Energy Band Structure of Open Circuited PN Junction

Energy Band Structure of Open Circuited P‒N Junction

• The Fermi level in n type materal lies just below the conduction band while in p‒type material, it lies just above the valence band.

• When p‒n junction is formed, the diffuson starts. The charges get adjusted so as to equalise the Fermi level in the two parts of p‒n junction. This is similar to the adjustment of water levels in two tanks of unequal level, when connected to each other. The charges flow from p to n and n to p side till, the Fermi level on the two sides get lined up.

Key Point: The transfer of charges does not disturb the relative positions of conduction band, valence band and Fermi level in any region either p or n.

• The transfer of charges and energy band structure showing equalisation of Fermi levels in p and n regions is shown in Fig. 2.3.1.

• In p region, the Fermi level E_Fp is near E_vp just above edge of valence band.

• In n region, the Fermi level E_Fn is near E_Cn just below edge of conduction band.

• And there is difference between levels of E_Fn and E_FP.

• The transport of charges, the edge of conduction band E_cp in the p type material becomes higher than E_Cn in the n type material. Similarly the edge of valence band E_vp in p type is higher than E_Vn in n type material. Thus there is a shift of E₁ in the Fermi level on p side while there is a shift of E₂ in the Fermi level on n side from their intrinsic levels. (i.e. centre of E_C and E_V).

• This adjusts the Fermi level on n and p side to get equivalent Fermi level E_F for the p‒n junction.

• The total shift in energy E₀ is E₁ + E₂ which is responsible to produce contact difference of potential across the junction. This is nothing but barrier potential or junction potential or contact potential given by,

 V_J = V_T In (N_AN_D / n_i²)

where V_T= kT

Where

N_A = Concentration of acceptor impurity

N_D = Concentration of donor impurity

n_i = Intrinsic concentration

k = Boltzmann's constant = 8.62 × 10⁵ eV/°K

1. Width of Space Charge Region

• As the reverse bias voltage V_R increases, the width of the space charge region increase.

• Many times it is necessary to calculate the width of the space charge region denoted as W.

• While calculating W, it is necessary to consider total potential barrier instead of the built in potential V_J.

• The total potential barrier is the difference between the junction potential and applied reverse bias.

 V_B = V_J ‒ (V_R)

• But V_R is taken as negative as reverse bias hence we can write,

 V_B = V_J + V_R

= Total potential barrier

   ………..(2.3.1)

where

 V_R = Magnitude at the reverse bias voltage.

• The width of the space charge region is given by

           …………. (2.3.2)

Where

q = Charge on each electron

= 1.6×10^‒19 C

ε = ε₀ ε_r,

 ε_r = 11.7 for Si,

ε_r = 16 for G_e

Ex. 2.3.1 Consider a Si PN junction at T = 300° K with doping concentrations of N_A =10¹⁶ cm^‒3 and N_D = 10¹⁵ cm^-3. Assume that n_i = 1.5×10¹⁰ cm^-3. Calculate width of the space charge region in a PN junction, when a reverse bias voltage V_R = 5 V is applied.

Solution:

T = 300°K,

N_A =10¹⁶ /cm³, N_D = 10¹⁵ / cm³,

 n_i =1.5×10¹⁰ /cm³, V_R = 5V, ε_r = 11.7 for Si,

 q = 1.6×10^‒19

V_J = V_T In [N_AN_D / n_i²]

= 26×10^‒3 In [ (10¹⁶×10⁶×10¹⁵×10⁶) / (1.5×10¹⁰×10⁶)² ]

= 0.637 V

V_B = V_J+V_R = 5 + 0.637 = 5.637 V

= 2.833 μm

Note that N_A =10¹⁶/cm³ = 10¹⁶ × 10⁶/m³ =10²²/m³

2. Derivation of Potential Barrier

• Consider non uniformly doped p type bar as shown in Fig. 2.3.2 (a).

• Fig 2.3.2 (b) shows the graph of concentration of holes against distance x.

• There exists potential difference between any two points of such a bar.

• Let concentration at x = x₁ is P₁ and concentration at x=x₂ is P₂.

• Let potential at x₁ = V₁ and at x₂ = V₂.

• The net current through the bar is zero hence

• In p‒n junction, there is abrupt change in the concentration of holes from Pp to Pn.

 P₁ ≈ P_P ≈ N_A and P₂ =P_n

while

V₂ ‒V₁ = V_J

• Using in equation (2.3.3),

 V_J = V_T In (N_A/P_n) = Potential barrier

• According to law of mass action, n_n ×P_n = n_i²

 P_n = n_i²/n_n but n_n = N_D hence P_n = n_i²/N_D

• This is the required expression for the built in potential barrier.

Ex. 2.3.2 Calculate the built in potential barrier in a PN junction. Consider a silicon PN junction at 300 °K with doping densities N_A =1×10¹⁸ cm^-3 and N_D =1×10¹⁵ cm^-3. Assume n_i =1.5×10¹⁰ cm^-3.

Solution:

T = 300 °K, N_A =1×10¹⁸ / cm³ = 1×10²⁴ /m³,

N_D =1×10¹⁵ / cm³ =1×10²¹/m³, n_i n_i =1×10¹⁵ / cm³ = 1×10²¹/m³

V_T = 0.0259 V at T = 300 °K

V_J = V_T In[N_AN_D /n_i²]

= 0.178 V

Review Questions

1. Draw and explain the energy band structure for open circuited p‒n junction.

2. Derive the expression for the built in potential barrier.

3. Consider a step graded germanium p‒n junction. It has N_D=10³ N_A and N_A corresponds to 1 atom per 10⁸ germanium atoms. Calculate the junction potential.

Assume n_i = 2.5×10³ cm³, atom density of Ge = 4.4 × 10²² atoms per cm³. [Ans. : 0.328 V]