Question 1

Draw the necessary energy band diagrams to distinguish between conductors, semiconductors and insulators. How does the change in temperature affect the behaviour of these materials ? Explain briefly.

Accepted Answer

Energy bands : In a solid , the energy of electrons lie within certain range. The energy levels of allowed energy are in the form of bands, these bands are separated by regions of forbidden energy called band gaps.Distinguish features : (a) In conductors : valence band and conduction band overlap each other.In semiconductors : Valence band and conduction band are separated by a small energy gap.In insulators : They are separated by a large energy gap.(b) In conductors : Large number of free electrons are available in conduction band.In semiconductor : A very small number of electrons are available for electrical conduction. In insulators : Conduction band is almost empty i.e., no electron is available for conduction.Effect of temperature :(i) In conductors : At high temperature , the collisions of electrons become more frequent with the atoms / molecules at lattice site in the metals as a result the conductivity decreases (or resistivity increases).(ii) In semiconductors : As the temperature of the semiconducting material increases, more electrons hole pairs becomes available in the conduction band and valence band , and hence the conductivity increases or the resistivity decreases. (ii) In insulators : The energy band between conduction band and valence band is very large , so it is unsurpassable for small temperature rise. So , there is no change in their behaviour.

Question 2

Write a short note on N type semiconductor.

Accepted Answer

When a small amount of  \mathbf{Pentavalent}   \mathbf{Impurities} is added to a pure semiconductor providing a large number of free electrons in it, the extrinsic semiconductor thus formed is known as  \mathbf{n-Type}   \mathbf{Semi}   \mathbf{Conductor.}  The conduction in the n-type semiconductor is because of the free electrons denoted by the pentavalent impurity atoms. These electrons are the excess free electrons with regards to the number of free electrons required to fill the covalent bonds in the semiconductors.

Question 3

Pure silicon at room temperature has an electron number density in the conduction band of about  5 	imes 10^{15}m^{-3}  and an equal density of holes in the valence band. Suppose that one of every  10^7  silicon atoms is replaced by a phosphorus atom.  (a)  Which type will thedoped semiconductor be,  n  or  p?   (b)  What charge carrier number density will the phosphorus add?  (c)  What is the ratio of the charge carrier number density (electrons in the conduction band and holesin the valence band) in the doped silicon to that in pure silicon?

Accepted Answer

(a)  The semiconductor is  n-type , since each phosphorus atom has one more valenceelectron than a silicon atom. (b)  The added charge carrier density is n_P=10^{-7}n_{Si}=10^{-7}(5	imes 10^{28}m^{-3})=5	imes 10^{21}m^{-3} (c)  The ratio is (5	imes 10^{21}m^{-3})/\left [ 2(5	imes 10^{15}m^{-3}) ight ]=5	imes 10^5 Here the factor of  2  in the denominator reflects the contribution to the charge carrierdensity from both the electrons in the conduction band and the holes in the valence band.

Question 4

Suppose a n-type wafer is created by doping  Si  crystal having  5	imes {10}^{28}  atoms /{m}^{3}  with  1ppm  concentration of  As . On the surface  200ppm  Boron is added to create  P  region in this wafer. Considering  {n}_{i}=1.5	imes {10}^{16}{m}^{-3} , (i) Calculate the densities of the charge carriers in the n and p regions. (ii) Comment which charge carriers would contribute largely for the reverse saturation. Current when diode is reverse biased.

Accepted Answer

When  As  (pentavalent) is added to  Si  the n-type water is createdSo the number of majority carrier in n-type water; {N}_{e}={N}_{D}  and  {D}_{Si}=\cfrac{1}{{10}^{6}}	imes 5	imes {10}^{28}=5	imes {10}^{22}/{m}^{3} For number of minority carriers  {n}_{h} {n}_{e}.{n}_{h}={n}_{i}^{2} {n}_{h}=\cfrac{{n}_{i}^{2}}{{n}_{e}}=\cfrac{1.5	imes {10}^{16}	imes 1.5	imes {10}^{16}}{5	imes {10}^{22}}({n}_{i}=1.5	imes {10}^{16}/{m}^{3})  (Given) =0.3	imes 1.5	imes {10}^{32-22}=0.45	imes {10}^{+10}  per  {m}^{3} When boron (trivalent) is implanted in  Si  crystal, p-type water is formed with number of holes {n}_{h}=({N}_{D}	imes n  of  Si ) =\cfrac{200}{{10}^{6}}	imes 5	imes {10}^{28}=1000	imes {10}^{28-6} {n}_{h}=1	imes {10}^{25}  per  {m}^{3} Minority carrier in p type water {n}_{e}.{n}_{h}={n}_{i}^{2} {n}_{e}=\cfrac{{n}_{i}^{2}}{{n}_{h}}=\cfrac{1.5	imes {10}^{16}	imes 1.5	imes {10}^{16}}{ {10}^{25}}=2.25	imes {10}^{32-25} =2.25	imes {10}^{7}  electrons per  {m}^{3} When reversed bias is appled on p-n junction then the minority charge carrier moves toward depletion layer ie holes  n_h=(0.45	imes {10}^{10}  per  {m}^{3})  from  n  side and  {n}_{e}=2.25	imes {10}^{7}/{m}^{3}  from p side moves towards junction and make the depletion layer thicker .

