Contact Potential and Band Scheme of a P-N Junction

Contact Potential (Definition)

When a p-type and an n-type semiconductor are joined together, diffusion of charge carriers occurs across the junction:

• Holes from the p-side diffuse into the n-side
• Electrons from the n-side diffuse into the p-side

At the junction, recombination takes place. This leaves behind:

• Negative acceptor ions on the p-side (immobile, fixed)
• Positive donor ions on the n-side (immobile, fixed)

This region of uncovered fixed ions is called the depletion layer. The separation of these fixed charges creates a built-in electric field directed from the n-side to the p-side. The potential difference established across this depletion layer due to these uncovered ions is called the contact potential (also called the built-in potential or barrier potential), denoted V₀.

The contact potential prevents further net diffusion of majority carriers across the junction at equilibrium. Typical values are ~0.3 V for Ge and ~0.7 V for Si.

Band Scheme of a P-N Junction

The energy band diagram of a p-n junction at equilibrium is drawn below, illustrating all three required aspects.

n-side          Depletion Layer          p-side
                |                |
                |                |
  Ec __________|                 |__________  Ec
               |  eV₀ (barrier) |
               |                 \
               |                  \________  Ei
  Ei _________ |                            
               |                  ________  Ev
  Ev __________|                 |
                |                |
                |                |
  (+) donor ions|                |(-) acceptor ions

A cleaner schematic representation:

ENERGY
  ^
  |   n-side                          p-side
  |
  |  Ec(n)----\                              /---- Ec(p)
  |            \          eV₀              /
  |             \    <------------->      /
  |              \                       /
  |  Ev(n)--------\                     /---------- Ev(p)
  |                \___________________/
  |
  |<--- n-region --->|<-depletion->|<--- p-region --->
  +---------------------------------------------------------> x

(a) Potential Difference V₀ Due to Fixed Ions in the Depletion Layer

When p-type and n-type semiconductors are joined:

• On the n-side of the depletion layer: electrons have left, exposing positive donor ions (+)
• On the p-side of the depletion layer: holes have left, exposing negative acceptor ions (-)

This charge separation creates an internal electric field E pointing from n-side (+) to p-side (-), which gives rise to a potential difference V₀ across the depletion layer.

p-side                              n-side
  |  (-)  (-)  (-)  | (-)  (+)  (+)  (+)  (+)  |
  |  Acceptor ions  |      Donor ions           |
  |                 |                           |
  |<---- p -------->|<------ depletion -------->|
  
  Electric field E:  <========================
                     (from n-side to p-side)

  Potential:
  V
  |                                    _____ V₀
  |                                   /
  |_________________________________ /
  |
  +-----------------------------------------> x
       p-side          depletion      n-side

The potential rises from p-side to n-side by the amount V₀ (contact potential). This is the built-in potential barrier.

(b) Potential Energy Barrier for Majority Electrons (n-side)

Electrons are the majority carriers in the n-side. As they attempt to cross the junction from n-side to p-side, they face the conduction band energy barrier.

Since the n-side is at higher potential (V₀), the potential energy of electrons (charge = -e) on the n-side is lower than on the p-side. Therefore, electrons moving from n to p must climb an energy hill of magnitude:

$$\boxed{E_{barrier} = eV_0}$$

ENERGY
  ^
  |
  |  Ec(n)                              Ec(p)
  |   ___                               ___
  |      \                             /
  |       \       eV₀ barrier         /
  |        \    <----------->        /
  |         \__________________________/
  |
  |  Electrons in n-side must climb UP by eV₀ to reach p-side
  |
  |         n-side -----> junction -----> p-side
  +---------------------------------------------------------> x

• The conduction band edge Ec(p) is higher than Ec(n) by eV₀
• Majority electrons in the n-side do not have sufficient thermal energy to surmount this barrier at equilibrium
• This barrier prevents spontaneous flow of electrons from n to p

(c) Potential Energy Barrier for Majority Holes (p-side)

Holes are the majority carriers in the p-side. As they attempt to cross the junction from p-side to n-side, they face the valence band energy barrier.

