Download presentation

1
TRANSISTOR BJT : DC BIASING

2
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)

3
**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)

4
**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)

5
**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)

6
**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.

7
**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)

8
**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)

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 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

12
**DC LOAD LINE 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 Saturation Region Q-Point DC Load Line Cutoff Region

13
**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

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 w/o capacitor

16
**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

17
**DC Analysis of Amplifier Circuit**

Thevenin Theorem; Amplifier Circuit w/o capacitor Simplified Circuit

18
**DC Analysis of Amplifier Circuit**

Important equation for DC Analysis From Thevenin Theorem; 2 From HVK; 1 1 2

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**

2nd step: Locate 2 main loops. BE Loop CE Loop 1 2 1 2

24
**FIXED BASE BIAS CIRCUIT**

BE Loop Analysis From HVK; 1 A IB

25
**FIXED BASE BIAS CIRCUIT**

CE Loop Analysis From HVK; As we known; Subtituting with 2 IC B B A

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**

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

28
**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

29
**FIXED BIAS WITH EMITTER RESISTOR**

1st step: Locate capacitors and replace them with an open circuit

30
**FIXED BIAS WITH EMITTER RESISTOR**

2nd step: Locate 2 main loops. BE Loop CE Loop 1 2 2 1

31
**FIXED BIAS WITH EMITTER RESISTOR**

BE Loop Analysis From HVK; Recall; Subtitute for IE 1

32
**FIXED BIAS WITH EMITTER RESISTOR**

CE Loop Analysis From HVK; Assume; Therefore; 2

33
**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

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**

2nd step: Locate 2 main loops. BE Loop CE Loop 2 1 2 1

38
**VOLTAGE DIVIDER BIAS CIRCUIT**

BE Loop Analysis From HVK; Recall; Subtitute for IE 1

39
**VOLTAGE DIVIDER BIAS CIRCUIT**

CE Loop Analysis From HVK; Assume; Therefore; 2

40
**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

Similar presentations

OK

DC Biasing - BJTs Chapter 4 Boylestad Electronic Devices and Circuit Theory.

DC Biasing - BJTs Chapter 4 Boylestad Electronic Devices and Circuit Theory.

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google