The whole Electronic Devices formula sheet, to print

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 $f(E)$.

Valid in thermal equilibrium for a non-degenerate semiconductor ($E_F$ at least about $3kT$ from $E_c$ and from $E_v$), whatever the doping. $N_c$ and $N_v$ are the effective densities of states of the conduction and valence bands.

In SI units these come out in m${}^{-3}$; divide by $10^6$ for cm${}^{-3}$. Both scale as $(m^{\ast }{)}^{3/2}{T}^{3/2}$, so they depend on the carrier effective masses and rise with temperature. They are material properties, not constants: in a numerical problem, use the values given.

Holds in thermal equilibrium, non-degenerate; $E_g=E_c-E_v$. For silicon at $T=300\text{K}$ the commonly used value is $n_i\approx 1.5\times 10^{10}{\text{cm}}^{-3}$ (some texts use $1.0\times 10^{10}{\text{cm}}^{-3}$). $n_i$ depends on the material and the temperature, so always use the value the problem states.

This follows from $N_c{N}_v\propto T^3$, and the exponential factor dominates. Taking $E_g$ as constant, two temperatures are related by $\frac{n_i(T_2)}{n_i(T_1)}={\left(\frac{T_2}{T_1}\right)}^{3/2}\exp \left[\frac{E_g}{2k}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\right]$.

Obtained by setting $n_0=p_0$ in the Boltzmann expressions. If $m_p^{\ast }>m_n^{\ast }$, $E_i$ lies slightly above mid-gap. The shift is of order $kT$, small compared with $E_g$.

Equivalent to the $N_c$, $N_v$ forms and valid under the same conditions (equilibrium, non-degenerate, any doping). Multiplying the two gives $n_0{p}_0=n_i^2$ directly.

From the Boltzmann expressions; non-degenerate, equilibrium. $E_F$ lies above $E_i$ when $n_0>n_i$.

From the Boltzmann expressions; non-degenerate, equilibrium. $E_F$ lies below $E_i$ when $p_0>n_i$.

Holds in thermal equilibrium in a non-degenerate semiconductor, doped or undoped, with $n_i$ evaluated at the same temperature. It does not hold under illumination or carrier injection (non-equilibrium).

Total negative charge equals total positive charge in the bulk, where $N_a^{-}$ and $N_d^{+}$ are the ionised acceptor and donor concentrations. With full ionisation, $N_a^{-}=N_a$ and $N_d^{+}=N_d$.

Combine with $n_0{p}_0=n_i^2$ to solve for $n_0$ and $p_0$. If $|N_d-N_a|\gg n_i$, the material is n-type when $N_d>N_a$ (then $n_0\approx N_d-N_a$) and p-type when $N_a>N_d$ (then $p_0\approx N_a-N_d$); the minority carrier follows from mass action.

This is the positive root of $n_0^2-(N_d-N_a)n_0-n_i^2=0$ (full ionisation, non-degenerate). For $N_a>N_d$ swap the roles: $p_0=\frac{N_a-N_d}{2}+\sqrt{{\left(\frac{N_a-N_d}{2}\right)}^2+n_i^2}$. Needed whenever $|N_d-N_a|$ is not much larger than $n_i$. With $N_d=N_a$ it gives $n_0=p_0=n_i$.

$V_T\approx 25.9\text{mV}$ at $T=300\text{K}$, with $k=1.38\times 10^{-23}\text{J/K}$ and $q=1.6\times 10^{-19}\text{C}$. Equivalently $kT\approx 0.0259\text{eV}$ at 300 K, so an energy difference in eV divided by $kT$ in eV gives the dimensionless exponent in the Boltzmann expressions directly.

Evaluated at the band edge, with $\hslash =h/2\pi$. A sharply curved band gives a small effective mass. For a parabolic conduction band, $E=E_c+\frac{{\hslash }^2{k}^2}{2m_n^{\ast }}$.

The wavelength of a photon of energy $E_g$. It gives the approximate emission wavelength of band-to-band recombination and the longest wavelength (cut-off) the material absorbs band-to-band. Uses $hc\approx 1.24\text{eV}\mu \text{m}$.

