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6.101 Course Roadmap Time Domain Analysis

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6.101 Course Roadmap Time Domain Analysis 6.101 Course Roadmap • Passive components: RLC –with RF • Diodes • Transistors: BJT, MOSFET, antennas • Op‐amps, 555 timer, ECG • • Course Roadmap Switch Mode Power Supplies • Rectification • • Bipolar Junction Transistor Fiber optics, PPG • Applications Acnowledgements: Neamen, Donald: Microelectronics Circuit Analysis and Design, 3rd Edition The Art Of Electronics by Horowitz and Hill 6.101 Spring 2020 Lecture 3 1 6.101 Spring 2020 Lecture 3 2 Time Domain Analysis Fourier Series ‐ Ramp v (Ac KAm cosmt)*cosct KA v A cos t m [cos( )t cos( )t] c c 2 c m c m function [ t, sum ] = ramp(number) %generate a ramp based on fixed number of terms % t = 0:.1:pi*4; % display two full cycles with 0.1 spacing sum = 0 for n=1:number sum = sum + sin(n*t)*(-1)^(n+1)/(n*pi); end plot(t, sum) shg end 6.101 Spring 2020 Lecture 3 3 6.101 Spring 2020 Lecture 3 4 CT: center tap Rectifier Circuits Full Wave Bridge vs Center Tapped + 1N4001 + V = 120 V 60 Hz 12.6 VCT RMS C F R L v OUT out - Pri Sec 3a) Half-wave rectifier circuit diagram 1N4001 + + 120 V 60 Hz 12.6 VCT RMS C F R L v OUT Vout = - Pri Sec Center tapped advantages: 1N4001 3b) Full-wave rectifier circuit diagram • Lower diode voltage drop (high efficiency) 4x 1N4001 + • Secondary windings carries ½ average + 12.6 VCT RMS 120 V 60 Hz CF R v L OUT Vout = current (thinner windings, easier to wind) - Pri Sec • Used in computer power supplies 3c) Bridge rectifier circuit diagram RC >> 16.6ms why? 6.101 Spring 2020 Lecture 3 5 6.101 Spring 2020 Lecture 3 6 Power Supply Ripple Voltage Calculation Physical Wiring Matters D2 conduction angle in degrees 6.101 Spring 2020 Lecture 3 7 6.101 Spring 2020 Lecture 3 8 5 V Adapters Diode AC Resistance 500 ma 1000 ma 300 ma 6.101 Spring 2020 Lecture 3 9 6.101 Spring 2020 Lecture 3 10 Log Amplifier Bipolar Junction Transistors bypass caps 0.1uf caps (2) NPN collector • BJT can operate in a linear ic = βib mode (amplifier) or can ID i IR = - ID b operate as a digital switch. 1N914 • IR Current controlled device Vout = - VD base • Two families: npn and pnp. 0.1 F 1.5k +15 i + i 2 b c • BJT’s are current controlled - 7 LF356 6 qV qV vout D D emitter devices 4 kT kT + I I (e 1) I e v 3 + D S S • _ in 0.1F -15 NPN – 2N2222 • PNP – 2N2907 • VCE ~30V, 500 mw power PNP 6.101 Spring 2020 Lecture 3 11 6.101 Spring 2020 Lecture 3 12 Why BJT’s ? • Preferred device for demanding analog application, both integrated and discrete (lower noise) • Great for high frequency applications; characteristics well understood. • High reliability makes it a key device in automotive applications. • Lower output resistance at emitter vs source • Larger gm compared to FET 6.101 Spring 2020 Lecture 3 13 6.101 Spring 2020 Lecture 3 14 BJT Symbols Packaging TO-18 