* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download ppt - EC - Unit 1 - Transistor, UJT, SCR
Electrical substation wikipedia , lookup
Negative feedback wikipedia , lookup
Mains electricity wikipedia , lookup
Alternating current wikipedia , lookup
Resistive opto-isolator wikipedia , lookup
Immunity-aware programming wikipedia , lookup
Buck converter wikipedia , lookup
Current source wikipedia , lookup
Wien bridge oscillator wikipedia , lookup
Switched-mode power supply wikipedia , lookup
Opto-isolator wikipedia , lookup
Regenerative circuit wikipedia , lookup
Two-port network wikipedia , lookup
Rectiverter wikipedia , lookup
Power MOSFET wikipedia , lookup
Semiconductor device wikipedia , lookup
History of the transistor wikipedia , lookup
Current mirror wikipedia , lookup
Unit 1 Transistors, UJTs and Thyristors Unit 1: Transistors, UJTs and Thyristors • • • • • • Objectives: Operating Point CE configuration Thermal runaway UJT SCR 1.1 Operating (Q) Point To design an amplifier: 1. DC & AC analysis 2. Operating Point in active region by biasing circuits 1. 2. 3. 4. Fixed bias Emitter bias Collector-to-base bias Voltage divider bias with emitter bias …selecting suitable Q point …selecting suitable Q point • Point A no bias Cut-off region • Point B near to knee portion Do not allow more output swing • Point C close to PD(max) curve Output’s +ve swing is limited • Point D middle of active region Allows +ve & -ve excursions of output 1.2 CE configuration • Popular as provides considerable AV & AI To learn: • Biasing Circuits: 1. Fixed bias 2. Emitter bias (self bias) 3. Voltage divider bias with emitter bias • Analysis: DC, load-line • merits & demerits fixed bias circuit • Simplest • Biasing components: – 2 resistors (RB & RC) – Vcc • BE junction FB by Vcc through RB (100s of kΩ) • CB junction RB by Vcc through Rc (few k Ω) 5/10/2017 4:19:34 PM Unit 1: Transistors, UJTs and SCR 7 DC analysis To get DC equivalent circuit: Open Circuit all capacitors, and redraw the circuit 5/10/2017 4:19:34 PM Unit 1: Transistors, UJTs and SCR 8 … From fundamentals • VBE = VB – VE Similarly • VCE = VC – VE Next, by KVL to BE loop VCC – IBRB – VBE = 0 VCC – VBE IB = ----------------RB 5/10/2017 4:19:34 PM Unit 1: Transistors, UJTs and SCR 9 … But IC = β IB where β is transistor gain VCC - VBE IC = β ( ---------------- ) RB VCC IC = β ( ---------- ) RB 5/10/2017 4:19:34 PM Unit 1: Transistors, UJTs and SCR -----(1) 10 … β VCC IC ≈ -----------RB IC Where β is transistor current gain = -----IB 5/10/2017 4:19:34 PM Unit 1: Transistors, UJTs and SCR 11 … Next, apply KVL to output loop, VCC = ICRC + VCE VCE = VCC – ICRC 5/10/2017 4:19:34 PM ----(2) Unit 1: Transistors, UJTs and SCR 12 Put Ic = 0 in eqn (2) ∴ VCE = Vcc Put VCE = 0 in eqn (2) ∴ Ic = Vcc / Ic 5/10/2017 4:19:34 PM Unit 1: Transistors, UJTs and SCR 13 5/10/2017 4:19:34 PM Unit 1: Transistors, UJTs and SCR 14 … • Q point is influenced by – IB – base current – RC – collector resistance – VCC – supply voltage 5/10/2017 4:19:34 PM Unit 1: Transistors, UJTs and SCR 15 5/10/2017 4:19:34 PM Unit 1: Transistors, UJTs and SCR 16 5/10/2017 4:19:34 PM Unit 1: Transistors, UJTs and SCR 17 5/10/2017 4:19:34 PM Unit 1: Transistors, UJTs and SCR 18 Numerical example: Find values of ICQ & VCQ Find values of IC & VCE, forcing