Electronic Devices Test 2

The number of negative charges per unit area in the p-region is equal to the number of positive charges per unit area in the n-region.

i.e. N_ax_p = N_dx_n

where, x_p and x_n­ are the width of the depletion region extending to the ‘p’ and ‘n’-side, respectively.

Calculation:

Given, N_d = 3 × 10¹⁶ cm^-3, N_a = 4 × 10¹⁵ cm^-3

x_p = 0.447 μm

with ⇒ N_ax_p = N_dx_n

\({x_n} = \frac{{{N_a}{x_p}}}{{{N_d}}}\)

\({x_n} = \frac{{4 \times {{10}^{15}} \times 0.447 \times {{10}^{ - 4}}}}{{3 \times {{10}^{16}}}}\)

= 5.96 × 10^-6

= 0.0596 μm

Concept:
Internal quantum efficiency η is the ratio of the number of electrons injected into the depletion region to the number of photons produced.

Ep = Energy in photon (in eV) given by:

\(E_p=\frac{hc}{λ.e}\)

c = Velocity of light 

λ = wavelength of LED

Calculation:

Given, the wavelength of LED light λ = 900 nm 

Forward current through LED I = 20 mA

η = 0.1

Output optical power will be:

P = η Ep I

The Energy is calculated as:

\(= \frac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{1.6 \times {{10}^{ - 19}} \times 9 \times {{10}^{ - 7}}}}\)

= 1.38 eV

∴ P = 0.1 × 1.38 × 20 × 10-3

= 2.76 mW

Explanation:

The minority current is given by:

\({I_0} = Aqn_i^2\left[ {\frac{{{D_p}}}{{{L_p}{N_D}}} + \frac{{{D_n}}}{{{L_n}{N_A}}}} \right]\)

σ ∝ Doping concentration

Hence For diffusion current

\({I_p} ∝ \frac{1}{L_p\sigma_p}\)

\({I_n} ∝ \frac{1}{L_n\sigma_n}\)

\(\frac{{{I_p}}}{{{I_n}}} = \left( {\frac{{{L_n}}}{{{L_p}}}} \right)\left( {\frac{{{σ _p}}}{{{σ _n}}}} \right)\)

\(4 = \left( {\frac{{{L_n}}}{{{L_p}}}} \right)\left( {\frac{{1.4}}{{2.8}}} \right)\)

Given: V_EC = 5V. This means that the pnp transistor is operating in the active node

\({I_{{E_1}}} = 10\;A\)

\({I_{{B_1}}} = \frac{{10}}{{1 + \beta }} = \frac{{10}}{{16}} = 0.625\;A\)

\({I_{{C_1}}} = \beta {I_{{B_1}}} = 15 \times \frac{{10}}{{16}} = 9.375\;A\)

\({I_{{C_1}}} = {I_{{S_1}}}{e^{{V_{EB}}/{V_T}}}\)

\(\Rightarrow {I_{S1}} = {I_{C1}}{e^{ - {V_{EB}}/{V_T}}}\)

\({I_{{S_1}}} = 9.375 \times {e^{ - \frac{{0.85}}{{26 \times {{10}^{ - 3}}}}}} \)

\(I_{S1}= 59.4 \times {10^{ - 15}}\;A\)

\(\frac{{{I_{{S_1}}}}}{{{I_{{S_2}}}}} = \frac{{{A_{EBJ1}}}}{{{A_{EBJ2}}}} = \frac{{59.4 \times {{10}^{ - 15}}A}}{{{I_{{C_2}}}{e^{ - {V_{BE}}/{V_T}}}}} \)

\(\frac{I_{S_{1}}}{I_{S_{2}}}= \frac{{59.4 \times {{10}^{ - 15}}}}{{2.03 \times {{10}^{ - 15}}}}= 29.3\)

The Relation between bandgap and intrinsic carrier concentration is

\(n_i^2 \propto {e^{ - {E_g}/KT}}\)

When E_g increase there is decrease in \(n_i^2\)

Now factor affecting BJT gain are

1) Emitter injection efficiency

2) Recombination in base

Due to decrease inn \(n_i^2\)( hence n_i ) the emitter injection efficiency increase. Hence there is decrease in current gain.

Concept:

The conductivity for a doped semiconductor is obtained as:

σ = q n0 μn = q p0 μp

where,

ni = Intrinsic carrier concentration.

μn = electron mobility

μp = Hole mobility

When a semiconductor crystal is optically excited, one hole is created for every electron, i.e. electron-hole pass are created.  

