        Nref = Nref0(k)*(T/300).^etaNref;
        mmin = mmin0(k)*(T/300).^etamin;
        m0   = m00(k)  *(T/300).^eta0(k);
        al   = al0(k)  *(T/300).^etaal;
        mu = mmin + m0./(1 + (N./Nref).^al);
        loglog(T, mu); hold on
    end
    grid; axis([200 500 100 1e4]);
    xlabel('T (K)'); ylabel('\mu (cm^2/V-sec)');
    if k==1, title('Electrons'), else, title('Holes'), end
end

Expected result: the lowest-doping curve is a straight line on log-log axes with slope about $-2.3$ (electrons) / $-2.2$ (holes); heavier-doping curves bend and flatten at low T, as in Figs. 3.7(a) and (b).

3.1.4 Resistivity

Resistivity $\rho$ is a “normalized resistance” that does not depend on the sample’s shape. Operationally it is the proportionality between the applied field and the resulting current density:

\mathcal{E} = \rho\,J \tag{3.5a}

J = \sigma\mathcal{E} = \frac{1}{\rho}\mathcal{E} \tag{3.5b}

with $\sigma =1/\rho$ the conductivity. In a homogeneous material all the current is drift current, so using (3.4):

J_{drift} = J_{N|drift} + J_{P|drift} = q(\mu_n n + \mu_p p)\,\mathcal{E} \tag{3.6}

Comparing with (3.5b):

\boxed{\rho = \frac{1}{q(\mu_n n + \mu_p p)}} \tag{3.7}

Simplification in doped material. In a non-degenerate n-type sample in the extrinsic region ($N_D\gg n_i$): $n=N_D$ and $p=n_i^2/N_D\ll n$ (Section 2.5.5). Then ${\mu }_{n}n+{\mu }_{p}p\approx {\mu }_n{N}_D$. Similarly for p-type, ${\mu }_p{N}_A$. Hence

\boxed{\rho \approx \frac{1}{q\mu_n N_D}\quad\text{(n-type)}} \tag{3.8a}

\boxed{\rho \approx \frac{1}{q\mu_p N_A}\quad\text{(p-type)}} \tag{3.8b}

Combined with the mobility-versus-doping data these give a one-to-one relation between a directly measurable quantity ($\rho$) and the doping (Fig. 3.8). Measuring $\rho$ is a routine way to find $N_A$ or $N_D$.

How to measure resistivity.

1. Straightforward bar method. Cut a bar, apply a voltage, measure the current, use $R=V/I=\rho \ell /A$ ($\ell$ = length, $A$ = cross section). It looks easy but is deceptively difficult (contact resistance), destroys good material, and does not suit wafers.

2. Four-point probe (Fig. 3.9a). Four collinear, equally spaced probes touch the wafer surface. Pass a known current $I$ through the outer two probes and measure the voltage $V$ between the inner two:

\rho = 2\pi s\,\frac{V}{I}\,\Gamma \tag{3.9}

where $s$ is the probe spacing and $\Gamma$ is a documented correction factor that depends on sample thickness and on whether the wafer’s bottom touches an insulator or a metal (commercial instruments compute it automatically from the thickness you enter). Why four probes? Because the voltmeter draws almost no current, the voltage drop across the contact resistances at the two voltage probes is negligible, which removes the contact-resistance headache. The method is quick and ideal for wafers, but it leaves slight surface damage, so it is not used on wafers destined for high-yield devices.

3. Eddy-current method (Fig. 3.9b), non-contacting. An RF-excited ferrite core produces fringing fields; a nearby conducting wafer develops eddy currents which absorb RF power. The power absorbed (calibrated) gives the sheet resistivity. Built-in ultrasound transducers measure the wafer thickness from the phase of echoes off the top and bottom surfaces. Then ${\rho }_{wafer}={\rho }_{sheet}\times \text{thickness}$.

3.1.5 Band bending

So far we have always drawn $E_C$ and $E_V$ as flat lines. When an electric field exists in the material, the band edges become functions of position; this is called band bending (Fig. 3.10a).

Meaning of the diagram. The energy band diagram plots allowed total electron energy vs. position. Bond-breaking with energy exactly $E_G$ creates an electron at $E_C$ and a hole at $E_V$, both essentially motionless. Any extra energy makes them move. Therefore

$\text{K.E. of electrons}=E-E_C,\text{K.E. of holes}=E_V-E$

(for holes, energy increases downward on the diagram, so a hole deep in the valence band has more kinetic energy). Total = K.E. + P.E., so the electron potential energy is $E_C-E_{ref}$ (Fig. 3.10c). $E_{ref}$ is an arbitrary constant.