Since holes carry positive charge (+e), and the p-side is at lower potential, holes moving from p to n must also climb an energy hill of magnitude:

$$\boxed{E_{barrier} = eV_0}$$

ENERGY
  ^
  |
  |         p-side -----> junction -----> n-side
  |
  |   Ev(p)                              Ev(n)
  |    ___                               ___
  |       \                             /
  |        \      eV₀ barrier          /
  |         \   <----------->         /
  |          \_______________________/
  |
  |  Holes in p-side must climb DOWN (in energy) by the same amount $eV_0$ to move into the n-side.

• The valence band edge Ev(p) is offset from Ev(n) by the same barrier height eV₀ as the conduction band offset in part (b), because the conduction and valence bands bend together across the depletion region.
• Majority holes in the p-side do not have sufficient thermal energy to surmount this barrier at equilibrium, exactly as electrons in the n-side could not surmount theirs.
• This barrier prevents spontaneous flow of holes from p to n.

Summary

At thermal equilibrium, a single built-in potential $V_0$ arises from the fixed donor and acceptor ions exposed in the depletion layer (part a). This potential presents an equal energy barrier of $eV_0$ to both kinds of majority carriers trying to diffuse further across the junction: electrons on the n-side (part b) and holes on the p-side (part c). It is precisely because both barriers have the same height $eV_0$, and both currents they suppress are balanced by an equal and opposite minority-carrier drift current, that the net current across an unbiased p-n junction is zero and the junction remains in stable equilibrium.

Torque on a Current-Carrying Rectangular Loop in a Magnetic Field

Setup and Description

Consider a rectangular loop of wire ABCD of dimensions a × b (width × length), carrying current I, mounted on a pivot rod (axis of rotation), and placed in a uniform external magnetic field B.

Let the plane of the loop make an angle φ with the magnetic field direction, so the normal to the loop makes an angle θ with B, where θ + φ = 90°.

Forces on Each Side of the Loop

Using the force law on a current-carrying conductor:

$$\vec{F} = I(\vec{L} \times \vec{B})$$

or magnitude: $F = BIL\sin\alpha$, where $\alpha$ is the angle between the current direction and B.

Side AB and Side CD (parallel to the pivot axis, length = b)

• The current in AB and CD flows parallel to the pivot rod axis.
• The force on each is:

$$F_{AB} = BIb\sin 90° = BIb$$ $$F_{CD} = BIb\sin 90° = BIb$$

• These two forces are equal in magnitude but opposite in direction, forming a couple (they do not cancel because they act on opposite sides).

Side BC and Side DA (perpendicular to the pivot axis, length = a)

• The forces on BC and DA are equal, opposite, and collinear along the pivot axis.
• They produce no net torque (they cancel each other).

Calculation of Torque

The two forces on AB and CD (each of magnitude $F = BIb$) act at perpendicular distances from the pivot axis.

The perpendicular distance between the two forces depends on the orientation of the loop. If the normal to the loop makes angle θ with B, the moment arm for each force is $\frac{a}{2}\sin\theta$.

Therefore, the net torque is:

$$\tau = F \cdot a\sin\theta$$

$$\tau = BIb \cdot a\sin\theta$$

$$\boxed{\tau = BIA\sin\theta}$$

where $A = ab$ is the area of the loop.

Expression in Terms of Magnetic Dipole Moment

The magnetic dipole moment of the current loop is defined as:

$$\vec{\mu} = I\vec{A} = IA\hat{n}$$

where $\hat{n}$ is the unit normal to the loop (direction given by the right-hand rule), and $A$ is the area of the loop.

The torque can then be written in vector form as:

$$\boxed{\vec{\tau} = \vec{\mu} \times \vec{B}}$$

with magnitude:

$$\tau = \mu B\sin\theta = IAB\sin\theta$$

Special Cases

Potential Energy of the Dipole

The potential energy of the magnetic dipole in the field is:

$$E_p = -\vec{\mu} \cdot \vec{B} = -\mu B\cos\theta$$

• Minimum energy (stable) at θ = 0°: $E_p = -\mu B$
• Maximum energy (unstable) at θ = 180°: $E_p = +\mu B$

Alternative Ways to Increase the Torque on the Coil

From the expression $\tau = NIAB\sin\theta$, the torque can be increased by:

1. Increasing the number of turns N: For a coil of N turns, $\tau = NIAB\sin\theta$. More turns means greater torque for the same current and field.

