Reverse Saturation Current MCQ Quiz - Objective Question with Answer for Reverse Saturation Current - Download Free PDF

Explanation:

Incorrect Statement Analysis for a Diode Operating in Forward Bias

In a forward-biased diode, the following processes occur that affect the depletion region and the potential barrier:

Correct Option:

Option 4: The reduction in the depletion region causes a heavy flow of minority carriers across the junction.

This statement is incorrect because, in a forward-biased diode, the reduction in the depletion region primarily facilitates the flow of majority carriers (electrons in the n-region and holes in the p-region) across the junction. The minority carriers (holes in the n-region and electrons in the p-region) do not play a significant role in the current flow in a forward-biased condition. The primary current flow in a forward-biased diode is due to the majority carriers overcoming the potential barrier and recombining at the junction.

Explanation of Correct Option:

When a diode is forward-biased, the external voltage applied across the diode reduces the potential barrier at the p-n junction. This reduction in the potential barrier is due to the narrowing of the depletion region, which occurs as the majority carriers are pushed towards the junction. The applied forward voltage causes the electrons in the n-region to move towards the p-region, and the holes in the p-region to move towards the n-region. As a result, the width of the depletion region decreases, allowing more majority carriers to cross the junction and recombine, leading to an increase in the current flow through the diode.

The key points to understand here are:

• The forward bias reduces the potential barrier, making it easier for majority carriers to cross the junction.
• The depletion region narrows, allowing for the increased flow of majority carriers.
• The current in a forward-biased diode is primarily due to the flow of majority carriers, not minority carriers.

Analysis of Other Options:

Option 1: The reduction in the width of the depletion region is due to the recombination of charge carriers and immobile ions near the junction.

This statement is correct. In forward bias, as the majority carriers move towards the junction, they recombine with the oppositely charged immobile ions (donors in the n-region and acceptors in the p-region), reducing the width of the depletion region.

Option 2: The reduction in the potential barrier occurs due to the narrowing of the depletion region.

This statement is correct. The applied forward voltage reduces the potential barrier at the junction by narrowing the depletion region, allowing more majority carriers to cross the junction.

Option 3: The reduction in the depletion region allows a majority carrier flow across the junction.

This statement is correct. The narrowing of the depletion region in forward bias facilitates the flow of majority carriers (electrons and holes) across the junction, leading to an increase in current.

In summary, the incorrect statement is option 4, as it incorrectly attributes the increased current flow in a forward-biased diode to the flow of minority carriers. The correct understanding is that the majority carriers are responsible for the increased current in a forward-biased diode.

Given circuit:

Here, \(I=I_{Ge}=I_{Si}\) & V = V_G + V_s

\(I_{Ge}=I_{Si}\) take the approximate current equation

\(I=I_0e^{V/\eta{V_T}}\)

⇒ \(I_{G_0}e^{V_{G_0}/\eta{V_T}}=I_{s_0}e^{V_{s_0}/\eta{V_T}}\)

\(I_{G_0}e^{V_{G_0}/\eta{V_T}}=I_{s_0}e^{\left(\frac{V-V_{G_0}}{2V_T}\right)}\)

⇒ \(I_{G_0}e^{V_{G_0}/\eta{V_T}}=I_{s_0}\left[e^{\frac{V}{2V_T}}e^{\frac{-V_{G_0}}{2V_T}}\right]\)

⇒ \(\dfrac{I_{G_0}}{I_{s_0}}e^{\left(\frac{V_{G_0}}{V_T}+\frac{V_{{G_0}}}{2V_T}\right)}=e^{\frac{V}{2V_T}}\)

⇒ \(I_{G_0}e^{\left(\frac{3V_{G_0}}{2V_T}\right)}=I_{s_0}e^{\frac{V}{2V_T}}\)

\(e^{\frac{3V_{G_0}}{2V_T}}=\dfrac{I_{s_0}}{I_{G_0}}e^{\frac{V}{2V_T}}\)

\(e^{\frac{V_{G_0}}{V_T}}=\left(\frac{I_{s_0}}{I_{G_0}}\right)^{2/3}e^{\frac{V}{3V_T}}\) ---- (1)

and,

 \(I=I_{G_0}e^{\frac{V_{G_0}}{V_T}}\) ---- (2)

From 1 and 2,

⇒ \(I=I_{G_0}\left(\frac{I_{s_0}}{I_{G_0}}\right)^{2/3}e^{\frac{V}{3V_T}}\)

⇒ \(I=\left(I_{G_0}I_{s_0}^2\right)^{1/3}e^{\frac{V}{3V_T}}\)

Explanation:

Reverse Saturation Current:

• It is only due to minority carriers and the current is denoted by I_0.
• It is independent of reverse bias voltage. 
• For a 1^oC rise in temperature, it increases by approximately 70%.
• It is the drift current and it flows from N to P and therefore it is called reverse current.
• It doubles for every 10^o C rise in temperature in both Ge and Si diodes.

