SEMICONDUCTORS. Conductivity lies between conductors and insulators. The flow of charge in a metal results from the

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1 SEMICONDUCTORS Conductivity lies between conductors and insulators The flow of charge in a metal results from the movement of electrons Electros are negatively charged particles (q=1.60x10-19 C ) The outermost orbit electrons of an atom in the metal, which is also called as valence electrons, are free to move and constitute the current

2 2D Depiction of ions and valence electrons in the metal Free Electrons Ion

3 Due to the thermal energy the free electrons are in continuous motion These electrons will collide with the ions and direction of the motion changes If no voltage is applied to a metal, then the random motion of the electrons results in zero average current

4 Application of a voltage to a Metal will result in an electric field E measured in volts per meter (V/m). For this applied electric field electrons will move at a speed denoted by u, which is also called as drift speed This drift speed is directly proportional to the applied electric field u=µe, where proportionality constant is known as mobility of the electrons. (m 2 /Vs)

5 Let us consider a metallic conductor of length L and has a cross section area of A square meter A N Free electrons L Let N be the number of free electrons in this section of the conductor and moves to the left with the drift speed of u. If an electron takes T seconds to travel L meters, then drift speed u=l/t, we can write T=L/u.

6 Flow of N electrons to left in T seconds means a current i going to the right Where i Nq Nq T L / u L Nqu L The current density i J A Thus we have J Nqu AL LA is the volume of the conductor section and number of free electrons is N, density of the free electron is n=n/la, which is called as free electron concentration. (m -3 )

7 Then current density can be written as J nqu The product nq is known as charge density (C/m 3 ) Now substitute for the u in J we get J=nqµE=σE Where σ=nqµ is known as the conductivity of the 1 metal (Ωm) -1 Reciprocal of this is known as resistivity (ρ=1/ σ) ) The current i=ja= σea= EAL A L L EL

8 The product EL with the unit (V/m)(m)=V is v which is the voltage across the length L conductor. Then by Ohms law A i v L 1 R v Where the resistance of the conductor is R=L/ σa= ρl/a Example1:Apieceofaluminumwirehasacross sectional area of 10 6 m 2. The aluminum has a free electron concentration of 1.81x10 29 m 33 and the mobility of the free electron is 10 3 m 2 /Vs. a) Find conductivity of the aluminum wire b) Find the resistivity of the wire c) Find the resistance of a 1 meter length of wire

9 Solution: Given: A= 10-6 m 2, n= 1.81x10 29 m -3 and µ= 10-3 m 2 /Vs a) conductivity of the aluminum wire (σ=nqµ ) σ= (1.81x10 29 )(1.6x10-19 )(10-3 ) = 2.9x10 7 mho/m b) Find the resistivity of the wire (ρ=1/ σ) ρ=1/2.9x10 7 =3.345x10-8 Ωm c) resistance of a 1 meter length of wire R= ρl/a R=(3.345x10-8 )(1)/10-6 = 34.5 mω

10 Intrinsic (Pure) Semiconductor) Semiconductors such as and Ge have 4 electrons (tetravalent) in the outermost shell In crystal structure of these materials, atoms are arranged in tetrahedron structure with one atom at each vertex Each atom contributes 4 valence electrons to the crystal; Each atom shares one electron each from its 4 neighbors, thus forming covalent bond Because of covalent bonding, electrons are tightly bound to crystal not available for conduction

11 Bond structure of and Ge at 0 K Covalent bond Valence electrons

12 At room temperature, due to thermal energy one electron will be detached and makes an vacancy in that position which is called as hole. Covalent bond Hole Free electron Valence electrons

13 When one vacancy (hole) is created, that space will be filled by another electrons in terms this process will continue and there will be holes movement in the opposite direction that of the electron. The current due to the movement of holes is called as Holes current. There are two currents exist, one is due to the holes and another is due to the electrons, hence it is called as bipolar device

14 In pure semiconductor the number of holes concentration are equal to the electrons concentration i.e n=p=n i Now the current density J is given by J=nqµE=σE For the semiconductor Let µ n =mobility of the free electron µ p =mobility of the holes Under the influence of the electric field. The consequence of this is electron and hole currents in the same direction.