each value to zero, each time 5/10/2017 4:19:34 PM Unit 1: Transistors, UJTs and SCR 19 Solution: 5/10/2017 4:19:34 PM Unit 1: Transistors, UJTs and SCR 20 • Thus, ICQ = 1.43 mA & VCEQ = 9.28 V Which are the coordinates of Q point 5/10/2017 4:19:34 PM Unit 1: Transistors, UJTs and SCR 21 5/10/2017 4:19:34 PM Unit 1: Transistors, UJTs and SCR 22 5/10/2017 4:19:34 PM Unit 1: Transistors, UJTs and SCR 23 Emitter bias / self bias configuration • Additional resistor RE, improves stability, as it produces negative feedback. 5/10/2017 4:19:34 PM Unit 1: Transistors, UJTs and SCR 24 DC analysis • By KVL to input base loop, VCC – IB RB – VBE – IE RE= 0 VCC – VBE IB = -------------------- ----(1) RB + (β+1)RE … neglecting VBE VCC IB = -------------------RB + (β+1)RE Compare this equation with Iin = Vin -------Rin From basics, IC = IB + IE IB is << … IE = IC We get input resistance = RB + (β+1)RE Next, applying KVL to output collector loop, VCC – IC RC – VCE – IE RE = 0 VCE = VCC – IC (RC + RE) ------(2) … Also VE = IE RE = IC RE From eqn. (1) & (2) VCC – VBE ICQ = β -------------------RB + (β+1)RE VCEQ = VCC – IC (RC + RE) The coordinates of Q point Load line analysis We have VCE = VCC – IC (RC + RE) Put IC = 0 VCE = VCC│IC = 0 is point of load line on x – axis (VCC, 0) … For second point, Put VCE = 0 VCC IC = --------------(RC + RE) VCE = 0 … thus we have on y- axis (0, VCC/(RC + RE)) as another point on load line. How stability increased: ICQ VCC – VBE = β -------------------RB + (β+1)RE VCEQ = VCC – IC (RC + RE) RE introduces negative (ac emitter) feedback, due to which • Stability of Av increases How stability is improved by RE: IE ≈ Ic IE = VE / RE IE VE Due to temperature or else, IC VCC = IB RB + VBE + VE IC Initial increase and later decrease proves that RE improves stability IB = β IB VBE IB VBE = IB RB AC and DC Currents in an Amplifier: Small Signal Operation: ICQ IC = ICQ + ic ic IBQ ib IE = IEQ + ie IB = IBQ + ib IEQ ie Numerical example: • Refer to figure. Find the values of RB, RC & RB, given that IB=40µA, VE=2V, IC=4mA, VCE=12V & supply voltage VCC=15V. Assume silicon transistor. Numerical: Numerical: 3. Voltage-divider-bias with emitter-bias configuration: • Stability further improved, ( but gain decreases) • Most commonly used configuration • Gain can be improved by bypass capacitor, later studied. DC equivalent circuit: To get DC equivalent circuit: Open Circuit all capacitors, and redraw the circuit DC analysis 2 methods: 1. Accurate method (uses Thevenin’s theorem) 2. Approximate method 1. Accurate Method: Using Thevenin’s Theorem: DC Equivalent Circuit Redrawn showing Vcc Thevenin’s Equivalent Circuit Calculating RTH & VTH: RTH: Identify the 2 points short circuit all DC voltages Little consideration will show that – RTH is = parallel combination of RB1 & RB2 RTH = RB1 ║ RB2 RTH VTH: Consider the circuit, again – VTH is nothing but – the voltage across the 2 points OR across RB2 VTH RB2 VCC VTH = ---------------(RB1 + RB2) Now replace the base circuit with Thevenin’s equivalent circuit IE IC + IB IE = IC + IB Dividing by IC, as it is to be eliminated, ---- = ----IC IC ----IC 1 = 1 + ---β β + 1 = ---------β IC (β + 1) = -------------IC / IB IE = IB (β + 1) • Applying KVL to the base loop – VTH – IB RTH – VBE – IE RE = 0 Substituting IE = (β+1)IB