Question 5

In the figure shown assuming the diodes to the ideal.

Accepted Answer

In circuit,  A  is at  -10V  and  B  is at  0V . So,  B  is positive than  A . So,  {D}_{2}  is in forward bias and  {D}_{1}  is in reverse bias so no current flows from  A  to  B  or  B  to  A

Question 6

When a p-n junction is reverse biased, the current becomes almost constant at  25\mu A . When it is forward biased at   200meV , a current of  75\mu A  is obtained. Find the magnitude of diffusion current when the diode is (a) unbiased (b) reverse-biased at  200mV  and (c) forward-biased at  200mV .

Accepted Answer

Given,p-n junction is reversed biased. i_1=25\mu A V=200 mA i^2=75\mu A (a) unbiased condition.Diffusion current=Drift current= i_1 	herefore  Diffusion Current =25\mu A (b) Reversed biased condition at  200 mV On the reverse biasing the diffusion current becomes zero.(c) Forward biased condition at  200 mV At,Forward biased, the actual current be  i So, i= Drift Current =  Forward biasing current. i-i_1=i_2 \Rightarrow i-25\mu A=75\mu A i=\left(75+25ight)\mu A=100 \mu A 	herefore  Diffusion current forward biasing becomes  100 \mu A  at  200 mV

Question 7

Explain the working of a half wave diode rectifier.

Accepted Answer

Half Wave Rectifier: A circuit which rectifies half of the a.c. wave is called half wave rectifier.Figure shows the circuit for half wave rectification. The a.c. voltage  \left( { V }_{ S } \right)   to be rectified is obtained across the secondary ends  of the transformer. The  P -end of the diode  D  is connected to  { S }_{ 1 }  of the secondary coil of the transformer.The  N -end of the diode is connected to the other end  { S }_{ 2 }  of the secondary coil of the transformer, through a load resistance  { R }_{ L } . The rectified output voltage  { V }_{ dc }  appears across the load resistance  { R }_{ L } .During the positive half cycle of the input a.c. voltage  { V }_{ S } ,  { S }_{ 1 }  will be positive and the diode is forward biased and hence it conducts.Therefore, current flows through the circuit and there is a voltage drop across  { R }_{ L } . This gives the output voltage as shown in Figure.During the negative half cycle of the input a.c. voltage  \left( { V }_{ S } \right) ,  { S }_{ 1 }  will be negative and the diode  D  is reverse biased. Hence the diode does not conduct. No current flows through the circuit and the voltage drop across  { R }_{ L }  will be zero. Hence no output voltage is obtained. Thus corresponding to an alternating input signal, unidirectional pulsating output is obtained.The ratio of d.c. power output to the a.c. power input is known as rectifier efficiency. The efficiency of half wave rectifier is approximately  40.6 %.

Question 8

Draw the circuit diagram of a full wave rectifier using two p-n junction diodes and explain its working. Show the input and output waveforms.

Accepted Answer

Working:- The AC input voltage across secondary  s_1  and  s_2  changes polarity after each half cycle. Suppose during the first half cycle of input AC signal, the terminal  s_1  is positive relative to centre tap O and  s_2  is negative relative to O. then diode  D_1  is forward bias and diode  D_2  is reverse biased. Therefore diode  D_1  conduct while diode  D_2  does not. In next half cycle,the terminal  s_1  is negative relative to centre tap O and  s_2  is positive relative to O. then diode  D_2  is forward bias and diode  D_1  is reverse biased. Therefore diode  D_2  conduct while diode  D_1  does not. The direction of current in resistance  R_L  is same from A to B for both half cycles. hence for input AC signal the output current is a continuous series of unidirectional pulses.

