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P-N JUNCTION.  Single piece of SC material with half n-tpye and half p-type  The plane dividing the two zones is called junction (plane lies where density.

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Presentation on theme: "P-N JUNCTION.  Single piece of SC material with half n-tpye and half p-type  The plane dividing the two zones is called junction (plane lies where density."— Presentation transcript:


2  Single piece of SC material with half n-tpye and half p-type  The plane dividing the two zones is called junction (plane lies where density of donors and acceptors is equal) + + + - - - + + + + + + - - - - - - PN Junction  Three phenomenas take place at the junction  Depletion layer  Barrier potential  Diffusion capacitance P-N JUNCTION

3  Formation of depletion layer  Also called Transition region  Both sides of the junction  Depleted of free charge carriers  Density gradient across junction (due to greater difference in number of electrons and holes)-Results into carrier diffusion  Diffusion of holes (from p to n) and electrons (from n to p)  Diffusion current is established  Devoid of free and mobile charge carriers (depletion region)  It seems that all holes and electrons would diffuse!!! P-N JUNCTION

4  But this does not happen  There is formation of ions on both sides of the junction  Formation of fixed +ve and –ve ions- parallel rows of ions  Any free charge carrier is either  Diffused by fixed ions on own side  Repelled by fixed ions of opposite side  Ultimately depletion layer widens and equilibrium condition reached + + + - - + + + + - - - - PN + + - - -

5 BARRIER VOLTAGE  Inspite of the fact that depletion region is cleared of charge carriers, there is establishment of electric potential difference or Barrier potential (V B ) due to immobile ions + + + - - + + + + - - - - PN + + - - - VBVB

6  V B for Ge is 0.3eV and 0.7eV for Si  Barrier voltage depends on temperature  V B for both Ge and Si decreases by about 2 mV/ 0 C Therefore  V B = -0.002  t where  t is the rise in temperature  V B causes the drift of carriers through depletion layer. Hence barrier potential causes the drift current which is equal and opposite to diffusion current when final equilibrium is reached- Net current through the crystal is zero PROBLEM  Calculate the barrier potential for Si junction at 100 0 C and 0 0 C if its value at 25 0 C is 0.7 V

7  Operation of P-N junction in terms of energy bands  Energy bands of trivalent impurity atoms in the P-region is at higher level than penta-valent impurity atoms in N- region (why???)  However, some overlap between respective bands  Process of diffusion and formation of depletion region  High energy electrons near the top of N-region conduction band diffuse into the lower part of the P-region conduction band  Then recombine with the holes in the valence band  Depletion layer begins to form  Energy bands in N-region shifts downward due to loss of high energy electrons  Equilibrium condition- When top of conduction band reaches at same level as bottom of conduction band in P-region- formation of steep energy hill Explanation of P-N junction on the basis of Energy band theory

8 VB CB VB CB P-RegionN-Region P-Region

9  Forward Biasing  Positive terminal of Battery is connected with P-region and negative terminal with N-region  Can be explained by two ways. One way is  Holes in P-region are repelled by +ve terminal of the battery and electrons in N-region by –ve terminal  Recombination of electrons and holes at the junction  Injection of new free electrons from negative terminal  Movement of holes continue due to breaking of more covalent bonds- keep continuous supply of holes  But electron are attracted to +ve terminal of battery  Only electrons will flow in external circuit Biasing of P-N junction

10 + + + - - + + + + - - - - PN + + - - -

11  Another way to explain conduction  Forward bias of V volts lowers the barrier potential (V-V B )  Thickness of depletion layer is reduced  Energy hill in energy band diagram is reduced  V-I Graph for Ge and Si  Threshold or knee voltage (practically same as barrier voltage)  Static (straight forward calculation) and dynamic resistance (reciprocal of the slope of the forward characteristics)

12  Reverse Biasing  Battery connections opposite  Electrons and holes move towards negative and positive terminals of the battery, respectively  So there is no electron-hole combination  Another way to explain this process is  The applied voltage increases the barrier potential (V+V B )- blocks the flow of majority carriers  Therefore width of depletion layer is increased  Practically no current flows  Small amount of current flows due to minority carriers (generated thermally)  Also called as leakage current  V-I curve and saturation

13 + + + - - + + + + - - - - PN + + - - -

14  Compute the intrinsic conductivity of a specimen of pure silicon at room temperature given that, and. Also calculate the individual contributions from electrons and holes. PROBLEMS  Find conductivity and resistance of a bar of pure silicon of length 1 cm and cross sectional area at 300 0 k. Given  A specimen of silicon is doped with acceptor impurity to a density of 10 22 per cubic cm. Given that to be ionized All impurity atoms may be assumed  Calculate the conductivity of a specimen of pure Si at room temperature of 300 0 k for which The Si specimen is now doped 2 parts per 10 8 of a donor impurity. If there are 5x10 28 Si atoms/m 3, calculate its conductivity.

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