AcademyCurrent and Resistance
Academy
Microscopic Model of Conduction
Level 1 - Physics topic page in Current and Resistance.
Principle
The microscopic model connects macroscopic current to charge-carrier density, carrier charge, cross-sectional area, and drift velocity.
Notation
\(n\)
mobile charge-carrier number density
\(\mathrm{m^{-3}}\)
\(q\)
charge of one carrier
\(\mathrm{C}\)
\(\vec v_d\)
drift velocity of carriers
\(\mathrm{m\,s^{-1}}\)
\(A\)
wire cross-sectional area
\(\mathrm{m^{2}}\)
\(I\)
conventional current
\(\mathrm{A}\)
\(\vec J\)
current density
\(\mathrm{A\,m^{-2}}\)
\(\mu\)
carrier mobility
\(\mathrm{m^{2}\,V^{-1}\,s^{-1}}\)
Method
Derivation 1: Count carriers crossing a surface
In time \(\Delta t\), carriers with drift speed \(v_d\) move through length \(v_d\Delta t\). The volume swept through area \(A\) is \(Av_d\Delta t\).
Swept volume
\[\Delta V_{\mathrm{vol}}=A v_d\Delta t\]
Number of carriers
\[\Delta N=nA v_d\Delta t\]
Charge crossing
\[\Delta Q=|q|nA v_d\Delta t\]
Current magnitude
\[I=n|q|A v_d\]
Derivation 2: Vector current density
Current density includes the sign of the carrier charge. Negative carriers drifting one way produce conventional current the other way.
Vector current density
\[\vec J=nq\vec v_d\]
Current through area
\[I=JA\quad\text{for uniform flow normal to area}\]
Electron drift
\[q=-e\Rightarrow \vec J\ \text{opposite }\vec v_d\]
Derivation 3: Mobility and conductivity
In a simple ohmic model, drift velocity is proportional to electric field. That proportionality leads to conductivity.
Drift response
\[\vec v_d=\mu\vec E\quad\text{for positive carriers}\]
Current density
\[\vec J=nq\mu\vec E\]
Conductivity
\[\sigma=nq\mu\quad\text{for positive carriers}\]
Rules
These are the compact microscopic conduction relations.
Current magnitude
\[I=n|q|A v_d\]
Current density
\[\vec J=nq\vec v_d\]
Uniform flow
\[I=JA\]
Ohmic mobility
\[\vec J=\sigma\vec E\]
Examples
Question
A wire has
\[n=8.0\times10^{28}\,\mathrm{m^{-3}}\]
area \[1.0\times10^{-6}\,\mathrm{m^2}\]
and current \[1.6\,\mathrm{A}\]
Estimate electron drift speed.Answer
\[v_d=\frac{I}{n e A}=\frac{1.6}{(8.0\times10^{28})(1.60\times10^{-19})(1.0\times10^{-6})}=1.25\times10^{-4}\,\mathrm{m\,s^{-1}}\]
Checks
- Drift speed is usually very small compared with random thermal speeds.
- Conventional current is opposite electron drift in metals.
- \(n\) counts mobile carriers per volume, not all particles in the material.
- Use the magnitude form \(I=n|q|Av_d\) when solving for speed.