note:  the derivations, equations and calculations given on this website are distorted(misarranged) which is due to transferring the text from Microsoft Word

          to Frontpage. PDF downloads will be soon available.

Fixed-Bias Circuit:

Parameters used:

                                                  V_CC = 9 volts

                                                     b = 150

                                                  V_CE = V_CC/2

                                                         = 4.5 volts

                                                    I_C  = 2.5 mA, since I_C must be greater than 2mA

Input (Base – Emitter Loop):

V_CC – I_BR_B – V_BE = 0

                                                                                                              I_B = V_CC – V_BE                                 ==>      Equation 1.1

       R_B

Output (Collector – Emitter Loop):

V_CC – I_CR_C – V_CE = 0

                                                                                                              I_C = V_CC – V_CE                                 ==>      Equation 1.2

R_C

b (expressing b as I_B  and I_C):

b =  I_C

         I_B

               V_CC – V_CE

                                                                                                             b    =            R_C

               V_CC – V_BE

                  R_B

                                                                                                                b =   (V_CC – V_CE) R_B                     ==>      Equation 1.3

                                                                                                                          (V_CC – V_BE) R_C

R_E is first solve so that I_C = 2.5mA:

   I_C  =  V_CC – V_CE                                 

         R_C

…substituting these parameters to equation 1.2:

                                                              V_CC = 9 volts                                               b = 150

                                                              V_CE = 4.5 volts                                           I_C  = 2.5 mA

            … we obtain the value of R_C which is 1.8kΩ.

Solving for R_B:

                                                                                                                b =   (V_CC – V_CE) R_B                    

                                                                                                                          (V_CC – V_BE) R_C

     …substituting the values above, we then obtain R_B = 495kΩ

Solving for I_B:

                                                                                          I_B =   V_CC – V_BE                                

                                                                                                         R_B

                     ….we obtain I_B = 16.67µA.

The computed value closely matched to the result of the simulated circuit.                                                                           

Fixed-Bias with Emitter Resistor:

Input (Base – Emitter Loop):

                                                                                                V_CC – I_BR_B – V_BE  - I_ER_E = 0

V_CC – I_BR_B – V_BE  - I_B(b  + 1)R_E = 0

V_CC – V_BE  =  I_BR_B + I_B(b  + 1)R_E

V_CC – V_BE  =  I_B[ R_B + (b  + 1)R_E]

  I_B  =           V_CC – V_BE  

                                                                    R_B + (b  + 1)R_E

Derivation of I_E to b - I_C Relationship:

I_E = I_C + I_B

since I_E = (b + 1)I_B                                                                                        I_E = I_C   +     I_E                                             

                                                                                                                 b + 1

I_E   -    I_E         =    I_C                                             

                                                                                                         b + 1

                                                                                               I_E  [1 - 1/(b + 1)] = I_C   

                                                                                               I_E = I_C  /  [1 - 1/(b + 1)]   

                                                                                               I_E = I_C  /  [b/(b + 1)]   

                                                                                               I_E = (b + 1) I_C  / b

 Output (Collector – Emitter Loop):

                                                                                                V_CC – I_CR_C – V_CE  - I_ER_E = 0

V_CC – I_CR_C – V_CE  - I_C (b + 1) R_E = 0

                                                                                                                               b

I_CR_C +  I_C (b + 1) R_E = V_CC - V_CE              

                                                                                                            b

                                                                                                I_C [R_C +  (b + 1) R_E] = V_CC - V_CE              

                                                                                                                  b

I_C    =          V_CC – V_CE 

                                                                                                                                                                                                                 R_C +  (b + 1) R_E

                                                                                                                      b

I_C    =          bV_CC – V_CE 

                                                                                                                                                                                                                 bR_C +  (b + 1) R_E

b (expressing b as I_B  and I_C):

                                                                                                                     b =  I_C

                                                                                                                              I_B

                          bV_CC – V_CE 

                                                                                                                                                                                                        b   =         bR_C +  (b + 1) R_E

                                                                                                                           V_CC – V_BE

                                                                                                                        R_B + (b  + 1)R_E

        b   =     b(V_CC – V_CE)        [R_B + (b  + 1)R_E]

          (V_CC – V_BE)      [bR_C +  (b + 1) R_E]

        1   =     (V_CC – V_CE)        [R_B + (b  + 1)R_E]

          (V_CC – V_BE)      [bR_C +  (b + 1) R_E]

                                                                                                       R_B  =     (V_CC – V_BE)      [bR_C +  (b + 1) R_E]    -  (b  + 1)R_E

                                                                                                                                    (V_CC – V_CE)

R_E is first solve so that I_C = 2.5mA and assigning a value of R_C which is equal to 1kΩ. Again, substituting these parameters:

  V_CC = 9 volts                                               b = 150

                                                              V_CE = 4.5 volts                                           I_C  = 2.5 mA

I_C    =          b(V_CC – V_CE) 

                                                                                                                                                                                                                 bR_C +  (b + 1) R_E

                  ...we obtain R_E = 794.7Ω.

