BIPOLAR JUNCTION TRANSISTOR(BJT) Introduction-1.pptx

Bipolar Junction Transistor (BJT) Introduction

1. Introduction to TRANSISTOR BIPOLAR JUNCTION TRANSISTOR (BJT)

2. TRANSISTOR Background and Introduction Beside diodes, the most popular semiconductor devices is transistors. Eg : Bipolar Junction Transistor (BJT) Transistors are more complex and can be used in many ways Most important feature: can amplify signals and as switch to Switches the flow of current between two terminal Amplification can make weak signal strong (make sounds louder and signal levels greater), in general, provide function called Gain.

3. BIPOLAR JUNCTION TRANSISTOR (BJT) STRUCTURE The BJT is constructed with three doped semiconductor regions separated by two pn Junctions. as shown in the Figure (a). The three regions are called emitter, base, and collector. Physical representations of the two types of BJTs are shown in Figure b) and (c). One type consists of two n regions separated by a p region ( npn ), and the other type consists of two p regions separated by an n region ( pnp ). The term bipolar refers to the use of both holes and electrons as current carriers in the transistor structure.

4. Transistor Material The transistor is an arrangement of semiconductor materials that share common physical boundaries. Materials most commonly used are Silicon gallium-arsenide and germanium Impurities have been introduced by a process called “doping”

5. Types Of BJT Two basic types of bipolar junction transistor construction, PNP NPN, Transistors have three terminals : Emitter ( E ) Base ( B ) Collector ( C )

6. Emitter Base Collector Memory aid: NPN means N ot P ointing i N . E B C NPN Schematic Symbol

7. Collector Base Emitter E B C PNP Schematic Symbol

8. Collector Emitter Base

9. EMITTER It is highly doped To inject a large number of charge carriers Main function is to supply the majority carriers to the base. It is always forward biased with respect to base

10. BASE It is a middle section of a transistor It is lightly doped So that most of the charge carriers pass to the collector. It controls flow of charges It forms two PN junctions with Emitter and Collector

11. COLLECTOR IT IS situated opposite to the emitter It is always reversed biased So that it can collect the majority carriers Size of collector is larger than emitter Its doping level is in the middle of base and emitter

12. P-type N-type N-type Collector (c) Base (B) Emitter (E )

13. P-type N-type N-type Collector (c) Base (B) Emitter (E )

14. N-type P-type P-type Collector (c) Base (B) Emiter (E )

15. N P N The collector is lightly doped. C The base is thin and is lightly doped. B The emitter is heavily doped. E NPN Transistor Structure

16. The C-B junction is reverse biased. N P N NPN Transistor Bias C B E No current flows.

17. The B-E junction is forward biased. N P N NPN Transistor Bias C B E Current flows.

18. When both junctions are biased.... N P N NPN Transistor Bias C B E Current flows everywhere. Note that I B is smaller than I E or I C . I C I B I E

19. N P N C B E Although I B is smaller it controls I E and I C . I C I B I E Note: when the switch opens, all currents go to zero. Gain is something small controlling something large (I B is small).

20. N P C B E I C = 99 mA I B = 1 mA I E = 100 mA b = I C I B The current gain from base to collector is called b. 99 mA 1 mA = 99

21. N P C B E I C = 99 mA I B = 1 mA I E = 100 mA I E = I B + I C 99 mA = 1 mA + = 100 mA Kirchhoff’s current law:

22. C B E I C = 99 mA I B = 1 mA I E = 100 mA In a PNP transistor, holes flow from emitter to collector. Notice the PNP bias voltages.

23. P P P N N N I B ( µA) I B ( µA) I c (m A ) I E ( m A ) I E ( m A ) I C ( mA ) Arrow shows the current flows Transistor Currents

24. Transistor Currents Emitter current (I E ) is the sum of the collector current (I C ) and the base current (I B ) . Kirchhoff’s current law;

25. Vce Vcb Vbe Vec Veb Vbc

26. Base Current (I B ) I B is very small compared to I C ; The ratio of I C to I B is the dc current gain of a transistor, called beta ( β ) The level of beta typically ranges from about 50 to over 400

27. Collector Current (I C ) The ratio of I C to I E is called alpha ( α ), values typically range from 0.95 to 0.99.

28. Forward Common emitter current transfer ratio 

29. Current Relationships Relations between I C and I E : α = I C I E Value of α usually 0.95 to 0.9999 , α ≈ 1 Relations between I C and I B : β = I C @ I C = βI B I B Value of β usually in range of 50  400 The equation, I E =I C + I B can also written in β I C = βI B I E = βI B + I B => I E = (β + 1)I B The current gain factor , α and β is: α = β @ β = α . β + 1 1- α