\(I_{0_{T2}}=I_{0_{T1}}[2^\frac{T_2-T_1}{10}]\)

So from the above equation, we can say that the given Assertion is true.

As shown in the graph above the curves at different temperatures are shown far apart just for illustration purposes. The curve shifts to the left at the rate of -2.5 mV per degree centigrade change in temperature.

\(\frac{dV}{dT}=-2.5 \ mV\) ------(2)

Similarly, it shifts to the right when the temperature decreases.

From equation (2) we can say that the given reason is true.

But the given reason does not satisfy the given assertion,

Hence option (2) is correct.

The very small current flows through the diode when the diode is in the reverse-biased state is called the reverse current of the diode.

The reverse saturation current of a silicon diode is of the order of nano amperes (nA). ∴ The silicon diode is operated as a good switch.

The reverse saturation current of a Germanium diode is of the orders of micro-amperes (μA).

Note:

Reverse Bias: When the case is opposite and n-type side is kept at a higher potential than the p-type side, only a small amount of reverse saturation current flows through the diode, referring to this condition as reverse biased.

Forward Bias: When the p-type side of the diode is connected to a higher potential than the n-type side, the diode is said to be forward-biased, because it enhances the capability of the diode to conduct forward current.

Concept:

The reverse saturation current of the diode increases with an increase in the temperature.

Mathematically, if the reverse saturation current is I01 at a temperature T1 and I02 at temperature T2 then:

\(\)I02 = I01 2(T2-T1)/10

Application:

Given:

I₀₁ = 3 nA at T₁ = 25°

At T2 = 45°,

the reverse saturation current will be:

\(\)I02 = 3 × 2(45-25)/10 nA

\(\)I02 = 3 × 22 nA

I₀₂ = 12 nA

The diode current equation is given as \(I = {I_o}\left( {{e^{\left( {\frac{V}{{\eta {V_T}}}} \right)}} - 1} \right)\)

given η = 1 (Ideality factor)

\({V_T} = \frac{T}{{11600}} = \frac{{300}}{{11600}} = 0.0258V\)

= 25.8 mV

Given: I = -0.75 I_o

\(- 0.75{I_o} = {I_o}\left( {{e^{\frac{V}{{{V_T}}}}} - 1} \right)\)

\(1 - 0.75 = {e^{V/{V_T}}}\)

\(0.25 = {e^{V/{V_T}}}\)

V = V_T ln(0.25)

V = -0.0357 Volts

V = -35.7 mV

Explanation:

Reverse Saturation Current:

• It is only due to minority carriers and the current is denoted by I_0.
• It is independent of reverse bias voltage. 
• For a 1^oC rise in temperature, it increases by approximately 70%.
• It is the drift current and it flows from N to P and therefore it is called reverse current.
• It doubles for every 10^o C rise in temperature in both Ge and Si diodes.

\(I_{0_{T2}}=I_{0_{T1}}[2^\frac{T_2-T_1}{10}]\)

So from the above equation, we can say that the given Assertion is true.

As shown in the graph above the curves at different temperatures are shown far apart just for illustration purposes. The curve shifts to the left at the rate of -2.5 mV per degree centigrade change in temperature.

\(\frac{dV}{dT}=-2.5 \ mV\) ------(2)

Similarly, it shifts to the right when the temperature decreases.

From equation (2) we can say that the given reason is true.

But the given reason does not satisfy the given assertion,

Hence option (2) is correct.

Concept:

The reverse saturation hole current is given by,

\({{\rm{J}}_{{\rm{po}}}} = {\rm{q\eta }}_{\rm{i}}^2 \cdot \left( {\frac{{{{\rm{D}}_{\rm{p}}}}}{{{{\rm{L}}_{\rm{p}}}{{\rm{N}}_{\rm{D}}}}}} \right)\). 

