Transistor-2

 

Transistor Operation

 

As we have studied that to operate the transistor  it needs to be biased properly.   Even if it is biased but not carefully  output signal  will erroneous and deformed.  We must use the Transistor  within its linear operation region.  Along with  certain other parameters,  care should be taken to take care of  Operating Point,  Cut-off Point,  Saturation Point  and Load-Line.

Also remember that  β(dc),  the Current  Gain  of a Transistor depends upon Four factors:

 

1. Transistor itself (mA)
2. Collector Current  IC
3. Temperature

 

Gain varies from Transistor to Transistor even having same type.  Temperature can affect the Collector current.

Variation of Current gain with Temperature

 

                       

Operating Point and Load Line

 

For proper operation of Transistor,  certain current and voltage values are set.  These values define a point,  called “Operating Point”,  also called Quiescent Point, or Q-Point.

Since the levels of currents and voltages are fixed,  so it is called DC Operating Point.

 

Keeping in view the circuit diagram,  Transistor is biased with DC power supplies, and  Base current IB, Collector current IC  and Collector to Emitter Voltage has been set.

the equation for the Collector- Emitter circuit is:

 

 

 

 

 

 

 

 

 

4

 

Now consider two cases:

 

1. When  IC = 0,  Then  VCE = VCC.   It is the 

Cut-off  point-A.

1. When VCE = 0,  then  IC = VCC / RC    

It is the  Saturation point-B

 

 

We rewrite the equation as 

 

IC = -VCE / RC   +  VCC/ RC.

 

It is similar to the equation for Straight Line, 

 Y = -mX + C.

 

All  points  on this  between A  and  B  form the  active region of Transistor.

Points “A”  and “B”  have been marked.   A straight joining both points have been  drawn.  This line  is called  “Load Line”. 

The Load Line intersects the Vertical axis at IC = 50 mA, and  Horizontal axis at VCE = 10V.

 

      Above discussions indicate that:

1.  when IB  increase,  IC also Increases.  
2. With the increase in IC,  VCE  Decreases.  

 

     This shows  that  as VBB  is increased or decreased,  the Q-Point moves up or down on the Load Line.

 

 

 

Examples:

 

For the Circuit shown, draw load line  and mark DC  Q-point.

 

Solution:

 

IC = -VCE / RC   +  VCC/ RC.

 

Cut-off point is where:   VCB = VCC = 20V. (Point-A)

And   IC(SAT)  = VCC / RC =  20 / 5K = 4mA.   (Point-B)

Line  joining A and B  point is Load Line.

For Q-Point,  IE = VEE / RE = 30/15K  = 2mA.

                      IC = α IE ≈ IE = 2mA.

                      So   VCB = VCC – ICRC = 20 – 2 x 5 = 10V

                      Therefore  Q-Point = (10V, 2mA).

 

 

 

 

 

 

Example:

For  Common-Emitter  circuit shown,  draw Load Line and make the  DC  Working Point.   Assume β = 100.

Neglect VBE. 

 

 

 

 

 

 

 

 

 

Solution:

 

Cutt-off point A  is located where   IC = 0,   and VCE = VCC = 30V.

Saturation point B is Given  where   VCE = 0,  and  IC(SAT) = 30 / 5K  = 6mA.

Line joining points A  and B  is Load Line.

For DC working Point

IB = 30 / 1.5M = 20uA20uA,    IC = βx IB = 100 x 20 =  2000uA = 2mA.

VCE = VCC – IC x RL = 30 – 2 x 5 = 20V.

So  Q-Point = (20V, 2mA)       

 

 

Q -Point

Position of Q-Point  on the DC Load Line  tells a lot about  the output signal. 

Q-Point tells about the Maximum signal we get  without Clipping(Cut-off Clipping)   versus saturation Clipping.

 

 

 

 

 

 

 

 

 

 

 

 

Q1.    If Q-point is near the Cut-off point, Signal will  get Clipped at Point-A.

It is called Cut-Off Clipping,    as negative  swing of I/P signal drives the Transistor  to Cut-off.

