SEU2012 BJT Biasing DC analysis by fd5565

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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;

IE  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
ICminority  ICBO
■ 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.
Collector Current (IC)
■ Then;
IC
                                   …(Eq. 3.3)
IE
■ 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
                              …(Eq. 3.4)
IB
■ 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 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)
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
VBB  VBE
IB                           …(Eq. 3.6)
RB
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  I C 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
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).
Q-Point
 Determined by using transistor
line
 A straight line intersecting the                  Saturation Region
vertical axis at approximately
IC(sat) and the horizontal axis at                                    Q-Point
VCE(off).
 IC(sat) occurs when transistor
operating in saturation region

VCC
I Csat 
RC transistor
 VCE(off) occurs whenV  0
CE

operating in cut-off region

VCE( off )  VCC  I C RC   I C 0   Cutoff Region
VCC = 8V               Draw DC Load Line and Find Q-point.

RC = 2 kΩ              VCC                  8V
RB = 360 kΩ                          I Csat                            4mA
RC    VCE  0
2k

VCE( off )  VCC  I C RC       I C 0

VCE( off )  VCC  8V
Draw DC Load Line and Find Q-point.

   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        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;
R1  R2
RTH  R1 // R2 
R1  R2
R2
VTH            VCC
R1  R2                               2

From HVK;
1
VTH  VBE
1   IB                    ;     I C  I B
RTH  (   1) RE

2   VCE  VCC  I C ( RC  RE )
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                    VCC  I B R B  VBE  0
VCC  VBE
IB                           IB                      A
RB
FIXED BASE BIAS CIRCUIT
■ CE Loop Analysis
■ From HVK;
VCC  IC R C  VCE  0
IC                    VCE  VCC  I C R C
■ As we known;
2              I C  I B     B
■   Subtituting      A with       B

 V  VBE   
I C   DC  CC
           

   RB      
FIXED BASE BIAS CIRCUIT
 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);

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;
 BE loop
Resistor, RE added              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

2
1                                                      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
VCC  VBE
IB 
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);

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

From Thevenin Theorem;
Thevenin Theorem;
R1  R2
RTH  R1 // R2 
R1  R2
R2
VTH            VCC
R1  R2

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

2
2

1                              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
Subtitute for IE
1
VTH  I B RTH  VBE  (   1) I B RE  0
VTH  VBE
IB 
RRTH  (   1) RE
VOLTAGE DIVIDER BIAS CIRCUIT
■ CE Loop Analysis
■ From HVK;
VCC  I C RC  VCE  I E RE  0
■ Assume;
I E  IC
2
■ 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);