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TRANSISTOR BJT : DC BIASING

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Transistor Currents ■ Emitter current (I E ) is the sum of the collector current (I C ) and the base current (I B ). ■ Kirchhoff’s current law; …(Eq. 3.1)

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

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Collector Current (I C ) ■ Then; ■ The ratio of I C to I E is called alpha (α), values typically range from 0.95 to …(Eq. 3.3)

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Base Current (I B ) ■ I B is very small compared to I C ; ■ The ratio of I C to I B 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)

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Current & Voltage Analysis ■ Consider below figure. Three dc currents and three dc voltages can be identified I B : dc base current I E : dc emitter current I C : dc collector current V BE : dc voltage across base- emitter junction V CB : dc voltage across collector-base junction V CE : dc voltage from collector to emitter Transistor bias circuit.

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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 R B is ■ By Ohm’s law; ■ Solving for I B …(Eq. 3.5) …(Eq. 3.6)

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Current & Voltage Analysis ■ The voltage at the collector is; ■ The voltage drop across R C is ■ V CE can be rewritten as ■ The voltage across the reverse-biased CB junction is …(Eq. 3.8) …(Eq. 3.7)

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

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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)

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Transistor Amplifier Circuit Analysis ■ Q-Point Operating point of an amplifier to state the values of collector current (I CQ ) and collector-emitter voltage (V CEQ ). Determined by using transistor output characteristic and DC load line Q-Point

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

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DC LOAD LINE (Example) V CC = 8V R B = 360 kΩ R C = 2 kΩ Draw DC Load Line and Find Q-point. Answers;

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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 I C and V CE 4V 2 mA

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DC Analysis of Amplifier Circuit Amplifier CircuitAmplifier Circuit w/o capacitor

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DC Analysis of Amplifier Circuit ■ Refer to the figure, for DC analysis: 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: I C V CE Amplifier Circuit w/o capacitor

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DC Analysis of Amplifier Circuit Amplifier Circuit w/o capacitorSimplified Circuit Thevenin Theorem;

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DC Analysis of Amplifier Circuit ■ Important equation for DC Analysis From HVK; From Thevenin Theorem;

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TRANSISTOR BJT BIASING CIRCUIT

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■ 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)

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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; BE loop CE loop

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

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FIXED BASE BIAS CIRCUIT ■ 2 nd step: Locate 2 main loops BE LoopCE Loop

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

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

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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.

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FIXED BASE BIAS CIRCUIT ■ Example 1 ■ Find I C, I B, V CE, V B, V C, V BC ? (Silikon transistor); ■ Answers; I C = 2.35 mA I B = μA V CE = 6.83V V B = 0.7V V C = 6.83V V BC = -6.13V

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FIXED BIAS WITH EMITTER RESISTOR ■ An emitter resistor, R E is 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; BE loop CE loop Resistor, R E added

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

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

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

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

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FIXED BIAS WITH EMITTER RESISTOR ■ Example 2 ■ Find I C, I B, V CE, V B, V C, V E & V BC ? (Silikon transistor); ■ Answers; I C = 2.01 mA I B = 40.1 μA V CE = 13.97V V B = 2.71V V E = 2.01V V C = 15.98V V BC = V

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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; BE loop CE loop

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

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

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

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

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

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VOLTAGE DIVIDER BIAS CIRCUIT ■ Example 3 ■ Find R TH, V TH, I C, I B, V CE, V B, V C, V E & V BC ? (Silikon transistor); ■ Answers; R TH = 3.55 kΩ V TH = 2V I C = 0.85 mA I B = 6.05 μA V CE = 12.22V V B = 1.978V V E = 1.275V V C = 13.5V V BC = V

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