• Course Roadmap • Rectification • Bipolar Junction Transistor 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 Course Roadmap • Passive components: RLC – with RF • Diodes • Transistors: BJT, MOSFET, antennas • Op‐amps, 555 timer, ECG • Switch Mode Power Supplies • Fiber optics, PPG • Applications 6.101 Spring 2020 Lecture 3 2 Time Domain Analysis v (Ac KAm cosmt)*cosct KA v A cos t m [cos( )t cos( )t] c c 2 c m c m 6.101 Spring 2020 Lecture 3 3 Fourier Series ‐ Ramp 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 4 CT: center tap Rectifier Circuits + 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 1N4001 3b) Full-wave rectifier circuit diagram 4x 1N4001 + + 12.6 VCT RMS 120 V 60 Hz CF R v L OUT Vout = - Pri Sec 3c) Bridge rectifier circuit diagram RC >> 16.6ms why? 6.101 Spring 2020 Lecture 3 5 Full Wave Bridge vs Center Tapped Center tapped advantages: • Lower diode voltage drop (high efficiency) • Secondary windings carries ½ average current (thinner windings, easier to wind) • Used in computer power supplies 6.101 Spring 2020 Lecture 3 6 Physical Wiring Matters 6.101 Spring 2020 Lecture 3 7 Power Supply Ripple Voltage Calculation D2 conduction angle in degrees 6.101 Spring 2020 Lecture 3 8 5 V Adapters 500 ma 1000 ma 300 ma 6.101 Spring 2020 Lecture 3 9 Diode AC Resistance 6.101 Spring 2020 Lecture 3 10 Log Amplifier bypass caps 0.1uf caps (2) ID IR = - ID 1N914 IR Vout = - VD 0.1 F 1.5k +15 2 - 7 LF356 6 qV qV vout D D 4 kT kT + I I (e 1) I e v 3 + D S S _ in 0.1F -15 6.101 Spring 2020 Lecture 3 11 Bipolar Junction Transistors NPN collector • BJT can operate in a linear ic = βib mode (amplifier) or can ib operate as a digital switch. • Current controlled device base • Two families: npn and pnp. i + i b c • BJT’s are current controlled emitter devices • NPN – 2N2222 • PNP – 2N2907 • VCE ~30V, 500 mw power PNP 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 2N2222 2N3904 2N3906 1 P2N2222 pinout reversed 2 3 6.101 Spring 2020 Lecture 3 15 Packaging TO-18 TO-220 TO-3 6.101 Spring 2020 Lecture 3 16 BJT Current Relationship NPN collector iE iC iB ic = βib iC iB iE ( 1)iB base ib + ic iC iE emitter hFE = β = large signal (DC) gain at fixed current 1 hFE < hfe 6.101 Spring 2020 Lecture 3 17 max voltage max continuous current max power at 25o C 6.101 Spring 2020 Lecture 3 18 hFE = f(Ic) peaks at ~ 0.5-10ma β hFE @1.0ma < hfe @1.0ma 6.101 Spring 2020 Lecture 3 19 hFE & Current & Temperature Characteristics 6.101 Spring 2020 Lecture 3 20 NPN Common Emitter V‐I Relationship β= ? 6.101 Spring 2020 Lecture 3 21 (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- width modulation. 6.101 Spring 2020 Lecture 3 22 Tek 575 Curve Tracer • Vertical axis: current • Horizontal axis: voltage • Voltage sweep: positive and negative with resistor current limit 0‐20v; 0‐200v! • Input: fixed current steps (0.001‐200ma); 240 steps • Tests: diodes, BJT, MOSFETs • Calibrate zero current step 6.101 Spring 2020 Lecture 3 23 Mcube • Tests: – Diodes (forward drop) – BJT (type, beta) – MOSFET (type, VTH and more) • Auto terminal identification 6.101 Spring 2020 Lecture 3 24 RLC – BJT MOSFET Testor 6.101 Spring 2020 Lecture 3 25 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 26 Circuit analysis by inspection 6.101 Spring 2020 Lecture 3 27 General