VIEWS: 106 PAGES: 40 POSTED ON: 5/19/2012
TRANSISTOR BJT : DC BIASING Transistor Currents ■ Emitter current (IE) is the sum of the collector current (IC) and the base current (IB) . ■ Kirchhoff’s current law; IE IC IB …(Eq. 3.1) Collector Current (IC) ■ Collector current (IC) comprises two components; majority carriers (electrons) from the emitter I Cmajority I E minority carriers (holes) from reverse-biased BC junction → leakage current, ICBO ICminority ICBO ■ Total collector current (IC); I C I E I CBO …(Eq. 3.2) ■ Since leakage current ICBO is usually so small that it can be ignored. Collector Current (IC) ■ Then; IC …(Eq. 3.3) IE ■ The ratio of IC to IE is called alpha (α), values typically range from 0.95 to 0.99. Base Current (IB) ■ IB is very small compared to IC; ■ The ratio of IC to IB is the dc current gain of a transistor, called beta (β) IC …(Eq. 3.4) IB ■ The level of beta typically ranges from about 50 to over 400 Current & Voltage Analysis ■ Consider below figure. Three dc currents and three dc voltages can be identified IB: dc base current IE: dc emitter current IC: dc collector current VBE: dc voltage across base- emitter junction VCB: dc voltage across collector-base junction VCE: dc voltage from collector to emitter Transistor bias circuit. Current & Voltage Analysis ■ When the BE junction is forward-biased, it is like a forward- biased diode. Thus; (Si = 0.7, Ge = 0.3) VBE 0.7V …(Eq. 3.5) ■ From HVK, the voltage across RB is VR B VBB VBE ■ By Ohm’s law; VR B I B R B ■ Solving for IB VBB VBE IB …(Eq. 3.6) RB Current & Voltage Analysis ■ The voltage at the collector is; VCE VCC VR C ■ The voltage drop across RC is VR C I C R C ■ VCE can be rewritten as VCE VCC I C R C …(Eq. 3.7) ■ The voltage across the reverse-biased CB junction is VCB VCE VBE …(Eq. 3.8) Transistor as Amplifier ■ Transistor is capable to amplify AC signal : (output signal > input signal) ■ Eg: Audio amplifier that amplify the sound of a radio Transistor Amplifier Circuit Analysis ■ There are 2 analysis; DC Analysis AC Analysis ■ Transistor will operate when DC voltage source is applied to the amplifier circuit ■ Q-point must be determined so that the transistor will operate in active region (can operate as an amplifier) Transistor Amplifier Circuit Analysis ■ Q-Point Operating point of an amplifier to state the values of collector current (ICQ) and collector-emitter voltage (VCEQ). Q-Point Determined by using transistor output characteristic and DC load line DC LOAD LINE ■ DC Load Line A straight line intersecting the Saturation Region vertical axis at approximately IC(sat) and the horizontal axis at Q-Point VCE(off). DC Load Line IC(sat) occurs when transistor operating in saturation region VCC I Csat RC transistor VCE(off) occurs whenV 0 CE operating in cut-off region VCE( off ) VCC I C RC I C 0 Cutoff Region DC LOAD LINE (Example) VCC = 8V Draw DC Load Line and Find Q-point. Answers; RC = 2 kΩ VCC 8V RB = 360 kΩ I Csat 4mA RC VCE 0 2k VCE( off ) VCC I C RC I C 0 VCE( off ) VCC 8V DC LOAD LINE (Example) Draw DC Load Line and Find Q-point. Answers; Q-point can be obtained by calculate the half values of maximum IC and VCE 2 mA 4V DC Analysis of Amplifier Circuit Amplifier Circuit Amplifier Circuit w/o capacitor DC Analysis of Amplifier Circuit ■ Refer to the figure, for DC analysis: Replace capacitor with an open-circuit ■ R1 and R2 create a voltage-divider circuit that connect to the base ■ Therefore, from DC analysis, you can find: IC VCE Amplifier Circuit w/o capacitor DC Analysis of Amplifier Circuit Thevenin Theorem; Amplifier Circuit w/o capacitor Simplified Circuit DC Analysis of Amplifier Circuit ■ Important