Drift current and electric field: Worked examples — Hardware deep dive

Everything here rests on two formulas from the parent note:

The scenario matrix

Before solving anything, here is the full space of cases this topic can throw at you. Every example below is tagged with the cell it fills.

Cell	Case class	What's special / where it bites
A	Single carrier, plain numbers	baseline: $v_{d} = μ E$ , unit bookkeeping
B	Field sign / direction reversed	electrons drift opposite to $E$ ; current still along $E$
C	Both carriers active (intrinsic)	$n$ and $p$ both matter, $n = p = n_{i}$
D	Mixed doping ( $n \neq = p$ , both non-negligible)	when the "ignore holes" 80/20 shortcut is illegal
E	Zero / degenerate input	$E = 0$ , or $τ \to 0$ , or $n = 0$ — what current do you get?
F	Limiting behaviour	$E$ so large that $v_{d} = μ E$ breaks (velocity saturation)
G	Real-world word problem	resistor made of silicon: find measurable current $I$
H	Exam twist	given $J$ and geometry, work backwards to find $μ$ or $τ$

We now cover cells A → H in order.

Cell A — Single carrier, plain numbers

Cell B — Field direction reversed

Worked example B · Which way does the electron actually move?

Take the Ge sample from A but reverse the field: $E = - 200 V/cm$ (pointing in the $- x$ direction). Describe the electron velocity direction and the conventional current direction.

Forecast: If the field flips, does the current flip too? Does the electron move with or against $E$ ?

Step 1. Force on an electron: $F = q E = (- e) E$ . Why this step? The electron's charge is negative, so its force is opposite to $E$ . With $E$ pointing $- x$ , force is $+ x$ , so the electron drifts in $+ x$ .

Step 2. Conventional current follows positive-charge flow, i.e. along $E$ ( $- x$ here), which is opposite to the electron motion. Why this step? Conventional current was historically defined as the direction a positive charge would move; electrons (negative) moving $+ x$ = current in $- x$ .

Step 3. Magnitude is unchanged: $∣ v_{d} ∣ = 7.8 \times 1 0^{5}$ cm/s. Why this step? Reversing sign of $E$ reverses direction but not magnitude — $∣ v_{d} ∣ = μ ∣ E ∣$ .

Verify: Two facts agree: (1) $J = σ E$ always points along $E$ , and (2) electrons move opposite to $J$ . Both satisfied ✓. See the figure below.

Figure — Drift current and electric field

Look at the amber field arrow: it points left. The cyan electron drifts right (against the field), yet the white conventional-current arrow points left (with the field). Two opposite objects, one consistent current.

Cell C — Both carriers active (intrinsic material)

Worked example C · Conductivity of intrinsic silicon

Pure (intrinsic) Si at room temperature: $n = p = n_{i} = 1.5 \times 1 0^{10} cm^{- 3}$ , $μ_{n} = 1350$ , $μ_{p} = 480 cm^{2} / (V\cdots)$ . Find $σ$ .

Forecast: In intrinsic material electrons and holes are equal in number. Do their currents cancel or add?

Step 1. Use the full formula — you may not drop $p$ here: $σ = e (n μ_{n} + p μ_{p}) = e n_{i} (μ_{n} + μ_{p})$ Why this step? Since $n = p = n_{i}$ , both terms are the same size order; ignoring holes would throw away a real chunk (here $μ_{p} / μ_{n} \approx 0.36$ ).

Step 2. Substitute: $σ = (1.6 \times 1 0^{- 19}) (1.5 \times 1 0^{10}) (1350 + 480)$ $= (1.6 \times 1 0^{- 19}) (1.5 \times 1 0^{10}) (1830) = 4.39 \times 1 0^{- 6} (Ω \cdotcm)^{- 1}$ Why this step? Add the mobilities first because $n_{i}$ multiplies both — one multiplication instead of two.

Verify: This tiny $σ$ (huge resistivity $ρ = 1/ σ \approx 2.3 \times 1 0^{5} Ω \cdotcm$ ) matches the known fact that pure silicon is a poor conductor — that's why we dope it ✓. Currents add (parent's mistake #2): both point along $E$ ✓.

Cell D — Mixed doping (the 80/20 shortcut is illegal)

Worked example D · When you MUST keep both terms

A silicon sample is lightly n-doped: $n = 5 \times 1 0^{15} cm^{- 3}$ . It is not strongly extrinsic, so holes are not negligible: $p = 4.5 \times 1 0^{4} cm^{- 3}$ … wait — check whether we can drop $p$ . Use $μ_{n} = 1350$ , $μ_{p} = 480$ .

Forecast: Is dropping $p$ safe here, or does it introduce error?

