GATE EC Electronic Devices Formula Sheet

Energy bands and carrier concentration

Syllabus: Formation of energy bands in solids, energy bands in intrinsic and extrinsic semiconductors, equilibrium carrier concentration, direct and indirect band-gap semiconductors.

Mass action law

n_0\, p_0 = n_i^2

Symbols

$n_0, p_0$ : equilibrium electron and hole concentrations (cm⁻³)
$n_i$ : intrinsic concentration (cm⁻³)

Applies when

Thermal equilibrium, non-degenerate doping. With

N_{D}

≫

n_{i}

, n₀ ≈

N_{D}

.

Common mistake

Applying it under bias or illumination, where np is no longer

n_{i}

².

Worked example

Si with

N_{D}

= 10¹⁶ cm⁻³: the hole concentration: 22500 cm⁻³ (2.25 × 10⁴)

Carrier concentration and the Fermi level

n_0 = N_C\, e^{-(E_C - E_F)/kT} = n_i\, e^{(E_F - E_i)/kT}, \qquad p_0 = N_V\, e^{-(E_F - E_V)/kT}

\begin{aligned} n_0 &= N_C\, e^{-(E_C - E_F)/kT} \\ &= n_i\, e^{(E_F - E_i)/kT}, \\ p_0 &= N_V\, e^{-(E_F - E_V)/kT} \end{aligned}

Symbols

$N_C, N_V$ : effective densities of states (cm⁻³)
$E_F$ : Fermi level
$E_i$ : intrinsic level

Applies when

Non-degenerate:

E_{F}

at least about 3kT away from both band edges.

N_{C}

and

N_{V}

grow as T^(3/2).

Common mistake

Changing the temperature without scaling both kT and

N_{C}

or

N_{V}

.

Worked example

N_{D}

= 10¹⁶ cm⁻³ in Si:

E_{F}

−

E_{i}

: 0.347 eV (kT ln(

N_{D}

/

n_{i}

))

Compensated doping

n_0 = \frac{N_D - N_A}{2} + \sqrt{\left(\frac{N_D - N_A}{2}\right)^2 + n_i^2}

\begin{aligned} n_0 &= \frac{N_D - N_A}{2} \\ &\quad + \sqrt{\left(\frac{N_D - N_A}{2}\right)^2 + n_i^2} \end{aligned}

Symbols

$N_D, N_A$ : donor and acceptor concentrations (cm⁻³)

Applies when

Complete ionisation and equilibrium. For

N_{D}

−

N_{A}

≫

n_{i}

it reduces to n₀ =

N_{D}

−

N_{A}

.

Common mistake

Using

N_{D}

alone when both dopants are present: the acceptors cancel an equal number of donors.

Worked example

N_{D}

= 10¹⁶,

N_{A}

= 4 × 10¹⁵ cm⁻³: 6000000000000000 cm⁻³

Band gap and emitted wavelength

\lambda = \frac{hc}{E_g} \;\Rightarrow\; \lambda\,(\mu\text{m}) = \frac{1.24}{E_g\,(\text{eV})}

\begin{aligned} \lambda &= \frac{hc}{E_g} \\ &\Rightarrow\; \lambda\,(\mu\text{m}) \\ &\quad = \frac{1.24}{E_g\,(\text{eV})} \end{aligned}

Symbols

$E_g$ : band gap (eV)
$\lambda$ : wavelength (µm)

Applies when

Light emitted or absorbed across the gap. Efficient emission (LEDs, lasers) needs a direct band gap, as in GaAs; Si and Ge are indirect.

Common mistake

Using 1.24 with the wavelength in nm: it is 1240 for nm.

Worked example

GaAs,

E_{g}

= 1.42 eV: 0.873 µm

Carrier transport

Syllabus: Diffusion current, drift current, mobility and resistivity, generation and recombination of carriers, Poisson and continuity equations.

Drift current and resistivity

J = q(n\mu_n + p\mu_p)\,\mathcal{E}, \qquad \rho = \frac{1}{q(n\mu_n + p\mu_p)}

\begin{aligned} J &= q(n\mu_n + p\mu_p)\,\mathcal{E}, \\ \rho &= \frac{1}{q(n\mu_n + p\mu_p)} \end{aligned}

Symbols

$\mu_n, \mu_p$ : electron and hole mobilities (cm²/V s)
$\mathcal{E}$ : electric field (V/cm)
$\rho$ : resistivity (Ω cm)

Applies when

Low fields, where drift velocity is proportional to field. At high fields the velocity saturates.

Common mistake

Adding the hole term with a minus sign: electrons and holes drift opposite ways but their currents add.

