Barrier Potential

Main ideas / concepts taught

• PN junction at zero applied bias: The lecture revisits how a p-n junction forms a depletion region even when no external voltage is applied.
• Built-in (barrier) potential:
  • Caused by the potential difference across the depletion region created by fixed (immobile) charges.
  • This potential opposes further diffusion of charge carriers, acting like a barrier.
• Diffusion and charge formation:
  • Diffusion is described as recombination between mobile charge carriers.
  • When carriers recombine, immobile ions are left behind, producing:
    • On the P side: negatively charged immobile ions
    • On the N side: positively charged immobile ions
• Depletion region meaning:
  • The region is called depleted because it lacks mobile charge carriers, leaving only fixed immobile ions.
• Barrier potential formula:
  • The lecture derives/mentions the standard expression and emphasizes you mainly need to remember and use the formula.

• Thermal voltage (V_T):

  • Introduced as: [ V_T = \frac{kT}{e} ]

  • At room temperature ((27^\circ\text{C}\approx 300\,\text{K})), it is commonly taken as:

    • (V_T \approx 0.026\,\text{V})
  • A generalized method to compute (V_T) for any temperature is also given.

Method / instruction steps (for computing barrier potential)

1. Use the barrier potential expression: [ V_B = V_T \ln\left(\frac{N_A N_D}{n_i^2}\right) ]

2. Compute thermal voltage (V_T):

   • Thermal voltage: [ V_T = \frac{kT}{e} ]

   • Constants:

     • (k) = Boltzmann constant (= 1.38\times 10^{-23}\,\text{J/K})
     • (e) = electron charge (= 1.6\times 10^{-19}\,\text{C})
     • (T) must be in Kelvin
   • Convert Celsius to Kelvin:
     • If (T’ ) is in (^{\circ}\text{C}), then (T = 273 + T’)
   • Room-temperature shortcut:
     • For (T’ = 27^\circ\text{C}\Rightarrow T\approx 300\,\text{K}\Rightarrow V_T \approx 0.026\,\text{V}) 3. Substitute values:
   • (N_A) = acceptor concentration
   • (N_D) = donor concentration
   • (n_i) = intrinsic carrier density 4. Evaluate (V_B):
   • Plug into the logarithmic expression and compute.

Numerical example results (Silicon PN junction, room temperature)

Given:

• Room temperature (\Rightarrow V_T = 0.026\,\text{V})
• (N_A = 10^{16}\ \text{cm}^{-3})
• (N_D = 10^{17}\ \text{cm}^{-3})
• (n_i = 1.5\times 10^{10}\ \text{cm}^{-3})

Computation (as presented):

[ V_B = 0.026 \ln\left(\frac{(10^{16})(10^{17})}{(1.5\times 10^{10})^2}\right) ]

Final stated answer:

• (V_B \approx 0.757\,\text{V}) (barrier/built-in potential for silicon at room temperature)

The lecture also states commonly used room-temperature values:

• Silicon: barrier potential (\approx 0.7\,\text{V})
• Germanium: barrier potential (\approx 0.3\,\text{V})

What’s coming next

• Next lecture topics:
  • Width of the depletion region
  • Another numerical problem on depletion width

Speakers / sources featured

• Unidentified single speaker (course instructor / lecturer) — the subtitles are from one person speaking.
• No external sources or named references are featured in the provided subtitles.