Electrical Circuits and Electronics
Part 5
                          Bipolar Junction Transistors
Pearson Education, Inc.
       Part 5: Bipolar Junction Transistors
1. Understand bipolar junction transistor operation in amplifier circuits.
2. Analyze simple amplifiers using the load-line technique.
3. Use large-signal equivalent circuits to analyze BJT circuits.
4. Analyze bias circuits.
5. Use small-signal equivalent circuits to analyze BJT amplifiers.
6. Compute performance of several important amplifier configurations.
                        Introduction
• Brattain, Bardeen, and Shockley invented the bipolar junction
transistor (BJT) in 1949, while working for Bell Telephone
Laboratories.
• This revolutionary invention changed the world.
• BJT is a three-terminal device.
• BJT is used as amplifier and switch.
• Nature of bipolar transistor :
        Ø BJT is a current amplifier (not a voltage amplifier)
        Ø BJT is current-controlled current source.
        Ø BJT base current controls the emitter current and thereby
        the collector current.
           The Bipolar Junction Transistor
• The term Bipolar is because two types of charge (electrons
and holes) are involved in the flow of electricity
• The term Junction is because there are two pn junctions
• There are two configurations for this device
                    NPN and PNP Transistors
• NPN is more widely used
       Ø Majority carriers are electrons so it operates more quickly
• PNP is used for special applications
       Ø Majority carriers are holes
• The terminals of the transistor are labelled Base, Emitter, and Collector
       Ø The emitter is always drawn with the arrow.
               Asymmetry of Transistor
• Although in the NPN schematic, it looks like the
collector and emitter can be reversed, in reality the
device is very inefficient in reverse connection and has
very little amplification (gain).
            Operation of NPN Transistor
• In normal operation, the EB junction is forward biased and
the BC junction is reverse biased
• The base region is very thin so the ratio L1:L2 is typically
about 150 :1
               L1 L2
      The npn BJT                   Bias conditions for pn junctions
Bipolar junction transistors: Animation
 http://www.learnabout-electronics.org/Semiconductors/bjt_04.php
                                Behaviour
• Forward biasing of the EB junction causes a heavy flow of majority carriers
(electrons) from the n-type material into the base junction and also majority
carriers (holes) from the base region into the emitter region. We denote this
current Ie.
• The base region is very thin so that most of the electrons attracted to this
region pass straight through it (attracted by the collector which is positive
relative to the base).
• Because of this, the collector current is very nearly equal in value to Ie.
Thus with the current directions shown :          iC
where α is close to unity (α=0.98)           α=
                                               iE
                                Behaviour
• The current that does not go through the collector forms the base current so that we
have iB = (1-α) iE.
• From this :                            iC     α
                                    β=
                                         iB
                                              =
                                                  1−α                 iC = βiB
 Typically β =50 to 200
• The parameter β is called the DC current gain and represents the current
amplification of the transistor.
• Indeed the use of the transistor as an amplifier is one of its main applications
                                        Only a small fraction of the emitter current
                                        flows into the base (provided that the
                                        collector-base junction is reverse biased and
                                        the base-emitter junction is forward biased)
               Equations of Operation
          ⎡ ⎛ v BE   ⎞ ⎤                    ⎛ v BE   ⎞
iE = I ES ⎢exp⎜⎜     ⎟⎟ − 1⎥   iC ≅ I s exp⎜⎜        ⎟⎟      I s = αI ES
          ⎣ ⎝ VT      ⎠ ⎦                   ⎝ VT      ⎠
                                                      iB = (1 − α )iE
     iE = iC + iB
        iC                                                  iC   α
     α=                                                   β= =
        iE                                                  iB 1 − α
Summary of the BJT operation
Summary of the BJT operation
         Basic amplifier circuits
• Common-base configuration
• Common-collector configuration
• Common-emitter configuration
                                     VBC = VBE − VCE
                                       iC   α
                                     β= =
                                       iB 1 − α
                                     β>100 State of the
                                       art transistor
     Common-emitter Characteristic
           Common-emitter Input
              Characteristic
          ⎡ ⎛ v BE   ⎞ ⎤
iE = I ES ⎢exp⎜⎜     ⎟⎟ − 1⎥   and   iB = (1 − α )iE
          ⎣ ⎝ VT      ⎠ ⎦
                  Common-emitter Output
                     Characteristic
                                             The transistor illustrated has :
                                                          iC
                                                       β = = 100
                                                          iB
• Output characteristics are plots of iC versus VCE for constant values of iB
Example: Determining β from the Characteristic Curve
• Find the value of β for the transistor with the characteristics below.