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.

For a fully ionized, uncompensated extrinsic sample with doping much larger than $n_i$, one term dominates: $\sigma \approx q{N}_D{\mu }_n$ (n-type) or $\sigma \approx q{N}_A{\mu }_p$ (p-type). In a compensated sample, use the net doping, e.g. $n\approx N_D-N_A$ when $N_D-N_A\gg n_i$. The mobility is still reduced by all $N_D+N_A$ ionized impurities. For an intrinsic sample, ${\sigma }_i=q{n}_i({\mu }_n+{\mu }_p)$. $L$ is the length along the current and $A$ the cross-section.

This uses $np=n_i^2$ (thermal equilibrium, non-degenerate). Since ${\mu }_n>{\mu }_p$ in Si, the minimum lies at $n<n_i$, i.e. in slightly p-type material, not at the intrinsic point. ${\sigma }_{min}<{\sigma }_i$ because $2\sqrt{{\mu }_n{\mu }_p}\le {\mu }_n+{\mu }_p$.

$D_n,D_p$ are diffusion coefficients in ${\text{cm}}^2/\text{s}$. Both carriers diffuse from high to low concentration. Electrons are negative, so the electron diffusion current points up the electron gradient (it has the sign of $dn/dx$). Holes are positive, so the hole diffusion current points down the hole gradient (it has the sign of $-dp/dx$).

The drift-diffusion equations for each carrier type. They are the inputs to the continuity equation.

Holds for a non-degenerate semiconductor. It is derived in thermal equilibrium and used at low fields in the drift-diffusion model. $V_T\approx 0.02585$ V at $T=300$ K, often rounded to $0.0259$ V or $0.026$ V. State the value you use, because the rounding shifts the third significant figure.

Conditions: thermal equilibrium (zero net current), full ionization, and quasi-neutrality ($n\approx N_D(x)$ or $p\approx N_A(x)$). Drift must cancel diffusion, so a doping gradient produces a field. Example: $N_D=N_0{e}^{-x/\lambda }$ gives a uniform field $E=V_T/\lambda$ along $+x$.

$I$ flows along the sample length and $B$ is perpendicular to $I$. $t$ is the sample thickness measured along $B$, $W$ is the width across which $V_H$ is measured, and $J=I/(Wt)$. $V_H$ does not depend on $W$. For a fixed geometry, $V_H$ has opposite polarity for holes and for electrons, so the sign of $V_H$ (or $R_H$) identifies the majority carrier. These simple forms assume one dominant carrier type and take the Hall scattering factor as 1. From a measured $V_H$: $p=IB/(q{V}_{H}t)$ or $n=IB/(q|V_H|t)$.

Valid when one carrier type dominates. It combines the measured Hall coefficient and the measured conductivity to give the majority-carrier mobility.

Valid for low-level injection, where the excess carrier density is much smaller than the majority-carrier density. $\delta p=p-p_0$ and $\delta n=n-n_0$ are the excess concentrations. In a neutral region with pair generation, $\delta n=\delta p$. ${\tau }_p,{\tau }_n$ are the minority-carrier lifetimes; they depend on the material and its defects and are given as constants in problems.

Conditions: a uniformly illuminated n-type sample, low-level injection, and no concentration gradients (so no diffusion). The continuity equation then reduces to $d(\delta p)/dt=G_L-\delta p/{\tau }_p$. $G_L$ is the optical generation rate in ${\text{cm}}^{-3}{\text{s}}^{-1}$. Since $\delta n=\delta p$, both carriers add to the conductivity.

The characteristic distance a minority carrier diffuses before it recombines. In a long sample it is the $1/e$ decay length of the steady-state excess-carrier profile (see the continuity-equation section).

$G$ and $R$ are the generation and recombination rates. The opposite signs come from the charge of each carrier: $J_n=-q\times$(electron flux) and $J_p=+q\times$(hole flux). Substituting the drift-diffusion expressions for $J_n$ and $J_p$ gives the minority-carrier equation below.

Assumes uniform doping, a uniform field $E$ and low-level injection. $G_{ext}$ is any externally driven generation, e.g. optical. With $E=0$ the drift term drops out. Setting the time derivative to zero gives the steady-state form.

