Transcript depletion region

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
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Formation of depletion layer
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
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


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
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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



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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

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
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
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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 1019 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 m3 e  0.13m2 / V  s h  0.05m / V  s e  1.6 1019 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 1019 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 1019 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.