Electronic Devices Test 3

Concept:

For a forward bias applied at the PN-junction, the total currently density is given as:

\(J = {J_s}\left( {{e^{\frac{{{V_a}}}{{{V_T}}}}} - 1} \right)\)

J_s = saturation current density calculated as:

\({J_s} = \left[ {\frac{{e{D_p}{p_{n0}}}}{{{L_p}}} + \frac{{e{D_n}{n_{p0}}}}{{{L_n}}}} \right]\)

p_n0 = minority carrier holes on n-side

n_p0 = minority carrier electrons at the p-side.

This can also be written as:

\({J_s} = \left( {\frac{{e{D_p} \cdot n_i^2}}{{{L_p} \cdot {N_d}}} + \frac{{e\;{D_n} \cdot n_i^2}}{{{L_n}{N_a}}}} \right)\)     ---(1)

The above equation can be written as:

J_s = J_sp + J_sn

J_sp = contribution of current because of holes.

J_sn = contribution of current because of electrons.

Now, the contribution of hole current (J_sp) from Equation (1) can increase in one of the two ways:

1) Reducing n-donor concentration (N_d).

2) Increasing the acceptor donor concentration (N_a). This will reduce the minority carrier electrons, which will result in a decrease in current contribution because of electrons

∴ Both (1) and (2) are correct.

Concept:

The current-voltage Relationship in a diode is:

\(i = {I_o}{e^{{V^{BE}}/{V_T}}}\)

where I_o is the reverse saturation current

and I_o ∝ Area of the cross-section

I_o ∝ width

Calculation:

At a bias of 0.6V in D₁:

\(1mA = {I_{01}}{e^{0.6/{V_T}}}\)   ---(1)

Since the Width of D₂ is double the width of D₁:

I₀₂ = 2I₀₁ 

Now, for a bias of 0.625 V in D2

\({I_2} = 2{I_{01}}{e^{\frac{{0.625}}{{{V_T}}}}}\)

The above equation can be written as:

\({I_2} = 2\left[ {{I_{01}}{e^{0.60/{V_T}}}.{e^{0.025/{V_T}}}} \right]\)

Using Equation (1), the above equation becomes:

\({I_2} = 2\left( {1mA} \right){e^{0.025/{V_T}}}\)

I₂ = 2mA (e¹) = 5.436 mA 

Diffusion capacitance:

• Diffusion capacitance is due to the transfer of minority carries during forward-bias.
• The minority carries diffuse from one end of the junction to another, causing variation of charge with applied voltage.
• This leads to capacitance, it is present only in forward bias and is significantly higher than depletion capacitance in the forward bias.

Depletion Capacitance:

Depletion capacitance is calculated as:

\({C_D} = \frac{{\varepsilon A}}{W}\)    ---(1)

W = Width of the depletion region given by:

\(W = {\left[ {\left[ {\frac{{2\varepsilon }}{q}\left( {\frac{1}{{{N_A}}} + \frac{1}{{{N_D}}}} \right)} \right]\left( {{V_{bi}} - V} \right)} \right]^{1/2}}\)      ---(2)

Vbi = built-in potential

V = Anode to cathode applied potential

• Depletion capacitance is due to the storage of charges in reverse bias junction, which acts like parallel plate capacitance.
• With forward-bias voltage, the depletion width decreases, which increases the depletion capacitance.
• It is less than diffusion capacitance in forward-bias mode.

Concept:

Junction/depletion capacitance is calculated as:

\({C_j} = \frac{{\varepsilon A}}{W}\)    ---(1)

W = Width of the depletion region given by:

\(W = {\left[ {\left[ {\frac{{2\varepsilon }}{q}\left( {\frac{1}{{{N_A}}} + \frac{1}{{{N_D}}}} \right)} \right]\left( {{V_{bi}} - V} \right)} \right]^{1/2}}\)      ---(2)

Vbi = built-in potential

V_a = Applied Voltage from anode to cathode

Substituting W from Equation (2) to Equation (1), we get:

\({C_j} = \frac{{\varepsilon A}}{{{{\left[ {\frac{{2\varepsilon }}{q}\left( {\frac{1}{{{N_A}}} + \frac{1}{{{N_D}}}} \right)\left( {{V_{bi}} - V} \right)} \right]}^{1/2}}}}\)

Application:

Since we have n⁺p junction, we have:

N_d ≫ N_a

The junction capacitance is approximated as:

\({{C}_{j}}=\sqrt{\frac{e~\epsilon {{N}_{a}}{{N}_{d}}}{2\left( {{V}_{bi}}+{{V}_{R}} \right)\left( {{N}_{d}} \right)}}\)

\({{C}_{j}}=\sqrt{\frac{e\epsilon {{N}_{a}}}{2\left( {{V}_{bi}}+{{V}_{R}} \right)}}\)     ---(1)

For N_a’ = 2 N_a, neglecting the change in built-in potential, the new depletion capacitance can be approximated as:

\({{C}_{j}}\left( 2{{N}_{a}} \right)=\sqrt{\frac{e\epsilon \left( 2{{N}_{a}} \right)}{e\left( {{V}_{bi}}+{{V}_{R}} \right)}}\)       ---(2)

Diving Cj(2N_a) with C_j, we get:

\(\frac{{{C}_{j}}\left( 2{{N}_{a}} \right)}{{{C}_{j}}}=\sqrt{2}\)

\(\frac{{{C}_{j}}\left( 2{{N}_{a}} \right)}{{{C}_{j}}}=1.424\)

Now, the percentage change in junction capacitance will be:

\(\frac{{{C}_{j}}\left( 2{{N}_{a}} \right)-{{C}_{j}}}{{{C}_{j}}}\times 100\)

\(=\frac{1.414-1}{1}\times 100\)

= 41.4%

Concept:

The reverse saturation current is given by:

\({I_o} = A.q.n_i^2\left[ {\frac{{{D_p}}}{{{L_P}{N_D}}} + \frac{{{D_n}}}{{{L_n}{N_A}}}} \right]\)

\({I_o} \propto n_i^2\;\)

Also,

\(n_i^2=N_cN_ve^{-{\frac{E_g}{KT}}}\)

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

Diode for which E_g/KT will be more will have less n_i and subsequently will have less reverse saturation current.

Observation:

Value of E_g/KT for:

1) D₁ at 100°C  = E_g1/(0.032)

2) D₁ at 30°C = E_g1/(0.026)

3) D₂ at 100°C = E_g2/(0.032)

4) D₁ at 30°C = E_g2/(0.026)

Given that E_g1 > E_g2

Hence the reverse saturation current I₀ will be minimum for D₁ operating at 30°C