This is the exact value of R_E without any approximation, which is used by Robert Boylestad in his book. If I_E and I_C are assumed to be equal, we will obtain a value of 800Ω but there will be more discrepancies between the computed value and simulated value.

Standard resistance are not used in the simulation since it would defeat the purpose and effort of deriving the formulas above and also, the comparison of the output here is important in which a slight change of value would cause much difference since the EWB software does not approximate in its calculation. But in actual design, we could then use the standard resistance since approximation is used most of the time.

Solving for R_B:

                                                                                   R_B  =     (V_CC – V_BE)      [bR_C +  (b + 1) R_E]    -  (b  + 1)R_E

                                                                                                                   (V_CC – V_CE)

   …substituing the values taken above, we obtain R_B = 375 kΩ.

Solving for I_B:

I_B  =           V_CC – V_BE  

                                                                    R_B + (b  + 1)R_E

                     ….we obtain I_B = 16.67µA.

Solving for I_E:

                                                                                                    I_E  = I_B + I_C

                                                                                                    I_E = 2.52mA

The computed value closely matched to the result of the simulated circuit.  

Voltage Divider Circuit:

The formulas of voltage divider is closely similar to emitter stablized only that R_B will then be R_TH (Thevenin Equivalent Resistor) which is the parallel equivalent of R₁ and R₂. The two also differ in the input loop, instead of using V_CC, the V_TH(Thevenin Equivalent Voltage) will be equal to the voltage divider of the resistors with respect to R₂.

                     1   =     (V_CC – V_CE)        [R_B + (b  + 1)R_E]

                                                                                             (V_CC – V_BE)      [bR_C +  (b + 1) R_E]

Above is an expression taken from emitter stabilized derivation, in order to transform it into a voltage divider equation, the R_B and V_cc  (with respect to the input loop) will be replaced with its Thevenin equivalents.

         1   =      (V_CC – V_CE)        [(R₁R₂/R₁+R₂) +(b  + 1)R_E]

                                                                                            [(R₂V_CC/R₁+R₂) – V_BE]      [bR_C +  (b + 1) R_E]

         1   =      (V_CC – V_CE)        [R₁R₂+(b  + 1) (R₁+R₂)R_E]

                                                                                            [R₂V_CC– V_BE (R₁+R₂)]      [bR_C +  (b + 1) R_E]

  1   =      (V_CC – V_CE)        [R₁R₂+(b  + 1) R_E R₁+(b  + 1)R_E R₂]

                                                                                           R₂V_CC – V_BE R₁-V_BE R₂)]      [bR_C +  (b + 1) R_E]

                1   =      (V_CC – V_CE) R₁R₂+(V_CC – V_CE) (b  + 1) R_E R₁+(V_CC – V_CE) (b  + 1)R_E R₂

                                                                                [R₂V_CC -V_BE R₂)]      [bR_C +  (b + 1) R_E] – V_BE R₁[bR_C +  (b + 1) R_E]

[R₂V_CC -V_BE R₂)]      [bR_C +  (b + 1) R_E] – V_BE R₁[bR_C +  (b + 1) R_E]   =     (V_CC – V_CE) R₁R₂+(V_CC – V_CE) (b  + 1) R_E R₁+(V_CC – V_CE) (b  + 1)R_E R₂

V_BE R₁[bR_C +  (b + 1) R_E]  +  (V_CC – V_CE) R₁R₂+(V_CC – V_CE) (b  + 1) R_E R₁   =     [R₂V_CC -V_BE R₂)]      [bR_C +  (b + 1) R_E] - (V_CC – V_CE) (b  + 1)R_E R₂

                                                      R₁       =              [R₂V_CC -V_BE R₂)]      [bR_C +  (b + 1) R_E] - (V_CC – V_CE) (b  + 1)R_E R_{2                       è}     equation 3.1

                                                                                                                                           V_BE [bR_C +  (b + 1) R_E]  +  (V_CC – V_CE) R₂+(V_CC – V_CE) (b  + 1) R_E    

The output loop is unaffected by R1 and R2, therefore, same parameters will be used (I_C, R_C and R_E) in the emitter stabilized which is already computed.