VTH - VBE IB = ---------------------RTH + (β+1)RE … VTH - VBE IBQ = β --------------------RTH + (β+1)RE -----(1) By KVL to output collector loop, VCC –IC RC – VCE – IE VE = 0 VCE = VCC – IC RC – IC RE VCEQ = VCC – ICQ (RC + RE) as IC = IE -----(2) Hence equations (1) & (2) represent Q point. Load line analysis: • Same as that of emitter bias: Advantages & disadvantages of VDB: • Excellent stability (against temperature & β), because of –ve feedback introduced by RE. • But RE reduces gain AV. • Again CE bypasses ac signal, increasing AV, also maintaining stability. DC do not pass through CE as f=0 in XC = 1 / (2πfC)= resistance IE ie Numerical: Solution: 4. Collector-to-base bias configuration: Little bit looking back: We have studied – 1. base bias 2. emitter (self) bias 3. voltage-divider-bias with emitter-bias with, • DC analysis (to know Q point/ICQ & VCEQ) • load-line analysis (points A & B of loadline) • RB provides negative feedback, voltage-shunt feedback • Better (not excellent, as VDB with EB) stability DC analysis: IB + IC IB IC • KVL to base-emitter loop, VCC – (IB + IC) RC – IB RB – VBE = 0 • Substitute IC = β IB VCC – VBE IB = --------------------RB + (β + 1) RC VCC – VBE ICQ = β --------------------RB + (β + 1) RC -----(1) • KVL to collector-emitter loop, VCC – (IB + IC) RC –VCE = 0 VCE = VCC – (IB + IC) RC • Ignoring IB, VCE = VCC – IC RC At quiescent conditions, VCEQ = VCC – ICQ RC -----(2) Advantages & disadvantages: • Provides stability • Reduced gain, due to –ve feedback, can be reduced by bypassing capacitor, as shown below, We are not studying … • • • • 4.3 Common base circuit 4.4 Common collector circuit 4.5 Bias stabilization 4.6 Bias compensation But let us now study … • 4.7 Thermal Runaway 4.7 Thermal Runaway • Power dissipation depends on • Physical size • Its construction • Mounting arrangement • Maximum power rating is limited by • Temperature that CB junction can withstand • Ambient temperature • When ambient temperature , power rating , is known as “power derating” … • Power dissipation capability range – 100s milliwatts to 250 Watts • Max. CB temperature – Si – 150 °C to 225 °C – Ge – 60 °C to 100 °C Power dissipation PD PD(MAX) 100 Typical power derating curve Maximum operating temperature 200 Case temperature °C Defn.: when the transistor is in operation it dissipates power & its junction temperature rises, which in turn causes collector current to increase. This may lead to more power dissipation & further increase in temperature & subsequent increase in collector current. If this cycle continues, it may result in permanent damage to the transistor. This phenomenon is known as “Thermal Runaway”. PD Tj IC … • Steady state junction temperature, Tj – TA = θ PD Where Tj – junction temperature TA – ambient temperature PD – power dissipated in transistor θ – thermal resistance °C/W Thermal Resistance: ratio of the rise in transistor junction temperature to the amount of power dissipated. Which depends on – • Transistor size • Size of heat sink • Other cooling method, such as forced air Thermal resistance (θ): Tj – TA dTj θ = ----------- = -------PD dPD dPD 1 ----------- = -------dTj θ Operating Point considerations against thermal run-away: At Q2: IC ↑ Q2 to move on 100mW curve PD ↓ thermally stable At Q3: IC ↑ Q3 to move