GATE Electronic Devices formula sheet

Probability that an allowed state at energy $E$ is occupied by an electron, in thermal equilibrium. $k=8.62\times 10^{-5}\text{eV/K}$. For any $T>0$, $f(E_F)=\frac{1}{2}$. At $T=0\text{K}$, $f=1$ for $E<E_F$ and $f=0$ for $E>E_F$.

Used for holes in the valence band; follows directly from $f(E)$.

Electrons drift opposite to $E$ (hence the minus sign) and holes drift along $E$. ${\mu }_n,{\mu }_p$ are the electron and hole mobilities in ${\text{cm}}^2/\text{V-s}$. ${\tau }_m$ is the mean time between scattering events and $m^{\ast }$ is the conductivity effective mass. The linear relation $v_d=\mu E$ holds only at low fields (see velocity saturation below).

$q=1.6\times 10^{-19}$ C. With $n,p$ in ${\text{cm}}^{-3}$ and $E$ in V/cm, $J$ is in ${\text{A/cm}}^2$. Electron and hole drift currents both point along $E$ (conventional current), so they add.

Holds at thermal equilibrium with complete ionization and non-degenerate doping ($N_A,N_D$ well below the effective densities of states). $N_A$ and $N_D$ are the net acceptor and donor concentrations on the two sides. $V_{bi}$ is positive: the n-side sits $V_{bi}$ above the p-side. It cannot be read with a voltmeter across the diode, because the contact potentials at the metal contacts cancel it.

Equilibrium majority and minority concentrations on the two sides, with $n_{n0}\approx N_D$ and $p_{p0}\approx N_A$. This is the same result as above, written with the mass-action law.

This applies in active mode with no avalanche multiplication at the CBJ (multiplication factor $M=1$). When $I_{CBO}$ is negligible, $\alpha =I_C/I_E$. For a real device $\alpha <1$, and it is close to 1.

When $I_{CBO}$ is negligible, $\beta =I_C/I_B$. The second form holds when $\gamma$ and ${\alpha }_T$ are both close to 1, and it shows how the two losses add. Because $\alpha$ is close to 1, $\beta$ is very sensitive to small changes in $\alpha$. Carry enough significant figures in $1-\alpha$ before dividing.

${\epsilon }_{ox}=3.9{\epsilon }_0$ for SiO${}_2$; $t_{ox}$ is the oxide thickness. This is the parallel-plate result: it ignores polysilicon-gate depletion and the finite thickness of the inversion layer. Example: $t_{ox}=10\text{ nm}$ gives $C_{ox}=3.45\times 10^{-7}{\text{ F/cm}}^2$.

Valid for a non-degenerate, fully ionised body with $N_A\gg n_i$; ${\varphi }_F>0$ here. For an n-type body (PMOS) the magnitude is $|{\varphi }_F|=V_T\ln (N_D/n_i)$ and ${\varphi }_F$ is negative. Example: $N_A=10^{16}{\text{ cm}}^{-3}$ gives ${\varphi }_F=0.0259\ln (6.67\times 10^5)\approx 0.347\text{ V}$.

Holds for a photon of free-space wavelength $\lambda$. Numeric form: $E(\text{eV})=\frac{1240}{\lambda (\text{nm})}=\frac{1.24}{\lambda (\mu \text{m})}$.

Number of photons per second carried by monochromatic light of optical power $P_{opt}$ (W). With $h\nu$ in eV, $\Phi =P_{opt}/(qh\nu [\text{eV}])$.