equation for DC Analysis From Thevenin Theorem; R1 R2 RTH R1 // R2 R1 R2 R2 VTH VCC R1 R2 2 From HVK; 1 VTH VBE 1 IB ; I C I B RTH ( 1) RE 2 VCE VCC I C ( RC RE ) TRANSISTOR BJT BIASING CIRCUIT BJT BIASING CIRCUIT ■ Fixed Base Bias Circuit (Litar Pincangan Tetap) ■ Fixed Bias with Emitter Resistor Circuit (Litar Pincangan Pemancar Terstabil) ■ Voltage-Divider Bias Circuit (Litar Pincangan Pembahagi Voltan) ■ Feedback Bias Circuit (Litar Pincangan Suap-Balik Voltan) FIXED BASE BIAS CIRCUIT ■ This is common emitter (CE) configuration ■ Solve the circuit using HVK ■ 1st step: Locate capacitors and replace them with an open circuit ■ 2nd step: Locate 2 main loops which; BE loop CE loop FIXED BASE BIAS CIRCUIT ■ 1st step: Locate capacitors and replace them with an open circuit FIXED BASE BIAS CIRCUIT ■ 2nd step: Locate 2 main loops. BE Loop CE Loop 1 2 1 2 FIXED BASE BIAS CIRCUIT ■ BE Loop Analysis ■ From HVK; 1 VCC I B R B VBE 0 VCC VBE IB IB A RB FIXED BASE BIAS CIRCUIT ■ CE Loop Analysis ■ From HVK; VCC IC R C VCE 0 IC VCE VCC I C R C ■ As we known; 2 I C I B B ■ Subtituting A with B V VBE I C DC CC RB FIXED BASE BIAS CIRCUIT ■ DISADVANTAGE Unstable – because it is too dependent on β and produce width change of Q-point For improved bias stability , add emitter resistor to dc bias. FIXED BASE BIAS CIRCUIT ■ Example 1 ■ Find IC, IB, VCE, VB, VC, VBC? (Silikon transistor); ■ Answers; IC = 2.35 mA IB = 47.08 μA VCE = 6.83V VB = 0.7V VC = 6.83V VBC = -6.13V FIXED BIAS WITH EMITTER RESISTOR ■ An emitter resistor, RE is added to improve stability ■ Solve the circuit using HVK ■ 1st step: Locate capacitors and replace them with an open circuit ■ 2nd step: Locate 2 main loops which; BE loop Resistor, RE added CE loop FIXED BIAS WITH EMITTER RESISTOR ■ 1st step: Locate capacitors and replace them with an open circuit FIXED BIAS WITH EMITTER RESISTOR ■ 2nd step: Locate 2 main loops. BE Loop CE Loop 1 2 1 2 FIXED BIAS WITH EMITTER RESISTOR ■ BE Loop Analysis 1 ■ From HVK; VCC I B RB VBE I E RE 0 Recall; I E ( 1) I B Subtitute for IE VCC I B RB VBE ( 1) I B RE 0 VCC VBE IB RB ( 1) RE FIXED BIAS WITH EMITTER RESISTOR ■ CE Loop Analysis ■ From HVK; VCC I C RC VCE I E RE 0 ■ Assume; 2 I E IC ■ Therefore; VCE VCC I C ( RC RE ) FIXED BIAS WITH EMITTER RESISTOR ■ Example 2 ■ Find IC, IB, VCE, VB, VC, VE & VBC? (Silikon transistor); ■ Answers; IC = 2.01 mA IB = 40.1 μA VCE = 13.97V VB = 2.71V VE = 2.01V VC = 15.98V VBC = -13.27V VOLTAGE DIVIDER BIAS CIRCUIT ■ Provides good Q-point stability with a single polarity supply voltage ■ Solve the circuit using HVK ■ 1st step: Locate capacitors and replace them with an open circuit ■ 2nd step: Simplified circuit using Thevenin Theorem ■ 3rd step: Locate 2 main loops which; BE loop CE loop VOLTAGE DIVIDER BIAS CIRCUIT ■ 1st step: Locate capacitors and replace them with an open circuit VOLTAGE DIVIDER BIAS CIRCUIT ■ 2nd step: : Simplified circuit using Thevenin Theorem From Thevenin Theorem; Thevenin Theorem; R1 R2 RTH R1 // R2 R1 R2 R2 VTH VCC R1 R2 Simplified Circuit VOLTAGE DIVIDER BIAS CIRCUIT ■ 2nd step: Locate 2 main loops. BE Loop CE Loop 2 2 1 1 VOLTAGE DIVIDER BIAS CIRCUIT ■ BE Loop Analysis ■ From HVK; VTH I B RTH VBE I E RE 0 Recall; I E ( 1) I B Subtitute for IE 1 VTH I B RTH VBE ( 1) I B RE 0 VTH VBE IB RRTH ( 1) RE VOLTAGE DIVIDER BIAS CIRCUIT ■ CE Loop Analysis ■ From HVK; VCC I C RC VCE I E RE 0 ■ Assume; I E IC 2 ■ Therefore; VCE VCC I C ( RC RE ) VOLTAGE DIVIDER BIAS CIRCUIT ■ Example 3 ■ Find RTH, VTH, IC, IB, VCE, VB, VC, VE & VBC? (Silikon transistor); ■ Answers; RTH = 3.55 kΩ VTH = 2V IC = 0.85 mA IB = 6.05 μA VCE = 12.22V VB = 1.978V VE = 1.275V VC = 13.5V VBC = -11.522V