on 200mW curve PD ↑ thermally instable checked for thermal runaway 4.8 Transistor switch • Another major application of transistor is Switch … While designing, Transistor is heavily saturated IC(sat) = VCC / RC IB(sat) = IC(sat)/ β Usually IB(max) = IB(sat) x 0.25 The minimum i/p voltage to drive transistor in saturation is VIH = IB(max) RB + VBE Rsat = VCE(sat) / IC(sat) ≈ few tens of Ohms Transistor Switching Delay: During ON: tD=delay time=to 10% tr=rise time=10% to 90% turn-ON time=tD+tr During OFF: ts=storage time=fall to 10% tf=fall time=90% to 10% toff=turn-OFF time =ts+tf Numerical: An input pulse is applied to the transistor switch shown in figure. What is the minimum input voltage required to make the LED glow? It is given that the minimum current required by the LED to glow is 10mA, voltage drop across LED is 1.5V, BE voltage 0.7V, CE voltage at saturation is 0.5V. 2..IC(sat) ? VCE(sat) = 0.5V 1..Vp ? IB=IC / β =10mA / 100 VBE = 0.7V = 1µA IE=IC= 10mA 3..IB(sat) ? 4..IB(max) ? =β x IC(sat) =1.25 x IB(sat) 1.5V UJT (unijunction transistor) • • • • Only 1 pn junction, unlike 2 (BE & CB) Current controlled Negative resistance exists May be used as switch Construction: Equivalent Circuit: Intrinsic stand-off ratio: = RBB1 / (RBB1 + RBB2) VI characteristics: VP = VBB + V Operation: when VE ↑ -IEO reaches 0, then ↑ to IP at VP, IE ↑ with VE ↓ up to IV & VV known as negative resistance, further behaves as a normal resistance. UJT Relaxation Oscillator: SCR (silicon controlled rectifier): Operation: A momentary pulse applied to the gate increases the base current of npn transistor initiating regenerative feedback action;. This action ultimately drives both transistors to saturation causing switching-ON by conducting heavily. Characteristics: BEFB CB CB BE FB Questions: 1. Explain selecting a suitable operating point (104) 2. Explain DC & load-line analysis of fixed bias (105) 3. What are the parameters that vary the Q – point (ans:- IB, RC, VCC -108) 4. Problem 4.1 (109) 5. Explain DC & load-line analysis of emitter / self bias (110) 6. Problem 4.2 & 4.3 (113) 7. Explain DC & load-line analysis of VDB with EB circuit (115) 8. Problem 4.4 (119) 9. Explain DC & load-line analysis of collector-tobase bias (124) 10. Problem 4.7 (127) 11. Define thermal runaway. Explain operating point considerations in thermal runaway (147) 12. Explain transistor as a switch. Brief about switching delays (152) 13. Explain UJT, relaxation oscillator (215) 14. Explain SCR (227) End of Unit 1: Transistors, UJT & Thyristors IC (Integrated Circuit) Overview 5/10/2017 4:19:35 PM Unit 1: Transistors, UJTs and SCR 87 5/10/2017 4:19:35 PM Unit 1: Transistors, UJTs and SCR 88 5/10/2017 4:19:35 PM Unit 1: Transistors, UJTs and SCR 89 5/10/2017 4:19:35 PM Unit 1: Transistors, UJTs and SCR 90 5/10/2017 4:19:35 PM Unit 1: Transistors, UJTs and SCR 91 5/10/2017 4:19:35 PM Unit 1: Transistors, UJTs and SCR 92 5/10/2017 4:19:35 PM Unit 1: Transistors, UJTs and SCR 93 5/10/2017 4:19:35 PM Unit 1: Transistors, UJTs and SCR 94 5/10/2017 4:19:35 PM Unit 1: Transistors, UJTs and SCR 95 Vin = HIGH (≈5Volts) Saturation region, IC(sat)=VCC / RC VCE = VCC - IC(sat) RC = 0 Volt = logic LOW 5/10/2017 4:19:35 PM Cut-off region, IC ≈ 0 VCE = VCC - IC RC = logic LOW Unit 1: Transistors, UJTs and SCR 96