1. Transistors Transfer Resistor Chapter 9

2. Bipolar Transistors Two PN junctions joined together Two types available – NPN and PNP The regions (from top to bottom) are called the collector (C), the base (B), and the emitter (E) Base Collector Emitter

3. Operation  Begin by reverse biasing the CB junction  Here we are showing an NPN transistor as an example  Now we apply a small forward bias on the emitter-base junction  Electrons are pushed into the base, which then quickly flow to the collector  The result is a large emitter-collector electron current (conventional current is C-E) which is maintained by a small E-B voltage  Some of the electrons pushed into the base by the forward bias E-B voltage end up depleting holes in that junction  This would eventually destroy the junction if we didn’t replenish the holes  The electrons that might do this are drawn off as a base current

4. Currents

5. Conventional View

6. Origin of the names  the Emitter 'emits' the electrons which pass through the device  the Collector 'collects' them again once they've passed through the Base ...and the Base?...

7. Original Manufacture

8. Base Thickness  The thickness of the unmodified Base region has to be just right.  Too thin, and the Base would essentially vanish. The Emitter and Collector would then form a continuous piece of semiconductor, so current would flow between them whatever the base potential.  Too thick, and electrons entering the Base from the Emitter wouldn't notice the Collector as it would be too far away. So then, the current would all be between the Emitter and the Base, and there'd be no Emitter-Collector current.

9. Amplification Properties  The C-B voltage junction operates near breakdown.  This ensures that a small E-B voltage causes avalanche  Large current through the device

10. Common Base NPN

11. Common Emitter NPN

12. Common Collector NPN How does I C vary with V CE for various I B ? Note that both dc sources are variable Set V BB to establish a certain I B

13. Collector Characteristic Curve  If V CC = 0, then I C = 0 and V CE = 0  As V CC ↑ both V CE and I C ↑  When V CE  0.7 V, base-collector becomes reverse-biased and I C reaches full value (I C =  I B )  I C ~ constant as V CE ↑. There is a slight increase of I C due to the widening of the depletion zone (BC) giving fewer holes for recombinations with e¯ in base.  Since I C =  I B, different base currents produce different I C plateaus.

14. NPN Characteristic Curves

15. PNP Characteristic Curves

16. Load Line For a constant load, stepping I B gives different currents (I C ) predicted by where the load line crosses the characteristic curve. I C =  I B works so long as the load line intersects on the plateau region of the curve. Slope of the load line is 1/R L

17. Saturation and Cut-off Note that the load line intersects the 75 mA curve below the plateau region. This is saturation and I C =  I B doesn’t work in this region. Cut-off

18. Example  We adjust the base current to 200  A and note that this transistor has a  = 100  Then I C =  I B = 100(200 X 10 -6 A) = 20 mA  Notice that we can use Kirchhoff’s voltage law around the right side of the circuit  V CE = V CC – I C R C = 10 V – (20 mA)(220  ) = 10 V – 4.4 V = 5.6 V

19. Example  Now adjust I B to 300  A  Now we get I C = 30 mA  And V CE = 10 V – (30 mA)(220  ) = 3.4 V  Finally, adjust I B = 400  A  I B = 40 mA and V CE = 1.2 V

20. Plot the load line V CE ICIC 5.6 V20 mA 3.4 V30 mA 1.2 V40 mA

21. Gain as a function of I C As temperature increases, the gain increases for all current values.

22. Operating Limits  There will be a limit on the dissipated power  P D(max) = V CE I C  V CE and I C were the parameters plotted on the characteristic curve.  If there is a voltage limit (V CE(max) ), then you can compute the I C that results  If there is a current limit (I C(max) ), then you can compute the V CE that results

23. Example  Assume P D(max) = 0.5 W V CE(max) = 20 V I C(max) = 50 mA P D(max) V CE I C 0.5 W5 V100 mA 1050 1533 2025

24. Operating Range Operating Range

25. Voltage Amplifiers Common Base PNP Now we have added an ac source The biasing of the junctions are: BE is forward biased by V BB - thus a small resistance BC is reverse biased by V CC – and a large resistance Since I B is small, I C  I E

26. Equivalent ac Circuit r E = internal ac emitter resistance I E = V in /r E (Ohm’s Law) V out = I C R C  I E R C Recall the name – transfer resistor

27. Current Gains  Common Base   = I C /I E < 1  Common Emitter   = I C /I B

28. Example  If  = 50, then  = 50/51 = 0.98  Recall  < 1  Rearranging,  =  +   (1-  ) =   =  /(1-  )

29. Transistors as Switches

30. The operating points We can control the base current using V BB (we don’t actually use a physical switch). The circuit then acts as a high speed switch.

