non_equil - Current Flow in PN Diodes S Tewksbury The...

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Current Flow in PN Diodes S. Tewksbury The conditions leading to the flow of current in a PN diode under forward bias are rather complex. These slides have been developed to help you “see” the underlying things that are happening. I recommend that you go through the material carefully, keeping track of the sequence of steps along the way. If you can understand the concepts that lead to the actual way a PN diode works, you will be far more able to visualize other kinds of devices. I have tried to show systematically how the math is applied as we proceed through the various steps. I recommend that you spend some time understanding the math that is used. It is not difficult, simply easy to get lost in.
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P-Type semiconductor: N a N-Type semiconductor: N d E C E V E F E i E i - E F E F - E V E C E V E F E i E F - E i E C - E F The majority carrier density is n n = n i exp E F E i kT = N d , the donor density. From this we obtain E F E i , namely E F E i = kT ln N d n i The minority carrier density is given by p n = n i exp E i E F kT The majority carrier density is p p = n i exp E i E F kT = N a , the donor density. From this we obtain E i E F , namely E i E F = kT ln N a n i The minority carrier density is given by n p = n i exp E F E i kT
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P-Type semiconductor N-Type semiconductor Depletion Layer p p ,0 = n i exp E i E F kT = N a E i E F P side = kT ln N a n i Equations for carrier concentrations n n ,0 = n i exp E F E i kT = N d E F E i N side = kT ln N d n i Equations for carrier concentrations 0 volts 0 volts n p,o p p,o p n,o n n,o E V E F E C E C E V E F E F - E i E i - E F E i E i Δ E i = E i E F ( ) P side + E F E i ( ) N side [ ] Total energy drop across depletion layer
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P-Type semiconductor N-Type semiconductor Depletion Layer p p ,0 = n i exp E i E F kT = N a E i E F P side = kT ln N a n i Equations for carrier concentrations n n ,0 = n i exp E F E i kT = N d E F E i N side = kT ln N d n i Equations for carrier concentrations 0 volts 0 volts Δ E i = E i E F ( ) P side + E F E i ( ) N side [ ] = kT ln N a n i + kT ln N d n i = kT ln N a N d n i ( ) 2 E = qV relates energy to voltage. The built - in voltage is therefore V bi = kT q ln N a N d n i ( ) 2 .
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