2. Increasing the current I: A larger current directly increases the magnetic dipole moment $\mu = NIA$, hence increasing torque.

3. Increasing the area A of the loop: A larger loop area (larger $a \times b$) increases the torque proportionally.

4. Increasing the magnetic field strength B: Using a stronger external magnetic field increases the torque directly.

5. Orienting the loop so that θ = 90°: The torque is maximum when the plane of the loop is parallel to B (i.e., the normal to the loop is perpendicular to B), giving $\tau_{max} = NIAB$.

6. Using a ferromagnetic core inside the coil: This concentrates the magnetic field lines and effectively increases the flux, thereby increasing the torque (used in practical galvanometers and motors).

Summary

The torque on a current-carrying rectangular loop in a magnetic field is:

$$\tau = NIAB\sin\theta$$

This principle is the fundamental operating principle of electric motors and galvanometers, where electrical energy is converted to mechanical rotation through the torque produced on a current loop in a magnetic field.

Solution: Oscillating Block on a Spring

STEP 1 - Given Data

Key interpretation: Moving between the two endpoints (extreme to extreme) is half of a full oscillation.

$$\frac{T}{2} = 0.15 \text{ s} \implies T = 0.30 \text{ s}$$

STEP 2 - Solve

(a) Frequency

$$f = \frac{1}{T} = \frac{1}{0.30} = 3.33 \text{ Hz}$$

(b) Amplitude

The endpoints are separated by $2A$:

$$2A = 0.40 \text{ m} \implies A = 0.20 \text{ m}$$

(c) Force Constant

$$T = 2\pi\sqrt{\frac{m}{k}} \implies k = \frac{4\pi^2 m}{T^2}$$

$$k = \frac{4 \times (3.14159)^2 \times 0.25}{(0.30)^2} = \frac{9.8696}{0.09} = 109.66 \text{ N/m}$$

Summary

Force on a Proton in a Magnetic Field

Given Data

Step 1: Velocity Gained by the Proton

Energy conservation:

$$qV = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2qV}{m}}$$

$$v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 200}{1.67 \times 10^{-27}}}$$

Numerator: $2 \times 1.6 \times 10^{-19} \times 200 = 6.4 \times 10^{-17}$

$$v = \sqrt{\frac{6.4 \times 10^{-17}}{1.67 \times 10^{-27}}} = \sqrt{3.832 \times 10^{10}}$$

$$v \approx 1.958 \times 10^{5} \text{ m/s}$$

Step 2: Magnetic Force

$$F = qvB\sin\theta, \qquad \theta = 90° \implies \sin\theta = 1$$

$$F = qvB = (1.6 \times 10^{-19})(1.958 \times 10^{5})(0.5)$$

Step-by-step:

• $1.6 \times 10^{-19} \times 1.958 \times 10^{5} = 3.133 \times 10^{-14}$
• $\times 0.5 = 1.566 \times 10^{-14}$

$$\boxed{F \approx 1.57 \times 10^{-14} \text{ N}}$$

Result

The force experienced by the proton is approximately $1.57 \times 10^{-14}$ N, directed perpendicular to both $\vec{v}$ and $\vec{B}$ (given by the right-hand rule). This force provides the centripetal force causing circular motion.

Digital Circuit: $y = (AB + \bar{C}BA)(B + C)$

Step 1: Extract Given Data

Boolean expression: $$y = (AB + \bar{C}BA)(B + C)$$

Inputs: $A$, $B$, $C$ Gates allowed: OR, AND, NOT

Step 2: Simplify the Expression

The first bracket: $$AB + \bar{C}BA = AB + AB\bar{C} = AB(1 + \bar{C}) = AB$$

(using $1 + \bar{C} = 1$)

So: $$y = AB(B + C)$$

Expand: $$y = AB\cdot B + AB\cdot C = ABB + ABC$$

Since $B\cdot B = B$: $$y = AB + ABC = AB(1 + C) = AB$$

Therefore: $$\boxed{y = AB}$$

Step 3: Circuit Diagram (drawn as per given expression)

                        AB
A ──┬──────────[AND1]──────────┐
    │            │             │
B ──┼──┬─────────┘             ├─[OR1]─┐
    │  │                       │       │
    │  │   ┌─[AND2]────────────┘       │
    │  │   │  ABC̄                      ├─[AND3]── y
C ──┼──┼─[NOT]                          │
    │  │   (C̄)                          │
    │  └───────────────[OR2]────────────┘
    │                   │  (B+C)
C ──┴───────────────────┘

Gate list:

Step 4: Truth Table

Conclusion

The circuit simplifies to $y = AB$. Output $y = 1$ only when both $A = 1$ and $B = 1$, independent of $C$.

