# TRANSISTOR BJT : DC BIASING.

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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; 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. …(Eq. 3.2)

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

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) The level of beta typically ranges from about 50 to over 400 …(Eq. 3.4)

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) From HVK, the voltage across RB is By Ohm’s law; Solving for IB …(Eq. 3.5) …(Eq. 3.6)

Current & Voltage Analysis
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 …(Eq. 3.7) …(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). Determined by using transistor output characteristic and DC load line Q-Point

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 Saturation Region Q-Point DC Load Line Cutoff Region

DC LOAD LINE (Example) Draw DC Load Line and Find Q-point. Answers;
VCC = 8V 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;
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 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; 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 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 A IB

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

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 = 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 CE loop Resistor, RE added

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 2 1

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

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

FIXED BIAS WITH EMITTER RESISTOR
Find IC, IB, VCE, VB, VC, VE & VBC? (Silikon transistor); Example 2 Answers; IC = 2.01 mA IB = 40.1 μA VCE = 13.97V VB = 2.71V VE = 2.01V VC = 15.98V VBC = V

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; Simplified Circuit

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

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

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

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 = V

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