Mobility & Limitations of Ohm's Law

What is Mobility?

We already saw that under an electric field $\vec{E}$ , electrons in a metal acquire a drift velocity

$\vec{v}_d = -\dfrac{e\tau}{m}\vec{E}$

whose magnitude is proportional to the field strength. So a natural quantity to define is: how much drift speed do you get per unit of applied field? That quantity is called the mobility $\mu$ .

$\boxed{\; \mu = \dfrac{|v_d|}{|E|} \;}$

For electrons in a metal:

$\mu_e = \dfrac{e\tau}{m}$

SI unit

$[\mu] = \dfrac{\text{m/s}}{\text{V/m}} = \dfrac{\text{m}^2}{\text{V}\cdot\text{s}}$

So mobility is measured in m $^2$ V $^{-1}$ s $^{-1}$ . Some texts use the smaller practical unit cm $^2$ V $^{-1}$ s $^{-1}$ (1 m $^2$ /V s = $10^4$ cm $^2$ /V s).

Typical values

Carrier	Mobility $\mu$ (m $^2$ V $^{-1}$ s $^{-1}$ )
Electrons in copper	$\sim 4 \times 10^{-3}$
Electrons in silicon (pure)	$\sim 0.15$
Holes in silicon	$\sim 0.05$
Electrons in GaAs	$\sim 0.85$
Ions in water	$\sim 10^{-7}$

The huge gap between metal and semiconductor mobilities is what makes transistor electronics possible.

Mobility and conductivity

Since $\sigma = 1/\rho = n e^2 \tau / m$ , and $\mu_e = e\tau/m$ :

$\boxed{\; \sigma = n e \mu_e \;}$

In a semiconductor that has both electrons and holes (concentrations $n_e$ , $n_h$ and mobilities $\mu_e$ , $\mu_h$ ):

$\sigma = e(n_e \mu_e + n_h \mu_h)$

Key point: mobility is a material property (depends on $\tau$ and the carrier mass), while drift velocity depends on both mobility and the applied field. Two different metals in the same field have different drift velocities because they have different mobilities.

[JEE Tip] You will meet $\sigma = ne\mu$ everywhere in Class 12 — semiconductors, plasmas, electrolytes. Memorise it separately from $\sigma = ne^2\tau/m$ ; they are the same formula in different dress.

Ohmic vs Non-Ohmic Devices

Ohmic Conductors: These obey Ohm's law strictly, so the $V-I$ graph is a straight line. Examples: metals at low currents, copper, silver.
Non-Ohmic Devices: These do not obey Ohm's law, so the $V-I$ graph is non-linear. Examples: semiconductor diodes, transistors, vacuum tubes, electrolytes.

Limitations of Ohm's Law

Ohm’s law is not a universal law for all materials under all conditions. It is an experimentally observed law valid only for certain materials and only under fixed conditions. It fails when:

High current: Joule heating increases temperature, which changes resistance.
Non-linear relation: The relation between $V$ and $I$ depends on the sign or magnitude of $V$ (for example, diodes).
Non-unique relation: A single current $I$ may correspond to multiple values of $V$ in special materials such as gallium arsenide.

Rectifier: the V-I Curve of a Diode

A p-n junction diode is a semiconductor device that lets current flow in essentially one direction only. Its V-I characteristic has two very distinct regions:

Forward bias ( $V > 0$ ): once $V$ exceeds a turn-on threshold (~0.3 V for Ge, ~0.7 V for Si), the current rises rapidly — almost exponentially.
Reverse bias ( $V < 0$ ): only a tiny leakage current ( $\sim \mu$ A) flows. The diode effectively blocks current.

This is asymmetric, so the diode violates Ohm's law in a qualitative way: reversing $V$ does not simply reverse $I$ .

Resistance is not a constant for a diode

If you compute $R = V/I$ on different points of the forward-bias curve, you get very different values. For a diode the useful notion is the dynamic resistance:

$r_d = \dfrac{dV}{dI}$

which also varies with operating point. So saying the resistance of a diode is X ohms is meaningless without specifying where on the curve.

[NEET Important] Never apply $V = IR$ to a diode unless you are told it has been linearised around a small signal at an operating point. Instead, use the given V-I graph — draw a load line if asked.

GaAs and Negative Differential Resistance

Gallium arsenide (GaAs) and certain semiconductor devices such as the tunnel diode show a curious V-I curve: as you increase the voltage the current at first rises, then falls, then rises again.

In the falling region:

$\dfrac{dI}{dV} < 0 \;\text{(locally)}$

This is called negative differential resistance (NDR). It is a serious violation of Ohm's law: a single value of $V$ can correspond to several possible currents.