 

Q2.  If  Q-point is located on the Load Line  near Saturation Point,  Then Clipping starts at Point-B  due to saturation.     As  Positive  swing of I/P signal drives the Transistor  to  Saturation.

 

Q3. If Q-point is located at the Centre  of the Load Line.   We  get maximum output signal.

Maximum un-distorted signal.

 

So during the Biasing  of Transistor it is required to establish the Q-Point  which is optimum.

Q-Point  should be stable.   It should not change or drift with temperature change.

 

Factors effecting the Biasing

 

Even after setting of Q-Point  carefully, it could shift due to thermal instability.   This thermal change occurs due to IC.   For common-Emitter mode,

IC  =  β . IB +  ICEO  =  β . IB  +  (1+ β) ICO.

In this equation there are three variables,   β,  IC,  and   ICO.  All of these variables found to increase with Temperature. 

Increase in the ICO (reverse saturation current)   produces significant increase in the  IC.    Increase in  IC  produces  more  power dissipation.

Increase in the temperature can cause  thermal runaway of Transistor.

 

Therefore   some techniques are used to overcome this problem.

If  circuit is modified in such away  which stabilizes  IC.

 

 

Transistor Biasing Methods

One more it is stated that for normal operation of Transistor it is essential that:

1. Forward Bias the Emitter – Base Junction.
2. Reverse Bias the Collector- Base Junction.
3. Establishing a proper Q-Point.

 

 

1. Base Bias

 

Figure  shows a Base Bias circuit.  It is known as  Fixed Bias  circuit.

 

Consider the Base-Emitter loop.  Apply Kirchhoff’s voltage law,

      IB . RB + VBE – VBB  =  0.

      IB  =      

 

VBB and VCC are fixed voltages.   Value of RB set the value of  IB.

 

Now consider the Collector-Emitter circuit.   Apply again Kirchhoff’s voltage Law,

       VCE   =  VCC  -   IC. RC

      IC  =          

       as  IC  =  β . IB,     so   IC  =  β .     

These equations show that  IC  and VCE  are dependent upon β.    But β  is strongly dependent upon Temperature.  

So if β  changes,  then  IC,  and VCE  will also change.  IC,  VCE  sets the  Q-Point.   This mean  Q-Point in Base-Bias  circuit will not be stable.   Therefore  Base-Bias  is never used in amplifier circuits.

 

 

Example:

A circuit is shown in the Figure.  Find whether transistor is operating in Saturation,  Active  or Cut-off region.

 

Solution:    

It is Given that: VBB = 5V,    RB  =  100K,    RC  =  5K, And  hFE ( β)  = 30.  Transistor is NPN and  Silicon   As     IB =     =    =   43 x 10-6A              = 43uA. And   IC = β . IB  =  30 x (43 x 10-6) = 1290 x 10-6A               = 1.29mA

 

For transistor to operate in saturation the IC  should  at least be,    IC =    =   =  1mA

Since calculate current in the circuit is 1.29mA,   which is more than the minimum requirement(1mA),

Therefore  the Transistor is  operating in  Saturation.

 

2. Emitter Bias  with Emitter Feedback

 

A Base Bias with a Emitter resistor RE  circuit is  shown

 here.     An Emitter resistor provides  better stability

than the Bas Bias only.    

       With the increase in temperature  Current Gain(β)

       Due to this Collector Current  IC  also increases. 

This increase in the IC,  produces a voltage drop across

The Emitter resistor  RE,  which reduces  the voltage

drop across Base resistor RB. 

This causes Base Current IB to decrease.  As a result  IC

  Also  decreases.    So a compensation in the increase in the IC takes place.

         For this  reason the  Emitter resistor is called  Emitter Feedback resistor.

        

         But !!!!!       The Emitter feedback Resistor is useful only  if the Emitter Resistor is large as possible.

         But in  actual practice  the Emitter resistor has to be kept relatively small to avoid  Saturation 

        

         Consider the Base-Emitter circuit loop.   Applying Kirchhoff Voltage law  to solve it for IB.