30. Transistor configuration Transistor configuration –is a connection of transistor to get variety operation. 3 types of configuration: Common Collector . Common Base . Common Emitter

31. Common-Collector Configuration The input signal is applied to the base terminal and the output is taken from the emitter terminal. Collector terminal is common to the input and output of the circuit Input – BC Output – EC Input = Output

32. Common-Base Configuration Base terminal is a common point for input and output. Input – EB Output – CB Not applicable as an amplifier because the relation between input current gain (I E ) and output current gain (I C ) is approximately 1

33. Common-Emitter Configuration Emitter terminal is common for input and output circuit Input – BE Output – CE Mostly applied in practical amplifier circuits, since it provides good voltage, current and power gain

34. Principle of operation of the two transistor types PNP and NPN Biasing polarity of the power supply for each type

35. Current & Voltage Analysis Consider the figure below . Three dc currents and three dc voltages can be identified I B : dc base current I E : dc emitter current I C : dc collector current V BE : dc voltage across base-emitter junction V CB : dc voltage across collector-base junction V CE : dc voltage from collector to emitter Transistor bias circuit.

36. Current & Voltage Analysis When the BE junction is forward-biased, it is like a forward-biased diode. Thus; (Si = 0.7, Ge = 0.3) From KVL, the voltage across R B is By Ohm’s law; Solving for I B

37. Current & Voltage Analysis The voltage at the collector is; The voltage drop across R C is V CE can be rewritten as The voltage across the reverse-biased CB junction is …( Eq. 3.8) …( Eq. 3.7)

38. BJT common emitter characteristics

39. Example 4-3 page (174)

40. I-V Characteristic for CE configuration : Output characteristic Output characteristic: output current (I C ) against output voltage (V CE ) for several input current (I B ) 3 operating regions: Saturation region Cut-off region Active region

41. Common-Emitter Kristin Ackerson, Virginia Tech EE Spring 2002 Circuit Diagram + _ V CC I C V CE I B Collector-Current Curves V CE I C Active Region I B Saturation Region Cutoff Region I B = 0 Region of Operation Description Active Small base current controls a large collector current Saturation V CE(sat) ~ 0.2V, V CE increases with I C Cutoff Achieved by reducing I B to 0, Ideally, I C will also equal 0.

42. 46 Hydraulic analogy of a transistor Analogy of technical direction of current in an NPN transistor – to a water flow in a pipe system

43. Recall: NPN and PNP Bias Fundamental operation of pnp transistor and npn transistor is similar except for: role of electron and hole, voltage bias polarity, and Current direction

44. I-V Characteristic for CE configuration : Input characteristic Input characteristic: input current (I B ) against input voltage (V BE ) for several output voltage (V CE ) From the graph I B = 0 A V BE < 0.7V (Si) I B = value V BE > 0.7V (Si) The transistor turned on when V BE = 0.7V

45. I-V Characteristic for CE configuration : Output characteristic Output characteristic: output current (I C ) against output voltage (V CE ) for several input current (I B ) 3 operating regions: Saturation region Cut-off region Active region

46. Modes of Operation Forward-Active B-E junction is forward biased B-C junction is reverse biased Saturation B-E and B-C junctions are forward biased Cut-Off B-E and B-C junctions are reverse biased

47. Saturation region – in which both junctions are forward-biased and I C increase linearly with V CE Cut-off region – where both junctions are reverse-biased, the I B is very small, and essentially no I C flows, I C is essentially zero with increasing V CE Active region – in which the transistor can act as a linear amplifier, where the BE junction is forward-biased and BC junction is reverse-biased. I C increases drastically although only small changes of I B. Saturation and cut-off regions – areas where the transistor can operate as a switch Active region – area where transistor operates as an amplifier I-V Characteristic for CE configuration : Output characteristic

48. Current Relationships Relations between I C and I E : α = I C I E Value of α usually 0.9998 to 0.9999 , α ≈ 1 Relations between I C and I B : β = I C @ I C = βI B I B Value of β usually in range of 50  400 The equation, I E =I C + I B can also written in β I C = βI B I E = βI B + I B => I E = (β + 1)I B The current gain factor , α and β is: α = β @ β = α . β + 1 α - 1

49. TRANSISTOR BJT : DC BIASING

50. Transistor Currents Emitter current (I E ) is the sum of the collector current (I C ) and the base current (I B ) . Kirchhoff’s current law; …( Eq. 3.1)

51. Collector Current (I C ) Collector current (I C ) comprises two components ; majority carriers (electrons) from the emitter minority carriers (holes) from reverse-biased BC junction → leakage current, I CBO Total collector current (I C ); Since leakage current I CBO is usually so small that it can be ignored. …( Eq. 3.2)