Why does this happen?

Without going too deep, in GaAs there are two conduction-band valleys with different mobilities. As the applied field grows, electrons transfer from the high-mobility valley to the low-mobility valley, reducing conductivity. Hence the region of negative slope.

NDR is actually useful: it is the basis of microwave oscillators (Gunn diodes) and certain memory devices.

[JEE Tip] Negative differential resistance does not mean the current flows backwards. It just means $I$ decreases as $V$ increases in that region. The resistance $R = V/I$ stays positive everywhere; only the derivative $dV/dI$ is negative.

More Examples of Non-Ohmic Behaviour

Incandescent filament (light bulb)

A tungsten filament at room temperature has resistivity much smaller than when white-hot. As current rises, Joule heating raises $T$ , raising $\rho$ . So V vs I is sub-linear — the curve bends toward the V-axis (panel c of the figure).

Practical consequence: a 100 W bulb draws a big inrush current when you first switch it on (cold, low resistance), then settles to its rated current once hot.

Electrolytes

In a salt solution the moving charges are ions (both positive and negative). At low voltages the V-I relation is roughly linear, but at higher voltages electrolysis kicks in, bubbles form, and the effective resistance changes. Again: not ohmic.

Thermistors

Thermistors are ceramic semiconductor resistors engineered to change resistance strongly with temperature. They are useful precisely because they violate Ohm's law — small temperature changes produce large resistance changes, letting them act as temperature sensors.

Vacuum tube (diode valve)

In a vacuum diode, current flows only when a hot cathode emits electrons. The V-I curve follows the three-halves power law:

$I \propto V^{3/2}$

Definitely not Ohm's law.

Quick sanity check

Plot V against I. If it is a straight line through the origin at a fixed temperature, the device is ohmic. If the line:

is curved → non-ohmic (filament, thermistor, vacuum tube)
is not symmetric around origin → non-ohmic (diode)
has a negative slope somewhere → non-ohmic with NDR (GaAs, tunnel diode)

Key point: Ohm's law is a beautifully simple relation, but real physics is full of counter-examples. Learn to recognise when it applies — and when you must fall back to the given V-I curve.

Solved Examples

Example 1: Calculating Mobility

In copper, the drift velocity of electrons is $1.2 \times 10^{-4}$ m/s when an electric field of 0.03 V/m is applied. Find the mobility of electrons in copper.

Solution: $\mu = \dfrac{v_d}{E} = \dfrac{1.2 \times 10^{-4}}{0.03} = 4 \times 10^{-3} \text{ m}^2\text{/V}\cdot\text{s}$

Answer: $\mu = 4 \times 10^{-3}$ m $^2$ V $^{-1}$ s $^{-1}$

Example 2: Conductivity from $n$ and $\mu$

A semiconductor has electron concentration $n_e = 10^{20}$ m $^{-3}$ and mobility $\mu_e = 0.15$ m $^2$ /V s. Compute its conductivity, assuming only electrons contribute.

Solution: Using $\sigma = ne\mu,$ we get $\sigma = (10^{20})(1.6 \times 10^{-19})(0.15) = 2.4 \text{ S/m}$

Answer: $\sigma = 2.4$ S/m

Example 3: Two-Carrier Semiconductor

A silicon sample has $n_e = 3 \times 10^{18}$ m $^{-3}$ with $\mu_e = 0.13$ , and $n_h = 2 \times 10^{19}$ m $^{-3}$ with $\mu_h = 0.05$ . Find its total conductivity.

Solution: $\sigma = e(n_e\mu_e + n_h\mu_h)$ $\sigma = (1.6 \times 10^{-19})\left[(3 \times 10^{18})(0.13) + (2 \times 10^{19})(0.05)\right]$ $= (1.6 \times 10^{-19})(3.9 \times 10^{17} + 10^{18})$ $= (1.6 \times 10^{-19})(1.39 \times 10^{18}) \approx 0.22 \text{ S/m}$

Answer: $\sigma \approx 0.22$ S/m

Example 4: Mobility and Relaxation Time

Electrons in a metal have mobility $\mu = 5 \times 10^{-3}$ m $^2$ /V s. Find the relaxation time $\tau$ (use $m = 9.11 \times 10^{-31}$ kg, $e = 1.6 \times 10^{-19}$ C).