TO-220 TO-3 2N2222 2N3904 2N3906 1 P2N2222 pinout reversed 2 3 6.101 Spring 2020 Lecture 3 15 6.101 Spring 2020 Lecture 3 16 BJT Current Relationship max voltage NPN collector iE iC iB ic = βib iC iB max continuous current iE ( 1)iB base max power at 25o C ib + ic iC iE emitter hFE = β = large signal (DC) gain at fixed current 1 hFE < hfe 6.101 Spring 2020 Lecture 3 17 6.101 Spring 2020 Lecture 3 18 hFE & Current & Temperature hFE = f(Ic) peaks at ~ 0.5-10ma Characteristics β hFE @1.0ma < hfe @1.0ma 6.101 Spring 2020 Lecture 3 19 6.101 Spring 2020 Lecture 3 20 NPN Common Emitter V‐I Relationship (James) Early Voltage A large VA is desirable for high voltage gains ~ 30-50v. VA is determined by transistor design and varies with base width, base and collector doping concentration. Early effect: the rise of Ic due to base- β = ? width modulation. 6.101 Spring 2020 Lecture 3 21 6.101 Spring 2020 Lecture 3 22 Tek 575 Curve Tracer Mcube • Vertical axis: current • Tests: • Horizontal axis: voltage – Diodes (forward • Voltage sweep: positive drop) and negative with resistor – BJT (type, beta) current limit 0‐20v; 0‐200v! – MOSFET (type, VTH and more) • Input: fixed current steps (0.001‐200ma); 240 steps • Auto terminal • Tests: diodes, BJT, MOSFETs identification • Calibrate zero current step 6.101 Spring 2020 Lecture 3 23 6.101 Spring 2020 Lecture 3 24 RLC –BJT MOSFET Testor BJT Configurations Voltage Current Power Gain Gain Gain Common Emitter X X X Common Collector X X Common Base X X Common emitter: hgh input impedance, for general amplification of voltage, current and power from low power, high impedance sources. Common collector: aka "emitter follower" for high input impedance and current gain without voltage gain, as in an amplifier output stage. Common base: low input impedance for low impedance sources, for high frequency response. Grounding the base short circuits the Miller capacitance from collector to base and makes possible much higher frequency response. 6.101 Spring 2020 Lecture 3 25 6.101 Spring 2020 Lecture 3 26 General Configuration Transistor Configurations TRANSISTOR AMPLIFIER CONFIGURATIONS Common Emitter +15V +15V +15V RL RL R2 R2 R2 + + + + + + + + + + + + R + 1 VOUT VOUT V Vin R 1 RE V V in R OUT in R1 E RE - - Common - - - - Common Base Collector [a] Common Emitter Amplifier [b] Common Collector [Emitter Follower] Amplifier [c] Common Base Amplifier 6.101 Spring 2020 Lecture 3 27 6.101 Spring 2020 Lecture 3 28 Common Emitter Operation –Quiescent Point Load Line – Operating Point +20 V 910 I R CQ 2 • Find Vout open circuit voltage: 20V + • Find I max = 20/(910 +91) = ~20ma 2N3904 CQ • Draw load line. vout R1 91 BFC - • For RE = 0, just choose Q at ½ VCC for maximum swing. • For RE > 0, set Q at ½ [VCC –VRE]. • For ICQ = 10 mA, VRL = 9.1V, VRE = 0.91V, VCE = 10V. For ICQ = 10.5mA, VRL = 9.6V, VRE = 0.96V, VCE = 9.5V 6.101 Spring 2020 Lecture 3 29 6.101 Spring 2020 Lecture 3 30 Variation of Collector Current with β Transistor Bias Instability One Resistor +15V IR07. V IR V BB CE CC +15V IRBB07. V FBE IR V CC F VCC 0.7V IRBB FE R V CC07. V IC 2 IC = 4 mA RB RB F RE VVCC 07. I 1 IC = 4 mA B RB RRBFE 2N3904 8.8V VV 07. Variation of Collector Current with Beta FCC 2N3904 IB IC 2 RRBFE IC F IB R = 2200 IE = 4 mA E 2.9 mA 50 VV 07. FCC R = 2200 IE = 4 mA RRBFE E 4.0 mA 100 IC 100 15VV 0. 