Diodes

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Why

Diodes are a magical [^1] semiconductor device that conduct current in one direction. It's one of the fundamental electronics components, and it's a good idea to understand how they work.

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Intrinsic carrier concentration

The intrinsic carrier concentration of silicon, or the density of free electrons and holes at a given temperature, is given by

\(n_i = \sqrt{N_c N_v} e^{\frac{-E_g}{2 k T}}\)

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The density of states are

N_c = 2 \left[\frac{2 \pi k T m_n^*}{h^2}\right]^{3/2} \text{ } N_v = 2 \left[\frac{2 \pi k T m_p^*}{h^2}\right]^{3/2}

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In BSIM 4.8 [@bsim] the intrinsic carrier concentration is

n_{i} = 1.45e10 \frac{TNOM}{300.15} \sqrt{\frac{T}{300.15} \exp^{21.5565981 - \frac{E_g}{2kT}}}

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Density of states

-\frac{\hbar^2}{2m}\nabla^2\psi = E\psi

N(dk) = \frac{2}{(2 \pi)^p} dk

E(k) = \frac{\hbar^2 k^2}{2 m^*}

m^* = \frac{\hbar^2}{\frac{d^2 E}{dk^2}}

N(E)dE = \frac{\sqrt{2}}{\pi^2}\left(\frac{m^*}{\hbar^2}\right)^{3/2} E^{1/2}dE

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f(E) = \frac{1}{e^{(E - E_F)/kT} + 1}

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\[N_e dE = N(E)f(E)dE\]

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n_e = 2\left( \frac{2 \pi m^\ast k T}{h^2}\right)^{3/2} e^{(E_F - E_C)/kT}

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For intrinsic silicon at thermal equilibrium, we could write

n_0 = 2\left( \frac{2 \pi m^\ast k T}{h^2}\right)^{3/2} e^{-E_g/(2kT)}

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Doping

The number of electrons and holes in a n-type material is

n_n = N_D \text{ , } p_n = \frac{n_i^2}{N_D}

and in a p-type material

p_p = N_A \text{ , } n_p = \frac{n_i^2}{N_A}

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PN junctions

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Built-in voltage

n = \int_{E_C}^{\infty} N(E) f(E) dE

n_n = e^{E_{F_n}/kT} \int_{E_C}^{\infty} N_n(E) e^{-E/kT}dE

n_p = e^{E_{F_p}/kT} \int_{E_C}^{\infty} N_p(E) e^{-E/kT}dE

\frac{n_n}{n_p} = \frac{ e^{E_{F_n}/kT}}{e^{E_{F_p}/kT}} = e^{(E_{F_n} - E_{F_p})/kT}

E_{F_n} - E_{F_p} = q\Phi

\frac{N_A N_D}{n_i^2} = e^{q\Phi_0/kT}

or rearranged to

\Phi_0 = \frac{kT}{q} ln\left( \frac{N_A N_D}{n_i^2} \right)

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Current

\frac{p_p}{p_n} = e^{-q\Phi_0/kT}

\frac{p(-x_{p0})}{p(x_{n0})} = e^{q(V-\Phi_0)/kT}

\frac{p(x_{n0})}{p_n} = e^{qV/kT}

\Delta p_n = p(x_{n0}) - p_n = p_n\left( e^{qV/kT} -1 \right)

J(x_n) = -q D_p \frac{\partial \rho}{\partial x}

\partial \rho(x_n) = \Delta p_n e^{-x_n/L_p}

J(0) = q\frac{D_p}{L_p} p_n \left( e^{qV/kT} - 1\right)

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I = q A n_i^2 \left( \frac{1}{N_A}\frac{D_n}{L_n} + \frac{1}{N_D}\frac{D_p}{L_p} \right)\left[ e^{qV/kT} - 1 \right]

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Forward voltage temperature dependence

V_D = V_T \ln\left(\frac{I_D}{I_S}\right)

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V_D = V_T \ln{I_D} - V_T \ln{I_S}

\ln{I_S} = 2 \ln{n_i} + \ln{Aq\left (\frac{D_n}{L_n N_A} + \frac{D_p}{L_p N_D}\right)}

n_i = \sqrt{B_c B_v} T^{3/2} e^\frac{-E_g}{2 kT}

B_c = 2 \left[\frac{2 \pi k m_n^*}{h^2}\right]^{3/2} \text{ } B_v = 2 \left[\frac{2 \pi k m_p^*}{h^2}\right]^{3/2}

2 \ln{n_i} = 2\ln{\sqrt{B_c B_v}} + 3 \ln T - \frac{V_G}{V_T}

\(V_D = \frac{kT}{q}(\ell - 3 \ln T) + V_G\)

\ell= \ln{I_D} - \ln{\left (Aq\frac{D_n}{L_n N_A} + \frac{D_p}{L_p N_D}\right)} - 2 \ln{\sqrt{B_c B_v}}

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From equations above we can see that at 0 K, we expect the diode voltage to be equal to the bandgap of silicon. Diodes don't work at 0 K though.

\frac{dV_D}{dT} = \frac{k}{q}\bigl(\ell - 3 \ln T - 3\bigr).

The slope of the diode voltage can be seen to depend on the area, the current, doping, diffusion constant, diffusion length and the effective masses.

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Current proportional to temperature

I_S e^\frac{qV_{D1}}{kT} = N I_S e^\frac{qV_{D2}}{kT}

Taking logarithm of both sides, and rearranging, we see that

V_{D1} - V_{D2} = \frac{kT}{q}\ln{N}

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Reverse leakage

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Diffusion leakage

I_S = q A n_i^2 \left( \frac{1}{N_A}\sqrt{\frac{D_n}{\tau_n}} + \frac{1}{N_D}\sqrt{\frac{D_p}{\tau_p}} \right)

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Generation in the depletion region

I_{gen} = \frac{q A n_i W}{\tau_g}

W = \sqrt{\frac{2 \varepsilon_{si} (\Phi_0 + V_R)}{q} \cdot \frac{N_A + N_D}{N_A N_D}}

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Plasma charging currents

J_{plasma} \sim 1\text{-}10\, \mathrm{mA/cm^2}

I_{ant} = J_{net} \cdot A_{antenna}

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Antenna ndiode, 200 K to 1000 K

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Wires below 1 mm

I_{ant} = J_{net} \cdot W_{wire} \cdot L_{wire}

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Thanks!

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Summary

Silicon's bandgap makes it a semiconductor: few free carriers at room temperature, exponentially more when heated
Doping moves the Fermi level - towards the conduction band with donors, towards the valence band with acceptors
Join n to p and diffusion fights drift until the depletion field balances: the built-in voltage
Forward bias lowers the barrier exponentially: \(I = I_S(e^{V/V_T} - 1)\)
The diode voltage falls roughly 2 mV/K and extrapolates to the bandgap at 0 K - the seed of every bandgap reference
Reverse leakage grows exponentially with temperature: an antenna diode sized at etch temperature is a different diode at deposition

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Would you like to know more?

[^1]: It doesn't stop being magic just because you know how it works. Terry Pratchett, The Wee Free Men

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