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TRANSISTOR

TRANSISTOR. BJT : DC BIASING. Transistor Currents. …( Eq. 3.1). Emitter current (I E ) is the sum of the collector current (I C ) and the base current (I B ) . Kirchhoff’s current law;. Collector Current (I C ). …( Eq. 3.2). Collector current (I C ) comprises two components ;

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TRANSISTOR

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  1. TRANSISTOR BJT : DC BIASING

  2. Transistor Currents …(Eq. 3.1) Emitter current (IE) is the sum of the collector current (IC) and the base current (IB) . Kirchhoff’s current law;

  3. Collector Current (IC) …(Eq. 3.2) • Collector current (IC) comprises two components; • majority carriers (electrons) from the emitter • minority carriers (holes) from reverse-biased BC junction → leakage current, ICBO • Total collector current (IC); • Since leakage current ICBO is usually so small that it can be ignored.

  4. Collector Current (IC) …(Eq. 3.3) Then; The ratio of IC to IE is called alpha (α), values typically range from 0.95 to 0.99.

  5. Base Current (IB) …(Eq. 3.4) IB is very small compared to IC; The ratio of IC to IB is the dc current gain of a transistor, called beta (β) The level of beta typically ranges from about 50 to over 400

  6. Current & Voltage Analysis 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. Consider below figure. Three dc currents and three dc voltages can be identified

  7. Current & Voltage Analysis …(Eq. 3.5) …(Eq. 3.6) When the BE junction is forward-biased, it is like a forward-biased diode. Thus; (Si = 0.7, Ge = 0.3) From HVK, the voltage across RB is By Ohm’s law; Solving for IB

  8. Current & Voltage Analysis …(Eq. 3.7) …(Eq. 3.8) The voltage at the collector is; The voltage drop across RC is VCE can be rewritten as The voltage across the reverse-biased CB junction is

  9. Transistor as Amplifier Transistor is capable to amplify AC signal : (output signal > input signal) Eg: Audio amplifier that amplify the sound of a radio

  10. 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)

  11. Transistor Amplifier Circuit Analysis Q-Point • Q-Point • Operating point of an amplifier to state the values of collector current (ICQ) and collector-emitter voltage (VCEQ). • Determined by using transistor output characteristic and DC load line

  12. DC LOAD LINE Saturation Region Q-Point DC Load Line Cutoff Region • DC Load Line • A straight line intersecting the vertical axis at approximately IC(sat) and the horizontal axis at VCE(off). • IC(sat) occurs when transistor operating in saturation region • VCE(off) occurs when transistor operating in cut-off region

  13. DC LOAD LINE (Example) VCC = 8V Draw DC Load Line and Find Q-point. Answers; RC = 2 kΩ RB = 360 kΩ

  14. 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

  15. DC Analysis of Amplifier Circuit Amplifier Circuit Amplifier Circuit w/o capacitor

  16. DC Analysis of Amplifier Circuit Amplifier Circuit w/o capacitor • 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

  17. DC Analysis of Amplifier Circuit Thevenin Theorem; Amplifier Circuit w/o capacitor Simplified Circuit

  18. DC Analysis of Amplifier Circuit From Thevenin Theorem; 2 1 From HVK; 1 2 Important equation for DC Analysis

  19. TRANSISTOR BJT BIASING CIRCUIT

  20. 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)

  21. 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

  22. FIXED BASE BIAS CIRCUIT 1st step: Locate capacitors and replace them with an open circuit

  23. FIXED BASE BIAS CIRCUIT BE Loop CE Loop 1 1 2 2 2nd step: Locate 2 main loops.

  24. FIXED BASE BIAS CIRCUIT • From HVK; IB 1 A BE Loop Analysis

  25. FIXED BASE BIAS CIRCUIT • From HVK; • As we known; • Subtituting with IC A B B 2 CE Loop Analysis

  26. 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.

  27. FIXED BASE BIAS CIRCUIT • 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 Example 1

  28. FIXED BIAS WITH EMITTER RESISTOR Resistor, RE added • 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 • CE loop

  29. FIXED BIAS WITH EMITTER RESISTOR 1st step: Locate capacitors and replace them with an open circuit

  30. FIXED BIAS WITH EMITTER RESISTOR CE Loop BE Loop 1 1 2 2 2nd step: Locate 2 main loops.

  31. FIXED BIAS WITH EMITTER RESISTOR • From HVK; • Recall; • Subtitute for IE 1 BE Loop Analysis

  32. FIXED BIAS WITH EMITTER RESISTOR • From HVK; • Assume; • Therefore; 2 CE Loop Analysis

  33. FIXED BIAS WITH EMITTER RESISTOR • 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 Example 2

  34. 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

  35. VOLTAGE DIVIDER BIAS CIRCUIT 1st step: Locate capacitors and replace them with an open circuit

  36. VOLTAGE DIVIDER BIAS CIRCUIT • 2nd step: : Simplified circuit using Thevenin Theorem From Thevenin Theorem; Thevenin Theorem; Simplified Circuit

  37. VOLTAGE DIVIDER BIAS CIRCUIT CE Loop BE Loop 1 1 2 2 2nd step: Locate 2 main loops.

  38. VOLTAGE DIVIDER BIAS CIRCUIT • From HVK; • Recall; • Subtitute for IE 1 BE Loop Analysis

  39. VOLTAGE DIVIDER BIAS CIRCUIT • From HVK; • Assume; • Therefore; 2 CE Loop Analysis

  40. VOLTAGE DIVIDER BIAS CIRCUIT • 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 Example 3

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