Worked example

n-Si with n = 10¹⁶ cm⁻³ and

\mu_n

= 1350 cm²/V s: 0.462 Ω cm

Diffusion current and the Einstein relation

J_n = qD_n\frac{dn}{dx}, \quad J_p = -qD_p\frac{dp}{dx}, \qquad \frac{D}{\mu} = \frac{kT}{q}

\begin{aligned} J_n &= qD_n\frac{dn}{dx}, \\ J_p &= -qD_p\frac{dp}{dx}, \\ \frac{D}{\mu} &= \frac{kT}{q} \end{aligned}

Symbols

$D_n, D_p$ : diffusion coefficients (cm²/s)
$kT/q$ : thermal voltage, 0.0259 V at 300 K

Applies when

Non-degenerate semiconductors in near-equilibrium.

Common mistake

Getting the sign of the hole diffusion current wrong: holes flow down their gradient, so

J_{p}

carries a minus sign.

Worked example

\mu_n

= 1350 cm²/V s at 300 K:

D_{n}

: 34.96 cm²/s

Diffusion length and the continuity equation

\frac{d^2 \delta p}{dx^2} = \frac{\delta p}{L_p^2} \;\Rightarrow\; \delta p(x) = \delta p(0)\, e^{-x/L_p}, \qquad L_p = \sqrt{D_p \tau_p}

\begin{aligned} \frac{d^2 \delta p}{dx^2} &= \frac{\delta p}{L_p^2} \\ \Rightarrow\; \delta p(x) &= \delta p(0)\, e^{-x/L_p}, \\ L_p &= \sqrt{D_p \tau_p} \end{aligned}

Symbols

$\delta p$ : excess minority carriers (cm⁻³)
$\tau_p$ : minority carrier lifetime (s)
$L_p$ : diffusion length (cm)

Applies when

Steady state, no field, low injection, a region much longer than

L_{p}

.

Common mistake

Multiplying D and τ without the square root, or mixing cm and µm.

Worked example

D_{p}

= 12 cm²/s,

τ_{p}

= 1 µs: 34.64 µm

Poisson's equation

\frac{d^2 V}{dx^2} = -\frac{\rho(x)}{\varepsilon_s}, \qquad \frac{d\mathcal{E}}{dx} = \frac{\rho(x)}{\varepsilon_s}

\begin{aligned} \frac{d^2 V}{dx^2} &= -\frac{\rho(x)}{\varepsilon_s}, \\ \frac{d\mathcal{E}}{dx} &= \frac{\rho(x)}{\varepsilon_s} \end{aligned}

Symbols

$\rho(x)$ : charge density (C/cm³)
$\varepsilon_s$ : permittivity of the semiconductor (F/cm)

Applies when

One dimension. In a depletion region ρ =

-qN_A

on the p side and

+qN_D

on the n side.

Common mistake

Using ε₀ instead of

\varepsilon_s

= 11.7ε₀ for silicon.

Worked example

Uniform

N_{D}

= 10¹⁶ cm⁻³ over 0.1 µm of depletion: the field change across it: 15.46 kV/cm

P-N junction and Zener diode

Syllabus: P-N junction, Zener diode.

Built-in potential

V_0 = \frac{kT}{q}\ln\frac{N_A N_D}{n_i^2}

Symbols

$N_A, N_D$ : doping on the p and n sides (cm⁻³)
$V_0$ : built-in potential (V)

Applies when

Step junction, equilibrium, non-degenerate doping.

Common mistake

Using

N_{A}

+

N_{D}

, or forgetting to square

n_{i}

.

Worked example

N_{A}

= 10¹⁷,

N_{D}

= 10¹⁶ cm⁻³, Si at 300 K: 0.754 V

Depletion width

W = \sqrt{\frac{2\varepsilon_s (V_0 - V)}{q}\left(\frac{1}{N_A} + \frac{1}{N_D}\right)}, \qquad N_A x_p = N_D x_n

\begin{gathered} W = \sqrt{\frac{2\varepsilon_s (V_0 - V)}{q}} \\ \times \sqrt{\frac{1}{N_A} + \frac{1}{N_D}} \\ N_A x_p = N_D x_n \end{gathered}

Symbols

$V$ : applied voltage, negative for reverse bias (V)
$x_p, x_n$ : widths on each side

Applies when

Abrupt junction, depletion approximation. Most of W lies on the lightly doped side.

Common mistake

Adding a reverse bias as V₀ − |V|: reverse bias widens the region, so the term is V₀ + |

V_{R}

|.

Worked example

The junction above at zero bias: 0.33 µm

Diode equation

I = I_S\left(e^{V/\eta V_T} - 1\right)

Symbols

$I_S$ : saturation current (A)
$\eta$ : ideality factor, 1 to 2
$V_T$ : thermal voltage (V)

Applies when

Forward or moderate reverse bias, below breakdown. At room temperature, every 60 mV (η = 1) multiplies the current by about 10.