Calculation:

Before Illumination:

NA = 1017 cm-3 and ni = 2.2 × 106 cm-3

Since NA ≫ ni, the excess hole concentration at thermal equilibrium will be:

p0 = NA = 1017 cm-3

The electron concentration using mass-action law is obtained as:

\({n_0} = \frac{{n_i^2}}{{{p_0}}}\)

\({n_0} = \frac{{{{\left( {2.2 \times {{10}^6}} \right)}^2}}}{{{{10}^{17}}}} = 4.8 \times {10^{ - 5}}c{m^{ - 3}}\)

The electron concentration before illumination is negligible.

Assuming room temperature, all dopants are ionized.

∴ p_o = N_A = 10¹⁷ cm^-3

Therefore, the electron concentration will be:

\({n_o} = \frac{{n_i^2}}{{{N_A}}} = \frac{{{{\left( {2.2 \times {{10}^6}} \right)}^2}}}{{{{10}^{17}}}} = 4.8 \times {10^{ - 5}}c{m^{ - 3}}\)

Here, we may notice that the electron contribution is negligible. So, we obtain the conductivity of the sample (in dark) as:

σ_o = q (μ_n n_o + μ_p p_o) ≅ qμ_pp_o

= (1.6 × 10^-19)(230)(10¹⁷)

σ₀ = 3.68 (Ω-cm)^-1

When electrons are generated optically, one hole is also created with the generation of an electron. Therefore, the excess hole concentration and excess electron concentration are the same for this case, i.e.

Δp = Δn = 10¹⁶ cm^-3

Now, the electron and hole concentration after illumination will be:

n = n0 + Δn

n = 4.8 × 10-5 + 1016

n = 1016

Similarly,

p = p0 + Δp

p = 1017 + 1016 cm-3

p = 11 × 1016 cm-3

∴ the conductivity of the given sample when the light is ON will be:

σ = q(n0 + Δn)μn + q(p0 + Δp)μp

σ = q[(1016)μn + (11 × 1016)μp]

= (1.6 × 10-19) (1016 × 5300 + (11 × 1016) (230))

= 1.6 × 10-19 (5.3 × 1019 + 2.53 × 1019)

= 1.6 × (5.3 + 2.53)

σ = 12.528 (Ω-cm)-1      

Concept:

For a field-effect transistor (FET) under certain operating conditions, the resistance of the drain-source channel is a function of the gate-source voltage alone and the JFET will behave as an almost pure ohmic resistor

For a MOSFET in the linear region:

VGS > Vth and

VDS > VGS – Vth

The current equation for a MOSFET in the saturation region is given by:

\({{I}_{D}}=\mu C_{ox} \frac{W}{2L}\left\{\left( {{V}_{GS}}-{{V}_{th}} \right)^2\right\}\)

Taking the derivative of the above w.r.t. V_DS, we get:

\(\frac{\partial I_D}{\partial V_{DS}}=\mu C_{ox}\left( {\frac{W}{L}} \right)\left( {{V_{GS}} - {V_{TH}}} \right)\)

Voltage-controlled Resistor (r_DS) is defined as:

\({r_{DS}} = \frac{{{V_{DS}}}}{{{I_{DS}}}} = \frac{1}{{{\mu _n}{C_{ox}}\left( {\frac{W}{L}} \right)\left( {{V_{GS}} - {V_{TH}}} \right)}}\)

Calculation:

Given:

V_GS = 2.0 V,

V_TH = -0.5 V,

\(\frac{W}{L} = 100\)

C_ox = 10^-8 f/cm²

μ_n = 800 cm²/V-s,

V_DS = 5 V

Putting on the respective values in Equation (1), we get:

\({r_{DS}} = \frac{1}{{800 \times {{10}^{ - 8}} \times 100 \times \left( {2 - \left( { - 0.5} \right)} \right)}}\)

\({r_{DS}} = 500\;{\rm{\Omega }}\)

The diffusion capacitance of a PN junction \({C_D} = \frac{{\tau I}}{{n{V_T}}}\)

τ = Carrier life time

I = Diode current

\(I = {I_0}\left( {{e^{\frac{v}{{\eta {V_T}}}}} - 1} \right)\)

\({I_1} = {I_0}\left( {{e^{\frac{{0.5}}{{{V_T}}}}} - 1} \right)\)

\({I_2} = {I_0}\left( {{e^{\frac{1}{{{V_T}}}}} - 1} \right)\)

\(\Rightarrow \frac{{{C_D}_1}}{{{C_D}_2}} = \frac{{{I_1}}}{{{I_2}}}\)

\({C_D}_2 = \frac{{{I_2}}}{{{I_1}}}{C_D}_1\)

\(= \frac{{{e^{\frac{1}{{{V_T}}}}}}}{{{e^{\frac{{0.5}}{{{V_T}}}}}}}{C_D}_1\left( {\therefore {e^{\frac{1}{{{V_T}}}}} > > 1} \right)\)

