TRANSISTOR BJT DC BIASING Transistor Currents Emitter current

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; …(Eq. 3. 1)

Collector Current (IC) ■ Collector current (IC) comprises two components; Ø majority carriers (electrons) from the emitter Ø minority carriers (holes) from reverse-biased BC junction → leakage current, ICBO ■ Total collector current (IC); …(Eq. 3. 2) ■ Since leakage current ICBO is usually so small that it can be ignored.

■ Then; Collector Current (IC) …(Eq. 3. 3) ■ 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) …(Eq. 3. 4) ■ 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 baseemitter 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 forwardbiased diode. Thus; (Si = 0. 7, Ge = 0. 3) …(Eq. 3. 5) ■ From HVK, the voltage across RB is ■ By Ohm’s law; ■ Solving for IB …(Eq. 3. 6)

Current & Voltage Analysis ■ The voltage at the collector is; ■ The voltage drop across RC is ■ VCE can be rewritten as …(Eq. 3. 7) ■ The voltage across the reverse-biased CB junction is …(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; Ø DC Analysis Ø 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 Ø Operating point of an amplifier to state the values of collector current (ICQ) and collector-emitter voltage (VCEQ). Ø Determined by using transistor output characteristic and DC load line Q-Point

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

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

DC LOAD LINE (Example) Draw DC Load Line and Find Q-point. Answers; ¢ 2 m. A 4 V Q-point can be obtained by calculate the half values of maximum IC and VCE

DC Analysis of Amplifier Circuit w/o capacitor

DC Analysis of Amplifier Circuit ■ Refer to the figure, for DC analysis: Ø Replace capacitor with an open-circuit ■ R 1 and R 2 create a voltage-divider circuit ■ that connect to the base Therefore, from DC analysis, you can find: Ø IC Ø 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; 2 From HVK; 1 1 2

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 1 st step: Locate capacitors and replace them with an open circuit 2 nd step: Locate 2 main loops which; Ø BE loop Ø CE loop

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

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

FIXED BASE BIAS CIRCUIT ■ BE Loop Analysis ■ From HVK; 1 IB A

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

FIXED BASE BIAS CIRCUIT ■ DISADVANTAGE Ø Unstable – because it is too dependent on β and produce width change of Q-point Ø 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 m. A IB = 47. 08 μA VCE = 6. 83 V VB = 0. 7 V VC = 6. 83 V VBC = -6. 13 V

FIXED BIAS WITH EMITTER RESISTOR ■ An emitter resistor, RE is added ■ ■ ■ Resistor, RE added to improve stability Solve the circuit using HVK 1 st step: Locate capacitors and replace them with an open circuit 2 nd step: Locate 2 main loops which; Ø BE loop Ø CE loop

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

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

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

FIXED BIAS WITH EMITTER RESISTOR ■ CE Loop Analysis ■ From HVK; ■ Assume; 2 ■ Therefore;

FIXED BIAS WITH EMITTER RESISTOR ■ Example 2 ■ Find IC, IB, VCE, VB, VC, VE & VBC? (Silikon transistor); ■ Answers; IC = 2. 01 m. A IB = 40. 1 μA VCE = 13. 97 V VB = 2. 71 V VE = 2. 01 V VC = 15. 98 V VBC = -13. 27 V

VOLTAGE DIVIDER BIAS CIRCUIT ■ Provides good Q-point stability with ■ ■ a single polarity supply voltage Solve the circuit using HVK 1 st step: Locate capacitors and replace them with an open circuit 2 nd step: Simplified circuit using Thevenin Theorem 3 rd step: Locate 2 main loops which; Ø BE loop Ø CE loop

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

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

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

VOLTAGE DIVIDER BIAS CIRCUIT ■ BE Loop Analysis ■ From HVK; Recall; 1 Subtitute for IE

VOLTAGE DIVIDER BIAS CIRCUIT ■ CE Loop Analysis ■ From HVK; ■ Assume; 2 ■ Therefore;

VOLTAGE DIVIDER BIAS CIRCUIT ■ Example 3 ■ Find RTH, VTH, IC, IB, VCE, VB, VC, VE & VBC? (Silikon transistor); ■ Answers; RTH = 3. 55 kΩ VTH = 2 V IC = 0. 85 m. A IB = 6. 05 μA VCE = 12. 22 V VB = 1. 978 V VE = 1. 275 V VC = 13. 5 V VBC = -11. 522 V