Step 1. Compare the two terms before deciding: $n μ_{n} = (5 \times 1 0^{15}) (1350) = 6.75 \times 1 0^{18}$ $p μ_{p} = (4.5 \times 1 0^{4}) (480) = 2.16 \times 1 0^{7}$ Why this step? The parent's "80/20" rule (drop $p$ ) is only legal when $p μ_{p} ≪ n μ_{n}$ . Always compute the ratio rather than assume.

Step 2. Ratio $= \frac{2.16 \times 1 0 ^{7}}{6.75 \times 1 0 ^{18}} = 3.2 \times 1 0^{- 12}$ — utterly negligible. So here dropping $p$ is legal: $σ \approx e n μ_{n} = (1.6 \times 1 0^{- 19}) (5 \times 1 0^{15}) (1350) = 1.08 (Ω \cdotcm)^{- 1}$ Why this step? We earned the right to simplify by showing the hole term is 12 orders of magnitude down.

Verify: Keeping $p$ would change $σ$ in the 12th significant figure — invisible. Decision justified by data, not habit ✓. (Compare with Cell C where the ratio was $\sim 0.36$ and dropping $p$ would have been wrong.)

Cell E — Zero and degenerate inputs

Worked example E · The three "nothing happens" cases

For the Si sample of Cell D, evaluate the drift current density in three degenerate scenarios: (E1) field off, $E = 0$ ; (E2) carriers frozen out, $n = 0$ (and $p = 0$ ); (E3) infinite scattering, $τ \to 0$ .

Forecast: Which of these gives exactly zero current, and why is each zero?

Step 1 — (E1) $E = 0$ . $J = σ E = σ \cdot 0 = 0$ . Why this step? No field ⇒ no systematic force ⇒ carriers keep only random thermal motion whose average is zero. This is the equilibrium baseline.

Step 2 — (E2) $n = p = 0$ . $σ = e (0 \cdot μ_{n} + 0 \cdot μ_{p}) = 0 \Rightarrow J = 0$ . Why this step? No mobile charge to carry current, no matter how strong the field. (This is a perfect insulator / fully carrier-frozen semiconductor.)

Step 3 — (E3) $τ \to 0$ . Since $μ = e τ / m^{*} \to 0$ , we get $v_{d} = μ E \to 0$ and $σ \to 0$ , so $J \to 0$ . Why this step? If a carrier collides instantly ( $τ = 0$ ) it can never accumulate any drift between collisions — it's reset before it moves.

Verify: All three give $J = 0$ but for three distinct physical reasons (no push / no carriers / no free flight). Each traces back to a different factor in $J = e n μ_{n} E$ hitting zero ✓. See Scattering Mechanisms and Mean Free Time for how $τ$ is set.

Cell F — Limiting behaviour ( $v_{d} = μ E$ breaks)

Worked example F · When the field is so strong the formula lies

For Si electrons, $μ_{n} = 1350$ . Predict $v_{d}$ at $E = 1 0^{5} V/cm$ using $v_{d} = μ E$ , then explain why the answer is wrong.

Forecast: Multiply it out — does the number stay physically believable?

Step 1. Naive substitution: $v_{d} = 1350 \times 1 0^{5} = 1.35 \times 1 0^{8} cm/s$ Why this step? We're deliberately pushing the linear law $v_{d} = μ E$ past where it's valid, to see it fail.

Step 2. Sanity check against thermal speed $\sim 1 0^{7}$ cm/s. Our answer $1.35 \times 1 0^{8}$ cm/s is 13× faster than thermal — impossible, because carriers cannot systematically exceed the speed they scatter at. Why this step? $v_{d} = μ E$ assumed $τ$ is constant. At high fields carriers gain so much energy that they scatter harder and more often, so $τ$ (and thus effective $μ$ ) drops. The real $v_{d}$ saturates near $1 0^{7}$ cm/s.

Verify: The linear formula gave a super-thermal speed — a red flag confirming we've left its domain ✓. The real physics is in Velocity Saturation and High-Field Effects; below, the plot shows the linear prediction diverging from the true saturating curve.

The cyan dashed line is the naive $v_{d} = μ E$ ; the white curve is reality. They agree at low field (Cell A regime) but the naive line rockets past the amber saturation ceiling $v_{s a t} \approx 1 0^{7}$ cm/s while the real carrier flattens.

Cell G — Real-world word problem

Worked example G · A silicon resistor

An n-type Si bar has $n = 1 0^{16} cm^{- 3}$ , $μ_{n} = 1350$ , length $L = 0.5 cm$ , cross-section $A = 2 \times 1 0^{- 4} cm^{2}$ . A voltage $V = 5 V$ is applied end to end. Find the current $I$ .