Common mistake

Forgetting η: with η = 2 a tenfold current needs 120 mV, not 60 mV.

Worked example

η = 1: the current ratio for 60 mV more forward bias: 10.14

Junction capacitance

C_j = \frac{\varepsilon_s A}{W} = \frac{C_{j0}}{\sqrt{1 - V/V_0}}

\begin{aligned} C_j &= \frac{\varepsilon_s A}{W} \\ &= \frac{C_{j0}}{\sqrt{1 - V/V_0}} \end{aligned}

Symbols

$C_{j0}$ : zero-bias capacitance (F)
$A$ : junction area (cm²)

Applies when

Abrupt junction under reverse or small forward bias. A linearly graded junction uses a cube root instead.

Common mistake

Taking the square root of the voltage alone instead of (1 − V/V₀).

Worked example

C_{j0}

= 10 pF,

V_{0}

= 0.75 V, reverse bias of 3 V: 4.47 pF

Zener regulator

I_Z = \frac{V_{in} - V_Z}{R_S} - I_L, \qquad I_Z \ge I_{Z,\min}

\begin{aligned} I_Z &= \frac{V_{in} - V_Z}{R_S} - I_L, \\ I_Z &\ge I_{Z,\min} \end{aligned}

Symbols

$V_Z$ : Zener voltage (V)
$R_S$ : series resistor (Ω)
$I_L$ : load current (A)

Applies when

The Zener stays in breakdown only while

I_{Z}

stays above its minimum: check the lowest

V_{in}

with the largest

I_{L}

.

Common mistake

Checking only the nominal input. Regulation fails at the lowest input and the heaviest load.

Worked example

V_{in}

= 12 V,

V_{Z}

= 5.1 V,

R_{S}

= 330 Ω,

I_{L}

= 10 mA: 10.91 mA

BJT, MOS capacitor and MOSFET

Syllabus: BJT, MOS capacitor, MOSFET, scaling in MOSFETs.

BJT current relations

I_C = \beta I_B, \qquad I_E = I_B + I_C = (\beta + 1) I_B, \qquad \alpha = \frac{\beta}{\beta + 1}

\begin{aligned} I_C &= \beta I_B, \\ I_E &= I_B + I_C \\ &= (\beta + 1) I_B, \\ \alpha &= \frac{\beta}{\beta + 1} \end{aligned}

Symbols

$\beta$ : common-emitter current gain
$\alpha$ : common-base current gain

Applies when

Forward-active region: base-emitter forward biased, base-collector reverse biased.

Common mistake

Using

I_{C}

= β

I_{B}

in saturation, where

I_{C}

is set by the circuit and is less than β

I_{B}

.

Worked example

β = 100: 0.9901 (α)

MOS capacitor threshold voltage

V_T = V_{FB} + 2\phi_F + \frac{\sqrt{2\varepsilon_s q N_A (2\phi_F)}}{C_{ox}}, \qquad \phi_F = \frac{kT}{q}\ln\frac{N_A}{n_i}, \qquad C_{ox} = \frac{\varepsilon_{ox}}{t_{ox}}

\begin{aligned} V_T &= V_{FB} \\ &\quad + 2\phi_F \\ &\quad + \frac{\sqrt{2\varepsilon_s q N_A (2\phi_F)}}{C_{ox}}, \\ \phi_F &= \frac{kT}{q}\ln\frac{N_A}{n_i}, \\ C_{ox} &= \frac{\varepsilon_{ox}}{t_{ox}} \end{aligned}

Symbols

$V_{FB}$ : flat-band voltage (V)
$\phi_F$ : bulk Fermi potential (V)
$C_{ox}$ : oxide capacitance per area (F/cm²)

Applies when

p-type substrate (n-channel). Strong inversion when the surface potential reaches

2\phi_F

; the depletion width is then at its maximum.

Common mistake

Using

\phi_F

instead of

2\phi_F

, both in the middle term and under the root.

Worked example

N_{A}

= 10¹⁷ cm⁻³,

t_{ox}

= 10 nm,

V_{FB}

= −0.9 V: 0.39 V

MOSFET drain current

I_D = \tfrac{1}{2}\mu_n C_{ox}\frac{W}{L}(V_{GS} - V_T)^2\ (\text{saturation}), \qquad I_D = \mu_n C_{ox}\frac{W}{L}\left[(V_{GS} - V_T)V_{DS} - \tfrac{1}{2}V_{DS}^2\right]\ (\text{triode})

\begin{gathered} \text{Saturation:} \\ I_D = \tfrac{1}{2}\mu_n C_{ox}\frac{W}{L}(V_{GS} - V_T)^2 \\ \text{Triode:} \\ I_D = \mu_n C_{ox}\frac{W}{L} \times {} \\ \left[(V_{GS} - V_T)V_{DS} - \tfrac{1}{2}V_{DS}^2\right] \end{gathered}

Symbols

$\mu_n C_{ox}$ : process transconductance parameter (A/V²)
$W/L$ : aspect ratio
$V_T$ : threshold voltage (V)

Applies when

Saturation when

V_{DS}

≥

V_{GS}

−

V_{T}

, triode below that. Channel-length modulation adds a factor (1 + λ

V_{DS}

) in saturation.

