Mechanism of electric conductance in crystals

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Brooke Lawson
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1 Mechanism of electric conductance in crystals 1

2 Electric current in conductors A When the conductor is in electric field, the field accelerates free electrons Electrons moving (drifting) in electric field transfer the electric charge. Moving electric charges create electric current through the conductor 2

3 Electric current in crystalline conductors Electric Field I = Q t Factors determining the electric current in crystalline conductors: 1. How many free electrons the crystal has 2. How fast the charge carriers (i.e. electrons) can move in the electric field 3

4 How many free electrons are there in the conductor? Metalshave ~ atoms per 1 cm 3 Every atom donates 1 free electron: Therefore, every 1 cm 3 of the metal contains ~ free electrons n ~ cm -3 = (1/cm 3 ) n is the electron concentration The concentration shows the number of particles per unit volume 4

5 Conductors, Insulators and Semiconductors Metal: ~ atoms per 1 cm 3 Every atom donates 1 free electron: n ~ cm -3 Insulator: ~ atoms per 1 cm 3; No free electrons: n ~ 0 Semiconductor: ~ atoms per 1 cm 3; Some atoms donate free electrons n ~ cm -3 5

6 Example problem 1 The metal bar dimensions are 2mm x 3 mm x 0.5 mm; Free electron concentration in the metal is 2x10 23 cm -3. How many free electrons the bar contains? 0 of Timed response 6

7 Example problem 2 Thin-film wire is 2 µm thick, 5 mm long and 20 µm wide. Free electron concentration in the metal is 5x10 22 cm -3. How many free electrons the wire contains? 0 of Timed response 7

8 How fast the electrons can move: electron mobility in crystals Equilibrium condition, no electric field (voltage) applied Free electron experiences very frequent collisions with atoms inthe metal. As a result it moves randomly (with the velocity of around 10 5 m/s). On average, the electron does not go anywhere! Average electric current is equal to zero (There is a flicker charge transfer, or the NOISE current though) 8

9 Electron in motion in conductors Electric field applied: There is an electric force, F = e Eexerting on any free electron. Electron still experiences very frequent random collisions. However, after each collision the electron s velocity has a component toward the positive electrode (against the field direction) On average, the electron drifts from negative electrode toward positive. There is a current flowing through the metal. 9

10 Electron mobility in crystals Let us ignore the random collisions and only monitor the drift in the electric field Ignoring the collisions, which are completely random, we can say that average electron velocity (drift velocity) is proportional to the electric field applied: v~ E v= µe µis called the electron mobility: µ = v/e[(m/s)/(v/m) = m 2 /(V. s)] 10

11 Electric current in conductors A Electric current Electron drift velocity increases with electric field: v= µe Higher electron concentration Electric field 11

Mechanism of electric conductance E L L = v. t A Gnome is counting the electrons A = wire cross-section area n= free electron concentration, i.e. the number of electrons per unit volume If the conductor volume is A L, the total number of electrons, N = n A L In the ttime interval, how many electrons will cross the dashed area A?

12 Mechanism of electric conductance E L L = v. t A Gnome is counting the electrons A = wire cross-section area n= free electron concentration, i.e. the number of electrons per unit volume If the conductor volume is A L, the total number of electrons, N = n A L In the ttime interval, how many electrons will cross the dashed area A? Ν= n (v t) A How much charge will pass through? Q = e Ν= e n (v t) A; where e = abs (electron charge) The electric current Q I = = env A t 12

13 Mechanism of electric conductance v. E A A = wire cross-section area Q The electric current I = = env A t The electron drift velocity, v= µe, from this, I = ( en µ A)E Electric current is proportional to the electric field (external parameter) Electric current is proportional to n, µ and A (conductor parameters) 13

14 Mechanism of electric conductance E v. A 1. The external power source (the battery) creates a potential difference along the conductor: V; 2. The potential difference (the voltage) creates an electric field across the conductor E = V/L; 3. Electric field accelerates free electrons and produces the current I = ( en µ A)E Using E = V/L: I A A = en µ V = en µ V L L 14

15 Mechanism of electric conductance I A A = en µ V = en µ V L L This product contains only the material parameters, does not depend on applied voltage or on the geometrical shape of the conductor. It is called conductivity: σ =en µ Material properties are also often characterized by resistivity: 1 1 = = en ρ σ µ 15

16 Using conductivity or resistivity the current voltage relationship is: The term I A = σ V L 1 A ρ L or I = ρ 1 A L V does not depend on the applied voltage but only on the material properties and the sample dimensions. R = ρ L A I is called resistance. Using the resistance, the I-V relationship is: 1 = V or V = I R R These two relationships are called the Ohm s law 16

17 The Ohm s Law George Ohm has established experimentally in 1827 the following law 1 V is the voltage across a conductor I = V I is the current thought a conductor R 1/R is a proportionality factor; R is called resistance - + I V Electric current Less resistance Voltage 17

18 Ohm s Law and electron charge transfer rate comparison v= µe 1 I = V R Voltage 18 Less resistance Electric current Higher electron concentration Electric field Electric current

19 Experimental observations of Ohm s Law I = 1 R V Electric current Voltage Conductor length decreases Current increases, i.e. R decreases 19

20 Experimental observations of Ohm s Law I = 1 R V Electric current Voltage Conductor area increases Current increases, i.e. R decreases 20

21 Ohm s law, resistance and resistivity - summary I = V R V = I R R = L ρ A Ohm s law in the form of I(V): the current through the conductor = applied voltage divided by the resistance Ohm s law in the form of V(I): the voltage needed to maintain the current I = current times resistance. L is the length of the conductor in the direction of the current flow; A is the cross-section area; L R = ρ A ρ describes the material ability to conduct the current. The higher ρ is, the higher R and the lower the current ρ = 1 en µ 21

