PN Junction: Definition, Formula & Applications

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PN Junction is formed at the interface or boundary between a p-type and the n-type semiconductor. The positive side of the semiconductor (p side) has an excess of holes. The negative side of the semiconductor (n side) has an excess of electrons. The PN junction is created by the process of doping. Doping is done to provide a seamless movement of electrons between the p side and n side of the semiconductor. 

Key Terms: Semiconductors, PN Junction, Biasing, Forward Bias, Reverse Bias, Diode, Junction, Battery, Electron, Equilibrium


What is PN Junction Diode?

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  • Fusing a p-type semiconductor with a n-type semiconductor creates a potential barrier voltage across the diode junction.
  • This interface creates a two terminal device better known as the PN Junction.
  • The p and the n region are separated by the charge rich depletion region.
  • The junction stays in equilibrium without the application of any external voltage.
  • The potential barrier can be overcomed by establishing a connection between the p and n type and connecting it to a battery.
  • The electrons thus cross the depletion region. 

PN Junction Diode

PN Junction Diode

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Formation of PN Junction

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  • The p-n junction in a semiconductor is formed by the doping method.
  • The method of doping is used because if the semiconductor is formed by combining different semiconductor materials, it will result in the creation of a grain boundary.
  • Grain boundary will scatter the holes and electrons and will stop the movement of electrons from one side to another side.

Drift current and Diffusion

Drift Current and Diffusion


Doping in Semiconductors

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Let us look at the step-by-step process of doping:

  • Consider there is a thin Silicon semiconductor of p-type.
  • Now, as a small amount of pentavalent impurity is added to the semiconductor wafer, it turns to be n-type silicon.
  • The wafer now has both p and n-type regions which are separated by an interface or a junction.
  • After that, the processes of diffusion and drift take place. 
  • Due to the difference of electrons at one side of the junction and the concentration of holes at another, the electrons from the n side diffuse to the p side.
  • Similarly, the holes that are in the p-side diffuse to the n-side.
  • Therefore, across the junction, there is a diffusion current.
  • As the holes from the p side diffuse to the n side,  there is an ionized acceptor that is left behind on the p side.
  • Hence, a layer of negative charge develops on the p side of the junction.

Doping in Semiconductors

Doping in Semiconductors

  • Similarly, as the electrons from the n side diffuse to the p side, there is an immobile ionized donor which is left behind on the n side.
  • Due to the continuation of the process, on the n side of the junction, a positive charge layer is developed.
  • The region with both negative and positive that is on both sides of the junction is called the depletion region.
  • Therefore, it leads to the development of an electric field due to which an electron that is on the p side of the junction moves to the n side.
  • This leads to a drift current and is employed in making PN junction diodes.
  • The direction of diffusion and drift current is opposite to each other.

Biasing Conditions of PN Junction

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Based on the voltage applied, there are 3 biasing conditions of a pn junction 

Forward Bias

  • When the positive terminal of the battery is connected to the p side of the junction and the negative terminal is connected to the n side of the junction, the pn junction is forward biased.
  • Here, the applied electric field and the built-in electric field, both are in opposite directions.
  • The resulting field is thus less than the built- in electric field.
  • The depletion region thus becomes thin and less resistive. 

Reverse Bias

  • When the positive terminal of the battery is connected to the n side of the junction and the negative terminal is connected to the p side of the junction, the pn junction is reverse biased.
  • Here, the applied electric field and the built-in electric field, both are in the same direction.
  • The resulting field is thus in the same direction as the built-in electric field.
  • The depletion region thus becomes thicker and more resistive.

Forward and Reverse Bias

Forward and Reverse Bias

Zero Bias

When no external voltage is applied to the pn junction, it is zero biased

Zero Bias


Formula of PN Junction

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The formula of PN junction is,

E\(V_T \ ln \frac{N_D.N_A}{n_i^2}\)

Here,

E0 is the junction voltage at zero bias

VT is the thermal voltage at room temperature of 26mV

ND and NA are the concentrations of impurity

ni is the intrinsic concentration


Applications of PN Junction

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  • PN junction is used in many semiconductor electronics.
  • PN junction can be used as a solar cell.
  • When forward biased, PN junctions is can be used in LED lights.
  • In LED, the PN junction can be used when it's forward biased.
  • When reverse biased, the pn junction is used as a photodiode.

