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
    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).
                                           Q-Point
    Determined by using transistor
     output characteristic and DC load
     line
                                DC LOAD LINE
■ DC Load Line
    A straight line intersecting the                  Saturation Region
     vertical axis at approximately
     IC(sat) and the horizontal axis at                                    Q-Point
     VCE(off).
                                                                             DC Load Line
    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
                    DC LOAD LINE (Example)
              VCC = 8V               Draw DC Load Line and Find Q-point.
                                     Answers;

                         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
                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
 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
■ 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;
                                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);

                          ■ 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


                                               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);
                           ■ 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

								
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