Application:

We have \({{\rm{L}}_{\rm{p}}} = \sqrt {{{\rm{D}}_{\rm{p}}}{{\rm{\tau }}_{\rm{p}}} = \sqrt {36} \times 100 \times {{10}^{ - 6}}}\)

\(\begin{array}{l} \Rightarrow {{\rm{L}}_{\rm{p}}} = 60 \times {10^{ - 3}}{\rm{cm}}\\ {{\rm{J}}_{{\rm{po}}}} = 1.6 \times {10^{ - 19}} \times {\left( {{{10}^{10}}} \right)^2} \times \left[ {\frac{{36}}{{60 \times {{10}^{ - 3}} \times {{10}^{15}}}}} \right]\\ \Rightarrow {{\rm{J}}_{{\rm{po}}}} = 9.6\frac{{{\rm{pA}}}}{{{\rm{c}}{{\rm{m}}^2}}} \end{array}\)

Now, current density at V = 208 mV

\(\begin{array}{l} {{\rm{J}}_{\rm{p}}} = {{\rm{J}}_{{\rm{po}}}}\left( {{{\rm{e}}^{\frac{{\rm{V}}}{{{{\rm{V}}_{\rm{T}}}}}}} - 1} \right)\\ = 9.6 \times {10^{ - 12}}\left( {{{\rm{e}}^{\left( {\frac{{208}}{{26}}} \right)}} - 1} \right)\\ \Rightarrow {{\rm{J}}_{\rm{p}}} = 28.61\frac{{{\rm{nA}}}}{{{\rm{c}}{{\rm{m}}^2}}} \end{array}\)

Explanation:

Incorrect Statement Analysis for a Diode Operating in Forward Bias

In a forward-biased diode, the following processes occur that affect the depletion region and the potential barrier:

Correct Option:

Option 4: The reduction in the depletion region causes a heavy flow of minority carriers across the junction.

This statement is incorrect because, in a forward-biased diode, the reduction in the depletion region primarily facilitates the flow of majority carriers (electrons in the n-region and holes in the p-region) across the junction. The minority carriers (holes in the n-region and electrons in the p-region) do not play a significant role in the current flow in a forward-biased condition. The primary current flow in a forward-biased diode is due to the majority carriers overcoming the potential barrier and recombining at the junction.

Explanation of Correct Option:

When a diode is forward-biased, the external voltage applied across the diode reduces the potential barrier at the p-n junction. This reduction in the potential barrier is due to the narrowing of the depletion region, which occurs as the majority carriers are pushed towards the junction. The applied forward voltage causes the electrons in the n-region to move towards the p-region, and the holes in the p-region to move towards the n-region. As a result, the width of the depletion region decreases, allowing more majority carriers to cross the junction and recombine, leading to an increase in the current flow through the diode.

The key points to understand here are:

• The forward bias reduces the potential barrier, making it easier for majority carriers to cross the junction.
• The depletion region narrows, allowing for the increased flow of majority carriers.
• The current in a forward-biased diode is primarily due to the flow of majority carriers, not minority carriers.

Analysis of Other Options:

Option 1: The reduction in the width of the depletion region is due to the recombination of charge carriers and immobile ions near the junction.

This statement is correct. In forward bias, as the majority carriers move towards the junction, they recombine with the oppositely charged immobile ions (donors in the n-region and acceptors in the p-region), reducing the width of the depletion region.

Option 2: The reduction in the potential barrier occurs due to the narrowing of the depletion region.

This statement is correct. The applied forward voltage reduces the potential barrier at the junction by narrowing the depletion region, allowing more majority carriers to cross the junction.

Option 3: The reduction in the depletion region allows a majority carrier flow across the junction.

This statement is correct. The narrowing of the depletion region in forward bias facilitates the flow of majority carriers (electrons and holes) across the junction, leading to an increase in current.

In summary, the incorrect statement is option 4, as it incorrectly attributes the increased current flow in a forward-biased diode to the flow of minority carriers. The correct understanding is that the majority carriers are responsible for the increased current in a forward-biased diode.

Reverse Saturation Current:

The reverse saturation current is the part of the reverse current in a semiconductor diode that is caused by the diffusion of minority carriers from the neutral regions to the depletion region.

