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

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!!!

P-N JUNCTION

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

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

VBVB

CBCB

VBVB

CBCB

P-Region

N-Region

N-Region

P-Region

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

+

+ + - -+ ++ +

- -- -

P N

++

---

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)

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

++

---

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

36104.1 mni sVme /145.0 2sVmh /05.0 2 Ce 19106.1

Find conductivity and resistance of a bar of pure silicon of length 1 cm and cross sectional area at 3000k. Given 21mm

316105.1 mni sVme /13.0 2 sVmh /05.0 2 Ce 19106.1 A specimen of silicon is doped with acceptor impurity to a density

of 1022 per cubic cm. Given that 316104.1 mni sVme /145.0 2sVmh /05.0 2 Ce 19106.1

to be ionized All impurity atoms may be assumed

Calculate the conductivity of a specimen of pure Si at room temperature of 3000k for which 316105.1 mni sVme /13.0 2

sVmh /05.0 2 Ce 19106.1 The Si specimen is now doped 2 parts per 108 of a donor impurity. If there are 5x1028 Si atoms/m3, calculate its conductivity.