Carrier Transport: Drift and Diffusion

Six worked problems on how carriers actually move in a semiconductor: drift velocity and drift current in an applied field, the Einstein relation linking diffusivity to mobility, diffusion current from a concentration gradient, diffusion length and the decay of injected carriers, drift and diffusion acting together, and a base-width design set by transit time.

Each problem below is solved in full — the given data is listed first, the governing relation is stated, and the arithmetic is carried through to a boxed answer. Work through the statement yourself before opening the solution. The underlying theory is covered in the Electronic Devices lecture series.

A bar of n-type silicon is \( 1~\text{cm} \) long with a square cross-section of \( 1~\text{mm}\times1~\text{mm} \). It is doped with \( N_D = 1\times10^{16}~\text{cm}^{-3} \); the mobilities are \( \mu_n = 1300 \) and \( \mu_p = 450~\text{cm}^2/\text{V}\cdot\text{s} \), and \( n_i = 1.5\times10^{10}~\text{cm}^{-3} \). A battery of \( 5~\text{V} \) is applied end to end.

1. Find the electric field and the electron drift velocity.

2. Find the drift current density and show that the hole drift current is negligible.

3. Find the resistance of the bar and the total current, and check it against part (b).

4. How long does an electron take to traverse the bar?

• \( N_D = 1\times10^{16}~\text{cm}^{-3} \Rightarrow n \approx 1\times10^{16} \), \( p = n_i^2/n = 2.25\times10^{4}~\text{cm}^{-3} \)

• \( L = 1~\text{cm} \), \( A = 0.1\times0.1 = 0.010~\text{cm}^2 \), \( V = 5~\text{V} \)

• Uniform field along the bar; low-field (constant-mobility) transport assumed.

Part (a) — Field and drift velocity

\( E = 5.0~\text{V/cm} \), \( v_d = 6.50\times10^{3}~\text{cm/s} \)

Part (b) — Drift current density

\( J = 10.41~\text{A/cm}^2 \), of which holes contribute only \( 8.1\times10^{-12}~\text{A/cm}^2 \)

Part (c) — Resistance and current

Cross-check against the current density: \( I = JA = (10.41)(0.010) = 0.1041~\text{A} \).

\( R = 48.02~\Omega \), \( I = 104.1~\text{mA} \)

Part (d) — Transit time

\( t \approx 154~\mu\text{s} \)

The drift velocity of \( 6.50\times10^{3}~\text{cm/s} \) is about \( 6.5\times10^{-4} \) of the random thermal velocity (\( \sim10^{7}~\text{cm/s} \)). Drift is a slow net crawl superposed on violent random motion, which is precisely why mobility is a constant and the bar obeys Ohm's law.

Silicon at 300 K has \( \mu_n = 1350 \) and \( \mu_p = 480~\text{cm}^2/\text{V}\cdot\text{s} \); take \( kT/q = 25.85~\text{mV} \).

1. Find the electron and hole diffusion constants.

2. A different sample is measured to have \( D_p = 12~\text{cm}^2/\text{s} \). What is its hole mobility?

3. Estimate \( D_n \) at 400 K if lattice scattering makes \( \mu_n \propto T^{-3/2} \).

• Einstein relation: \( \dfrac{D_n}{\mu_n} = \dfrac{D_p}{\mu_p} = \dfrac{kT}{q} = V_T \)

• \( V_T = 25.85~\text{mV} \) at 300 K

• Non-degenerate, near-equilibrium material (the relation fails at very heavy doping).

Part (a) — Diffusion constants at 300 K

\( D_n = 34.90~\text{cm}^2/\text{s} \), \( D_p = 12.41~\text{cm}^2/\text{s} \)

Part (b) — Mobility from a measured diffusivity

\( \mu_p = 464.2~\text{cm}^2/\text{V}\cdot\text{s} \)

Part (c) — Diffusivity at 400 K

Two things change in opposite directions: \( V_T \) rises linearly with \( T \) while lattice scattering pulls \( \mu_n \) down as \( T^{-3/2} \).