  V_CC = 9 volts                                               b = 150

                                                              V_CE = 4.5 volts                                           I_C  = 2.5 mA

                                                               R_C = 1KΩ                                                 R_E = 794.7Ω

For the RB values, we assign a value of R₂ since it usually lesser compared to R₁ since it would be difficult to adjust a greater value which in this case is R₁. In assigning values for R2, we must take note that bR_E≥10R₂ to make sure that our transistor is in god operating point. I then let R2 be equal to 710Ω which agree with the expression.

Substituting the values above to the equation 3.1, then we will obtain R1 = 1.59kΩ.

The computed value closely matched to the result of the simulated circuit.         

Voltage Feedback:

The transistor current I_C’ = I_C + I_B.

Input Loop:

      V_CC – (I_C+ I_B)R_C- I_BR_B- V_BE – I_ER_E = 0

                                                                             V_CC – bI_BR_C - I_BR_C - I_BR_B – V_BE  - I_B(b  + 1)R_E = 0

                                                                              bI_BR_C + I_BR_C + I_BR_B + I_B(b  + 1)R_E = V_CC – V_BE  

                                                                                    I_B  =                    V_CC – V_BE  

                                                                                                 bR_C + R_B + (b  + 1)R_E + R_C

Output Loop:

V_CC – I_CR_C – V_CE  - I_ER_E = 0

                                                                                     V_CC – (I_C + I_B)R_C – V_CE  - I_C (b + 1) R_E = 0

                                                                                                                               b

                                                                                               I_CR_C -  I_CR_C  +  I_C (b + 1) R_E  = V_CC - V_CE              

                                                                                                           b          b

                                                                                                I_C [R_C +  (b + 1) R_E + R_C] = V_CC - V_CE              

                                                                                                      b         b

I_C    =          V_CC – V_CE 

                                                                                                                                                                                                                 R_C +  (b + 1) R_E  + R_C

                                                                                                         b          b

I_C    =          b(V_CC – V_{CE )}

                                                                                                                                                                                                                 R_C +  (b + 1) R_E + bR_C

b (expressing b as I_B  and I_C):

                                                                                             b =   I_C

                                                                                                                              I_B

                          b(V_CC – V_{CE )}

                                                                                                                                                                                                        b   =            R_C +  (b + 1) R_E + bR_C

                                                                                                                           V_CC – V_BE

                                                                                                               bR_C + R_B + (b  + 1)R_E + R_C

      b   =     b(V_CC – V_CE)        [bR_C + R_B + (b  + 1)R_E + R_C]

 (V_CC – V_BE)      [R_C +  (b + 1) R_E + bR_C]

                                                                                           1   =     (V_CC – V_CE)        [bR_C + R_B + (b  + 1)R_E + R_C]

 (V_CC – V_BE)      [R_C +  (b + 1) R_E + bR_C]

Again with the same parameters used from the previous biasing, we can solve for R_B.

  V_CC = 9 volts                                               b = 150

                                                              V_CE = 4.5 volts                                           I_C’  = 2.5 mA

                                                               R_E = 794.7Ω                                    R_C = 1kΩ

    …substituting this parameters, we obtain R_B = 225kΩ.

Solving for IB:

I_B  =                    V_CC – V_BE  

                                                                                                 bR_C + R_B + (b  + 1)R_E + R_C

    ..we obtain I_B = 16.34µA

The computed value closely matched to the result of the simulated circuit.  

Conclusion:

The major difference between the four biasing is there temperature stability. b is sensitive to temperature. Therefore a change of temperature will then trigger a change of Ic so as to its output. Taking note that transistor is susceptible to heating effects.

            In order to make a good design, temperature stability in particular, it is important to add an emitter resistor which will lessen the effect of a change of b. This is observed in the equations of emitter stabilized. Adding more resistor in the base, which then be voltage divider, it will become more stable.

note:  the derivations, equations and calculations given on this website are distorted(misarranged) this is due to transferring the text from Microsoft Word

          to Frontpage. Anyways, soon, I'll provide PDF  format for this for you to download.