Electronic Devices Test 6

• The process of extension of a single crystal surface by growing a film in such a way that the added atoms form a continuation of the single-crystal structure is called epitaxy.
• Onto a substrate it is required to grow a uniform layer of one type of semiconductor.
• For example, one can grow a layer of silicon with one doping level into a substrate of another. We require that the new layer form a continuous crystal with the substrate, so that there are no defects or interface states that trap carriers. For this, epitaxial (epi) growth is used.
• In epitaxy, a thin layer of crystal is grown, rather than simply deposited onto the substrate wafer, and the substrate wafer acts as a seed crystal. 

The electron travelling from -15eV state to -5eV state radiate 10 eV energy as photon whose wavelength λ is given by

\( E = \frac{{hc}}{{\rm{\lambda }}}\)

\( \Rightarrow \lambda = \frac{E}{{hc}}\;\)

\( = \frac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{10 \times 1.6 \times {{10}^{ - 19}}}}\;\)

= 124.2 nm

Concept:

Dry oxidation is used to form thin oxide films

Wet oxidation proceeds at a faster rate and is used to form thick oxide films.

This is because water vapor diffuses through SiO2 faster than oxygen.

The Corresponding Reactions involved are

Si + O2 → SiO2  (Dry Oxidation)

Si + 2H2­O → SiO2 + H2 (Wet Oxidation)

Application:

Since the thickness given is 1 μm which is large, hence Wet Oxidation is preferred.

Dry Oxidation is used when the film thickness is about 100 nm 

Concept:

The photon energy corresponding to the incident wavelength is:

\({W_{ph}} = hf = \frac{{hc}}{\lambda } \)

\(W_{ph}= 497 \times {10^{ - 21}}J\)

Also, the light power density P is defined as:

P = ϕ.W_Ph

Where ϕ = Photon flux

Calculation:

Putting on the respective values, we get:

\(\phi = \frac{P}{{{W_{Ph}}}} = \frac{{1000}}{{497 \times {{10}^{ - 21}}}} \)

\(\phi = 2.0 \times {10^{21}}\frac{photons}{m^2- sec} \)

Each photon liberates 1 electron (100% Quantum efficiency), ∴  2 × 10²¹ electrons/ m-sec² are circulated.

 Since the area is 10^-2 m², the current is

I = 2 × 10²¹ × 10^-2 × 1.6 × 10^-15

= 3.22 A

The steps involved in Twin-tub or Twin-well process are

i) Epitaxial layer serving as base of device is chosen

ii) N-well and P-well formation

iii) Gate and Field Oxide

iv) Source and Drain implantation

v) Contact cut defined in both the Wells

Given g_op = 1 × 10¹⁸ cm^-3

The short circuit current generated I_SC = q. A. g_op (L_n + L_p + w)

A = area of junction

L_p = L_n = diffusion length

I_SC = 1.6 × 10^-19 × n × 10^-4 × 10¹⁸ × 10⁶ (2 + 2 + 1) × 10^-6

= 0.32 mA

The open-circuit voltage will be:

\({V_{OC}} = {V_T}\ln \left( {1 + \frac{{{I_{SC}}}}{{{I_{th}}}}} \right)\)

\( = 0.0254\ln \left( {1 + \frac{{0.32 \times {{10}^{ - 3}}}}{{32 \times {{10}^{ - 9}}}}} \right)\)

V_OC = 0.24 V

Concept:

The output flux, input flux and absorption coefficient and thickness are related using:

I_out = I_input e^-αd

Where, α is the absorption coefficient d is the thickness of slab.

Rearranging the equation:

\(d=\frac{-1}{\alpha }\ln \left( \frac{{{I}_{out}}}{{{I}_{in}}} \right)\)

\(\frac{{{I}_{out}}}{{{I}_{in}}}=0.1\) [since 90% is absorbed]

Calculations:

\(d=\frac{-1}{\alpha }\ln \left( 0.1 \right)\)

\(d=\frac{-1}{{{10}^{2}}}\ln \left( 0.1 \right)\)

d = 0.0230 cm

Concept:

To find the efficiency first find the minimum frequency that is required to excite electron across bandgap Eg. (f_min)

\(Efficiency\left( \eta \right) = \frac{{{f_{min}}}}{{{f_{input}}}} \times 100\% \) 

Calculation:

The frequency of photon whose energy is “Eg” eV is:

\({f_{min}} = \frac{{qEg}}{h} = \frac{{2.35 \times 1.6 \times {{10}^{ - 19}}}}{{6.62 \times {{10}^{ - 34}}}}\) 

= 568 × 10¹² Hz

Frequency that corresponds to light with 400 nm wavelength is:

\(f = \frac{c}{\lambda } = \frac{{3 \times {{10}^8}}}{{400 \times {{10}^{ - 9}}}} = 750 \times {10^{12}}Hz\) 

Efficiency:

\(\eta = \frac{{{f_{min}}}}{f} \times 100\) 

\(= \frac{{568}}{{750}} \times 100\) 

= 75.7%