Load-line analysis of a common-Emitter
        amplifier (Input Circuit)
             VBB + vin (t ) = RBiB (t ) + vBE (t )
• The quiescent operating point, or Q point, corresponds to :
                           vin (t ) = 0
Load-line analysis of a common-Emitter
       amplifier (Output Circuit)
             VCC = RC iC + vCE
Example : Load-line analysis of a BJT Amplifier
•Assume that the circuit below has VCC=10V, VBB =1.6V, RB=40kΩ,
and RC= 2 kΩ. The input signal is a 0.4V-peak 1-kHz sinusoid given
by vin(t) = 0.4 sin(2000πt). The common-emitter characteristics for
the transistor are shown below. Find the maximum, minimum, and
Q-point values for VCE
Example : Load-line analysis of a BJT
             Amplifier
                  BJT operation modes
• When iC becomes
zero, we say that the
transistor is
cutoff.
            ≅
• When vCE 0.2 V,        ≅
we say that the
transistor is in
saturation.
• Amplification occurs
only in the active
region
BJT operation modes : summary
Large-Signal DC Circuit Models
    (Active-Region Model)
Large-Signal DC Circuit Models
   (Saturation-Region Model)
Large-Signal DC Circuit Models
    (Cutoff-Region Model)
Example: Large-Signal DC Circuit
            Models
Example: Large-Signal DC Circuit Models
     Small-Signal Equivalent Circuit for the BJT
• For small-signal variations around the Q point, the base-to-emitter
junction of the transistor appears to be a resistance rπ :
           VT   βV
    rπ =       = T       With :   VT ≅ 0.026V At room temperature
           I BQ I CQ
• The signal components are related by :     ic (t ) = βib (t )
Example :Small-Signal Equivalent Circuit for the BJT
                         common-emitter amplifier
• The common-emitter amplifier is inverting and has large voltage
gain magnitude, large current gain, and hence large power gain.
• In small-signal ac analysis, we treat capacitors as short circuits.
• DC voltage sources are treated as a short circuit to ground for ac
signals.
 Example :Small-Signal Equivalent Circuit for the BJT
                    Small-signal ac equivalent circuit
                                                                    1
• RB is the parallel combination of R1 and R2 : RB = R1 R2 =
                                                               1 R1 + 1 R2
                                                                    1
• R’L is the parallel combination of RC and RL : RLʹ = RL RC = 1 R + 1 R
                                                                  L     C
  Example :Small-Signal Equivalent Circuit for the BJT
• Voltage gain of the amplifier :                      vo      RLʹ β
                                                 Av =       =−
                                                       vin      rπ
                                                        vo     RC β
• Open-circuit voltage gain of the amplifier :   Avo =      =−
                                                        vin      rπ
• Imput Impedance: is the impedance seen                  vin     1
                                                 Z in =       =
looking into the imput terminals :                        iin 1 RB + 1 rπ
                                                     io      Z
• Current gain of the amplifier:                 Ai =    = Av in
                                                     iin     RL
• Power gain of the amplifier:                   G = Ai Av
• Output Impedance: is the impedance seen
looking back from the load terminals, with
the source voltage Vs set to zero :               Z o = RC