31. Details  In Cut-off  All currents are zero and V CE = V CC  In Saturation  I B big enough to produce I C(sat)   I B  Using Kirchhoff’s Voltage Law through the ground loop  V CC = V CE(sat) + I C(sat) R C  but V CE(sat) is very small (few tenths), so  I C(sat)  V CC /R C

32. Example a)What is V CE when V in = 0 V? Ans. V CE = V CC = 10 V b) What minimum value of I B is required to saturate the transistor if  = 200? Take V CE(sat) = 0 V I C(sat)  V CC /R C = 10 V/1000  = 10 mA Then, I B = I C(sat) /  = 10 mA/200 = 0.05mA

33. Example LED If a square wave is input for V BB, then the LED will be on when the input is high, and off when the input is low.

34. Transistors with ac Input Assume that  is such that I C varies between 20 and 40 mA. The transistor is constantly changing curves along the load line.

35. Pt. A corresponds to the positive peak. Pt. B corresponds to the negative peak. This graph shows ideal operation.

36. Distortion  The location of the point Q (size of the dc source on input) may cause an operating point to lie outside of the active range. Driven to saturation Driven into Cutoff

37. Base Biasing  It is usually not necessary to provide two sources for biasing the transistor. The red arrows follow the base-emitter part of the circuit, which contains the resistor R B. The voltage drop across R B is V CC – V BE (Kirchhoff’s Voltage Law). The base current is then… and I C =  I B

38. Base Biasing  Use Kirchhoff’s Voltage Law on the black arrowed loop of the circuit V CC = I C R C + V CE So, V CE = V CC – I C R C V CE = V CC –  I B R C  Disadvantge   occurs in the equation for both V CE and I C  But  varies – thus so do V CE and I C  This shifts the Q-point (  -dpendent)

39. Example  Let R C = 560  @ 25 °C  = 100 R B = 100 k  @ 75 °C  = 150 V CC = +12 V @ 75 °C I B is the same I C = 16.95 mA V CE = 2.51 V I C increases by 50% V CE decreases by 56%

40. Transistor Amplifiers  Amplification  The process of increasing the strength of a signal.  The result of controlling a relatively large quantity of current (output) with a small quantity of current (input).  Amplifier  Device use to increase the current, voltage, or power of the input signal without appreciably altering the essential quality.

41. Class A  Entire input waveform is faithfully reproduced.  Transistor spends its entire time in the active mode  Never reaches either cutoff or saturation.  Drive the transistor exactly halfway between cutoff and saturation.  Transistor is always on – always dissipating power – can be quite inefficient

42. Class A

43. Class B  No DC bias voltage  The transistor spends half its time in active mode and the other half in cutoff

44. Push-pull Pair Transistor Q1 "pushes" (drives the output voltage in a positive direction with respect to ground), while transistor Q2 "pulls" the output voltage (in a negative direction, toward 0 volts with respect to ground). Individually, each of these transistors is operating in class B mode, active only for one-half of the input waveform cycle. Together, however, they function as a team to produce an output waveform identical in shape to the input waveform.

45. Class AB  Between Class A (100% operation) and Class B (50% operation).

46. Class C I C flows for less than half then cycle. Usually get more gain in Class B and C, but more distortion

47. Common Emitter Transistor Amplifier Notice that V BB forward biases the emitter-base junction and dc current flows through the circuit at all times The class of the amplifier is determined by V BB with respect to the input signal. Signal that adds to V BB causes transistor current to increase Signal that subtracts from V BB causes transistor current to decrease

48. Details  At positive peak of input, V BB is adding to the input  Resistance in the transistor is reduced  Current in the circuit increases  Larger current means more voltage drop across R C (V RC = IR C )  Larger voltage drop across R C leaves less voltage to be dropped across the transistor  We take the output V CE – as input increases, V CE decreases.

49. More details  As the input goes to the negative peak  Transistor resistance increases  Less current flows  Less voltage is dropped across R C  More voltage can be dropped across C-E  The result is a phase reversal  Feature of the common emitter amplifier  The closer V BB is to V CC, the larger the transistor current.

50. PNP Common Emitter Amplifier

51. NPN Common Base Transistor Amplifier Signal that adds to V BB causes transistor current to increase Signal that subtracts from V BB causes transistor current to decrease At positive peak of input, V BB is adding to the input Resistance in the transistor is reduced Current in the circuit increases Larger current means more voltage drop across R C (V RC = IR C ) Collector current increases No phase reversal

52. PNP Common Base Amplifier

53. NPN Common Collector Transistor Amplifier Also called an Emitter Follower circuit – output on emitter is almost a replica of the input Input is across the C-B junction – this is reversed biased and the impedance is high Output is across the B-E junction – this is forward biased and the impedance is low. Current gain is high but voltage gain is low.

