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P-N Junction MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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\mathrm{Ge} and \mathrm{Si} diodes conduct at \mathrm{0.3\ V} and \mathrm{0.7\ V} respectively. In the following figure if \mathrm{Ge} diode connection is reversed, the value of \mathrm{V_{0}} changes by

In the circuit, if the forward voltage drop for the diode is \mathrm{0.5\ V}, the current will be

The current in the circuit shown in the figure, considering ideal diode is

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Two identical capacitors \mathrm{A} and \mathrm{B} are charged to the same potential \mathrm{V}and are connected in two circuits at \mathrm{t = 0}, as shown in figure. The charge on the capacitors at time \mathrm{t = CR} are respectively

The diode used in the circuit shown in the figure has a constant voltage drop of \mathrm{0.5\ V} at all currents and a maximumpower rating of \mathrm{100\ milli-watt} . What should be the value of the resistance \mathrm{R}, connected in series with the diode, for obtaining maximum current?

In the following circuit of \mathrm{PN} junction diodes \mathrm{D_{1},D_{2}} and \mathrm{D_{3}} are ideal then i is

When a silicon \mathrm{PN} junction is in forward biased condition with series resistance, it has knee voltage of \mathrm{0.6\ V}. Current flow in it is \mathrm{5\ mA}, when PN junction is connected with \mathrm{2.6\ V} battery, the value of series resistance is

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In the following circuits \mathrm{PN} - junction diodes \mathrm{D_{1},D_{2}} and \mathrm{D_{3}} are ideal for the following potentials of \mathrm{A} and \mathrm{B}. The correct increasing order of resistance between \mathrm{A} and \mathrm{B} will be

(i) -10 \mathrm{~V},-5 \mathrm{~V}\\ (ii) -5 \mathrm{~V},-10 \mathrm{~V}\\ (ii) -4 \mathrm{~V},-12 \mathrm{~V}

A \mathrm{2\ V} battery is connected across the points \mathrm{A} and \mathrm{B} as shown in the figure given below. Assuming that the resistance of each diode is zero in forward bias and infinity in reverse bias, the current supplied by the battery when its positive terminal is connected to \mathrm{A} is

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A potential barrier of \mathrm{0.50\ V} exists across a \mathrm{P-N} junction. If the depletion region is \mathrm{5.0 \times 10^{-7} m} wide, the intensity of the electric field in this region is

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P-N Junction

P-N Junction-

A p-n junction is the basic building block of many semiconductor devices like diodes, transistors,etc.

When a P-type semiconductor is suitably joined to an N-type semiconductor, then the resulting arrangement is called P-N junction or P-N junction diode.

                                                                  

                                                                       

 

Formation of a p-n Junction

Let’s imagine thin p-type silicon (p-Si) semiconductor wafer. Now, we have to add a pentavalent impurity to it.  This result in a small part of the p-Si wafer to convert into an n-Si. Due to this two-region will get created one is the wafer contains a p-region and another is n-region with a metallurgical junction between the two. There are two important processes take place during the formation of a p-n Junction:

  1. Diffusion
  2. Drift

As we have learned that in an n-type semiconductor, the concentration of electrons is more than that of holes similarly in a p-type semiconductor, the concentration of holes is more than that of electrons.

When a p-n junction is being formed, diffusion of holes starts from the p-side to the n-side (p→n) while diffusion of hole occurs from the n-side to the p-side (n→p). The reason behind this diffusion is the concentration gradient across p and n sides. Due to this, a diffusion current generates across the junction. Let’s discuss both the scenarios one by one -

Electron diffusion from n→p -

Electron diffusion leaves an positive charge ( ionized donor ) on the n-side. This positive charge is bonded to the surrounding atoms and is not moveable. As diffusion is going on, more electrons start diffusing to the p-side, a layer of positive charge in the n-side of the junction is formed.

Hole Diffuses from p→n

Hole diffusion leaves an negative charge on the p-side. As the diffusion proceeds, holes start diffusing to the n-side, a layer of negative charge on the p-side of the junction is formed. Both the phenomens i.e., diffusion of electrons and holes across the junction depletes the region of its free charges, these space charge regions together are called the depletion region.

                                                                    

This process is shown in the above figure. The thickness of the depletion region very small and its thickness is around one-tenth of a micrometre. Since there is an electric field which is directed from the p-side to the n-side of the junction. Due to this electric field electrons moves from the p-side to the n-side and holes from the n-side to the p-side. This motion of charged carriers due to the electric field is called drift. From this we can conclude that the drift current direction is opposite to the direction to the diffusion current. This is also seen in the figure given above.

The Last Stages of Formation of a p-n Junction

When the diffusion starts, the diffusion current is large as compare to the drift current. As diffusion process continues, the space-charge regions on either side of the junction start extending. Due to this the electric field gets strengthen and same with the drift current. This process will continues till diffusion current = drift current. This is how a p-n junction is formed.

Barrier Potential

In the state of equilibrium, there will be no current in a p-n junction. Due to increase in potential difference across the junction of the two regions due to the loss of electrons by the n-region and the subsequent gain by the p-region. This potential opposes the further flow of carriers to maintain the state of equilibrium. This potential is called Barrier potential.

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P-N Junction

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