Heisenberg Uncertainty Principle

Step 1 - Given Data

• Mass of small particle: $m = 10^{-6}$ g $= 10^{-6} \times 10^{-3}$ kg $= 10^{-9}$ kg
• Uncertainty in speed: $\Delta v_x = 10^{-6}$ m/s
• Mass of electron: $m_e = 9.11 \times 10^{-31}$ kg
• Planck's constant: $h = 6.626 \times 10^{-34}$ J·s

Step 2 - Solution

The uncertainty principle:

$$\Delta x \cdot \Delta p_x \geq \frac{h}{4\pi}$$

With $\Delta p_x = m,\Delta v_x$, the minimum position uncertainty is:

$$\Delta x = \frac{h}{4\pi, m, \Delta v_x}$$

Part (a): Small particle

$$\Delta x = \frac{6.626 \times 10^{-34}}{4\pi \times 10^{-9} \times 10^{-6}}$$

Denominator:

$$4\pi \times 10^{-15} = 1.2566 \times 10^{-14}$$

$$\Delta x = \frac{6.626 \times 10^{-34}}{1.2566 \times 10^{-14}} = 5.27 \times 10^{-20}\ \text{m}$$

$$\boxed{\Delta x \approx 5.27 \times 10^{-20}\ \text{m}}$$

This is negligibly small for a macroscopic particle.

Part (b): Electron

$$\Delta x = \frac{6.626 \times 10^{-34}}{4\pi \times 9.11 \times 10^{-31} \times 10^{-6}}$$

Denominator:

$$4\pi \times 9.11 \times 10^{-37} = 12.566 \times 9.11 \times 10^{-37} = 1.1448 \times 10^{-35}$$

$$\Delta x = \frac{6.626 \times 10^{-34}}{1.1448 \times 10^{-35}} = 57.9\ \text{m}$$

$$\boxed{\Delta x \approx 57.9\ \text{m}}$$

Conclusion

The uncertainty principle is significant only at the microscopic scale. For the macroscopic particle the position uncertainty is unmeasurably small, while for the electron it is enormous, reflecting the fundamental quantum limitation on simultaneous knowledge of position and momentum.

(Note: If the simpler form $\Delta x,\Delta p \geq h/2\pi$ or $\hbar$ is used, results differ by a constant factor of 2, but the $h/4\pi$ minimum-uncertainty form used here is standard.)

Stern-Gerlach Experiment: Beam Separation

Given Data

Physics

Force on each spin component: $F = \mu_B \dfrac{dB}{dz}$ (magnitude), directed oppositely for $m_s = \pm\tfrac12$.

Step 1: Force

$$F = (9.274\times 10^{-24})(3\times 10^4) = 2.782\times 10^{-19}\text{ N}$$

Step 2: Acceleration

$$a = \frac{F}{m} = \frac{2.782\times 10^{-19}}{1.67\times 10^{-27}} = 1.666\times 10^{8}\text{ m/s}^2$$

Step 3: Transit time

$$t = \frac{L}{v} = \frac{0.20}{10^4} = 2\times 10^{-5}\text{ s}$$

Step 4: Deflection of one component

$$z = \tfrac12 a t^2 = \tfrac12 (1.666\times 10^{8})(2\times 10^{-5})^2$$ $$= \tfrac12(1.666\times 10^{8})(4\times 10^{-10}) = 3.33\times 10^{-2}\text{ m}$$

Step 5: Total separation

$$\Delta z = 2z = 6.66\times 10^{-2}\text{ m} \approx 6.66\text{ cm}$$

$$\boxed{\Delta z \approx 6.66\text{ cm}}$$

The two components are separated by about 6.66 cm on emerging from the magnet.