Forecast: Guess whether $I$ is microamps, milliamps, or amps.

Step 1 — Field from voltage. $E = V / L = 5/0.5 = 10 V/cm$ . Why this step? The field is what drives drift, and for a uniform bar $E = V / L$ (voltage per length).

Step 2 — Conductivity. $σ = e n μ_{n} = (1.6 \times 1 0^{- 19}) (1 0^{16}) (1350) = 2.16 (Ω \cdotcm)^{- 1}$ . Why this step? Same as parent Example 2 — strongly n-type, so holes safely dropped (ratio check from Cell D applies).

Step 3 — Current density, then current. $J = σ E = 2.16 \times 10 = 21.6 A/cm^{2}$ $I = J A = 21.6 \times 2 \times 1 0^{- 4} = 4.32 \times 1 0^{- 3} A = 4.32 mA$ Why this step? Compute the size-independent $J$ first, then multiply by area only at the end to reach the measurable current.

Verify (cross-check via resistance): $R = \frac{L}{σ A} = \frac{0.5}{2.16 \times 2 \times 1 0 ^{- 4}} = 1157 Ω$ . Then $I = V / R = 5/1157 = 4.32 mA$ ✓ — matches, confirming Ohm's law $J = σ E$ and the lumped $V = I R$ are the same physics (Conductivity and Resistivity of Semiconductors).

Cell H — Exam twist (work backwards)

Worked example H · Given the current, find the scattering time

A silicon sample carries $J = 100 A/cm^{2}$ under $E = 50 V/cm$ , with electron density $n = 1 0^{16} cm^{- 3}$ (n-type, drop holes). The electron effective mass is $m^{*} = 0.26 m_{0}$ with $m_{0} = 9.11 \times 1 0^{- 31}$ kg. Find the mean free time $τ$ .

Forecast: This runs the whole chain in reverse: $J \to σ \to μ \to τ$ . Guess if $τ$ lands near $1 0^{- 13}$ s (typical).

Step 1 — Back out conductivity. $σ = J / E = 100/50 = 2 (Ω \cdotcm)^{- 1}$ . Why this step? $J = σ E$ inverts directly; this isolates the material property from the drive.

Step 2 — Back out mobility. From $σ = e n μ_{n}$ : $μ_{n} = \frac{σ}{e n} = \frac{2}{( 1.6 \times 1 0 ^{- 19} ) ( 1 0 ^{16} )} = 1250 cm^{2} / (V\cdots)$ Why this step? We know $σ$ , $e$ , $n$ ; the only unknown left in $σ = e n μ$ is $μ$ .

Step 3 — Convert $μ$ to SI, then get $τ$ . From $μ = e τ / m^{*}$ ⇒ $τ = μ m^{*} / e$ . First convert $μ$ : $1250 cm^{2} / Vs = 1250 \times 1 0^{- 4} m^{2} / Vs = 0.125 m^{2} / Vs$ . $τ = \frac{μ m ^{*}}{e} = \frac{0.125 \times ( 0.26 \times 9.11 \times 1 0 ^{- 31} )}{1.6 \times 1 0 ^{- 19}}$ $= \frac{0.125 \times 2.369 \times 1 0 ^{- 31}}{1.6 \times 1 0 ^{- 19}} = 1.85 \times 1 0^{- 13} s$ Why this step? $μ = e τ / m^{*}$ is the microscopic definition (Effective Mass supplies $m^{*}$ ); it's the only equation linking $μ$ to $τ$ . Units MUST be SI here — this is where exams trap you (cm² vs m²).

Verify: $τ \approx 0.18 ps$ , squarely in the typical semiconductor range of $1 0^{- 13}$ s ✓. Chain closes: $τ \to μ \to σ \to J$ would reproduce the given $J = 100$ A/cm² ✓.

Matrix coverage check

Recall Did we hit every cell?

A ::: Example A (drift velocity) B ::: Example B (reversed field, current vs electron direction) C ::: Example C (intrinsic, both carriers) D ::: Example D (ratio check — when to drop holes) E ::: Example E (three zero-current cases) F ::: Example F (velocity saturation limit) G ::: Example G (silicon resistor word problem) H ::: Example H (backwards: J → τ)

Connections

Effective Mass — supplies $m^{*}$ used in Cell H's $τ = μ m^{*} / e$ .
Scattering Mechanisms and Mean Free Time — the $τ$ behind Cells E3 and H.
Conductivity and Resistivity of Semiconductors — $σ$ – $ρ$ used in Cells C, G.
Velocity Saturation and High-Field Effects — the real physics of Cell F.
Diffusion Current and Carrier Gradients · Drift-Diffusion Equation · Einstein Relation — the companion transport mechanisms.