Configuration Common Emitter Common Common Base Collector 6.101 Spring 2020 Lecture 3 28 Transistor Configurations TRANSISTOR AMPLIFIER CONFIGURATIONS +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 - - - - - - [a] Common Emitter Amplifier [b] Common Collector [Emitter Follower] Amplifier [c] Common Base Amplifier 6.101 Spring 2020 Lecture 3 29 Common Emitter Operation – Quiescent Point 6.101 Spring 2020 Lecture 3 30 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 31 Transistor Bias Instability +15V IRBB 07. V IR CE V CC IRBB07. V FBE IR V CC IRBB FE R V CC07. V IC = 4 mA RB VVCC 07. IB 1 RRBFE 2N3904 8.8V FCCVV 07. IB IC 2 RRBFE I = 4 mA RE = 2200 E FCCVV 07. RRBFE IC 100 15VV 0. 7 R 100 2200 B 4mA 100, I I 1430 F C F B Rk220 103 B 4 I E F 1 I B , I E IC RkB 220 358 k RkB 138 6.101 Spring 2020 Lecture 3 32 Variation of Collector Current with β One Resistor +15V F VCC 0.7V IC 2 RB F RE IC = 4 mA RB Variation of Collector Current with Beta 2N3904 IC F IB 2.9 mA 50 R = 2200 IE = 4 mA E 4.0 mA 100 5.0 mA 200 5.4 mA 300 IC=2.5 mA 6.101 Spring 2020 Lecture 3 33 Two Resistor Biasing +15V +15V IC = 4 mA IC = 4 mA R2 2N3904 IB 2N3904 RTH= RB RB R = E V = V R 1 2200 TH B R = 2200 IC = 4 mA E VB [b] [a] [c] R1 R1R2 VB Vcc 3 RB R1 //R2 4 R1 R2 R1 R2 6.101 Spring 2020 Lecture 3 34 Thevenin Circuit 6.101 Spring 2020 Lecture 3 35 Two Resistor Biasing VIRVIRBBB 07. CE 0 +15V +15V VIRVBBB07. FBE IR VVIRIRIRRBBBFBEBBFE07. IC = 4 mA IC = 4 mA R2 2N3904 IB 2N3904 RTH= RB R VVB 07. B IB 5 R = E V = V RR R 1 2200 TH B R = 2200 BFE IC = 4 mA E VB [b] [a] [c] FBVV 07. IC 6 RRBFE Assume RB = 22kΩ, 4mA22k 220k 100VB 0.7V 4mA 242k 100VB 70 βRE = 220kΩ and ignore RB 968 70 100VB VB 10.4V 6.101 Spring 2020 Lecture 3 36 Two Resistor Biasing RR12 RkB 22 Given VB= 10.4 V and RR12 R = 22kΩ, we can now RR 045. B 11 22k solve equations (3) and RR11 045. (4) for R1 and R2. 045. R2 1 22k 145. R1 0310. Rk1 22 Rkusek1 709. 68 RR210.... 45 0 45 70 9 k 319 kusek 33 R1 VBCC V RR12 VCC 15V RR121 R R 1 145. R 1 VB 10. 4V RR12145. R 1 045. RR12 6.101 Spring 2020 Lecture 3 37 Variation of Collector Current with β Two Resistor Biasing V 0.7V Variation of Collector Current with Beta I F B 6 C R R B F E Two Resistor One Resistor IC IC F F 10.. 4 0 7V 3.7 mA 2.9 mA 50 IC 22k F 2200 4.0 mA 4.0 mA 100 4.2 mA 5.0 mA 200 4.3 mA 5.4 mA 300 IC=0.6 mA IC=2.5 mA 6.101 Spring 2020 Lecture 3 38 Base Current – Resistor Divider IC F 3.7 mA 50 4.0 mA 100 68K 4.2 mA 200 4.3 mA 300 ib IC=0.6 mA 33K Make i b small compared to the current through R2 See handout: Transistor bias stability 6.101 Spring 2020 Lecture 3 39 Common Collector – Emitter Follower Biasing +15V • Β = 100, iB = 7.5ma/100 =‐ 75µa 7.5 mA • Using Thevenin equivalent, R2 R A 15 1 2N3904 RB = R1||R2, VB = R1 R2 R 1.0 k 1 7.5 mA VB = IBRB + 0.6V + 7.5V B VB = [75 µA x 10k] + 0.6V + 7.5V VB = 750 mV + 0.6V + 7.5V +15V VB = 8.9V 7.5 mA [15 R1] ÷ [R1 + R2] = 8.9V 15 R1 = 8.9 x [R1 + R2] 2N3904 [15−8.9] R = 8.9 R IB 1 2 RB R1 = 1.44 R2 7.5 V [R1 x R2] ÷ [R1 + R2] = 10 kΩ VB [1.44R2 x R2] ÷ [1.44 R2 + R2] = 10kΩ R2 = 16.9 kΩ (use 16 kΩ) R1 = 1.44 R2 = 24.4 kΩ (use 24 kΩ) 6.101 Spring 2020 Lecture 3 40 Common Collector – Emitter Follower Biasing • With R1 = 24kΩ, R2 = 16 kΩ, the +15V current through the voltage divider is 15 ÷ [40 kΩ] = 375 µA.