15 The current density J is given by J=(nµ n +pµ p )qe=σe σ=(nµ n +pµ p )q is conductivity of the semiconductor At 300 K, the intrinsic concentration of germanium (GE) is 2.5x10 19 m -3 Furthermore, the free-electron mobility is 0.38 m 2 /Vs and the hole mobility is 0.18 m 2 /Vs. Find the conductivity and resistivity of the pure germanium. Given n i =2.5x10 19, µ n = 0.38 µ p =0.18 Answers : Conductivity= 2.25 mho/m Resistivity = Ω m

16 Doped Semiconductor Conductivity of the silicon or Germanium increases as temperature increases Conductivity can also be increased by adding small amount of impurity atom to the intrinsic semiconductor. Addition of small percentage of foreign atoms into p g g the crystal lattice of silicon or germanium in order to change its electrical properties is called Doping

17 Atoms used for doping are called dopants Two types of dopants Donors and Acceptors Donor Pentavalent (5 electrons in outermost shell) Examples: Phosphorus (P), Arsenic (As), Antimony (Sb), and Bismuth (Bi) Donates one electron to the crystal lattice Onefreeelectronper donor atom Concentration of free electrons increases Concentration of holes decreases

18 N-type semiconductor Free electron P Electrons are majority charge carriers Holes are minority charge carriers

19 Acceptor Trivalent (3 electrons in outermost shell) Examples: Boron (B), Aluminum (Al), Gallium (Ga) and Indium (In) Accepts one electron from the crystal structure One hole per acceptor atom Concentration of holes increases Concentration of free electrons decreases Resulting semiconductor is called P-type semiconductor, because holes are in majority than free electrons (thermally generated)

20 P-type semiconductor B Hole Holes are majority charge carriers Elecrons are minority charge carriers

21 In intrinsic (pure) semiconductor, concentration of free electrons is equal to concentration of holes (n = p =n i i) ) In extrinsic N-type semiconductor, n>>p In extrinsic P-type semiconductor, p >> n Under thermal equilibrium, product of negative and positive charge concentrations is a constant, equal to square of intrinsic concentration called law of Mass Action np=n i 2

22 Suppose that a semiconductor o is doped with both donor and acceptor impurities Let donor atom concentration = N D Let acceptor atom concentration = N A After donating one free electron to the crystal structure, donor atom now has deficit of one negative charge (i.e., net positive) milarly, after accepting one electron from crystal structure, acceptor atom now has one extra electron (i.e., net negative) Total negative charge concentration = n+n A Total positive charge concentration = p + N D Under equilibrium condition, n+n A =p+n D

23 In N-type semiconductor, N A = 0. Hence n=p+n D But n >> p. Hence n N D Now, p=n i2 /n = n i2 /N D milarly in P-type semiconductor, N D = 0. Hence p=n+n A But p >> n. Hence p N A Now, n=n i2 /p = n i2 /N A Note: If N D = N A then, semiconductor behaves like intrinsic If N D >N A then semiconductor is N-type If N A >N D then semiconductor is P-type

24 Diffusion current Results due to flow of charge carriers from the region of higher concentration to the region of lower concentration Suppose that holeconcentration varies with distance x, then concentration gradient is dp/dx If dp/dx is negative, then it results in a current in positive x direction

25 PN-Junction P-N Junction diode is a 2-terminal, 2-layer, singlejunction semiconductor device made out of a single block of silicon or germanium, with one side doped with acceptor (p-type) impurity and the other side with donor (n-type) impurity Very important device with numerous applications like Switch, Rectifier, Regulator, Voltage multiplier, Clipping, Clamping, etc.

26 Anode P N Cathode The two terminals are called Anode and Cathode At the instant the two materials are joined, electrons and holes near the junction cross over and combine with each other Holes cross from P-side to N-side Free electrons cross from N-side to P-side At P-side of junction, negative ions are formed At N-side of junction, positive ions are formed

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28 P-N Junction Diode P-type N-type Holes Depletion region Free electrons

29 Depletion region is the region having no free carriers Further movement of electrons and holes across the junction stops due to formation of depletion region Depletion region acts as barrier opposing further diffusion of charge carriers. So diffusion stops within no time Current through the diode under no-bias condition is zero Bias Application of external voltage across the two terminals of the device

30 Reverse bias Positive of battery connected to n-type material (cathode) Negative of battery connected to p-type material (anode) P V D N I 0

31 Reverse bias Free electrons in n-region are drawn towards positive of battery, Holes in p-region are drawn towards negative of battery Depletion region widens, barrier increases for the flow of majority carriers Majority charge carrier flow reduces to zero Minority charge carriers generated thermally can cross the junction results in a current called reverse saturation current I s I s is in micro or nano amperes or less I s does not increase significantly with increase in the reverse bias voltage