54. PNP Common Collector Transistor Amplifier

55. Gain Factors Usually given for common base amplifier Usually given for common emitter amplifier Usually given for common collector amplifier

56. Gamma  Recall from Kirchhoff’s Current Law  I B + I C = I E Ex. For  = 100  =  /(1+  ) = 0.99  = 1 +  = 101

57. Bringing it Together TypeCommon Base Common Emitter Common Collector Relation between input/output phase 0°180°0° Voltage GainHighMediumLow Current Gain Low (  )Medium (  )High (  ) Power GainLowHighMedium Input ZLowMediumHigh Output ZHighMediumLow

58. Hybrid Parameters Condition hihi Input resistanceOutput shorted hrhr Voltage feedback ratioInput open hfhf Forward current gainOutput shorted hoho Output conductanceInput open Second subscript indicates common base (b), common emitter (e), or common collector (c)

59. Hybrid Parameters =  = Slope of curve

60. Hybrid Parameters h ie = V B /I B Ohm’s Law h ie =input impedance h re = V B /V C

61. Hybrid Parameters h fe = I C /I B Equivalent of  h oe = I C /V C

62. Various Forms Common Emitter (e) Common Base (b) Common Collector (c) h i (ohms)V B /I B V E /I B V B /I B h r (unitless)V B /V C V E /V C V B /V E h f (unitless)I C /I B I C /I E I E /I B h o (watts)ICVCICVC ICVCICVC IEVEIEVE

63. Pin-outs No standard – look at the spec sheet or the case

64. Loudness  When the energy (intensity) of the sound increases by a factor of 10, the loudness increases by 1 bel  Named for A. G. Bell  One bel is a large unit and we use 1/10 th bel, or decibels  When the energy (intensity) of the sound increases by a factor of 10, the loudness increases by 10 dB

65. Decibel Scale  For intensities  L = 10 log(I/I o )  For energies  L = 10 log(E/E o )  For amplitudes  L = 20 log(A/A o )

66. Threshold of Hearing  The I o or E o or A o refers to the intensity, energy, or amplitude of the sound wave for the threshold of hearing  I o = 10 -12 W/m 2  Loudness levels always compared to threshold  Relative measure

67. Common Loud Sounds 160 Jet engine - close up 150 Snare drums played hard at 6 inches away Trumpet peaks at 5 inches away 140 Rock singer screaming in microphone (lips on mic) 130 Pneumatic (jack) hammer Cymbal crash Planes on airport runway120 Threshold of pain - Piccolo strongly played Fender guitar amplifier, full volume at 10 inches away Power tools110 Subway (not the sandwich shop) 100 Flute in players right ear - Violin in players left ear

68. Common Quieter Sounds 90 Heavy truck traffic Chamber music80 Typical home stereo listening level Acoustic guitar, played with finger at 1 foot away Average factory 70 Busy street Small orchestra 60Conversational speech at 1 foot away Average office noise50 Quiet conversation40 Quiet office30 Quiet living room20 10Quiet recording studio 0Threshold of hearing for healthy youths

69. The Math l 1 = 10 log(I 1 /I o ) l 2 = 10 log(I 2 /I o ) l 2 – l 1 =  l = 10(log I 2 – log I o – log I 1 + log I o ) = 10(log I 2 – log I 1 ) l 2 – l 1 =  l = 10 log(I 2 /I 1 ) Threshold of Hearing when I = I o l = 0 dB Threshold of Pain when I  10 12 I o l = 120 dB

70. Example  A loudspeaker produces loudness rated at 90 dB ( l 1 ) at a distance of 4 ft (d 1 ). How far can the sound travel (d 2 ) and still give a loudness at the listener’s ear of 40 dB ( l 2 - conversation at 3 ft.)? Sound follows the inverse square law I 1 /I 2 = d 2 2 /d 1 2  l = 50 dB = 10 log(I 2 /I 1 ) log(I 2 /I 1 ) = 5 which means I 2 /I 1 = 10 5 If d 1 = 4 ft, then d 2 2 = (I 1 /I 2 ) d 1 2 = 10 5 (4 ft) 2 d 2 = 1260 ft (about ¼ mile)

71. Common Emitter Current Gain  For the -3 dB point   l = 3 dB = 10 log (I 1 /I 2 )  I 1 /I 2 = 2 = P 1 /P 2  so 3 dB below initial level mean half the power Frequency h fe 0 dB -3 dB

72. Why do Frequency limits occur?  It takes a certain time for e - to travel from emitter to collector (transit time)  If frequency is too high, applied current varies too rapidly  Electrons may be unable to dislodge rapidly enough to move from E to C before current surges in the other direction. Making the base thinner reduces transit time and improves frequency response

73. Interelement Capacitance  As reverse bias increases on the C-B junction, the depletion zone increases and C decreases (C =  A/d and d increasing).  As emitter current increases, C increases (d decreasing).  If capacitance changes, so does capacitive reactance  Increasing C decreases X C

74. Feedback  Small base current provides a path back to input  If the feedback voltage aids the input voltage, then it is positive (regenerative) feedback  If the feedback is too large, the amplifier will oscillate

75. Superheterodyne Receiver