Common mistake

Using the saturation formula without checking

V_{DS}

≥

V_{GS}

−

V_{T}

.

Worked example

\mu_n C_{ox}

= 200 µA/V², W/L = 10,

V_{GS}

−

V_{T}

= 0.5 V, saturation: 0.25 mA

MOSFET scaling (constant field)

L, W, t_{ox}, V \to \frac{1}{\kappa};\quad N_A \to \kappa N_A;\quad I \to \frac{I}{\kappa};\quad P \to \frac{P}{\kappa^2};\quad \tau \to \frac{\tau}{\kappa};\quad \frac{P}{\text{area}} \text{ unchanged}

\begin{aligned} L, W, t_{ox}, V &\to \frac{1}{\kappa}; \\ N_A &\to \kappa N_A; \\ I &\to \frac{I}{\kappa}; \\ P &\to \frac{P}{\kappa^2}; \\ \tau &\to \frac{\tau}{\kappa}; \\ \frac{P}{\text{area}} \text{ unchanged} \end{aligned}

Symbols

$\kappa$ : scaling factor, greater than 1
$\tau$ : gate delay

Applies when

Ideal constant-field (Dennard) scaling. Constant-voltage scaling instead raises the power density as κ³.

Common mistake

Assuming power density falls too: power per device falls as 1/κ², but so does the area.

Worked example

κ = 2: power per device as a fraction of before: 0.25

LED, photodiode and solar cell

Syllabus: LED, photo diode and solar cell.

Photodiode responsivity

\mathcal{R} = \frac{I_{ph}}{P_{opt}} = \frac{\eta q}{h\nu} = \frac{\eta\,\lambda\,(\mu\text{m})}{1.24}\ \text{A/W}

\begin{aligned} \mathcal{R} &= \frac{I_{ph}}{P_{opt}} \\ &= \frac{\eta q}{h\nu} \\ &= \frac{\eta\,\lambda\,(\mu\text{m})}{1.24}\ \text{A/W} \end{aligned}

Symbols

$\eta$ : quantum efficiency
$P_{opt}$ : incident optical power (W)
$I_{ph}$ : photocurrent (A)

Applies when

Photon energy above the band gap; no avalanche gain.

Common mistake

Expecting responsivity to fall at longer wavelengths below the cut-off: each photon carries less energy, so the same power brings more photons and it rises.

Worked example

η = 0.8 at λ = 0.85 µm: 0.548 A/W

Solar cell open-circuit voltage and fill factor

V_{oc} = \eta V_T \ln\!\left(1 + \frac{I_L}{I_S}\right), \qquad FF = \frac{V_m I_m}{V_{oc} I_{sc}}, \qquad P_{max} = FF \cdot V_{oc} I_{sc}

\begin{aligned} V_{oc} &= \eta V_T \ln\!\left(1 + \frac{I_L}{I_S}\right), \\ FF &= \frac{V_m I_m}{V_{oc} I_{sc}}, \\ P_{max} &= FF \cdot V_{oc} I_{sc} \end{aligned}

Symbols

$I_L$ : light-generated current, equal to $I_{sc}$ (A)
$V_m, I_m$ : voltage and current at maximum power

Applies when

Ideal diode model, I =

I_{L}

−

I_{S}

(e^(V/η

V_{T}

) − 1), no series resistance.

Common mistake

Taking

P_{max}

as

V_{oc}

×

I_{sc}

: the two never occur together, so the fill factor (below 1) is needed.

Worked example

I_{L}

= 30 mA,

I_{S}

= 10⁻¹² A, η = 1: 0.625 V

GATE EC Electronic Devices Formula Sheet

Energy bands and carrier concentration

Mass action law

Carrier concentration and the Fermi level

Compensated doping

Band gap and emitted wavelength

Carrier transport

Drift current and resistivity

Diffusion current and the Einstein relation

Diffusion length and the continuity equation

Poisson's equation

P-N junction and Zener diode

Built-in potential

Depletion width

Diode equation

Junction capacitance

Zener regulator

BJT, MOS capacitor and MOSFET

BJT current relations

MOS capacitor threshold voltage

MOSFET drain current

MOSFET scaling (constant field)

LED, photodiode and solar cell

Photodiode responsivity

Solar cell open-circuit voltage and fill factor

Other GATE EC formula sheets