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                                                       ELECTRONIC COMPONENTS FOR MECHATRONIC SYSTEMS  275
                                3. Ohmic Region: When V GS  is large enough (V GS  > V ) and V DS  < V GS  − V ,the i D
                                                                            T
                                                                                              T
                                  is determined by the source circuit connected to the D terminal and the transistor
                                  behaves like a closed switch between D and S terminals. V GS  ≫ V , then and i =
                                                                                                  D
                                                                                        T
                                  V DS ∕R ON (V GS ). The MOSFET acts like a nonlinear resistor where the value of the
                                  resistance is nonlinear function of the gate voltage.
                             It should be noted that the advantages of MOSFETs over BJTs are higher efficiency, higher
                             switching frequency, and better thermal stability. On the other hand, MOSFETs are more
                             sensitive to static voltage. In general, MOSFETs have been replacing BJTs in low voltage
                             (<500 V) applications.
                             Insulated Gate Bipolar Transistors (IGBT)     IGBTs are the new alternative to
                             BJTs in high voltage (>500 V) applications where they combine the advantages of BJTs and
                             MOSFETs. IGBT is a four layer device with three terminals similar to BJTs (Figure 5.15c).
                             IGBTs are widely used in motor drive applications and operate in PWM mode for high
                             efficiency.

                             Example     In Figure 5.17, let V CC1  = 5V, V CC2  = 25 V, R = 100 KΩ, R = 1KΩ.The
                                                                              1
                                                                                          2
                             base current is (assuming V BE  = 0.0 for simplicity)
                                                             5V
                                                       i =         = 0.05 mA                   (5.139)
                                                        B
                                                            100 KΩ
                             Let the gain of the transistor be    = 100. The collector current is
                                                          i = i (i , V  )                      (5.140)
                                                          C   C B   CE
                                                          i ≈    ⋅ i = 5 mA                    (5.141)
                                                          C      B
                             Hence, the voltage balance in the output circuit dictates that

                                                     25 V = 1000 ⋅ 5 ⋅ 10 −3  + V CE           (5.142)
                                                      V CE  = 25 − 5 = 20 VDC                  (5.143)

                             In general, the output voltage across the transistor for the common-emitter configuration
                             showninFigure5.17is(let V  = V  , V = V   , V = V  )
                                                     out   CE  in   CC1  s   CC2
                                                                  R 2
                                                        V out  = V −  ⋅    ⋅ V in              (5.144)
                                                               s
                                                                  R 1
                             and the voltage potential across the output circuit resistor (V )is
                                                                             R2
                                                                R 2
                                                          V   =   ⋅    ⋅ V                     (5.145)
                                                           R2          in
                                                                R
                                                                 1
                             The common-emmiter configuration of the transistor as shown in Figure 5.17 can be used
                             as a voltage amplifier. If the output voltage is taken as the voltage across the transistor,
                                                         V out  = V − K ⋅ V in                 (5.146)
                                                                s
                                                                    1
                             or if the output voltage is taken as the voltage across the resistor preceeding the collector,
                                                           V out  = K ⋅ V in                   (5.147)
                                                                  1
                             Example     In order to illustrate the fact that a transistor can be operated like a switch
                             (ON or OFF) as well as a proportional amplifier, let us consider the following cases of
                             input voltage in the same Figure 5.17. Furthermore, let us more accurately assume that
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