         VBE  +  IE . RE  -  VCC  +  IB . RB = 0        

          As  IE  =  (β + 1) IB   so

         IB  =    =        ……. Taking β + 1  ≈  β

         

          collector current IC:   IC  =  β . IB  =      =      …………    as VCC >> VBE

          This equation shows that if RE     is made very large  as compared to  RB/,  then  bias circuit become

          independent of .  But this  not possible without  Saturating the Transistor. 

 

          Consider the Collector-Emitter circuit.  Apply Kirchhoff’s voltage law.

          VCE  +  IE . RE  -  VCC  +  IC . RC  =  0         take  IC  ≈  IE  and solving it for VCE

          VCE   =   VCC -  IC (RC  +  RE) 

          VE  =  IE . RE  =  IC . RC          ………….. it is the voltage measured from Emitter to ground.

          VC  =  VCC  -  IC . RC                …………..  It is the voltage measured  from  collector to ground

          Also       VCE  =  VC  -   VE

          Saturation  value of the collector current  IC(SAT)  =      =     …..if VCC  >>  VCE

 

 

Example:     In the circuit shown,  β = 50  for a silicon Transistor.  Find  IC and VCE.   Draw DC Load Line  and indicate the Q-Point.   Take VBE = 0.7V.

 

Solution:

It is given that   VCC = 12V,     RC =  2.2K,      RB = 240K,     

 β =  50,   VBE = 0.7V.

 

Collector current: 

IC  =    =    =  2.35 mA.

Collector to emitter Voltage:

  VCE  =   VCC  IC . RC   =  12  -  (2.35 x 10^-3) x 2.2 x 10³) V

          =  12  -  5.17 V  = 6.83V

Collector Saturation current  IC(SAT)  =    =   =  5.45 mA.

Cut-off Voltage  VCE(Cut-Off)   =  12V

 

 

 

Using  IC(SAT),   IC,   VCC,  and  VCE   the Dc Load Line  is shown below.

 

                           

 

 

 

 

 

 

 

 

 

 

 

 

Base Bias with Collector feedback

 

In the Figure  RB is connected to VC, rather than  VCC  or VBB.

Here Collector Voltage provides the Bias to  Base-Emitter

Junction. 

Resistor RB acts as a Feedback resistor.  It provides a very

stable Q-Point   by reducing the effects of variations in β.

 

Apply Kirchhoff’s Voltage law on the Base-Emitter loop.

            VBE  -  VCC  +  (IB  +  IC)RC  +  IB . RB  =  0

As   IC  =  β.IB,  putting  this value in the equation and solving it for   IB  we get

            IB  =    =      …………..  Taking  β + 1 = β

 

Collector Current  IC  =  β . IB  =     =      …………    as  VCC  >>  VBE

 

Now Consider the Collector-Emitter loop,  and Apply Kirchhoff’s  Voltage  law

 

            VCE  -  VCC  +  (IC  =  IB)RC  =  0

 

Substituting  IC  +  IB  equal  to  IC  in the equation   and solving it for  VCE

 

            VCE  =  VCC  -  IC . RC

 

 

 

 

Example:     Calculate  the Q-Point (Ic,  VCE)   for DC  Biased circuit  shown.

                      Aso draw  the  Load Line  and  locate  Q-Point  on it.  Take VBE = 0.7V.

 

Solution:

It is given that   VCC = 10V,     RC =  10K,      RB = 100K,     

 βDC =  100,   VBE = 0.7V.

 

Value of collector Current  IC  = 

                      =    =  0.845 mA.

Collector-Emitter voltage VCE  =  VCC  -  IC . RC

                                =  10  -  {(0.845 x 10^-3) x (10 x 10³)]

                      =  10  -  8.45  =  1.55 V

 

Dc  Load Line: 

            Collector Saturation Current IC(sat)  +  VCC/RC  10/(10 x 10³)  =  1 x 10^-3 A  1 mA.

