PN Junction Diodes: Two Marks Important Questions and Answers

Electron Devices

Chapter 2: PN Junction Diodes

Two Marks Questions with Answers

1. What is peak inverse voltage?

In reverse biased, opposite polarity voltage across diode. The maximum reverse appears voltage which diode can withstand without breakdown is called peak inverse voltage.

2. Draw the energy band structure of Silicon and Germanium at room temperature indicating the energy gap (E_G) value in eV.

The energy band structure is shown in Fig. 2.16.1.

3. What is meant by zener breakdown?

When a p‒n junction is heavily doped the depletion region is very narrow. So under reverse bias conditions, the electric field across the depletion layer is very intense. Electric field is voltage per distance and due to narrow depletion region and high reverse voltage, it is intense. Such an intense field is enough to pull the electrons out of the valence bands of the stable atoms. Such a creation of free electrons is called zener effect. These minority carriers constitute very large current to cause the breakdown and the mechanism is called zener breakdown.

4.  Compare the Silicon and Germanium diodes with respect to cut‒in voltage and reverse saturation current.

Barrier potential:

Silicon (Si): 0.7 V

Germanium (Ge): 0.3 V

Reverse saturation Current:

Silicon (Si): Few nA

Germanium (Ge): Few μA

Reverse breakdown voltage:

Silicon (Si): Higher than Ge

Germanium (Ge): Lower than Si

5. Define barrier potential at the junction.

Due to immobile positive charges on n side and negative charges on p side, there exists an electric field across the unbiased p‒n junction. This creates potential difference across the junction which is called a barrier potential. Thus the voltage existing across the depletion region or barrier is called a barrier potential.

6. Give the diode current equation.

The diode current equation is,

 I = I₀[e^V/ηVT‒1] A

where

 I₀ = Reverse saturation current in amperes

 V = Applied voltage,

 V_T = Voltage equivalent of temperature in volts

 η = 1 for germanium diode and 2 for silicon diode

7. Define Knee voltage or a cut‒in voltage of a diode.

When diode is forward biased, some voltage is necessary to overcome barrier potential, to make diode conduct. This is called its cut‒in voltage. The minimum voltage at which the diode starts conducting and current starts increasing exponentially is called cut‒in voltage, offset voltage, break‒point voltage, threshold voltage or knee voltage. It is denoted as Vγ and its, value is 0.2 V for Germanium while 0.6 V for Silicon. Below this voltage, the diode current is very very small and practically considered to be zero.

8. What is meant by depletion region?

In a P‒N junction, the diffusion of holes and electrons start initially. Near the junction, holes recombine in n‒region to form immobile positive ions. Similarly electrons recombine in p‒region to form immobile negative ions. With sufficient accumulation of such immobile ions on both sides, the diffusion stops. So near junction, there exists a region in which immobile positive and negative charges reside while mobile charge carriers in this region get completely depleted. This region is called depletion region.

9. Define the transition capacitance of a diode.

A capacitance existing at the P‒N junction when the diode is reverse biased where the two regions act as the plates while the depletion region acts as dielectric is called a transition capacitance of a diode.

10. Draw the VI characteristics of a PN junction diode.

11. Why a series resistor is necessary when a diode is forward biased ?

For limiting the forward current which increases exponentially with voltage.

12. What is meant by breakdown voltage in PN junction diode ?

In reverse biased condition, as reverse voltage increases the reverse current is almost constant upto certain value of reverse voltage. At some reverse voltage, the diode breaks down and large reverse current flows through it. The voltage at which the diode breakdown occurs is called reverse breakdown voltage.

13. State the expression for the dynamic resistance of a diode.

The dynamic resistance of a diode is given by,

 r = ηV_T / I₀e^V/ηVT

where

I₀ = Reverse saturation current in amperes,

V = Applied voltage,

V_T = Voltage equivalent of temperature in volts,

 η = 1 for Ge diode and 2 for Si diode

14. What is p‒n junction diode ?

The two materials namely p‒type and n‒type are chemically combined with a special fabrication technique to form p‒n junction. There are two electrodes each from p‒type and n‒type materials and due to due to these two electrodes (di+electrode), the device is called a diode. It conducts only in one direction.