Conditions: no generation for $x>0$, the excess concentration held at $\delta p(0)$ (e.g. by an injecting junction), and $\delta p\to 0$ as $x\to \infty$ (sample length much greater than $L_p$). The diffusion current decays with the same length $L_p$ as the profile.

The same equation with $\delta p(W)=0$, as at an ohmic contact or a reverse-biased junction at $x=W$. When $W\ll L_p$, almost no carriers recombine in transit. The profile is then linear, and the diffusion current $q{D}_p\delta p(0)/W$ is nearly uniform. This is the short-base picture used for BJTs and short diodes.

$E=-dV/dx$. ${\rho }_{net}$ here is the space-charge density, not resistivity. ${ϵ}_s={ϵ}_r{ϵ}_0$ is the semiconductor permittivity, with ${ϵ}_r\approx 11.7\text{–}11.9$ for Si and ${ϵ}_0=8.85\times 10^{-14}$ F/cm. With the depletion approximation ($n,p\approx 0$ inside the depletion region), it gives the field and potential profiles and the depletion width of a junction. For example, on the n-side of a step junction $dE/dx=q{N}_D/{ϵ}_s$, so $E$ varies linearly with $x$.

Valid for a non-degenerate semiconductor. With no excess generation or injection, both reduce to the single equilibrium Fermi level ($E_{Fn}=E_{Fp}=E_F$, $np=n_i^2$). Quoted values of $n_i$ for Si at 300 K range from about $9.65\times 10^9$ to $1.5\times 10^{10}{\text{cm}}^{-3}$. Use the value the question gives, and state it whenever a numerical answer depends on it.

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.

For an abrupt (step) junction under the depletion approximation. It holds in reverse bias and in forward bias only while $V_A$ stays well below $V_{bi}$; the approximation fails as $V_A\to V_{bi}$. In reverse bias $V_A=-V_R$, so the bracket becomes $V_{bi}+V_R$ and $W$ grows. ${ϵ}_s={ϵ}_r{ϵ}_0$; for Si, ${ϵ}_r\approx 11.7$, giving ${ϵ}_s\approx 1.04\times 10^{-12}\text{F/cm}$.

Charge neutrality: the ionized charge on the two sides is equal and opposite. The more lightly doped side always holds most of $W$.

The peak is at the metallurgical junction ($x=0$). The field falls linearly to zero at each depletion edge, so the profile is a triangle whose area equals $V_{bi}-V_A$.

For a doping profile $N_D-N_A=ax$ that changes linearly through the junction, where $a$ is the grading constant in ${\text{cm}}^{-4}$. Here $W\propto (V_{bi}-V_A{)}^{1/3}$, compared with the square-root dependence of the step junction.

$A$ is the junction cross-sectional area and $W$ is the depletion width at the bias point. This holds for any doping profile as a small-signal capacitance.

$C_{j0}$ is the capacitance at zero bias. The grading exponent is $m=1/2$ for an abrupt junction and $m=1/3$ for a linearly graded one. In reverse bias $V_A=-V_R$, so the bracket is $1+V_R/V_{bi}$ and $C_j<C_{j0}$. The formula applies in reverse bias and at small forward bias only; it wrongly diverges as $V_A\to V_{bi}$.

$N_B$ is the doping of the lightly doped side. A plot of $1/C_j^2$ against $V_R$ is a straight line: its slope gives $N_B$, and its intercept on the voltage axis is at $V_R=-V_{bi}$.

Minority-carrier concentrations at the n-side and p-side depletion edges. Valid for low-level injection, meaning the injected minority density stays well below the majority density.

$I$ flows from p to n through the diode. $V_A>0$ is forward bias and $V_A<0$ is reverse bias. When $|V_A|\gg V_T$ in reverse bias, $I\to -I_0$. Assumes low-level injection and no generation or recombination inside the depletion region.

$\eta \approx 1$ when diffusion current dominates. $\eta$ approaches 2 when recombination in the depletion region dominates, which is typical of Si at low forward current. Use $\eta =1$ unless the question gives another value.