7 5.0 mA 200 R 100 2200 B 4mA 5.4 mA 300 F 100, IC F I B 1430 3 Rk220 10 IC=2.5 mA B 4 I E F 1 I B , I E IC RkB 220 358 k RkB 138 6.101 Spring 2020 Lecture 3 31 6.101 Spring 2020 Lecture 3 32 Two Resistor Biasing Two Resistor Biasing VIRVIRBBB07. CE 0 +15V +15V +15V +15V VIRVBBB07. FBE IR VVIRIRIRRBBBFBEBBFE07. IC = 4 mA I = 4 mA IC = 4 mA R C I = 4 mA 2 R C 2N3904 2 IB 2N3904 R = R TH B VV 07. I 2N3904 R B B B IB 5 2N3904 R = R = R E RR TH B R 2200 VTH= VB BFE 1 I = 4 mA RE = 2200 C V RB B [b] R = [a] [c] E V = V VV 07. R 1 2200 TH B R = 2200 FB IC = 4 mA E I 6 V C B RRBFE [b] [a] [c] Assume RB = 22kΩ, 4mA22k 220k 100 VB 0.7V 4mA 242k 100VB 70 R1 R1R2 βRE = 220kΩ and ignore RB VB Vcc 3 RB R1 //R2 4 968 70 100VB R1 R2 R1 R2 VB 10.4V 6.101 Spring 2020 Lecture 3 33 6.101 Spring 2020 Lecture 3 34 Two Resistor Biasing Variation of Collector Current with β Two Resistor Biasing RR12 RkB 22 Given VB= 10.4 V and RR12 R = 22kΩ, we can now RR 045. B 11 22k Variation of Collector Current with Beta solve equations (3) and RR11 045. F VB 0.7V IC 6 (4) for R1 and R2. 045. R2 R R 1 22k B F E Two Resistor One Resistor 145. R1 IC IC F 0310. Rk1 22 10.. 4 0 7V 3.7 mA 2.9 mA 50 Rkusek 709. 68 I F 1 C 22k 2200 RR0.... 45 0 45 70 9 k 319 kusek 33 F 4.0 mA 4.0 mA 100 R 21 V 1 V 4.2 mA 5.0 mA 200 BCCRR 12 4.3 mA 5.4 mA 300 V 15V CC I =0.6 mA I =2.5 mA RR121 R R 1 145. R 1 C C VB 10. 4V RR12145. R 1 045. RR12 6.101 Spring 2020 Lecture 3 35 6.101 Spring 2020 Lecture 3 36 Base Current –Resistor Divider Common Collector –Emitter Follower Biasing +15V • Β = 100, iB = 7.5ma/100 =‐ 75µa 7.5 mA • Using Thevenin equivalent, R I 2 C F R A 15 1 3.7 mA 50 2N3904 RB = R1||R2, VB = R1 R2 4.0 mA 100 68K R 1 1.0 k 7.5 mA 4.2 mA 200 VB = IBRB + 0.6V + 7.5V B 4.3 mA 300 VB = [75 µA x 10k] + 0.6V + 7.5V ib VB = 750 mV + 0.6V + 7.5V IC=0.6 mA +15V VB = 8.9V 7.5 mA [15 R1] ÷ [R1 + R2] = 8.9V 33K 15 R1 = 8.9 x [R1 + R2] Make small 2N3904 [15−8.9] R = 8.9 R ib IB 1 2 compared to the RB R1 = 1.44 R2 [R x R ] ÷ [R + R ] = 10 kΩ current through R2 7.5 V 1 2 1 2 VB [1.44R2 x R2] ÷ [1.44 R2 + R2] = 10kΩ R = 16.9 kΩ (use 16 kΩ) See handout: Transistor bias stability 2 R1 = 1.44 R2 = 24.4 kΩ (use 24 kΩ) 6.101 Spring 2020 Lecture 3 37 6.101 Spring 2020 Lecture 3 38 Common Collector –Emitter Follower Biasing Bootstrapping –Higher Input Impedance • With R1 = 24kΩ, R2 = 16 kΩ, the +15V current through the voltage divider is The base is connected to the emitter 15 ÷[40 kΩ] = 375 µA.
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