Question 9

Explain with the help of a circuit diagram, the working of a photo-diode. Write briefly how it is used to detect the optical signals.

Accepted Answer

A photo-diode is fabricated using photosensitive Semiconducting material with a transparent window to allow light to fall on the junction of the diode.Working : (Refer to Fig.1) In diode (any type of diode) , an electrical field 'E' exists across the junction from n-side to p-side , when light with energy   h 
u   greater than energy gap   E_g ( h 
u > E_g)   illuminates the junction, then electron-hole pairs are generated due top absorption of photons , in or near the depletion region of the diode . Due to existing electric field , electrons and holes get separated. The free electrons are collected on n-side and holes are collected on p-side , giving rise to emf. Due to the generated emf, an electric current of   \mu A   order flows the external resistance.  Detection of optical signals :(Refer to Fig.2) It is easier to observe the change in the current with changes in the light intensity if a reverse bias is applied. Thus , photodiode can be used as a photodetector to detect optical signals. The characteristics curves of a photodiodes for two different illuminations   I_1   and   I_2 (I_2 > I_1)   are shown.

Question 10

(i) Describe the working principle of a solar cell. Mention three basic processes involved in the generation of emf.(ii) Why are Si and GaAs preferred materials for solar cell ?

Accepted Answer

Principle : It is based on photovoltaic effect (generation of voltage due to bombardment of light photons). When solar cell is illuminated with light photons of energy   (h 
u)   greater than the energy gap   (E_g)   of the semiconductor , then electron-hole pairs are generated due to absorption of photons. The three basics processes involved are : generation , separation and collection a. generation of electron-hole pairs due to light (with   h 
u > E_g   ) close to the junctionb. separation of electrons and holes due to electric field of the depletion region. Electrons swept to n-side and holes p-side.c. the electrons reaching the n-side are collected by the front contact and holes reaching p-side are collected by the back contact. Thus , p-side becomes positive and n-side becomes negative giving rise to photo voltage. Important criteria for the selection of a material for solar cell fabrications are :i. band gap   (\sim 1.0   to   1.8 \,eV)   ,ii. high optical absorption   (-10^4 \, cm^{-1})  ,iii. electrical conductivity,iv. availability of the raw material , and v. cost Solar radiation has maximum intensity of photons of energy   = 1.5 \,eV   Hence semiconducting materials Si and GaAs with band gap   \approx 1.5\,eV   are , preferred materials for solar cells.

Question 11

Explain the working of a LED.

Accepted Answer

Working of LED:The working of the LED depends on quantum theory. The quantum theory states that when the energy of electrons decreases from the higher level to lower level, it emits energy in the form of photons. The energy of the photons is equal to the gap between the higher and lower level.The LED is connected in the forward-biased, which allows the current to flows in the forward direction. The flow of current is because of the movement of electrons in the opposite direction. The recombination shows that the electrons move from the conduction band to valence band and they emit electromagnetic energy in the form of photons. The energy of photons is equal to the gap between the valance and the conduction band.

Question 12

You are given two circuits as shown in Fig., which consist of NAND gates. Identify the logic operation carried out by the two circuits.

Accepted Answer

In both the given circuits,  A  and  B  are the inputs and  Y  is the output.For first circuit, The output of the left NAND gate will be  \overline{AB} .Hence, the output of the combination of the two NAND gates is given as: Y=\overline{\overline{A.B}.\overline{A.B}}=AB Hence, this circuit functions as an AND gate.For the second circuit,  \bar{A}  is the output of the upper left of the NAND gate and  \bar{B}  is the output of the lower half of the NAND gate, as shown in the following figure.Hence, the output of the combination of the NAND gates will be given as: Y=\overline{\bar{A}+\bar{B}}=A+B Hence, this circuit functions as an OR gate.

Question 13

Write the truth table for circuit given in the figure, consisting of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing. (Hint: A = 0, B = 1 then A and B inputs of second NOR gate will be 0 and hence Y=1. Similarly work out the values of Y for other combinations of A and B. Compare with the truth table of OR, AND, NOT gates and find the correct one).

Accepted Answer

A  and  B  are the inputs of the given circuit. The output of the first NOR gate is  \overline{A+B} . It can be observed that the inputs of the second NOR gate become the output of the first one.Hence, the output of the combination is given as: Y=\overline{\overline{A+B}+\overline{A+B}}=A+B The truth table for this operation is given in the table. AB Y (=A+B) 00 0 011101111

Semiconductor Electronics: Materials, Devices And Simple Circuits