hapter 3 ipolar Junction Transistors (JT) 1 Transistors 1. ipolar Junction Transistor (JT) 2. Field ffect Transistor (FT) ipolar Junction Transistor (JT) Field ffect Transistor (FT) 2
Transistors 1. - KWatt - Mega Watt Milli Amp, - Micro Amp 2. Digital - Microprocessor -Memory - LD Display FT MOS 3 Transistor ias Transistor Transistor Digital Transistor Transistor ell Labs 1970 4
5 ipolar Junction Transistor (JT) PN Junction ipolar 2 + Transistor Transfer + esistor * nput Output * JT 2 NPN Transistor PNP Transistor 6
Transistors NPN Transistors PNP Transistors mitter ollector mitter ollector ase ase N P 7 N P ase, mitter ollector Dope JT ase 1:150, ase Dope mitter Dope 10 8
JT ase mitter ollector JT ertical ase N P N 9 Transistor D AS ias ase - mitter ias ase - ollector ias 3 mode» Active Mode» ut Off» Saturation 10
Active Mode = Active i o t t Active Transistor ircuit ut Off Mode = Transistor OFF ut OFF i o o = 0 t t utoff Transistor ircuit 11 Saturation Mode = Saturation o i o t t Transistor in Saturation Active utoff Saturation 12
Transistor ias mode ias Transistor 2 Junction 2 Junction» ase-mitter Junction ( Junction) ase- ollector Junction ( Junction) Junction Junction Supply Supply 13 ias Transistor Active Transistor Active Mode Junction Forward Junction everse Junction Forward Transistor P N P Junction everse, ommon ase 14
Depletion egion Forward ias Junction Depletion egion everse ias Junction Depletion egion P N P ase ollector mitter 15 1. open Open ase mitter = O everse ias = S = O O O O ollector ase Open A na 16
2. open = FWD Forward ias Open ase ollector = forward ma 0.7 17 1 2 1, 2?? Minority O (everse) Forward 2 1 2 18
2 > 1 1. ase ( ase ) 2. ase Dope ase ( ) 3. Depletion ase * * ase NPN Transistor 19 PNP Transistor O Majority O TOTAL O O 20
NPNTransistor ase mitter, Forward O ase ollector, everse Majority O TOTAL O O 21 Transistor NPN Transistors PNP Transistors mitter mitter 22
Transistor ommon Transistor ( common ) nput, Output ommon ommon 3 1. ommon ase ; 2. ommon mitter ; 3. ommon ollector ; 23 ommon ase ase nput Output D OUT N N nput D = ommon ase urrent Gain = ommon ase Output OUT ommon ase L 24
ommon-base pnp npn 25 D Transistor Active D = 1 D = 0.95-0.99 JT dc O O Open common base O O O D D A ac D constant 26
- curve common base nput = 0.7 27 - curve common base Output forward, reverse bias forward bias reverse bias 28
ommon mitter OUT mitter nput ase Output ollector N nput Output cc D OUT N D = ommon mitter urrent Gain = ommon mitter ommon mitter 29 ommon mitter npn pnp 30
- curve common emitter npn = 0.7 31 - curve common emitter forward, reverse bias 32
D Transistor Active 100 D D ( 1) >> D 50 500 D = 50 500 JT O Open common emitter O O 1 O O A constant 33 ac A ommon mitter A constant 34
ommon ollector OUT ollector nput ase Output mitter N nput Output cc ommon ollector OUT D 1 D 1 ommon ollector ( D +1) N 35 ommon ollector npn pnp 36
Transistor JT urrent Gain Parameters 2 D D Transistor D D 1 2 3 0.95 50-0.99 37-500 2 3 1 D D D D 1 D D 1 D D Transistor Ground,, Junction = to = - = to = = to = - + - + - + + - - + - + - Transistor = + 38
Operating egion Operating egion reakdown max = 20 39 4 4 1. Fixed ias = 2. Self ias = 3. oltage Divider ias = 4. ollector Feedback ias = 40
JT bias common-emitter npn transistor npn transistor nput Output 41 JT bias transistor 3modes Active ut-off Saturation 42