15. A silicon diode has a saturation current of 7.5 μA at room temperature 300 °K. Calculate the saturation current at 400 °K.

 T₁ = 300 °K, T₂ = 400 °K, I₀₁ = 7.5 μA, ΔT = T₂‒ T₁ = 100

 I₀₂ = (2^ΔT/10) (I₀₁)

= 2^(100/10) × 7.5

= 7680 μA

= 7.68 mA

16. A Ge diode has a saturation current of 10 μА at 300 °K. Find the saturation current at 400 °K.

 T₁= 300 °K,

T₂ = 400 °K,

I₀₁ = 10 μA

i.e. ΔT = T₂‒T₁=100

 I₀₂ = (2^ΔT/10) I₀₁

= (2^100/10) × 10×10^‒6

= 10.24 mA

17. What is is avalanche breakdown ?

When reverse voltage is increased, the velocity of minority charge carriers increases. In such highly accelerated charge carrier collides against electron involved in covalent bond, then it breaks the covalent bond and creats new electron‒hole pair. These secondary particles also accelerate and involve in the further collisions to produce more electron‒hole pairs. This is called carrier multiplication. Due to such large number of charge carriers breakdown of junction takes place. Such a breakdown is called avalanche breakdown.

18. Define storage time.

When the diode is switched from forward biased to reverse biased, the minority charge carriers remain stored and decrease slowly to zero. This time for which minority charge carriers remain stored is called storage time.

19. Define transition time.

The time required by the diode current to reduce to its reverse saturation value, after minority carrier concentration reduces to zero is called transition time.

20. Define reverse recovery time.

The total time required by a diode to switch from ON to OFF state which is addition of storage time and transition time is called reverse recovery time of a diode.

T_rr = t_s + t_t

21. What is the principle operation of a PN Junction diode in reverse bias condition ?

• When the p‒n junction is reverse biased the negative terminal attracts the holes in the p‒region, away from the junction.

• The positive terminal attracts the free electrons in the n‒region away from the junction.

• No charge carrier is able to cross the junction.

• As electrons and holes both move away from the junction, the depletion region widens.

• This creates more positive ions and hence more positive charge in the n‒region and more negative ions and hence more negative charge in the n‒region. This is because the applied voltage helps the barrier potential. This is shown in Fig. 2.6.2.

22. Sketch the forward bias characteristics of the pn Junction dlode.

23. What are the applications of PN diode ?

The various applications of p‒n diode are, different types of rectifiers, various protection circuits, multivibrators, in amplitude limiting circuits for oscillators, clipper circuits, clamper circuits, in various electronic and op‒amp based circuits, comparators. schemitt trigger circuits etc.

24. Find the voltage at which the reverse current in a germanium PN junction diode attains a value of 90% of its saturation value at room temperature.

 η = 1 for Ge,

V_T = 26 mV at room temperature

I = 0.9 I₀ but reverse hence take negative

I = I₀ [e^V/nVT ‒1]

‒0.9 I₀ = I₀[e^V/26×10‒3 ‒1]

 ‒0.91 + 1 = e ^V/26×10‒3

i.e. In 0.1 = V / 26×10^‒3

V = 2.3026×26×10^‒3 = ‒59.867 mV

The negative sign indicates the reverse voltage applied to the diode.

25. State the relationship between diffusion capacitance and diode current in a PN diode.

The diffusion capacitance is given by,

C_D = τI / ηV_T

where I is the diode current.

26. Determine the total forward bias current density in a PN junction diode under an applied forward bias voltage of 0.65 V at 300 K.

Assume J_S = 4.155 × 10^‒11 A/cm².

V= 0.65 V, T = 300 °K,

J_S = 4.155×10^‒11 A/cm²

The diode current equation is,

 I = I_S[e^V/ηVт −1] ...Divided by area

V_T = 26 mV at 300 °K, η = 1

J = 4.155×10^‒11 [e^{0.65/26x10‒3} ‒1]

J = 3 A/cm²

 ……. Forward bias current density

27. Define diffusion capacitance.

In forward bias condition, the width of depletion region decreases and more holes on p‒side get diffused in n‒side while electrons from n‒side move into p‒side. As the applied voltage increases, the concentration of the injected charged particles increases. This rate of change of the injected charge with the applied voltage is defined as diffusion capacitance.