52. Collector Current (I C ) Then; The ratio of I C to I E is called alpha ( α ), values typically range from 0.95 to 0.99. …( Eq. 3.3)

53. Base Current (I B ) I B is very small compared to I C ; The ratio of I C to I B is the dc current gain of a transistor, called beta ( β ) The level of beta typically ranges from about 50 to over 400 …( Eq. 3.4)

54. Current & Voltage Analysis Consider below figure. Three dc currents and three dc voltages can be identified I B : dc base current I E : dc emitter current I C : dc collector current V BE : dc voltage across base-emitter junction V CB : dc voltage across collector-base junction V CE : dc voltage from collector to emitter Transistor bias circuit.

55. Current & Voltage Analysis When the BE junction is forward-biased, it is like a forward-biased diode. Thus; (Si = 0.7, Ge = 0.3) From HVK, the voltage across R B is By Ohm’s law; Solving for I B …( Eq. 3.5) …( Eq. 3.6)

56. Current & Voltage Analysis The voltage at the collector is; The voltage drop across R C is V CE can be rewritten as The voltage across the reverse-biased CB junction is …( Eq. 3.8) …( Eq. 3.7)

57. Transistor as Amplifier Transistor is capable to amplify AC signal : (output signal > input signal) Eg: Audio amplifier that amplify the sound of a radio

58. Transistor Amplifier Circuit Analysis There are 2 analysis; DC Analysis AC Analysis Transistor will operate when DC voltage source is applied to the amplifier circuit Q-point must be determined so that the transistor will operate in active region (can operate as an amplifier)

59. Transistor Amplifier Circuit Analysis Q-Point Operating point of an amplifier to state the values of collector current (I CQ ) and collector-emitter voltage (V CEQ ). Determined by using transistor output characteristic and DC load line Q-Point

60. DC LOAD LINE DC Load Line A straight line intersecting the vertical axis at approximately I C(sat) and the horizontal axis at V CE (off) . I C(sat) occurs when transistor operating in saturation region V CE(off) occurs when transistor operating in cut-off region DC Load Line Cutoff Region Saturation Region Q-Point

61. DC LOAD LINE (Example) V CC = 8V R B = 360 k Ω R C = 2 k Ω Draw DC Load Line and Find Q-point. Answers;

62. DC LOAD LINE (Example) Draw DC Load Line and Find Q-point. Answers; Q-point can be obtained by calculate the half values of maximum I C and V CE 4V 2 mA

63. DC Analysis of Amplifier Circuit Amplifier Circuit Amplifier Circuit w/o capacitor

64. DC Analysis of Amplifier Circuit Refer to the figure, for DC analysis: Replace capacitor with an open-circuit R 1 and R 2 create a voltage-divider circuit that connect to the base Therefore, from DC analysis, you can find: I C V CE Amplifier Circuit w/o capacitor

65. DC Analysis of Amplifier Circuit Amplifier Circuit w/o capacitor Simplified Circuit Thevenin Theorem;

66. DC Analysis of Amplifier Circuit Important equation for DC Analysis 1 2 1 2 From HVK; From Thevenin Theorem;

67. TRANSISTOR BJT BIASING CIRCUIT

68. BJT BIASING CIRCUIT Fixed Base Bias Circuit (Litar Pincangan Tetap) Fixed Bias with Emitter Resistor Circuit (Litar Pincangan Pemancar Terstabil) Voltage-Divider Bias Circuit (Litar Pincangan Pembahagi Voltan) Feedback Bias Circuit (Litar Pincangan Suap-Balik Voltan)

69. FIXED BASE BIAS CIRCUIT This is common emitter (CE) configuration Solve the circuit using HVK 1 st step : Locate capacitors and replace them with an open circuit 2 nd step : Locate 2 main loops which; BE loop CE loop

70. FIXED BASE BIAS CIRCUIT 1 st step : Locate capacitors and replace them with an open circuit

71. FIXED BASE BIAS CIRCUIT 2 nd step : Locate 2 main loops. 1 2 1 2 BE Loop CE Loop

72. FIXED BASE BIAS CIRCUIT BE Loop Analysis 1 From HVK; I B A

73. FIXED BASE BIAS CIRCUIT CE Loop Analysis From HVK; As we known; Subtituting with 2 I C B A B

74. FIXED BASE BIAS CIRCUIT DISADVANTAGE Unstable – because it is too dependent on β and produce width change of Q-point For improved bias stability , add emitter resistor to dc bias.

