PN Junction Diode MCQ Quiz in मल्याळम - Objective Question with Answer for PN Junction Diode - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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 

Ideal diode:

• An ideal diode is a diode that is neither current-controlled nor voltage-controlled resistor
• An ideal diode is a diode that acts like a perfect conductor when a voltage is applied forward biased and like a perfect insulator when a voltage is applied reverse biased

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

  Here, n is the ideality factor, for the ideal diode is 1

• Ideal diodes do not have a threshold voltage
• Once any forward voltage is applied across the diode, it will conduct current instantly across its junctions

CONCEPT:

• The material which is not a good conductor or a good insulator is called as semiconductor. For example: Silicon, Germanium, etc.
• The semiconductor device which is used to control the flow of electric current is called as p-n junction diode.

• The depletion region/layer is an area at the junction where a field is formed because of formation of a negative charge layer on p side and positive charge layer on n side.

• In a p-n junction, holes from the p-side diffuse towards the n-side leaving behind the negative charges. Similarly, electrons from the n-side diffuse towards the p-side leaving behind the positive charges.
• These negative and positive charges thus formed are immobile.
• A layer of these negative and positive space-charge regions is formed. These two layers of space-charge carriers form the depletion region at the p-n junction.
• The movement of electrons from the n-side to the p-side gives rise to a potential, which prevents this movement. Hence it is termed as the barrier potential.
• Reverse bias is a situation where the n-side of the p-n junction is connected to the positive terminal of the battery.

EXPLANATION:

• In Reverse bias, the direction of the applied voltage and the barrier potential is the same.
• Due to this, the depletion layer widens as more charge carriers become immobile. So option 2 is correct.

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,

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

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

Depletion region:

When we join a p-type & n-type semiconductor, a p-n junction is formed.

In a p-n junction, holes are majority carriers in p-side and electrons are majority carriers in the n-side. This is as shown:

• As there is a high electron & hole concentration in n-type & p-type near the junction, diffusion of majority carriers will take place.
• When majority carriers of n-type & p-type cross the junction, they become minority carriesr. So this is called excess minority carrier injection (diffusion of excesses minority carriers).
• When diffusion takes place, positive ions in n-side & negative ions in the p-side remain static. This is as shown below:

Hence we can say that the depletion region is formed due to the diffusion of excess carriers.

And depletion region contains fixed donor & acceptor ions only.

The VI characteristic of a p-n junction diode:

Forward Bias:

• When forward biased, the applied voltage V of the battery mostly drops across the depletion region and the voltage drops across the p-side and n-side of the p-n junction is negligibly small.
• In forward biasing the forward voltage opposes the potential barrier Vbi. As a result, the potential barrier height is reduced and the width of the depletion layer decreases.
• As forward voltage is increased, at a particular value the depletion region becomes very much narrow such that a large number of majority charge carriers can cross the junction.
• The diode current increases exponentially with increase in voltage. 

Reversed Bias:

1) When reverse biased, more charge carriers are depleted, resulting in the widening of the depletion region. (Statement I is correct)

2) This increases the opposing electric field for the diffusion carriers and does not allow them to cross the junction, offering a high resistance.

3) If the reverse voltage increases beyond a certain level, the junction breakdown happens.

The correct option is 2

Concept:

• In diode terminology, PIV stands for Peak Inverse Voltage. 
• PIV rating indicates the maximum allowable reverse bias voltage which can be safely applied to a diode.
• If a reverse potential is greater than the PIV rating, then the diode will enter the reverse breakdown region.
• Silicon diodes have a forward voltage of approximately 0.7 volts.
• Germanium diodes have a forward voltage of approximately 0.3 volts.

There is a certain amount of minimum current flowing through the p-n junction under the reverse bias condition.

This current is referred to as the reverse saturation current (IS) and is due to the minority charge carriers in the semiconductor device.

The reverse saturation current is independent of reverse voltage but it doubles for every 10^o rise in temperature and it is given as:

\({{\rm{I}}_{{\rm{S}}2}} = {{\rm{I}}_{{\rm{S}}1}}{.2^{\frac{{{\rm{\Delta T}}}}{{10}}}}\)

The VI characteristic of a p-n junction diode:

Concept:

• An LED is a diode that conducts only when forward biased.
• Cut-in voltage is the minimum voltage necessary to switch ON the diode/LED.
• When forward-biased, an LED can be replaced with a voltage source equal to its cut-in voltage.

Application:

The given situation is drawn as shown:

Replacing the LED with a voltage source of value equal to the cut-in voltage, we get:

The current through the LED can now be obtained by applying KVL through the loop as:

12 - I ×1k - 1.8 = 0

10.2 = I × 1k

\(I=\frac{10.2}{1k}=10.2~mA\)

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 Germanium diode is of the orders of micro-amperes.

While the reverse saturation current of a silicon diode is of the order of nano amperes. so that the silicon diode is operated as a good switch.