32 Forward bias Positive of battery connected to p-type (anode) Negative of battery connected to n-type (cathode) P N V D I D

33 Forward bias Electrons in n-type are forced to recombine with positive ions near the boundary, similarly holes in p- type are forced to recombine with negative ions Depletion region width reduces An electron in n-region sees a reduced barrier at the junction and strong attraction for positive potential As forward bias is increased, depletion region narrows down and finally disappears leads to exponential rise in current Forward current is measured in milli amperes

34 Diode characteristics I (ma) Diode symbol P N Reverse saturation current (magnified) V γ 1 V (volts) Breakdown (μa) Cut-in or knee Vγ is ~ 03forGe 0.3 voltage 0.6 ~ 0.7 for

35 licon vs. Germanium ma G e V μa

36 I D Diode current equation I s ( V V D T e 1) I s e V V V D T I s I D is dioded current I s is reverse saturation current V D is voltage across diode V T is thermal voltage = T / η is a constant = 1 for Ge and 2 for

37 Diode current equation For positive values of V D (forward bias), the first term grows quickly and overpowers the second term. So, I I D V D V T se For large negative values of V D (reverse bias), the first term drops much below the second term. So, I D I s

38 Expression for the voltage across the diode V D V D i V ln 1 I T s volts Example : A licon diode has reverse saturation current of 1nA at 300 K. Find I D when V D is a) 0.7V, b) 0.1V c) 0V d) -0.1 V, e) -0.7V Answers: a) ma, b) 5.90nA, c) 0A d) na e) -1.0nA

39 Example : A licon diode has a saturation current of 5nA at 300 K a)find the current for the case that the forward bias voltage is 0.7V b)find the forward bias voltage that results in a current of 15 ma c) Find the value of Resistance R in the following circuit, when the is 15 ma Ans: a) 3.71 ma, b) V c) R=3.49 Ω ---V V D ---

40 DIODE BEHAVIOUR The reverse saturation current increases as the temperature is increased It approximately doubles for every 10 o C rise in temperature The reverse saturation current at temperature T b>t a is given by I s ( T b ) 2 ( T T b a )/10 I s ( T a )

41 i D T b >T a i 2 i 1 0 V γ v 1 v 2 v D Temperature dependence of diode i-v characteristics

42 Example: Given that the current through a IN4153 silicon diode is 10 ma, find the voltage across the diode when the temperature is a) 290 K b) 310 K c) 320 K Reverse saturation current at 300 k is 10 na Ans: a) V b) 0.70 V c) V

43 i D Diode Circuits ---V D --- i D V 1 ( V V D T I I e 11 ) R D o ( Applying KVL at above circuit, we get, V 1 =V D +Ri D I Q Q 1 V 1 id VD R R 0 V V V γ Q 1 v D

44 Assume that V 1 and R is known Here one of the equation is non-linear, so to find the unknowns, we cannot apply the simultaneous algebraic equation method. We can do some qualitative analysis by looking at the circuit Suppose the V 1 is positive, the current flows thoroughh the resistor it to alower potential ti, both i D and V D are positive (Forward bias)

45 We can get enough points to sketch on i D -V D, plot i D 1 V R D V1 R This is the equation of a straight line with slope -1/R and vertical axis intercept V 1 /R This straight line is called as Load line and it is drawn on the i-v characteristics curve of the Diode Any point on the diode d current equation curve satisfies I D I o ( V V D T e 1) Any point on the load line satisfies i D 1 V R D V1 R

46 The intersection of both the curve will satisfy both the equations simultaneously and that point is called as quiescent point (Operating point) Thus solution to the equations I Q and V Q is obtained graphically, instead of analytically

47 Example : Suppose that for the diode circuit shown V 1 =9V, R= 2KΩ, and the silicon diode has the saturation current of 5nA at 300 K. Determine I D and V D ( Begin by using V D =0.5 V)

48 Ideal Diode I D + ---V D I D > 0 A + V D =0V - FORWARD BIAS I D =0 A + V D 0 V - REVERSE BIAS

49 I D Reverse erse Bias (V D 0V) Forward Bias (I D >0A) 0 V D

50 Half wave rectifier + v - + v=0v - i=0 + v -

51 Output of a Half wave rectifier V s 0 π 2π ωt v o V s 0 π 2π ωt

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