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Things to Remember

  • An interface or a boundary that is between p-type and n-type, the two types of semiconductor material, is known as a p-n junction.
  • The positive side of the semiconductor, which is the p side, has an excess of holes while the negative side of the semiconductor, which is the n side, has an excess of electrons. 
  • The p-n junction is formed by doping.
  • Doping combines different semiconductor materials together.
  • Based on the voltage applied, there are 3 biasing conditions of a p-n junction Forward Bias, Reverse Bias, Zero Bias. 
  • The P-N junction can be used as a solar cell.
  • In LED, the PN junction can be used when it's forward biased.

Previous Year Questions

  1. A common emitter amplifier circuit, built using an npn transistor, is shown in the figure…? [JEE 2019]
  2. Identify the correct output signal Y in the given combination of gates…? [JEE 2020]
  3. An n−p−n transistor has three leads A,B and C. Connecting B and C by moist fingers…? [JEE 2014]
  4. The minimum base current and the input voltage at which the transistor will go to saturation…? [JEE 2019]
  5. Pure silicon doped with phosphorus is a…?
  6. If the rms value of sinusoidal input to a full wave rectifier is…? [KEAM]
  7. Identify the mismatched pair from the following…? [KEAM]
  8. A common emitter amplifier gives an output of 3V for an input of 0.01V . If β of the transistor…?
  9. The electrical conductivity of an intrinsic semiconductor at 0K is…? [KEAM]
  10. In n- p-n transistor, the collector current is 10 mA. If 90% of the electrons emitted…? [KEAM]
  11. The principle of LASER action involves…? [KCET 2007]
  12. The density of an electron-hole pair in a pure germanium is 3×1016m−3 at room temperature…? [KCET 2018]
  13. Which one of the following is the weakest kind of the bonding in solids? [NEET 1992]
  14. When using a triode, as an amplifier, the electrons are emitted by…? [NEET 1996]
  15. In forward bias, the width of potential barrier in a p-n junction diode…? [NEET 1999]
  16. When n type semiconductor is heated…? [NEET 1989]
  17. In the energy band diagram of a material shown below, the open circles and filled circles…? [NEET 2007]
  18. In the combination of the following gates the output Y can be written in terms of inputs…? [NEET 2018]
  19. In the circuit shown in the figure, the input voltage Vi is 20V…? [NEET 2018]
  20. In semiconductors at a room temperature…? [NEET 2004]

Sample Questions

Ques: What is a p-n junction? (1 mark)

Ans. A p-n junction is an interface or a boundary that is between the two types of semiconductor material p-type and n-type.

Ques: What are the two processes that take place in the formation of a p-n junction? (1 mark)

Ans. In the formation of a p-n junction, the processes of diffusion and drift take place.

Ques: When is a p-n junction said to be reverse biased? (1 mark)

Ans. When the positive terminal of the battery is connected to the n side of the junction and the negative terminal is connected to the p side of the junction, the p-n junction is said to be reverse biased. 

Ques. Give reason to explain why n and p regions of a Zener diode are heavily doped. Find the current through the Zener diode in the circuit given below. (3 marks)
(Zener breakdown voltage is 15 V) (2019 outside Delhi)
Circuit diagram

Ans. When both the sides of the junction, p and n are heavily doped the depletion region that forms is very thin i.e., <10? m. Therefore the electric field throughout the junction is very high Electrical Junction even when there is a little reverse bias voltage. This could be the cause of breakdown during the reverse bias. 

Current through Zener Diode

Calculation of Current

Ques. (a) A student wants to use two p-n junction diodes to convert alternating current into direct current. Draw the labeled circuit diagram she would use and explain how it works.
(b) Give the truth table and circuit symbol for the NAND gate. (2018)

Ans.

Circuit Diagram of p-n diode

NAND Gate

In the positive half cycle of the AC, end A changes to (+) and A1 changes to forward biased and D2 is reverse-biased, hence D1 conducts and D2 doesn't. So traditionally current flows through the D1, RL, and the top half of the secondary winding. Likewise, the (-) half cycle of the AC diode D2 changes to forward biased and D1 turns to reversed biased, current flows through D2, RL, and the bottom half of secondary winding. Thus the current flows in a similar direction in the two half cycles of input AC voltage. 

