Download presentation

Presentation is loading. Please wait.

Published byRussell Winbush Modified over 2 years ago

1
1 BJT, Bipolar Junction Transisor Bollen Base Current Controls Output current

2
2 AGENDA Bollen BJT transistorman Transistor types Bipolar Junction Transistor BJT models parameters water model NPN and PNP operation modes switch open switch closed BJT linear, controlled current source active operation characteristics DC input characteristics ac input characteristics BJT DC biasing circuits base bias base bias + collector feedback base bias + emitter feedback voltage divider

3
3 BJT, transistor man Bollen

4
4 Transistor Types Bollen Output current controlled by input current Output current controlled by input voltage = BJT = Bipolar Junction Transistor = FET = Field Effect Transistor

5
5 BJT, Bipolar Junction Transisor Bollen BE Forward bias, BC Reverse bias So low ohmichigh ohmic Transistor = Transfer Resistor

6
6 BJT, Bipolar Junction Transisor Bollen Emitter = Sent electrons Base = Base Collector = Get electrons

7
7 BJT, Models Bollen

8
8 BJT, parameters Bollen

9
9 BJT, Water model Bollen

10
10 BJT, Water model Bollen

11
11 BJT, NPN and PNP Bollen

12
12 BJT, Operation modes Bollen Cut-off and saturation; BJT is used as a switch Active operation Quiecent Point; BJT is used as a controlled current source, or analog amplifier

13
13 BJT, Switch open Bollen

14
14 BJT, Switch closed Bollen

15
15 BJT, Lineair, controlled current source Bollen

16
16 BJT, active operation Bollen

17
17 BJT, characteristics Bollen DC model ac model DC model; Vbe = 0V7 Ube, Uce, Ic, Ib, Ie Capitals ac model; re = 26mV/Ie ube, uce, ic, ib, ie Low cases

18
18 BJT, DC input characteristics Bollen Vbe = 0V7

19
19 BJT, AC input characteristics Bollen r e = 26mV/I c The dynamic resistor can be calculated by the DC current Ic

20
20 BJT, characteristics Bollen

21
21 BJT, DC biasing circuits Bollen A base bias B base bias + emitter feedback C base bias + collector feedback D voltage divider

22
22 BJT, base bias, introduction Bollen Base current determined by Vcc, Rb and Vbe

23
23 BJT, base bias Bollen Calculate Ib and then Ic Ic directly dependent on ß variation So, for stability a “bad” circuit

24
24 BJT, base bias load line Bollen Load line is the loading of the transistor seen from Uce (>0V7) Vcc and Rc determines the; “open voltage” and the “short circuit current” Q-point = Quiecient point = Working point

25
25 BJT, base bias load line Bollen Load line is the loading of the transistor seen from Uce (>0V7) Vcc and Rc determines the; “open voltage” and the “short circuit current” Reliable circuit = Q-point stability

26
26 BJT, base bias load line Bollen Vce always > 0V7 BC junction REVERSE If Rc too big, transistor in saturation; then;

27
27 BJT, base bias load line Bollen Vce always > 0V7 BC junction REVERSE If Vcc too small, transistor in saturation; then;

28
28 BJT, base bias example Bollen Calculate; I b, I c U Rc, U c, U ce Draw output caracteristic Calculate now; U ce if ß = 40 How many % did U ce Change I b = 47 uA, I c = 2,35 mA, U Rc = 5,17 V, U c = 6,83 V, U ce = 6,83 V U ce (for ß = 40) = 7,86 Ξ 15 %

29
29 BJT, base bias example Bollen I b = 33 uA, I c = 2,9 mA U Rc = 7,9 V, U c = 8,1 V R b = 282,5 kΩ, I c = 3,2 mA, R c = 1,855 kΩ

30
30 BJT, base bias example Bollen ß = 200, V Rc = 8,8 V V cc = 16 V Rb = 765 kΩ

31
31 BJT, base bias + emitter feedback Bollen Base current determined by Vcc, Rb, Vbe and Ve More stable for ß variations, than base bias.

32
32 BJT, base bias + emitter feedback Bollen Always calculate in the smallest current Ib !!

33
33 BJT, base bias + emitter feedback Bollen Load line is the loading of the transistor seen from Uce (>0V7) Vcc, Rc and Re determines the; “open voltage” and the “short circuit current”

34
34 BJT, base bias + emitter feedback example Bollen Calculate; I b, I c U Rc, U c, U e, U ce Draw output caracteristic I b = 6,2 uA, I c = 0,74 mA, U Rc = 8,9 V, U c = 7,1 V, U e =-0,9 V, U ce = 8,0 V

35
35 BJT, base bias + emitter feedback example Bollen Calculate; I b, I e U Re, U e, U ce Draw output caracteristic I b = 24 uA, I e = 2,9 mA, U Rc = 3,5 V, U e = -2,5 V, U ce = 2,5 V

36
36 BJT, base bias + collector/emitter feedback Bollen If Ic > then Uc < then Ib < If Ic > then Uc < and Ue > then Ib <

37
37 BJT, base bias + collector feedback Bollen Always calculate in the smallest current Ib !! The current through Rc is not Ic but Ic + Ib, so (β+1)Ib !!! If Ic rises for any reason, then Uc falls and also Ib decreases, so then Ic decreases

38
38 BJT, base bias collector feedback example Bollen Calculate; I b, ß, I c Draw output caracteristic I b = 13 uA, ß = 196, I c = 2,5 mA

39
39 BJT, base bias collector/emitter feedback Bollen Always calculate in the smallest current Ib !!

40
40 BJT, base bias collector/emitter feedback ex Bollen Calculate; I b, I e U Rc, U c, U e, U ce Draw output caracteristic I b = 11,8 uA, I e = 1,1 mA U Rc = 5,2 V, U c = 4,8 V U e = 1,3 V, U ce = 3,5 V

41
41 BJT, voltage divider Bollen Vb is a stable voltage - 0,7 V = so Ve is a stable voltage Ie is determined by Ve/ Re Ic = Ie. ß/(ß+1) Ic is very stable and nearly independent to ß variation, as long as ß is BIG in value 2 methods of calculating Ic - neglegting Ib, use voltage divider - not neglecting Ib and use thevenin

42
42 BJT, voltage divider, neglect Ib Bollen So neglegt Ib to R2, or in general Ri >> R2 In practice 10 times bigger

43
43 BJT, voltage divider, exact, thevenin Bollen Thevenin resistance R1 // R2 62k // 9k1= 7k9 Thevenin voltage

44
44 BJT, voltage divider, exact, thevenin Bollen 2V0 7k9 Ib = 20 uA

45
45 BJT, voltage divider, example Bollen Thevenin resistance = 6k8 Thevenin voltage = 3V1 I b = 18,8 uA I c = 2,25 mA r e = 11,5 Ω U Rc = 7V4 U c = 10V6 U e = 2V3 U ce = 5V1

46
46 BJT, voltage divider, example Bollen Thevenin resistance = 255k Thevenin voltage = 0V0 I b = 14,3 uA I c = 1,9 mA r e = 14 Ω U Rc = 17V3 U c = 0V7 U e = -3V7 U ce = 4V4

47
47 BJT Bollen

48
48 BJT Bollen

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google