Link to the electrostatic potential. For a particle of charge $-q$ in an electrostatic potential $V$ (when magnetic, thermal-gradient and stress effects are negligible),

\text{P.E.} = -qV \tag{3.10}

but we also found

\text{P.E.} = E_C - E_{ref} \tag{3.11}

Equate them and solve for $V$:

V = -\frac{1}{q}(E_C - E_{ref}) \tag{3.12}

By definition $\mathcal{E}=-\nabla V$ (3.13), which in one dimension reads

\mathcal{E} = -\frac{dV}{dx} \tag{3.14}

Differentiate (3.12):

$\mathcal{E}=-\frac{dV}{dx}=\frac{1}{q}\frac{d{E}_C}{dx}$

and since $E_C$, $E_V$, $E_i$ differ only by additive constants (their slopes are equal):

\boxed{\mathcal{E} = \frac{1}{q}\frac{dE_C}{dx} = \frac{1}{q}\frac{dE_V}{dx} = \frac{1}{q}\frac{dE_i}{dx}} \tag{3.15}

Reading a band diagram by inspection:

• To get the shape of $V(x)$: flip the band picture “upside down” (since $V=-(E_C-E_{ref})/q$). The vertical offset is arbitrary.
• To get the shape of $\mathcal{E}(x)$: look at the slope of $E_C$ (or $E_V$, $E_i$). Bands sloping up toward $+x$ mean $\mathcal{E}>0$.

Mountain-landscape analogy. Electrons roll downhill on the band diagram (toward lower $E_C$), holes “bubble” uphill (like air bubbles in water). A steep band means a strong field. Sign confusion disappears if you remember that the diagram is an electron potential-energy landscape.

Exercise 3.2: Reading a band diagram

Problem. The figure in the book is an energy band diagram for Si at 300 K with $E_F$ taken as the energy reference. The Fermi level sits $E_G/4$ below $E_i$ at $x=\pm L$ (p-type there, $E_i-E_F=E_G/4$) and $E_G/4$ above $E_i$ at $x=0$ ($E_F-E_i=E_G/4$, n-type there); the bands bend smoothly between. Three electrons and three holes (labelled 1, 2, 3) are marked on the diagram. (a) Sketch $V(x)$. (b) Sketch $\mathcal{E}(x)$. (c) Find the K.E. and P.E. of the marked carriers. (d) Determine the resistivity of the $x>L$ region.

Solution.

(a) $V(x)$ has the same form as the “upside-down” $E_C(x)$. Choosing $V=0$ at $x=L$ (arbitrary), the curve is the mirror image of the band edge: where the bands dip (near $x=0$) the potential has its peak, and it is flat for $|x|>L$.

(b) $\mathcal{E}(x)=(1/q)d{E}_C/dx$: proportional to the slope of the bands. It is zero where the bands are flat ($|x|>L$), positive where $E_C$ rises with $x$, negative where it falls, and it changes sign at $x=0$ where the slope reverses.

(c) For an electron: $\text{K.E.}=E-E_C$ and $\text{P.E.}=E_C-E_{ref}=E_C-E_F$. For a hole (energy increases downward): $\text{K.E.}=E_V-E$ and $\text{P.E.}=E_{ref}-E_V=E_F-E_V$. All differences are taken at the same $x$. With $E_G=1.12$ eV, $E_G/4=0.28$ eV and $E_G/2=0.56$ eV:

• at $x=0$ (n-type): $E_C-E_F=E_G/2-E_G/4=0.28$ eV, and $E_F-E_V=E_G/2+E_G/4=0.84$ eV;
• at $x=\pm L$ (p-type): $E_C-E_F=0.84$ eV, and $E_F-E_V=0.28$ eV.

Carrier	K.E. (eV)	P.E. (eV)
Electron 1	0	0.28
Electron 2	0.56	0.28
Electron 3	0	0.84
Hole 1	0	0.28
Hole 2	0.56	0.28
Hole 3	0	0.84

(Electron 1 sits at $E_C$ at $x=0$; electron 2 is $0.56$ eV above $E_C$ at $x=0$; electron 3 sits at $E_C$ at $x=\pm L$. Hole 1 sits at $E_V$ at $x=\pm L$; hole 2 is $0.56$ eV below $E_V$ there; hole 3 sits at $E_V$ at $x=0$.)

(d) In the $x>L$ region $E_i-E_F=E_G/4=0.28$ eV, so the material is p-type with

$N_A\approx p=n_i{e}^{(E_i-E_F)/kT}$

$=10^{10}{e}^{0.28/0.0259}=4.96\times 10^{14}{\text{cm}}^{-3}$

From Fig. 3.5(a), ${\mu }_p\approx 459$ cm${}^2$/V s. By (3.8b),

$\rho =\frac{1}{q{\mu }_p{N}_A}=\frac{1}{(1.6\times 10^{-19})(459)(4.96\times 10^{14})}=27.5\Omega \text{cm}$

Cross-check with Fig. 3.8(a): $\rho \approx 25\Omega$ cm.

---