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