Lecture5 - 2 Summary Inprevious device(pn junctiondiode...

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11/18/2010 1 LECTURE 5: Bipolar Junction Transistor (BJT) EECS 170A 11/18/2010 1 Summary In previous lectures we understood how the most basic electronic device (p n junction diode) works. We started basic assumptions to idealize the diode The diode is operating under steady state condition: There are no transient effects, d/dt 0 The semiconductor is a non degenerately doped semiconductor, |E C E F |>3kT, & |E V E F |>3kT, One dimensional diode structure, (variations are always along x direction) Low level injection: Carrier density diffusing from across the junction is much lower than the majority carriers, n<<p in p side, and p<<n in n side, The generation process in the quasi neutral regions is negligibly smal , G L =0 Total current flowing through the device is constant at any point along the device, J total (x)=FIXED We said that everything is triggered by diffusion of carriers from one side to 11/18/2010 2 -V A +V A q(Vbi) q(Vbi-VA) q(Vbi-VA) Holes with enough energy to overcome built- in potential electrons with enough energy to overcome built-in potential other. Applying voltage regulates the amount of carriers diffusing across the junction by lowering or increasing the potential barrier seen by carriers Summary (2) As a result we create a minority carrier gradient at quasineutral regions 11/18/2010 3 Under Forward Bias V A >0 Diffusion of these carrier will generate diffusion current V A <0 Summary (3) Mathematically you can link the total current passing through the diode by using continuity equations 11/18/2010 4 P Quasi NeutralRegion Space Charge (Depletion) Region N QuasiNeutral Region V A X=-xp X=0 X=xn  2 2 If you are after calculating electron or hole current at any point you can us 1 1 () @ ( : 1 ) e p n pN A A xx L qV kT i q Np P n N P xx x Vk T i x dn dp JJ x J n d J xq D q D dx n qD e e qD e e x x x d        N N p A P N AN dx N NL J  2 2 2 If you are after calculating the total current you can evaluate electron and hole currents at and : 1 1 1 1@ nP A P A n A qV kT i P n qV kT i P DP qV k pn T D i n n d qD n J n qD e e ee x dx N JJ x q e    2 NiP Dn D NL NL    J 0
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11/18/2010 2 Summary (4) The same calculation can be applied to any diode but we specifically introduced narrow base diode where N region is narrow Where the math is slightly different due to different boundary conditions We can apply the same continuity equation and we get solution as: 11/18/2010 5 N - P X=x n X=-x p X=x c  2 2 () ( ) cosh 1 A Pn N p cn P i qV kT Ni P Total current J J x J x xx L n J Dn D qq e A N D I A J A a r e a      If the diode is P + N diode, i.e. N A >>N D ' sinh .
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Lecture5 - 2 Summary Inprevious device(pn junctiondiode...

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