Communication Microelectronics (W17)

Transcription

1 Communication Microelectronics (W17) Lecture 4: Bipolar Junction Transistor Assistant Professor Office: C

2 Bipolar Junction Transistor (BJT) Physical Structure and I-V Characteristics 2

3 BJT Physical Structure Two back to back PN Junctions NPN or PNP Transistor Three terminal device, for NPN: Base (P-Type), Emitter and Collector (N-Type) Base-Emitter Junction Base-Collector Junction Emitter doping is higher than Collector doping NPN BJT Transistor PNP BJT Transistor 3

4 BJT NPN I-V characteristics 1. BJT NPN Transistor in Cutoff Mode Base-Emitter Junction is biased Base-Collector Junction is biased BEJ BCJ Cutoff Mode I I I 0 C B E 4

5 BJT NPN I-V characteristics 2. BJT NPN Transistor in Active Mode Base-Emitter Junction is biased Base-Collector Junction is biased BEJ Forwar d BCJ Revers e Active Mode 5

6 BJT NPN I-V characteristics 2. BJT NPN Transistor in Active Mode Base-Emitter Junction is biased and electrons pass through the Base to the collector due to the Base small area Electrons from the Emitter are collected at the Collector side The transistor s Collector current can be modeled by a current dependent current source I s depends on doping and width of the Base V I I exp V C s BE T I C I F B V BE 0.7V 6

7 BJT NPN I-V characteristics 3. NPN Transistor in Active mode Base-Collector Junction is biased Base-Emitter Junction is reverse biased Emitter and collector reverse their roles However, BJT has an asymmetrical physical structure The current gain from Base to Emitter is very small V BC 0.5V BEJ BCJ Active Mode R F I E I R B 7

8 BJT NPN I-V characteristics 4. NPN Transistor in Saturation Mode Both junctions are biased The total current is the EBJ diffusion current subtracted from CBJ diffusion current V BE = 0.7V V BC = 0.5V BEJ BCJ Saturation Mode BJT could be used as a closed switch in Saturation mode I C I F B V 0.2V CE 8

9 BJT NPN I-V characteristics I C = I s exp V BE V T For Active ONLY I C versus V BE and V CE 9

10 BJT NPN I-V characteristics V BE ic ISexp( ) 1 VT v V CE A r o i v C CE V I A C I C versus V CE The Early effect 10

11 BJT Large Signal Model in Active Mode 11

12 BJT PNP Physical Structure BJT PNP Transistor PNP is the NPN Complementary structure Two back to back PN Junctions Three terminal device: Base (Ntype), Emitter and Collector (Ptype) Emitter-Base Junction Collector-Base Junction Emitter doping is higher than Collector doping Same I-V Characteristics as NPN Transistor PNP BJT Transistor 12

13 BJT PNP Physical Structure BJT PNP Transistor in Active Mode EBJ Forwar d CBJ Active Mode 13

14 BJT Modes of Operation Electrical Equations of BJT 14

15 BJT NPN Modes of Operation Mode BEJ BCJ Equations Condition Cutoff I C = I E = I B =0 Active () Saturation Active V BE = 0.7 I E = I C + I B I C = β F I B = α F I E β F α F = 1 + β F V BE = 0.7 V BC = 0.5 V CE = 0.2 I E = I C + I B V BC = 0.5 I C = I E + I B I E = β R I B = α R I C β R α R = 1 + β R V BE < 0.7 V BC < 0.5 V BC < 0.5 Or V CE > 0.2 I C < β F I B V BE <

16 BJT PNP Modes of Operation Mode EBJ CBJ Equations Condition Cutoff I C = I E = I B =0 Active () Saturation Active V EB = 0.7 I E = I C + I B I C = β F I B = α F I E β F α F = 1 + β F V EB = 0.7 V CB = 0.5 V EC = 0.2 I E = I C + I B V CB = 0.5 I C = I E + I B I E = β R I B = α R I C β R α R = 1 + β R V EB < 0.7 V CB < 0.5 V CB < 0.5 Or V EC > 0.2 I C < β F I B V EB <

17 Calculating DC operating point Solved Exercise 17

18 Solved Example Find the DC Operating point of the Transistors? Given: V BE =0.7V,β=10 (Ans.: I B =0.023mA, I C =0.23mA, I E =0.253mA, V CE =9.54V, Active ) 18

19 Solution Steps: 1. Identify the BJT Type 2. Place the terminals name on the circuit 3. Write a KVL in the INPUT Loop Input loop for BJT is any loop containing V BE 4. Assume the BJT mode (most of the time active) 5. Calculate the currents and voltages 6. Write KVL in the OUTPUT loop Output loop for BJT is any loop containing V CE 7. Verify your assumption! 19

20 Example Find the DC Operating point of the Transistors? Given: V EB =0.7V,β=10 (Ans.: I B =93µA, I C =0.93mA, I E =1.023mA, V EC =8.14V, Active ) 20