JT bias curve 43 Active mode ondition : 0 A sat ( ) 0.7 on 44
Active mode dc ( 1) 45 ut-off mode ut-off condition: ase-mitter reverse bias = 0 A 46
Saturation mode Saturation condition: ase-mitter forward bias = (on) = 0.7 = sat = sat 47 Fixed-bias (Fixed Q ) Q Q Q + Q - + - cc 0.7 olts Q 48
Fixed-bias ase ollector Q Q + + - Q - Active 0.7 olts 49 Fixed-bias : Active mode 50
Fixed-bias : Active mode Set = (on) = 0.7 = 0.7 > (on) 51 Fixed-bias : Active mode Q sat 52
Fixed-bias : Active mode : xample Q, Q, Q 15 477 k 2.5 k 107 sat 1 53 Fixed-bias : Load line analysis (Q-point) - curve load line 54
Fixed-bias : Load line analysis load line : output Load line 55 Fixed-bias : Load line analysis load line output KL : 1 56
Fixed-bias : Load line analysis 1 y ( = 0 ) 0 x ( = 0 A ) 0 57 Fixed-bias : Load line analysis 58
Fixed-bias : Load line : xample x 15 y 15 3 2.510 6 ma 59 Fixed-bias : Load line : xample Q = 3.21 ma Q point = 7 Q > sat 60
Fixed-bias : Active mode : Q-point 61 Fixed-bias : Active mode : Q-point 62
Fixed-bias : Active mode : Q-point 63 Fixed-bias : Load line : xample (1) (2)? Q = 4.1 ma (1) (2) Q = 7.5 64
Fixed-bias : Load line : xample = 4.1 ma 8.4 Option 1 =2.5k 1.78k 477k 357k = 7.5 65 Fixed-bias : Load line : xample Option 2 = 4.1 ma =15 18 477k 432.5k 18 = 7.5 66
Fixed-bias : Saturation sat Q sat Q sat sat sat sat Saturation Saturated Q sat 67 Fixed-bias : xample Saturation? 2.5 k 15 80 sat 1 68
Fixed-bias : Load line : xample = 5.6 ma Q point 69 Self-bias 70
Self-bias Self-bias mitter stabilized bias i ( 1) ( on) ( 1) 71 Self-bias ( ) 72
Self-bias : xample x: Q-Point 73 x: Q-Point lock Q Q Solve: Q ( 1) 20 0.7 Q 430K (50 1) 1 K + - + Q - 40.1A Q Q 2.01mA Q Q ( ) 20 2.01mA(2K 1K ) 13.97 Active 74
Self-bias : xample x:,,, 75 x:,,, 2.01m 40.1 Solve: 20 2.01mA 2K + _ 2.7115.98 + 0.7-13.27 + 13.97 - + _ + _ everse ias 15.98 76 15.98 13. 97 2.01 0.7 2.01 2.71
Self-bias : xample 77 Self-bias : Load line X 0 ma Y 0 78
Self-bias : Saturation mode sat ( ) Q Q sat Q Q sat sat ( ) on Q sat sat 79 oltage-divider bias Self ias 80
oltage-divider bias: Thevenin nput 2 TH 1 2 TH 1 // 2 Q TH TH ( 1) + Q - Q Q Q ( ) 81 oltage-divider bias : xample x: Q-Point 82
oltage-divider bias x: Q-Point Solve: Thevenin nput Q Short Q + Q - Short Thevenin TH 1 2 8.2K 2. 2K 1.73K 83 Thevenin 20 20 ( 2 ) 20 1 20 2 3.85 ma 1 0 40 10.4 K TH 2 20 3.85 ma 2.2 K 20 11.53 84
Q Q 11.53 Q Q + Q - 20 20 11.53 0.7 Q 1.73K ( 1)1. 8 K 35.39 A 20 Q Q 120 35.39 A 4.25 ma Q 20 20 Q( ) 23 20 20 4.25 ma 2.7 K Q 8.53 Q 0.7 20 11.59 Q Q 85 oltage-divider : xample 86
oltage-divider : Load line Self bias X Y 0 ma 0 87 oltage-divider : Saturation sat ( ) Q Q Q sat sat Q sat sat 88
ollector Feedback ias ollector KL: nput ' ' Q ' ( ) KL: Output Q ( ), 89 ollector Feedback ias x: Q-Point Q Q + + - Q Q - Solve 90 Q 250K ( ) 11.91 A Q Q 10 0.7 90(4.7K 1.07 ma 1.2K) Q ( ) 10 1.07mA (4.7K 1.2K) 3.69 90
4.1 Find,,,,. (h4 Pb7) 91 4.2 Design a voltage-divider network given a transistor curve shown. (a) Determine and for the voltagedivider network having a Q-point of Q =5mA and Q =8. Use =24 and =3. (b) Find. (c) Determine. (d) Find 2 if 1 =24k assuming that 10 2. (e) alculate at the Q-point. (h4 Pb19) 92
93 4.3 Design the transistor inverter of the figure below to operate with a saturation current of 8mA using a transistor with =100. Use a level of equal to 120% of sat and standard resistor values. (h4 Pb37) 94