depletion region
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P-N JUNCTION 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 of donors and acceptors is equal) P + + + + + + + + + Junction N - - Three phenomenas take place at the junction Depletion layer Barrier potential Diffusion capacitance P-N JUNCTION 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!!! 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 P + + + + + + - N + + + - - BARRIER VOLTAGE Inspite of the fact that depletion region is cleared of charge carriers, there is establishment of electric potential difference or Barrier potential (VB) due to immobile ions P N + + - + + + - + + + - + - - - - - - VB VB for Ge is 0.3eV and 0.7eV for Si Barrier voltage depends on temperature VB for both Ge and Si decreases by about 2 mV/0C Therefore VB= -0.002 t where t is the rise in temperature VB 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 1000C and 00C if its value at 250C is 0.7 V Explanation of P-N junction on the basis of Energy band theory 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 Nregion (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 CB CB CB VB VB VB P-Region N-Region P-Region CB VB N-Region Biasing of P-N junction 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 P N + + - + + + - + + + - + - - - - - - Another Forward way to explain conduction bias of V volts lowers the barrier potential (V-VB) 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) 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+VB)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 P N + + - + + + - + + + - + - - - - - - PROBLEMS Compute the intrinsic conductivity of a specimen of pure silicon at 6 3 2 room temperature given that ni 1.4 10 m , e 0.145m / V s h 0.05m2 / V s and e 1.6 1019 C . Also calculate the individual contributions from electrons and holes. Find conductivity and resistance of a bar of pure silicon of length 1 cm and cross sectional area 1m m2 at 3000k. Given 2 ni 1.5 1016 m3 e 0.13m2 / V s h 0.05m / V s e 1.6 1019 C A specimen of silicon is doped with acceptor impurity to a density 2 16 3 of 1022 per cubic cm. Given that ni 1.4 10 m e 0.145m / V s h 0.05m2 / V s e 1.6 1019 C All impurity atoms may be assumed to be ionized Calculate the conductivity of a specimen of pure Si at room 16 3 2 temperature of 3000k for which ni 1.5 10 m e 0.13m / V s h 0.05m2 / V s e 1.6 1019 C The Si specimen is now doped 2 parts per 108 of a donor impurity. If there are 5x1028 Si atoms/m3, calculate its conductivity.