The reverse saturation current of the diode increases with an increase in the temperature.

The rise is 7% /°C for both germanium and silicon and approximately doubles for every 10°C rise in temperature.

Mathematically if reverse saturation current is I₀₁ at a temperature T₁ and I₀₂ at temperature T₂ then,

\(\large{I_{02}=I_{01}×2^{(\frac{Δ T}{10})}}\) .... (1)

Where ΔT = T₂ - T₁

If reverse saturation current doubles then,

I₀₂ = 2 × I_o1

From equation (1),

\(\large{2I_{01}=I_{01}×2^{(\frac{Δ T}{10})}}\)

or, \(2^{(\frac{Δ T}{10})}=2\) .... (2)

On solving equation (2),

Δ T = 10°C

Hence, The reverse saturation current doubles when the junction temperature increases by 10°C.

Concept:

The reverse saturation current of the diode increases with an increase in the temperature.

Mathematically, if the reverse saturation current is I01 at a temperature T1 and I02 at temperature T2 then:

\(\)I02 = I01 2(T2-T1)/10

Application:

Given:

I₀₁ = 3 nA at T₁ = 25°

At T2 = 45°,

the reverse saturation current will be:

\(\)I02 = 3 × 2(45-25)/10 nA

\(\)I02 = 3 × 22 nA

I₀₂ = 12 nA

The very small current flows through the diode when the diode is in the reverse-biased state is called the reverse current of the diode.

The reverse saturation current of a silicon diode is of the order of nano amperes (nA). ∴ The silicon diode is operated as a good switch.

The reverse saturation current of a Germanium diode is of the orders of micro-amperes (μA).

Note:

Reverse Bias: When the case is opposite and n-type side is kept at a higher potential than the p-type side, only a small amount of reverse saturation current flows through the diode, referring to this condition as reverse biased.

Forward Bias: When the p-type side of the diode is connected to a higher potential than the n-type side, the diode is said to be forward-biased, because it enhances the capability of the diode to conduct forward current.

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 

Given circuit:

Here, \(I=I_{Ge}=I_{Si}\) & V = V_G + V_s

\(I_{Ge}=I_{Si}\) take the approximate current equation

\(I=I_0e^{V/\eta{V_T}}\)

⇒ \(I_{G_0}e^{V_{G_0}/\eta{V_T}}=I_{s_0}e^{V_{s_0}/\eta{V_T}}\)

\(I_{G_0}e^{V_{G_0}/\eta{V_T}}=I_{s_0}e^{\left(\frac{V-V_{G_0}}{2V_T}\right)}\)

⇒ \(I_{G_0}e^{V_{G_0}/\eta{V_T}}=I_{s_0}\left[e^{\frac{V}{2V_T}}e^{\frac{-V_{G_0}}{2V_T}}\right]\)

⇒ \(\dfrac{I_{G_0}}{I_{s_0}}e^{\left(\frac{V_{G_0}}{V_T}+\frac{V_{{G_0}}}{2V_T}\right)}=e^{\frac{V}{2V_T}}\)

⇒ \(I_{G_0}e^{\left(\frac{3V_{G_0}}{2V_T}\right)}=I_{s_0}e^{\frac{V}{2V_T}}\)

\(e^{\frac{3V_{G_0}}{2V_T}}=\dfrac{I_{s_0}}{I_{G_0}}e^{\frac{V}{2V_T}}\)

\(e^{\frac{V_{G_0}}{V_T}}=\left(\frac{I_{s_0}}{I_{G_0}}\right)^{2/3}e^{\frac{V}{3V_T}}\) ---- (1)

and,

 \(I=I_{G_0}e^{\frac{V_{G_0}}{V_T}}\) ---- (2)

From 1 and 2,

⇒ \(I=I_{G_0}\left(\frac{I_{s_0}}{I_{G_0}}\right)^{2/3}e^{\frac{V}{3V_T}}\)

⇒ \(I=\left(I_{G_0}I_{s_0}^2\right)^{1/3}e^{\frac{V}{3V_T}}\)

Given η = 1 IS = 1 pA

Let \({I_{{S_1}}}\) = Reverse saturation current for D1

\({I_{{S_2}}}\) = Reverse saturation current for D2

When D1 is Forward Bias

Ieffective = smallest of two current = \({I_{{S_2}}}\)

I = IS = 1 pA