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Npn Transistor Pnp Transistor

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Npn Transistor Pnp Transistor ELL 100 - Introduction to Electrical Engineering LECTURE 23: BIPOLAR JUNCTION TRANSISTOR (BJT) 1 CONTENTS • Introduction • Construction & working principle of BJT • Common-base and common-emitter circuits 2 BJT INTRODUCTION • Bipolar: Both electrons and holes (positively charged quasi- particles, absence of electron) contribute to current flow • Junction: Consists of an n- (or p-) type silicon sandwiched between two p- (or n-) type silicon regions • Transistor: 3-terminal device (Base, Emitter, Collector) Two types: npn & pnp 3 Bipolar Junction Transistor (BJT) 4 Various Types of General-Purpose or Switching Transistors: (a) Low Power (b) Medium Power (c) Medium to high power 5 First Working Transistor, Invented in 1947. Earliest Raytheon Blue transistors, Replica of 1st device made by first appearing in 1955 Shockley, Bardeen & Brattain at Bell Labs 6 Applications: Transistor As Switch 7 Applications: Transistor As Relay (Switch) Single channel 5V relay breakout board 8 Applications: Transistor As Amplifier TDA7297 Power Amplifier & Audio amplifier Module 9 Applications: Transistors in Radio 10 Applications: Transistor in Oscillator Colpitts Oscillator 11 Applications: Transistor in Oscillator Astable Multivibrator 12 Transistors in Timer Circuits 13 Transistor As NOT Gate 14 Transistors in NAND Gate 15 Bipolar Junction Transistor (BJT) The three terminals of the BJT are called the Base (B), the Collector (C) and the Emitter (E). n p p n n p npn transistor pnp transistor 16 Bipolar Junction Transistor (BJT) npn pnp 17 p-n-p Transistor IE, IB, IC – Emitter, Base and Collector currents VEB, VCB, VCE – Emitter-Base, Collector-Base and Collector-Emitter voltages 18 n-p-n Transistor IE, IB, IC – Emitter, Base and Collector currents VEB, VCB, VCE – Emitter-Base, Collector-Base and Collector-Emitter voltages 19 BJT Construction 20 Bipolar Junction Transistor (BJT) Basic Principle The voltage between two terminals (B and E) controls the current through the third terminal (C). 21 Bipolar Junction Transistor (BJT) : Modes Mode VBE VBC Forward active Forward Bias Reverse Bias Amplifier Reverse active Reverse Bias Forward Bias Saturation Forward Bias Forward Bias Switch ON (VCE ~ 0) Cut off Reverse Bias Reverse Bias Switch OFF (IC,E ~ 0) 22 Biasing of transistors in “Forward Active” mode B-E forward-biased and B-C reverse-biased 23 (a) B-E forward-bias => large current (b) B-C reverse-bias => small current carried by majority carriers (IE) carried by minority carriers (ICBO) 24 BJT carrier flow in Forward-Active mode 25 Mode V ~ 0.7 V V < 0 Basic Working BE BC Forward Active Forward Bias Reverse Bias 26 BJT carrier flow in Forward-Active mode (IC = αIE, Diffusion Drift α ~ 0.95-0.99) Thus IC ~ IE and IB << IE,C Also, IC fairly independent of VCB 27 BJT Currents in Forward-Active Mode I I IE = IB+IC E C C IC ~ αIE 0.95 < α < 0.995 npn α is the emitter-to-collector gain B In addition, due to the reverse IB biased C-B junction, a small reverse saturation current flows, IC = αIE+ICBO = α(IC+IB) + ICBO ICBO (collector-to-base leakage current with emitter open) => (1- α)IC= αIB+ICBO => I = (α/1- α)I + (1/1-α)I β = α/(1-α) is the base-to-collector gain C B CBO 20 < β < 200 => IC = βIB + (β+1)ICBO 28 BJT Amplifier/Circuit Configurations Common-Base Configuration Common-Emitter Configuration Common-Collector Configuration 29 Common-Base Configuration Vin R E C RLL VVEB VCB BE B The base terminal is common to the input and output ports 30 The arrows in the graphics define the direction of current through the device. pnp transistor npn transistor 31 Input dc characteristics in Common-Base configuration V in RL E C RL VBE VEB VCB Notation: B Input current (IE) Input voltage (VBE) Output voltage (VCB) 32 Output dc characteristics in Common-Base configuration V in RL E C RL VBE VEB VCB B Notation: Output current (IC) Output voltage (VCB) Input current (IE) 33 Modes of operation of npn transistor in Common-Base configuration Cutoff region is defined as where the collector current IC = ICBO ~ 0 A The base–emitter and collector–base junctions are both reverse-biased. Saturation region is defined as where VCB < 0 V The base–emitter and collector–base junctions are both forward-biased. Active region is defined as where VCB > 0 V The base–emitter junction is forward-biased and the collector–base junction is reverse-biased. 