1. Review

Equilibrium

  • Open circuit

  • Depletion region forms with separated and fixed-in-place charges, setting up an internal E-field.

  • Built-in potential \(V_0\).

Reverse bias

  • Same direction as the built-in field and \(V_0\).

  • No real change in current (≈0).

  • Depletion width increases with more reverse bias.

  • Effect is a voltage-variable capacitance.

2. Forward bias




pn forward bias

Revisit the built-in potential:

\[\begin{align} V_0 &= \frac{k_B T}{q} \ln \left(\frac{p_p}{p_n}\right)\\ V_0 &= V_T \ln \left(\frac{n_n}{n_p}\right). \end{align}\]

Solve for the minority carrier concentrations in equilibrium:

  • \(p_n (\text{equ}) = \)

  • \(n_p (\text{equ}) = \)

The barrier (\(V_0\)) is reduced by the external applied voltage VF since they are in opposite directions.

We can compute the new minority carrier concentration by replacing \(V_0 \Rightarrow V_0 - V_F\).

  • \(p_n (\text{fwd}) = \)

  • \(n_p (\text{fwd}) = \)

Details

\(p_n (\text{fwd}) = \dfrac{p_p}{\exp\left(\dfrac{V_0 - V_F}{V_T}\right)}\)

pn forward bias

2.1. Towards current flow

ONLY the minority carriers are affected.

→ the pn junction is a minority carrier device.

The change in hole concentration is:

\[\begin{align} \Delta p_n &= p_{n,\text{fwd}} - p_{n,\text{equ}}\\ \nonumber\\ \Delta p_n & = \dfrac{p_p}{\exp\left(\dfrac{V_0 - V_F}{V_T}\right)} - \dfrac{p_p}{\exp\left(\dfrac{V_0}{V_T}\right)}\\ \end{align}\]

Collect the terms with V0

\[\Delta p_n = \dfrac{ }{\exp\left(\dfrac{V_0}{V_T}\right)}\left[ \phantom{\exp\left(\dfrac{V_F}{V_T}\right) - 1} \qquad \right]\]
Details
\[\Delta p_n = \dfrac{N_A}{\exp\left(\dfrac{V_0}{V_T}\right)}\left[ \exp\left(\dfrac{V_F}{V_T}\right) - 1 \right]\]

What we know:

  • np and pn increase exponentially lots,

  • diffusion currents therefore also increase lots.

{ …​ insert another day or so of semiconductor physics …​ }

We also know that:

  • Total current is proportional to this change in minority carrier concentration,

  • \(I_\text{total} \propto \left(\Delta p_n + \Delta n_p\right)\)

So:

\[I_\text{total} \left( v_D, T \right) = \underbrace{A \,q \, n_i^2 \left(\dfrac{D_n}{N_A L_n} + \dfrac{D_p}{N_D L_p}\right)}_{\displaystyle I_S} \left[ \exp \left( \dfrac{v_D}{V_T} \right) - 1 \right]\]

Circle all of the parts where temperature is buried.

2.2. Shockley diode equation

\[i_D = I_S \left[ \exp \left( \dfrac{v_D}{V_T}\right) - 1\right]\]

The −1 makes things difficult, can it be dropped? Continue here.