$D_p$ and $D_n$ are the minority-carrier diffusion coefficients, with $D=\mu V_T$ (Einstein relation). $L_p=\sqrt{D_p{\tau }_p}$ and $L_n=\sqrt{D_n{\tau }_n}$ are the diffusion lengths. This form holds when each neutral region is much longer than its diffusion length. In a short-base diode, replace $L_p$ and $L_n$ by the neutral-region widths.

$p_{n0}$ is the equilibrium hole concentration on the n-side and $n_{p0}$ the equilibrium electron concentration on the p-side. From the mass-action law, for non-degenerate doping.

$I$ is the DC forward current at the bias point, with $I\gg I_0$. For example, $r_d\approx 25.9\Omega$ at 1 mA with $\eta =1$ at 300 K.

Charge-control result for a one-sided long-base diode in forward bias ($I\gg I_0$). The stored charge is $Q=\tau I$, where $\tau$ is the minority-carrier lifetime on the lightly doped (injected) side. In a short-base diode, replace $\tau$ by the transit time $W_n^2/(2D_p)$, where $W_n$ is the neutral n-region width. A full small-signal analysis of a long-base diode at low frequency gives half this value, $\tau I/(2\eta V_T)$; use whichever form the question specifies.

$N_B$ is the doping of the lightly doped side. $E_{crit}$ is the critical field for avalanche, a few $\times 10^5$ V/cm in Si, and it changes only weakly with doping. $V_{bi}$ is neglected next to $V_{BR}$. So $V_{BR}\propto 1/N_B$: lighter doping gives a higher breakdown voltage.

$I_Z$ is the reverse current, taken positive, with $I_Z\ge I_{Z,min}$. $V_{Z0}$ is the knee voltage and $r_z$ the dynamic resistance. An ideal Zener has $r_z=0$.

Supply $V_S$ feeds the Zener in parallel with the load through a series resistor $R$. $R_{max}$ keeps the Zener in regulation in the worst case (lowest supply, highest load current). $R_{min}$ keeps the Zener current within its rating in the worst case (highest supply, lowest load current). Assumes $r_z=0$.

${\varphi }_M$ is the metal work function and $\chi$ the semiconductor's electron affinity. ${\varphi }_{Bn}$ is the barrier to electrons on n-type material and ${\varphi }_{Bp}$ the barrier to holes on p-type material. On n-type, the built-in potential of the contact is ${\varphi }_{Bn}-(E_C-E_F)$, in volts.

$A$ is the contact area, $A^{\ast \ast }$ the effective Richardson constant, and ${\varphi }_B$ the barrier height in volts (numerically equal to its value in eV). The prefactor $A{A}^{\ast \ast }{T}^2{e}^{-{\varphi }_B/V_T}$ is the saturation current, which replaces the diffusion-based $I_0$ of a p–n junction. For typical barrier heights it is several orders of magnitude larger than a comparable p–n junction's $I_0$.

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.

The first relation is KCL for the whole device (currents taken into C and B and out of E for an npn). The others convert between $\alpha$ and $\beta$ and give any one of $I_E$, $I_B$, $I_C$ from another. A useful identity is $\beta +1=\frac{1}{1-\alpha }$.

$I_{CBO}$ is the collector current with the emitter open and the CBJ reverse-biased. $I_{CEO}$ is the collector current with the base open. $I_{CEO}$ is larger by a factor of $(\beta +1)$ because the leakage current, flowing into the base, is itself amplified. Both relations hold in active mode.

This follows from the reciprocity of the device. It reduces the four Ebers–Moll parameters to three independent ones, and it defines the transport saturation current $I_S$.

These hold when the EBJ is forward-biased with $V_{BE}$ several times $V_T$, so the $-1$ is negligible, and $V_{BC}\le 0$, so the CBJ diode term contributes at most a small leakage. They follow from the Ebers–Moll equations using $I_B=I_E-I_C=(1-{\alpha }_F)I_{ES}{e}^{V_{BE}/V_T}$ and $I_S={\alpha }_F{I}_{ES}$. This is the exponential law used for the active-mode $I_C$–$V_{BE}$ characteristic.

