2P-N JUNCTIONSingle piece of SC material with half n-tpye and half p-typeThe plane dividing the two zones is called junction (plane lies where density of donors and acceptors is equal)+-PNJunctionThree phenomenas take place at the junctionDepletion layerBarrier potentialDiffusion capacitance
3It seems that all holes and electrons would diffuse!!! P-N JUNCTIONFormation of depletion layerAlso called Transition regionBoth sides of the junctionDepleted of free charge carriersDensity gradient across junction (due to greater difference in number of electrons and holes)-Results into carrier diffusionDiffusion of holes (from p to n) and electrons (from n to p)Diffusion current is establishedDevoid of free and mobile charge carriers (depletion region)It seems that all holes and electrons would diffuse!!!
4But this does not happen There is formation of ions on both sides of the junctionFormation of fixed +ve and –ve ions- parallel rows of ionsAny free charge carrier is eitherDiffused by fixed ions on own sideRepelled by fixed ions of opposite sideUltimately depletion layer widens and equilibrium condition reached+-PN
5BARRIER VOLTAGEInspite 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+-PNVB
6VB for Ge is 0.3eV and 0.7eV for Si Barrier voltage depends on temperatureVB for both Ge and Si decreases by about 2 mV/0CTherefore VB= twhere t is the rise in temperatureVB 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 zeroPROBLEMCalculate the barrier potential for Si junction at 1000C and 00C if its value at 250C is 0.7 V
7Explanation of P-N junction on the basis of Energy band theory Operation of P-N junction in terms of energy bandsEnergy 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 bandsProcess of diffusion and formation of depletion regionHigh energy electrons near the top of N-region conduction band diffuse into the lower part of the P-region conduction bandThen recombine with the holes in the valence bandDepletion layer begins to formEnergy bands in N-region shifts downward due to loss of high energy electronsEquilibrium condition- When top of conduction band reaches at same level as bottom of conduction band in P-region- formation of steep energy hill
9Biasing of P-N junction Forward BiasingPositive terminal of Battery is connected with P-region and negative terminal with N-regionCan be explained by two ways. One way isHoles in P-region are repelled by +ve terminal of the battery and electrons in N-region by –ve terminalRecombination of electrons and holes at the junctionInjection of new free electrons from negative terminalMovement of holes continue due to breaking of more covalent bonds- keep continuous supply of holesBut electron are attracted to +ve terminal of batteryOnly electrons will flow in external circuit
11Another way to explain conduction Forward bias of V volts lowers the barrier potential (V-VB)Thickness of depletion layer is reducedEnergy hill in energy band diagram is reducedV-I Graph for Ge and SiThreshold or knee voltage (practically same as barrier voltage)Static (straight forward calculation) and dynamic resistance (reciprocal of the slope of the forward characteristics)
12Reverse Biasing Battery connections opposite Electrons and holes move towards negative and positive terminals of the battery, respectivelySo there is no electron-hole combinationAnother way to explain this process isThe applied voltage increases the barrier potential (V+VB)- blocks the flow of majority carriersTherefore width of depletion layer is increasedPractically no current flowsSmall amount of current flows due to minority carriers (generated thermally)Also called as leakage currentV-I curve and saturation
14PROBLEMSCompute the intrinsic conductivity of a specimen of pure silicon at room temperature given that , and 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 at 3000k. GivenA specimen of silicon is doped with acceptor impurity to a density of 1022 per cubic cm. Given thatAll impurity atoms may be assumedto be ionizedCalculate the conductivity of a specimen of pure Si at room temperature of 3000k for whichThe Si specimen is now doped2 parts per 108 of a donor impurity. If there are 5x1028 Si atoms/m3,calculate its conductivity.