\( D_n(400~\text{K}) \approx 30.22~\text{cm}^2/\text{s} \)

Since \( D \propto T\mu \propto T^{-1/2} \) under pure lattice scattering, diffusivity is far less temperature-sensitive than either \( n_i \) or the mobility alone — it falls only about 13 per cent over a 100 K rise.

In a piece of silicon the hole concentration falls linearly from \( 1\times10^{16}~\text{cm}^{-3} \) at \( x=0 \) to \( 1\times10^{14}~\text{cm}^{-3} \) at \( x = 2~\mu\text{m} \). Use \( D_n = 34.90 \) and \( D_p = 12.41~\text{cm}^2/\text{s} \).

1. Find the hole diffusion current density and state its direction.

2. If instead electrons had exactly this profile, what would the diffusion current density be?

• \( p(0) = 1\times10^{16} \), \( p(2~\mu\text{m}) = 1\times10^{14}~\text{cm}^{-3} \)

• \( 2~\mu\text{m} = 2\times10^{-4}~\text{cm} \); the profile is linear, so the gradient is constant.

• No electric field is applied — the current here is pure diffusion.

Step 1 — The gradient

Step 2 — Hole diffusion current

Holes are positive, so their particle flux and their current point the same way; the minus sign in the law makes the current flow down the concentration slope:

\( J_p = 98.4~\text{A/cm}^2 \), directed along \( +x \) (from high to low concentration)

Step 3 — The same profile in electrons

\( J_n = -277~\text{A/cm}^2 \) directed along \( -x \) — electrons still move toward \( +x \), but being negative they carry current the other way

To push \( 98.4~\text{A/cm}^2 \) through the same material by drift alone, at the mean hole density of \( 5.05\times10^{15}~\text{cm}^{-3} \), would need a field of about \( 253~\text{V/cm} \). Steep gradients over micrometre distances are a very effective way to move charge — which is exactly the mechanism a forward-biased junction and a transistor base rely on.

Holes are injected steadily into a long n-type silicon bar at \( x = 0 \), maintaining an excess concentration \( \Delta p(0) = 1\times10^{14}~\text{cm}^{-3} \) there. The hole lifetime is \( \tau_p = 1~\mu\text{s} \) and \( D_p = 12.41~\text{cm}^2/\text{s} \). The bar is field-free beyond \( x=0 \).

1. Find the diffusion length \( L_p \).

2. Find the excess hole concentration at \( x = 50~\mu\text{m} \), and the distance at which it has decayed to 1 per cent of its injected value.

3. Find the injected hole current density at \( x = 0 \).

4. How would \( L_p \) change if the lifetime were improved to \( 10~\mu\text{s} \)?

• \( \tau_p = 1~\mu\text{s} = 1\times10^{-6}~\text{s} \), \( D_p = 12.41~\text{cm}^2/\text{s} \)

• \( \Delta p(0) = 1\times10^{14}~\text{cm}^{-3} \), low-level injection into a long bar

• Steady state, no field: the continuity equation reduces to \( D_p\dfrac{d^2\Delta p}{dx^2} = \dfrac{\Delta p}{\tau_p} \)

Step 1 — Diffusion length

The decaying solution of that equation is \( \Delta p(x) = \Delta p(0)e^{-x/L_p} \) with

\( L_p = 35.2~\mu\text{m} \)

Step 2 — Profile

\( \Delta p = 2.42\times10^{13}~\text{cm}^{-3} \) at \( 50~\mu\text{m} \); it drops to 1 per cent by \( x = 162.2~\mu\text{m} \)

Step 3 — Injected current density

\( J_p(0) = 56.4~\text{mA/cm}^2 \)

Step 4 — Longer lifetime

Ten times the lifetime gives only \( \sqrt{10} = 3.16 \) times the length, \( L_p' = 111.4~\mu\text{m} \)

\( L_p \) is the average distance a hole diffuses before it recombines, and it is what sets the minimum useful width of a diode's neutral region. A transistor base is made thin compared with \( L_p \) precisely so that carriers cross before recombining.