Solution: From $\mu = \frac{e\tau}{m},$ we get $\tau = \dfrac{m\mu}{e} = \dfrac{(9.11 \times 10^{-31})(5 \times 10^{-3})}{1.6 \times 10^{-19}}$ $\tau = \dfrac{4.555 \times 10^{-33}}{1.6 \times 10^{-19}} = 2.85 \times 10^{-14} \text{ s}$

Answer: $\tau \approx 2.85 \times 10^{-14}$ s

Example 5: Diode V-I Reading

A diode is found to pass 10 mA at $V = 0.7$ V (forward bias) and 5 $\mu$ A at $V = -5$ V (reverse bias). Compute the static resistance $V/I$ in each case and comment.

Solution: Forward: $R_f = \frac{0.7}{0.01} = 70\,\Omega$

Reverse: Using magnitudes, $R_r = \frac{5}{5 \times 10^{-6}} = 10^6\,\Omega = 1\,\text{M}\Omega$

Answer: $R_f = 70\,\Omega$ , $R_r = 1\,\text{M}\Omega$

The huge difference shows that the diode is strongly non-ohmic.

Example 6: Filament Lamp

A 100 W, 230 V bulb has a hot filament. At the operating point, compute its resistance. At room temperature, its resistance is measured as 40 $\Omega$ . What is the ratio of hot to cold resistance?

Solution: Hot resistance: $R_{\text{hot}} = \frac{V^2}{P} = \frac{230^2}{100} = 529\,\Omega$

Cold resistance: $R_{\text{cold}} = 40\,\Omega$

Ratio: $\frac{R_{\text{hot}}}{R_{\text{cold}}} = \frac{529}{40} \approx 13.2$

Answer: $R_{\text{hot}}/R_{\text{cold}} \approx 13.2$

Example 7: Conductivity Change with Temperature in a Metal

A metal has relaxation time $\tau = 3.0 \times 10^{-14}$ s at 300 K and $\tau = 1.5 \times 10^{-14}$ s at 600 K. Find the ratio of conductivities $\sigma_{300}/\sigma_{600}$ .

Solution: Since $\sigma = \frac{ne^2\tau}{m},$ for fixed $n$ , $e$ , and $m$ , $\sigma \propto \tau$

So, $\dfrac{\sigma_{300}}{\sigma_{600}} = \dfrac{\tau_{300}}{\tau_{600}} = \dfrac{3.0}{1.5} = 2$

Answer: $\sigma_{300}/\sigma_{600} = 2$

Example 8: Negative Differential Resistance

A GaAs sample has the following two operating points on its V-I curve:

$V_1 = 1.0$ V, $I_1 = 20$ mA
$V_2 = 1.2$ V, $I_2 = 15$ mA

Find: (a) the static resistance at the two points, (b) the differential resistance between them.

Solution: (a) Static resistances: $R_1 = \frac{V_1}{I_1} = \frac{1.0}{0.02} = 50\,\Omega$ $R_2 = \frac{V_2}{I_2} = \frac{1.2}{0.015} = 80\,\Omega$

(b) Differential resistance: $r_d = \dfrac{\Delta V}{\Delta I} = \dfrac{1.2 - 1.0}{0.015 - 0.020} = \dfrac{0.2}{-0.005} = -40\,\Omega$

Answer: $R_1 = 50\,\Omega$ , $R_2 = 80\,\Omega$ , $r_d = -40\,\Omega$

Example 9: Drift Velocity for Different Carriers

In an ionic solution, sodium ions have mobility $\mu_{Na^+} = 5.2 \times 10^{-8}$ m $^2$ /V s and chloride ions $\mu_{Cl^-} = 7.9 \times 10^{-8}$ m $^2$ /V s. In a field of 100 V/m, find the drift speeds.

Solution: Using $v_d = \mu E,$ we get $v_{Na^+} = (5.2 \times 10^{-8})(100) = 5.2 \times 10^{-6} \text{ m/s}$ $v_{Cl^-} = (7.9 \times 10^{-8})(100) = 7.9 \times 10^{-6} \text{ m/s}$

Answer:

Na $^+$ : $5.2 \times 10^{-6}$ m/s
Cl $^-$ : $7.9 \times 10^{-6}$ m/s

Example 10: Why Toasters Glow but Copper Wires Do Not

A toaster wire and the copper wires carrying current to it both carry the same current. The toaster wire has resistance $R_t = 60\,\Omega$ and copper wire resistance $R_c = 0.2\,\Omega$ . Current is $I = 4$ A. Find the power dissipated in each.

Solution: Using $P = I^2R,$ we get $P_t = 4^2 \times 60 = 960\,\text{W}$ $P_c = 4^2 \times 0.2 = 3.2\,\text{W}$