\(= {e^{\left( {\frac{{0.5}}{{0.0259}}} \right)}} \times {10^{ - 8}}F\)

= 2.42 F

Concept:

Energy of a photon emitted by a semiconductor is given by:

\(E = \frac{{hc}}{λ}\)

Rearranging the above, we get:

\(λ = \frac{{hc}}{E}\)

Differentiating this with respect to T:

\(\frac{{dλ }}{{dT}} = \frac{{ - hc}}{{E_g^2}}\frac{{d{E_g}}}{{dT}}\)    ----(1)

Calculation:

Given \({E_g} = \sqrt {1.24} ~eV\) , \(\frac{{d{E_g}}}{{dT}} = - 5 × {10^{ - 4}}~eV/K\) and hc = 1240 eV nm.

Putting these values in Equation- (1) we get:

\(\frac{{dλ }}{{dT}} = \frac{{ - 1240\left( {eV\;nm} \right)}}{{{{\left( {\sqrt(1.24)} \right)}^2}e{V^2}}}\left( { -~5 × {{10}^{ - 4}}} \right)\frac{{eV}}{K}\)

\(\frac{{dλ }}{{dT}} = 0.5\;nm/K\)

dλ = 0.5 × dT

Given dT = 10°C or 10 K, the required change in emitted wavelength will be:

dλ = 0.5 × 10 = 5 nm

At T = 300 K, we have:

n_i = 1.5 × 10¹⁰ cm^-3

N_d = 10¹⁶ cm^-3

Since N_d ≫ n_i, the electron concentration will be:

n₀ = N_d

= 10¹⁶ cm^-3

Using mass-action law, the hole concentration (minority concentration) is obtained as:

\({p_0} = \frac{{n_i^2}}{{{n_0}}}\)

\( = \frac{{{{\left( {1.5 \times {{10}^{10}}} \right)}^2}}}{{{{10}^{16}}}}\)

= 2.25 × 10⁴ cm^-3

Hence, the hole recombination rate in thermal equilibrium will be:

\({R_{p0}} = \frac{{{p_0}}}{{{\tau _{p0}}}}\)

\( = \frac{{2.25 \times {{10}^4}}}{{20 \times {{10}^{ - 6}}}} = 1.125 \times {10^9}c{m^{ - 3}}{s^{ - 1}}\)

Now, the recombination rate of the majority carrier electrons will be the same as that of the minority carrier holes, i.e.

R’_p = R’_n

\(\frac{{{p_0}}}{{{\tau _{p0}}}} = \frac{{{n_0}}}{{{\tau _{n0}}}}\)

Hence, the lifetime of the majority carrier electrons will be:

\({\tau _{n0}} = \frac{{{n_0}}}{{{p_0}}}{\tau _{p0}} = \frac{{{{10}^{16}}}}{{2.25 \times {{10}^4}}} \times 20 \times {10^{ - 6}}\)

= 8.89 × 10⁶ sec

Concept:

The built-in potential barrier for a PN junction with given doping concentration Na, Nd, is given by:

\({{V}_{bi}}=\frac{kT}{q}\ln \left( \frac{{{N}_{a}}{{N}_{d}}}{n_{i}^{2}} \right)\)

Also, the conductivity is defined as:

σ = eμn

n = carrier concentration

Calculation:

The conductivity of a p-type semiconductor is defined as:

σ_po = eμ_pp_po

So, the hole (majority) concentration in p-side is:

\({P_{po}} = \frac{{{\sigma _{po}}}}{{e{\mu _p}}}\)

\({P_{po}} = \frac{{\frac{1}{2}}}{{\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {1800} \right)}}\)

\({P_{po}} = 1.74 \times {10^{15}}c{m^{ - 3}}\)

Similarly, the electron (majority) concentration in n-side is obtained as:

\({n_{no}} = \frac{{{\sigma _{no}}}}{{e{\mu _n}}}\)

\( = \frac{{\frac{1}{1}}}{{\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {3800} \right)}}\)

= 1.64 × 10¹⁵ cm^-3

Hence, the built-in potential in the germanium diode is:

\({V_{bi}} = \frac{{KT}}{q}\ln \left( {\frac{{{N_A}{N_D}}}{{n_i^2}}} \right)\)

\( = \frac{{KT}}{q}\ln \left( {\frac{{{p_{po}} \times {n_{no}}}}{{n_i^2}}} \right)\)

\( = 0.0259\ln \left( {\frac{{1.74 \times {{10}^{15}} \times 1.64 \times {{10}^{15}}}}{{{{\left( {2.25 \times {{10}^{13}}} \right)}^2}}}} \right)\)

= 0.218 V