22 The units for resistance and resistivity V V = I R R = I Resistance R is measured in Ohms (Ohm, Ω) 1 Ohm is the resistance of the sample that passes the current of1a when the voltage of 1 V is applied across it. L R = ρ A 2 A m ρ = R Ohm = Ohm m L m Resistivity is measured in Ohm -meters (Ohm m) 22

23 Ohms law using conductance and conductivity Conductance G = 1/ R Using the conductance, the Ohm s law can be written as I = G V Also, from G = 1 / R and L R = ρ A, G = 1/ R = 1 A ρ L 1 σ = = en µ σ is the conductivity of the material ρ (does not depend on the conductor shape) Using conductivity, the conductance of the sample is given by G A = σ L 23

24 Ohm s Law using conductance - summary The expression that relates the electric current to the applied voltage is called the Ohm s Law (established experimentally in 1827) I = G V, A G = σ L σ = en µ where is the conductance of is the conductivity of the sample (wire), the material A is the conductor cross-section area; L is the conductor length along the current direction n is the concentration of free electrons µ is the electron mobility 24

25 The units for conductance and conductivity I I = G V G = V Conductance is measured in Siemens (S) 1 S is the conductance of the sample that passes the current of 1A when the voltage of 1 V is applied across it. G A = σ L L m σ = G S = S / m 2 A m Conductivity is measured in Siemens per meter (S/m) 25

26 Ohm s Law in the differential ( microscopic ) form: I = ( en µ A)E where A is the cross-section area, E is the electric field in the conductor I or = σ A E where σ= enµis the conductivity of the material I or V = σ A L where V is the voltage across the conductor and L is the conductor length in the direction of current flow 26

27 Problem 3. What is the current in the circuit below? 1. 3 A 2. 4 A A A 60 0% 0% 0% 0%

28 Resistivity of different materials Material Wires are used to connect different components in the network; Wires have very low resistance Resistors are used to dissipate the power and to change the voltage (potential). Aluminum[Al] 27 Aluminum Alloy 50 Brass Carbon[C] Copper[Cu] 17 Copper Alloy Gold[Au] 24 Iron[Fe] 97 Electric Resistivity ( 10-9 Ohm m) 28

29 Problem 4. Find the resistance of a wiremade of copper[cu] (ρ= 17*10-9 Ohm-m). The wire is 1 m long and is 1 mm in diameter Ohm *10-2 Ohm *10-2 Ohm S 180 0% 0% 0% 0%

30 Solution Find the resistance of a wiremade of copper[cu] (ρ= 17*10-9 Ohm-m). The wire is 1 m long and is 1 mm in diameter. R = ρ*l/a L = 1 m; D = 1mm = 10-3 m; The area, A = π*d 2 /4 = 3.14*(10-3 ) 2 /4 = 7.85*10-7 m 2 The resistance R = 17*10-9 Ohm*m*1m/ 7.85*10-7 m 2 = 2.17*10-2 Ohm = 21.7 mohm Find the voltage across this wire, in the current through it is 1 A V = I R = 1 A 21.7 m Ohm = 21.7 mv Conclusion: wires do not change the potentials in the circuit; Voltage across the wire can be taken as 0 in most problems 30

31 Problem 5. Find the resistance of a carbonresistor (ρ = Ohm*m), that is 1 cm long and 0.1 mm in diameter Ohm Ohm Ohm Ohm 180 0% 0% 0% 0%

32 Solution Find the resistance of a carbonresistor, which is 1 cm long and 0.1 mm in diameter R = ρ*l/a; ρ = * 10-9 Ohm*m = Ohm*m ; L = 1 cm = 10-2 m; A = 7.85*10-9 m 2 R = 1.4*10-5 Ohm*m*10-2 m/ 7.85*10-9 m 2 = 17.8 Ohm Compare: Cu wire, R 1 = 21.7 mohm 0.02 Ohm; Carbon-resistor, R 2 = 17.8 Ohm 18 Ohm; Given electric circuit with the current of 1 A, the voltage drop across the Cu-wire, V 1 = I*R 1 = 0.02 V; The voltage drop across the Carbon-resistor, V 2 = I*R 2 = 18 V; Resistors significantly change the potentials in the circuit. 32

33 Problem 6. Two wires are made of the same material. The lengths and the diameters are as follows: Wire 1 Wire 2 Length 10 cm 20 cm Diameter 1 mm 2 mm Which wire, #1 or #2, has higher resistance? 1. Wire #1 2. Wire #2 3. Equal resistances 0% 0% 0%

34 Problem 7. The bar made of carbon (ρ = Ohm*m) The current flows through the bar in the direction shown by a blue arrow. L = 10 cm; W=5 mm; d = 2 mm. What is the resistance of the bar? d I L W e-5 Ohm Ohm e-4 Ohm 0% 0% 0%

35 Problem 8. The bar made of carbon (ρ = Ohm*m) The current flows through the bar in the direction shown by a blue arrow. L = 10 cm; W=5 mm; d = 2 mm. What is the resistance of the bar? I d L W e-5 Ohm Ohm e-4 Ohm 0% 0% 0%

36 Problem 9. What voltage is needed to generate the current of 2A in a resistor of 100 Ohm? V 2. 2 V V V 60 0% 0% 0% 0%

37 Problem 10. What voltage is needed to generate the current of 2A in a resistor of 0.01 Ohm? V V V V 60 0% 0% 0% 0%

and in a simple circuit Part 2

and in a simple circuit Part 2 Current, Resistance, and Voltage in a simple circuit Part 2 Electric Current Whenever electric charges of like signs move, an electric current is said to exist. Look at the charges flowing perpendicularly