1. P-N JUNCTION

2. 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 N
Junction
Three phenomenas take place at the junction
Depletion layer
Barrier potential
Diffusion capacitance

3. It seems that all holes and electrons would diffuse!!!
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!!!

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 + - P N

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 (VB) due to immobile ions + - P N VB

6. 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= 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

7. 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 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

8. VB CB VB CB
P-Region
N-Region
P-Region
N-Region

9. 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

10. + - P N

11. Another way to explain conduction
Forward 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)

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+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

13. + - P N

14. PROBLEMS
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.
Find conductivity and resistance of a bar of pure silicon of length 1 cm and cross sectional area at 3000k. Given
A specimen of silicon is doped with acceptor impurity to a density of 1022 per cubic cm. Given that
All impurity atoms may be assumed
to be ionized
Calculate the conductivity of a specimen of pure Si at room temperature of 3000k for which
The Si specimen is now doped
2 parts per 108 of a donor impurity. If there are 5x1028 Si atoms/m3,
calculate its conductivity.