Ques. Name the junction diode whose I-V characteristics are drawn below. (2017)
Junction diode attribute to a solar cell

Ans. This is the attribute of the solar cell.

Ques. Write the two processes that take place in the formation of a p-n junction. Explain with the help of a diagram, the formation of depletion region and barrier potential in a p-n junction. (2017)

Ans. Diffusion and drift are the two processes that take place in the formation of a p-n junction. Diffusion of an electron:

Diffusion of electron

Development of the barrier potential and depletion region:

At this particular point of the p-n junction development the independent electrons closer to the junction diffuse throughout the junction into the p region and merge with holes. Therefore on merging with the hole, it forms a negative ion and leaves a positive ion on the n-side. These two particular strata of the immobile (+) and (-) charges form the depletion region. 

Furthermore, the negative charge helps in repelling any more diffusion of electrons when they reach a point through the continuous diffusion of electrons across the junction. This depletion region now forms the barrier and the energy from the exterior is provided to get the electrons to shift across the barrier of the electric field. This possible difference needed to shift the electrons through the electric field is known as the barrier potential.

Ques. (a) In the following diagram, is the junction diode forward biased or reverse biased?
reversed biased
(b) Draw the circuit diagram of a full-wave rectifier and state how it works. (2017 outside Delhi)

Ans.

(a) The junction diode is reverse biased.

junction diode is reverse biased

(b) When the junction can be stated as forward based only then diode D1 conducts. Therefore in the first half cycle of the input AC, we can see that D1 conducts and D2 doesn’t and the current will flow to B from A. D2 can be stated as reverse biased and in the second half cycle of input AC, diode D2 conducts while D1 doesn’t. The current in this case also flows from A to B. Therefore we can say that the full cycle will be unidirectional. 

circuit diagram of a full-wave rectifier

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CBSE CLASS XII Related Questions

  • 1.
    In the circuit, three ideal cells of e.m.f. \( V \), \( V \), and \( 2V \) are connected to a resistor of resistance \( R \), a capacitor of capacitance \( C \), and another resistor of resistance \( 2R \) as shown in the figure. In the steady state, find (i) the potential difference between P and Q, (ii) the potential difference across capacitor C.
    potential difference across capacitor C


      • 2.
        A vertically held bar magnet is dropped along the axis of a copper ring having a cut as shown in the diagram. The acceleration of the falling magnet is:
        vertically held bar magnet is dropped along the axis of a copper ring

          • zero
          • less than \( g \)
          • \( g \)
          • greater than \( g \)

        • 3.
          The resistance of a wire at 25°C is 10.0 \( \Omega \). When heated to 125°C, its resistance becomes 10.5 \( \Omega \). Find (i) the temperature coefficient of resistance of the wire, and (ii) the resistance of the wire at 425°C.


            • 4.
              A parallel plate capacitor has plate area \( A \) and plate separation \( d \). Half of the space between the plates is filled with a material of dielectric constant \( K \) in two ways as shown in the figure. Find the values of the capacitance of the capacitors in the two cases. parallel plate capacitor


                • 5.
                  Write the mathematical forms of three postulates of Bohr’s theory of the hydrogen atom. Using them prove that, for an electron revolving in the \( n \)-th orbit,
                  (a) the radius of the orbit is proportional to \( n^2 \), and
                  (b) the total energy of the atom is proportional to \( \frac{1}{n^2} \).


                    • 6.
                      (a) Consider the so-called ‘D-T reaction’ (Deuterium-Tritium reaction).
                      In a thermonuclear fusion reactor, the following nuclear reaction occurs: \[ \ ^{2}_1 \text{H} + \ ^{3}_1 \text{H} \longrightarrow \ ^{4}_2 \text{He} + \ ^{1}_0 \text{n} + Q \] Find the amount of energy released in the reaction.
                      % Given data Given:
                      \( m\left(^{2}_1 \text{H}\right) = 2.014102 \, \text{u} \)
                      \( m\left(^{3}_1 \text{H}\right) = 3.016049 \, \text{u} \)
                      \( m\left(^{4}_2 \text{He}\right) = 4.002603 \, \text{u} \)
                      \( m\left(^{1}_0 \text{n}\right) = 1.008665 \, \text{u} \)
                      \( 1 \, \text{u} = 931 \, \text{MeV}/c^2 \)

                        CBSE CLASS XII Previous Year Papers

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