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320 Lecture 24

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  Whites, EE 320 Lecture 24 Page 1 of 5 © 2009 Keith W. Whites Lecture 24: BJT as an Electronic Switch. The transistor can be used as an electronic switch, in addition to an amplifier. As a switch, we use the cutoff and saturation regions of BJT operation. (Fig. 5.74) ã   Cutoff Region. If 0.5  I  v    or so, the EBJ will conduct negligible current. Also, the CBJ will be reversed biased with a large V  CC . Consequently, 0  B i  ≈ , 0 C  i  ≈ , and 0  E  i  ≈  (1) which means O CC  v V  =  (2) These are the cutoff conditions and the BJT is in the “off ” state. ã   Saturation Region. For the “on” state of the switch, we increase v  I   until the BJT saturates. This occurs when the EBJ and the CBJ are both forward biased.  Whites, EE 320 Lecture 24 Page 2 of 5 Due to asymmetries in the device fabrication, the voltage drops are different for these two forward-biased junctions. sat 0.2 V CE  V   ≈   sat 0.2 V  EC  V   ≈  These are only approximate values for saturated BJTs. The actual values of sat CE  V   and sat  EC  V   depend heavily on i C  . Equivalent circuit models for these saturated BJTs are: 0.7 V  BE  V   ≈ sat 0.2 V CE  V   ≈   0.7 V  EB V   ≈ sat 0.2 V  EC  V   ≈  So, with v  I   “large,” then  R C  V  CC  i C  sat O CE  v V  ≈ i  E  v  I   R  B i  B +-    Whites, EE 320 Lecture 24 Page 3 of 5 With sat O CE  v V  =  (3) then 0.7  I  B B vi R −= , satsat CC CE C C  V V i R −= , sat  E B C  i i i = +  (4) Remember that because the BJT is no longer operating in the active region, C B i i  β  ≠ . Instead, if the BJT is operating in the saturation mode satforced  C  B ii  β β  ≡ <  (5) Example N24.1 . The BJT in the circuit below has 50150  β  ≤ ≤ . Find the  R  B  that saturates the BJT with a so-called overdrive factor of at least 10. 0.2 V O V   ≈  Designing at “electronic switch” has essentially two parts: cutoff and saturation. Cutoff is easy to design. Just make 0.5  I  v    V or so.  Whites, EE 320 Lecture 24 Page 4 of 5 Saturation is a bit more difficult to design. We need v  I   sufficiently large so that the collector current becomes large enough for the CBJ to become forward biased . For this problem, assume the BJT is saturated so that sat 0.2 CE  V   =  V. Therefore, sat 100.29.81,000 C C   I I   −= = =  mA. To saturate the BJT with the smallest  β   we need to provide satmin 9.8 mA0.19650 C  B  I  I   β  = = =  mA This is  I   B  just on the edge of saturation (EOS). For an “overdrive factor (ODF)” of 10 means we want to force 10 times this current into the base of the BJT: EOS ODF  B B  I I  = ⋅  (6) or 100.196 mA1.96  B  I   = ⋅ =  mA. Therefore, since 50.7  B B  I  R −=   ⇒  4.32.2  B B  R I  = =  k  Ω   Now, with this design and the transistor saturated, what is the “forced ”  β  ? forced  9.8 mA51.96 mA C sat  B  I  I   β   = = =  

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Jul 23, 2017
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