75. FIXED BASE BIAS CIRCUIT Example 1 Find I C , I B , V CE , V B , V C , V BC ? (Silikon transistor); Answers; I C = 2.35 mA I B = 47.08 μ A V CE = 6.83V V B = 0.7V V C = 6.83V V BC = -6.13V

76. FIXED BIAS WITH EMITTER RESISTOR An emitter resistor, R E is added to improve stability Solve the circuit using HVK 1 st step : Locate capacitors and replace them with an open circuit 2 nd step : Locate 2 main loops which; BE loop CE loop Resistor, R E added

77. FIXED BIAS WITH EMITTER RESISTOR 1 st step : Locate capacitors and replace them with an open circuit

78. FIXED BIAS WITH EMITTER RESISTOR 2 nd step : Locate 2 main loops. 1 2 2 BE Loop CE Loop 1

79. FIXED BIAS WITH EMITTER RESISTOR BE Loop Analysis From HVK; Recall; Subtitute for I E 1

80. FIXED BIAS WITH EMITTER RESISTOR CE Loop Analysis From HVK; Assume; Therefore; 2

81. FIXED BIAS WITH EMITTER RESISTOR Example 2 Find I C , I B , V CE , V B , V C , V E & V BC ? (Silikon transistor); Answers; I C = 2.01 mA I B = 40.1 μ A V CE = 13.97V V B = 2.71V V E = 2.01V V C = 15.98V V BC = -13.27V

82. VOLTAGE DIVIDER BIAS CIRCUIT Provides good Q-point stability with a single polarity supply voltage Solve the circuit using HVK 1 st step : Locate capacitors and replace them with an open circuit 2 nd step : Simplified circuit using Thevenin Theorem 3 rd step : Locate 2 main loops which; BE loop CE loop

83. VOLTAGE DIVIDER BIAS CIRCUIT 1 st step : Locate capacitors and replace them with an open circuit

84. VOLTAGE DIVIDER BIAS CIRCUIT Simplified Circuit Thevenin Theorem; 2 nd step : : Simplified circuit using Thevenin Theorem From Thevenin Theorem;

85. VOLTAGE DIVIDER BIAS CIRCUIT 2 nd step : Locate 2 main loops. 1 2 BE Loop CE Loop 1 2

86. VOLTAGE DIVIDER BIAS CIRCUIT BE Loop Analysis From HVK; Recall; Subtitute for I E 1

87. VOLTAGE DIVIDER BIAS CIRCUIT BE Loop Analysis From HVK; Recall; Subtitute for I E 1

88. VOLTAGE DIVIDER BIAS CIRCUIT CE Loop Analysis From HVK; Assume; Therefore; 2

89. VOLTAGE DIVIDER BIAS CIRCUIT Example 3 Find R TH , V TH , I C , I B , V CE , V B , V C , V E & V BC ? (Silikon transistor); Answers; R TH = 3.55 k Ω V TH = 2V I C = 0.85 mA I B = 6.05 μ A V CE = 12.22V V B = 1.978V V E = 1.275V V C = 13.5V V BC = -11.522V

90. BJT Operation Regions Operation Region I B or V CE Char. BC and BE Junctions Mode Cutoff I B = Very small Reverse & Reverse Open Switch Saturation V CE = Small Forward & Forward Closed Switch Active Linear V CE = Moderate Reverse & Forward Linear Amplifier Break-down V CE = Large Beyond Limits Overload

91. استخدام الترانزستور مضخم : لتيار الكهربائي ، للجهد الكهربائي أو للتيار و الجهد معا. هناك ثلاثة توصيلات :- 1- توصيلة الباعث مشترك : تضخيم فرق الجهد و التيار . 2- توصيلة القاعدة المشتركة : تضخيم فرق الجهد. 3- توصيلة المجمع المشتركة: تضخيم التيار.

92. 97 Measuring (testing) transistors For some transistors, the pin function can be identified from packaging:

93. 98 Measuring (testing) transistors But sometimes, we have to measure: Set a digital multimeter to diode test and an analogue multimeter to a low resistance range such as × 10, as described above for testing a diode. Test each pair of leads both ways (six tests in total): The base-emitter (BE) junction should behave like a diode and conduct one way only . The base-collector (BC) junction should behave like a diode and conduct one way only . The collector-emitter (CE) should not conduct either way .

94. BJT is bipolar because both holes (+) and electrons (-) will take part in the current flow through the device N-type regions contains free electrons (negative carriers) P-type regions contains free holes (positive carriers) 2 types of BJT NPN transistor PNP transistor The transistor regions are: Emitter (E) – send the carriers into the base region and then on to the collector Base (B) – acts as control region. It can allow none,some or many carriers to flow Collector (C) – collects the carriers Transistor Structure