            Transistor Cut-Off voltage  VCE(Cutt-Off)  =  VCC  =  10V

Load Line id drawn below.   Q-Point = (IC , VCE)  = (0.845,  1.55)

 

 

 

 

 

 

 

 

 

 

 

 

 

4. Voltage Divider Bias

It is always desirable  to provide a DC Bias circuit which is independent of transistor’s β.  The DC Bias circuit shown in the Figure  meets the requirement and is very popular.  Here Base voltage can be set by the ratio of two resistors.  Base current can be  set   by the values of the resistor. Voltage Divider (also called self bias circuit)  is most widely used Biasing technique.  This name is given to the circuit because  resistors R1 and R2 are being used as voltage divider. Base voltage  VB =   VCC  Voltage drop across R2, gives a forward bias to transistor.  Here assumption is that R1 much greater than the R2.  Then current passing through R1  almost completely passes through R2. Current passing through Base is negligible as compared to current through R1 and R2

 

 

 

 

 

 

Here          VB =   VCC

                   VE  =  VB  -  VBE     as  VBE  is very  small in forward Bias   so   VE  =  VB    

                   IE  = 

                   IC  ≈  IE

                   VC  =  VCC  -  ICRC

                   VCE  =  VC  -  VE  =  VCC  -  IC . RC  -  IE . RE  =  VCC  -  IE (RC  +  RE)

                   VRC = IC . RC    ……………………Voltage drop across collector resistor

These equations are based on Ohm’s and Kirchhoff’s laws.   These are widely used in circuit analysis.

 

Example:      What  is  the Collector-Emitter  voltage  of the circuit shown. Solution: Voltage divider produces unloaded base voltage  VBB  =   10V  = 1.8V VE  = 1.8V  -  0.7V  =  1.1V Emitter Current  IE  =     1.1 mA.  It is an approximation as  VBB will slightly change  due to RE. Since IC ≈ IE,     so   VC  =  10V  -  (1.1 mA)(3K6)  =  6.04 V The Collector-Emitter voltage  is:    VCE  = VC – VE  =  6.04  -  1.1 V  =  4.94 V.                       0  =  6.04V

 

 

 

As IC  =  1.1 mA,   and  VCE  =  4.94 V.   So  plot Q-Point  is given below. 

 

 

 

Example:      What  is  the Collector current and Collector-Emitter voltage for voltage divider bias circuit.

                       VBE  =  0.7V  and   β  = 100.

Solution:

            It is Given that :  Vcc  =  10V,  RC  =  1K,  R1  ==10K,  R2  =  5K,

                                            RE  =  500 Ω,  VBE  =  0.7V  and   β  =  100.

 

Find  Base voltage                VB =   VCC   =     x 10    =   3.33 V

Similarly Emitter Voltage    VE  =  VB  -  VBE     =  3.33  -  0.7  =  5.26 mA

Emitter Current                      IE  =        =           =  5.26 x 10^-3  =  5.26mA

And Collector Current           IC  ≈   IE  5.26 mA  ……Answer

Collector-Emitter Voltage   VCE  =  10 V  -  (RC  +  RE) x IE 

   =   10  -   (1500) . 5.26 x 10^-3 V  =  10  -  7.89 

   =  2.11 V  …..Answer

 

 

 

 

 

 

 

 

 

Emitter  Bias

 

Emitter Bias circuit is shown.   It uses both Positive and Negative

Power  Supplies.   If  here  Base resistor RB,  is small enough  in the

Emitter Bias circuit   then  Base voltage VBE  is approximately zero.

 

       The Base-Emitter junction is forward Biased.  

       DC Bias current is calculated  as,

            If  RB  is very high,  so current flow through it is very small.

              Therefore                              VE  =  -VBE

                        Emitter Current                      IE  =       =         

       Since VEE  supply is Negative,  there  IE  will be   

                                                                IE  =        =                

We know that                                  IC  ≈  IE     

So     Collector voltage                  VC  =  VCC  -  IC . RC

And  Collector-emitter voltage  VCE  =  VC  -  VE