34 Common-Emitter Configuration RRLL B C VCE VBE E The emitter terminal is common to the input and output ports 35 The arrows in the graphics define the direction of current through the device. npn transistor pnp transistor 36 Input dc characteristics in Common-Emitter configuration RL RL B C VCE VBE E Notation: Input current (IB) Input voltage (VBE) Output voltage (VCE) 37 Output dc characteristics in Common-Emitter configuration RL RL B C VCE VBE E Notation: Output current (IC) Output voltage (VCE) Input current (IB) 38 Saturation region is defined as where VCE < VCE(sat) ~ 0.7-1 V (i.e. VCB < 0 V) The base–emitter and collector–base junctions are both forward-biased. Active region is defined as where VCE > VCE(sat) ~ 0.7-1 V (i.e. VCB > 0 V) The base–emitter junction is forward-biased and the collector–base junction is reverse-biased. Cutoff region is defined as where the collector current IC = ICEO = βICBO The base–emitter and collector–base junctions are both reverse-biased. 39 Common-Collector Configuration RLRL CE B - - + V VCEEC VBECB + EC The common-collector configuration is used primarily for impedance- matching purposes: High input impedance and low output impedance 40 The arrows in the graphics define the direction of current through the device. pnp transistor npn transistor 41 • The collector is tied to ground even though the transistor is connected in a manner similar to the common-emitter configuration • The output characteristics of the common-collector configuration are same as for the common-emitter configuration Common-collector configuration used for impedance-matching purposes. 42 BJT Configurations in pnp Transistor 43 BJT amplifier configurations : Summary Characteristics Common Base Common Emitter Common Collector Input Impedance Low Medium High Output Impedance Very high High Low Phase shift 00 1800 00 Voltage gain High Medium Low Current Gain Low Medium High Power gain Low Very High Medium Application Preamplifier Most Common Buffer 44 Numerical Problems 45 Q1: In an npn transistor, 108 holes/µs move from the base to the emitter region while 1010 electrons/µs move from the emitter to the base region. An ammeter reads the base current as IB = 16 µA. Determine the emitter current IE and the collector current IC 46 Solution: 8 IB = 16 μA (corresponding to 10 positive electronic charges/sec) 10 IE = (100 × 16) μA (corresponding to 10 negative electronic charges/sec) + 16 μA (corresponding to 108 positive electronic charges/sec) = (101 x 16) μA = 1616 μA = 1.616 mA IC = IE − IB = 1.6 mA 47 Q2: For the previous question (data given below), find the transistor α and β values, if the reverse collector-base leakage current is considered negligible Transistor: npn transistor Base current: IB = 16 µA Emitter current: IE = 1.616 mA Collector current: IC = 1.6 mA 48 Solution: If we assume ICBO = ICEO = 0, then Emitter-collector current gain α = IC / IE = 1.6/1.616 = 0.99 Base-collector current gain β = IC / IB = 1.6/0.016 = 100 49 Q3: A BJT has α = 0.99, IB = 25 µA, ICBO = 200 nA. Find the following parameters: a) the dc collector current, b) the dc emitter current, and c) the percentage error in the emitter current when the leakage current is neglected. 50 Solution: 0.99 a) 99 1 1 0.99 69 IC I B ( 1) I CBO 99(25 10 ) (99 1)(200 10 ) 2.495 mA b) IE = IC + IB = 2.495 + 0.025 = 2.52 mA c) Neglecting the leakage current, we have giving an emitter current error of 6 ICB I 99(25 10 ) 2.475 mA 2.52 2.5 100% 0.793% I 2.475 IC 2.5 mA 2.52 E 0.99 51 Q4: A Si transistor connected in the common-emitter configuration (see fig.) has base current of 40 µA and ICBO = 0. If VBB = 6 V, RE = 1 kΩ, and β = 80, find (a) IE, (b) RB , (c) If VCC = 15 V and RC = 3 kΩ, find VCE . RE 52 Solution: a) IE = (β+1)IB = 81 × 0.04 = 3.24 mA b) KVL in input (B-E) loop: VBB = IBRB + VBE + IERE RB = (VBB – VBE – IERE)/IB = (6 – 0.7 – 3.24×1)/(0.04) kΩ RE = 51.85 kΩ b) KVL in output (C-E) loop: VCC = ICRC + VCE + IERE (IC = βIB = 80 × 0.04 = 3.2 mA) => VCE = VCC – IERE – ICRC = 15 – (3.24×1) – (3.2×3) = 2.16 V 53 Q5: Using the characteristics shown below, determine a. the resulting collector current if IE = 3 mA and VCB =10 V. b. the resulting collector current if IE remains at 3 mA but VCB is reduced to 2 V. c. VBE if IC = 4 mA and VCB = 20 V. 54 Solution: a. The output characteristics clearly indicate that IC ~ IE = 3 mA. b. The effect of changing VCB is negligible and IC continues to be 3 mA. c. From output characteristics, IE = IC = 4 mA. From input characteristics, for IE = 4 mA and VCB = 20 V, the resulting value of VBE is ~0.75 V. 55 Q6: Using the characteristics shown below, determine a. IC at IB = 30 μA and VCE = 10 V.
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