This first-order model holds in active mode at constant $V_{BE}$. $I_{C0}=I_S{e}^{V_{BE}/V_T}$ is the current extrapolated to $V_{CE}=0$. $V_A$ is the Early voltage. Extending the straight-line $I_C$–$V_{CE}$ characteristic back to the $V_{CE}$ axis gives a crossing at $V_{CE}=-V_A$.

It is evaluated at the DC operating point, in active mode. The middle expression follows exactly from the model above. The approximation $V_A/I_C$ is the one normally used, and it holds when $V_A\gg V_{CE}$.

${\epsilon }_{ox}=3.9{\epsilon }_0$ 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 ${\varphi }_F=0.0259\ln (6.67\times 10^5)\approx 0.347\text{ V}$.

${\varphi }_m$ is the gate work function and $\chi$ the electron affinity of Si (about 4.05 V), both in volts. $E_g\approx 1.12\text{ eV}$ for Si at 300 K, and the intrinsic level is taken at midgap. The bracket is the work function of the p-type body. For an n${}^{+}$ polysilicon gate (${\varphi }_m\approx \chi$) this gives ${\varphi }_{ms}\approx -\left(\frac{E_g}{2q}+{\varphi }_F\right)$. For an n-type body, replace $+{\varphi }_F$ by $-|{\varphi }_F|$.

$Q_{ox}$ is the effective oxide charge per unit area (C/cm${}^2$), treated as a sheet at the oxide-silicon interface. Charge located deeper in the oxide, closer to the gate, shifts $V_{FB}$ less. Positive $Q_{ox}$ makes $V_{FB}$ more negative, and negative $Q_{ox}$ makes it more positive.

Depletion width at the onset of strong inversion (${\varphi }_s=2{\varphi }_F$), uniform doping, depletion approximation. In equilibrium (slow gate sweep), it barely grows beyond this, because the inversion charge supports any further gate charge. Under a fast ramp (deep depletion) it can exceed $W_{dm}$.

Charge per unit area of the ionised acceptors in the depletion layer at ${\varphi }_s=2{\varphi }_F$ and $V_{SB}=0$. It is negative for a p-type body.

Uniform body doping, with $Q_{ox}$ included through $V_{FB}$. Equivalently $V_{TH0}=V_{FB}+2{\varphi }_F+|Q_{d,max}|/C_{ox}$.

$|{\varphi }_F|=V_T\ln (N_D/n_i)$. The depletion charge is positive (ionised donors) and the inversion layer is made of holes, so both terms add negatively. $V_{FB}$ uses ${\varphi }_{ms}$ for the n-type body.

$V_{SB}\ge 0$ is the source-to-body reverse bias, and $\gamma$ is the body-effect coefficient (units $\sqrt{\text{V}}$). A larger $V_{SB}$ widens the depletion region and raises $V_{TH}$. So in a stack of NMOS transistors with the body grounded, an upper device (source above ground) has a higher threshold. For PMOS the same form holds with magnitudes: $|V_{TH}|$ grows with $V_{BS}\ge 0$, with $N_D$ and $|{\varphi }_F|$ in place of $N_A$ and ${\varphi }_F$.

$D_I$ is the implanted dose (ions/cm${}^2$), assumed to sit as a thin sheet at the Si surface. An acceptor implant into an NMOS channel raises $V_{TH}$, and a donor implant lowers it.

Charge-sheet approximation, per unit area, for $V_{GS}>V_{TH}$ with $V_{DS}=0$ (uniform channel). With $V_{DS}>0$ the local channel voltage is subtracted from $V_{GS}$, which is where the square-law I-V model comes from.

Isolated MOS capacitor, high test frequency, equilibrium (slow) DC sweep. $W_{dm}$ is the maximum depletion width at ${\varphi }_s=2{\varphi }_F$.

Valid for $V_{GS}>V_{TH}$ and $0\le V_{DS}\le V_{GS}-V_{TH}$. ${\mu }_n$ is the electron mobility in the channel.

For $V_{DS}\ll 2(V_{GS}-V_{TH})$ the $V_{DS}^2$ term is negligible, so the channel acts as a gate-controlled linear resistor.