In a silicon sample the electron concentration varies linearly from \( n = 1\times10^{17}~\text{cm}^{-3} \) at \( x=0 \) to \( n = 6\times10^{16}~\text{cm}^{-3} \) at \( x = 2~\mu\text{m} \). A uniform field \( E = 100~\text{V/cm} \) is applied in the \( +x \) direction. Take \( \mu_n = 1000~\text{cm}^2/\text{V}\cdot\text{s} \) and \( V_T = 25.85~\text{mV} \).

1. Find the drift, diffusion and total electron current densities at \( x = 1~\mu\text{m} \).

2. What field would make the total electron current zero at that point?

• Linear profile \(\Rightarrow\) \( \dfrac{dn}{dx} = \dfrac{6\times10^{16}-1\times10^{17}}{2\times10^{-4}} = -2.00\times10^{20}~\text{cm}^{-4} \)

• At \( x = 1~\mu\text{m} \) (midpoint), \( n = 8.00\times10^{16}~\text{cm}^{-3} \)

• \( D_n = V_T\mu_n = (0.02585)(1000) = 25.85~\text{cm}^2/\text{s} \) by the Einstein relation

Step 1 — Drift component

Step 2 — Diffusion component

It is negative: electrons diffuse toward \( +x \) down the slope, so the conventional current they carry points along \( -x \), opposing the drift term.

Step 3 — Total

\( J_{\text{drift}} = 1282 \), \( J_{\text{diff}} = -828.2 \), \( J_{\text{total}} = 453.4~\text{A/cm}^2 \) along \( +x \)

Step 4 — Field for zero net current

\( E = 64.62~\text{V/cm} \) exactly balances diffusion

That last result is the whole of junction physics in one line: whenever a doping gradient exists in equilibrium, the crystal builds its own field of just this size so that the net current is zero. Integrating \( E = -(V_T/n)(dn/dx) \) across a step junction gives the familiar \( V_{bi} = V_T\ln(N_AN_D/n_i^2) \).

Design / estimation problem. In an npn transistor, electrons injected into the p-type base cross it almost entirely by diffusion, giving a base transit time \( \tau_B = W^2/2D_n \), where \( W \) is the neutral base width. Take \( D_n = 25.85~\text{cm}^2/\text{s} \) in the base and a minority-electron lifetime there of \( \tau_n = 1~\mu\text{s} \).

1. Find \( \tau_B \) for a base width of \( 0.5~\mu\text{m} \), and the ceiling it puts on the transition frequency, \( f_T \approx 1/2\pi\tau_B \).

2. Choose the base width needed to bring \( \tau_B \) down to \( 10~\text{ps} \).

3. For the \( 0.5~\mu\text{m} \) base, estimate the base transport factor \( \alpha_T \approx 1 - \tau_B/\tau_n \) and the current gain it allows.

• \( \tau_B = \dfrac{W^2}{2D_n} \), \( D_n = 25.85~\text{cm}^2/\text{s} \), \( \tau_n = 1~\mu\text{s} \)

• \( 0.5~\mu\text{m} = 0.5\times10^{-4}~\text{cm} \)

• Uniformly doped base (pure diffusion, no built-in drift field); transit time is the only delay counted in \( f_T \).

Part (a) — Transit time of a \( 0.5~\mu\text{m} \) base

\( \tau_B = 48.4~\text{ps} \), \( f_T \lesssim 3.29~\text{GHz} \)

Part (b) — Base width for \( \tau_B = 10~\text{ps} \)

\( W \approx 0.227~\mu\text{m} \) — the base must be narrowed by a factor of \( \sqrt{4.84} = 2.20 \)

Part (c) — Base transport factor and gain

\( \alpha_T = 0.9999516 \), so recombination in the base alone would permit \( \beta \approx 2.1\times10^{4} \)

Transit time falls as \( W^2 \), so halving the base quarters the delay — the single strongest lever on transistor speed. The gain ceiling from part (c) is enormous, which tells you that real \( \beta \) values of 100 to 300 are set by emitter injection efficiency, not by recombination in the base.