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;
I E  IC  IB
…(Eq. 3.1)
Collector Current (IC)
■ Collector current (IC) comprises two components;
 majority carriers (electrons) from the emitter
I Cmajority   I E
 minority carriers (holes) from reverse-biased BC junction → leakage
current, ICBO
■
I Cminority  I CBO
Total collector current (IC);
I C  I E  I CBO
…(Eq. 3.2)
■ Since leakage current ICBO is usually so small that it can be
ignored.
■ Then;
Collector Current (IC)
IC

IE
…(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)
 
IC
IB
…(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)
VBE  0.7V
…(Eq. 3.5)
■ From HVK, the voltage across RB is
VR B  VBB  VBE
■ By Ohm’s law;
VR B  I B R B
■ Solving for IB
IB 
VBB  VBE
RB
…(Eq. 3.6)
Current & Voltage Analysis
■ The voltage at the collector is;
VCE  VCC  VR C
■ The voltage drop across RC is
VR C  I C R C
■ VCE can be rewritten as
VCE  VCC  IC R C
…(Eq. 3.7)
■ The voltage across the reverse-biased CB junction is
VCB  VCE  VBE
…(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
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
Saturation Region
Q-Point
DC Load Line
VCC
I Csa t 
RC Vtransistor
0
 VCE(off) occurs when
CE
operating in cut-off region
VCE( off )  VCC  I C RC
I C 0
Cutoff Region
DC LOAD LINE (Example)
VCC = 8V
Draw DC Load Line and Find Q-point.
Answers;
RC = 2 kΩ
RB = 360 kΩ
I Csa t 
VCC
RC

VCE  0
8V
 4mA
2k
VCE( off )  VCC  I C RC
VCE( off )  VCC  8V
I C 0
DC LOAD LINE (Example)
Draw DC Load Line and Find Q-point.
Answers;

2 mA
4V
Q-point can be obtained by
calculate the half values of
maximum IC and VCE
DC Analysis of Amplifier Circuit
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;
RTH  R1 // R2 
VTH
R1  R2
R1  R2
R2

VCC
R1  R2
2
From HVK;
1
2
VTH  VBE
IB 
;
RTH  (   1) RE
I C  I B
VCE  VCC  I C ( RC  RE )
1
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
1
2
2
FIXED BASE BIAS CIRCUIT
■ BE Loop Analysis
■ From HVK;
1
IB
VCC  I B R B  VBE  0
VCC  VBE
 IB 
RB
A
FIXED BASE BIAS CIRCUIT
■ CE Loop Analysis
■ From HVK;
VCC  I C R C  VCE  0
 VCE  VCC  I C R C
IC
■ As we known;
2
■
I C  I B
Subtituting
B
A with
 V  VBE 

IC   DC  CC
RB


B
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 = 47.08 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;
Resistor, RE added
 BE loop
 CE loop
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
1
2
2
FIXED BIAS WITH EMITTER RESISTOR
■ BE Loop Analysis
1
■ From HVK;
VCC  I B RB  VBE  I E RE  0
Recall; I E  (  1) I B
Subtitute for IE
VCC  I B RB  VBE  (   1) I B RE  0
 IB 
VCC  VBE
RB  (   1) RE
FIXED BIAS WITH EMITTER RESISTOR
■ CE Loop Analysis
■ From HVK;
VCC  I C RC  VCE  I E RE  0
■ Assume;
2
I E  IC
■ Therefore;
VCE  VCC  I C ( RC  RE )
FIXED BIAS WITH EMITTER RESISTOR
■ Example 2
■ Find IC, IB, VCE, VB,
VC, VE & VBC?
(Silikon transistor);
■ Answers;
IC = 2.01 mA
IB = 40.1 μA
VCE = 13.97V
VB = 2.71V
VE = 2.01V
VC = 15.98V
VBC = -13.27V
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
Thevenin Theorem;
From Thevenin Theorem;
RTH  R1 // R2 
VTH
Simplified Circuit
R1  R2
R1  R2
R2

VCC
R1  R2
VOLTAGE DIVIDER BIAS CIRCUIT
■ 2nd step: Locate 2 main loops.
BE Loop
2
1
CE Loop
2
1
VOLTAGE DIVIDER BIAS CIRCUIT
■ BE Loop Analysis
■ From HVK;
VTH  I B RTH  VBE  I E RE  0
Recall; I E  (  1) I B
1
Subtitute for IE
VTH  I B RTH  VBE  (   1) I B RE  0
 IB 
VTH  VBE
RRTH  (   1) RE
VOLTAGE DIVIDER BIAS CIRCUIT
■ CE Loop Analysis
■ From HVK;
VCC  I C RC  VCE  I E RE  0
■ Assume;
2
I E  IC
■ Therefore;
VCE  VCC  I C ( RC  RE )
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 = -11.522V