PN Junction Diodes: Important Example Solved Problems

Electron Devices

Chapter 2: PN Junction Diodes

Important Part B Questions

Energy Band Structure of Open Circuited P‒N Junction

Ex. 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

 

Ex. 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

 

Diode Current Equation and Nature of Characteristics

 

Ex: 3: The reverse saturation of a silicon PN junction diode is 10 μA Calculate the diode current for the forward bias voltage of 0.6 V at 25 °C.

Solution:

 I₀ = 10 μA, η = 2 for silicon, V = 0.6 V,

 T = 25°C = 25+ 273 °K = 298 °K

V_T = kT = 8.62×10^‒5×298 = 0.025 V

Using diode current equation,

I = I₀ [e^V/ηVT ‒1]

= 10×10^‒6 [e^0.6/2×0.025 −1] = 1.6275 A

 

Ex. 4: The voltage across a silicon diode at room temperature of 300 °K is 0.71 V when 2.5 mA current flows through it. If the voltage increases to 0.8 V, calculate the new diode current.

Solution:

The current equation of a diode is

I = I₀ [e^V/ηVT ‒1]

At 300 °K, V_T = 26 mV = 26×10^‒3 V

 V = 0.71 V for I = 2.5 mA

and

 η = 2 for silicon

2.5×10^‒3 = I₀ [e^{(0.71/2×26×10−3)} ‒ 1]

i.e., I₀ = 2.93 × 10^‒9 V

Now V = 0.8 V, I₀ remains same.

 I = 2.93 × 10^‒9 [e^{(0.8/2×26×10‒3)} ‒1]

 = 0.0141 A = 14.11 mA

 

Ex. 5: A germanium diode has a reverse saturation current of 3 μA. Calculate the voltage at which 1 % of the rated current will flow through the diode, at room temperature if diode is rated for 1 A.

Solution:

 η = 1 for germanium, I₀ = 3 μА = 3×10^‒6 A

Rated current is 1 A, and

 I = 1 % of rated current = 0.01 A.

 V_T = 26 mV at room temperature

Using current equation of a diode,

 I = I₀ [e^V/ηVт − 1]

i.e., 0.01= 3×10^‒6 [e^{V/1×26×10−3} ‒ 1]

 3333.33 = e^{V/1×26×10−3} ‒ 1

i.e. e^{V/1×26×10‒3} = 3334.33

V / 26×10^‒3 = 8.112

i.e. V = 0.2109 V

 

Ex.6: The diode current is 0.6 mA when the applied voltage is 400 mV and 20 mA when the applied voltage is 500 mV. Determine η. Assume kT/q = 25mV.

Solution:

At V₁ = 400 mV, I₁ = 0.6 mA

V₂ = 500 mV, I₂ = 20 mA,

V_T = 25 mV (given)

I = I₀ [e^V/ηVт − 1] 

       ….. Diode current equation

 0.6×10^‒3 = I₀ [e^{400×10‒3/η×25×10‒3} ‒ 1]

         ……….. (1)

 20×10^‒3 = I₀ [e^{500×10‒3/η×25×10−3} ‒ 1]

      ………..(2)

In forward biased condition, 1 <<< e^V/ηVT so neglecting 1,

 0.6×10^‒3 = I₀ e^16/η

and 20×10^‒3 = 1₀ e^20/η

Dividing the two equations,

0.6 / 20 = e^16/η / e^20/η

e^16/η = 0.03 e^20/η

Taking natural logarithm of both sides,

 16/η = In 0.03 + 20/η

i.e. 1/η(16‒20) = ‒ 3.50655

1/η = 0.87664

i.e. n = 1.1407

Solving equation (1)

I₀ = 4.8598 × 10^‒10 A

Derivation of V‒I Characteristics of P‒N Junction Diode (Diode Current Equation)

Ex. 7: Determine the ideal reverse saturation current density in a silicon pn junction at T = 300 °K. Consider the following parameters in the silicon pn junction: N_A = N_D = 10¹⁶ cm³, n_i = 1.5×10¹⁰ cm^-3, D_n =25 cm²/s. τ_p0 = τ_n0 =5×10^‒7 s, D_p = 10 cm² / s, ε_r = 117. Comment on the result.

Solution:

The reverse current density is given by,

The diffusion length for the holes and electrons is given by,

L_p = √(D_pτ_p)

L_n = √(D_nτ_n)

n_p0 = n_i² / N_A

P_no = n_i² / N_D

  ….. by law of mass action

L_p = √(D_pτ_p) = √[10×10^‒4×5×10^‒7] = 2.236×10^‒5

L_n = √(D_nτ_n) = √[25×10^‒4×5×10^‒7] = 3.535×10^‒5

n_p0 = n_i² / N_A = (1.5×10¹⁶)² / 10²² =2.25×10¹⁰

P_no = n_i² / N_D= (1.5×10¹⁶)² / 10²² =2.25×10¹⁰

q = 16×10^‒19 C and using in (1),

J₀ = 1.6×10^‒19 [ (10×10^‒4×2.25×10¹⁰ / 2.236×10^‒5 ) + (25×10^-4×2.25×10¹⁰ /3.535×10^‒5) ]

 = 4.15×10^‒7A/m²

i.e.4.15×10^‒11 A/cm²

 

Mathematical Expression for the Dynamic Resistance

 

Ex. 8: A p‒n junction diode has at a temperature of 125°C a reverse saturation current of 30 μ A. At a temperature of 125°C find the dynamic resistance for 0.2 V bias in forward and reverse direction.

Solution:

Now

T = 125 °C, V = ± 0.2 V, I₀ = 30 μА

V_T = kT at temperature T in °K

T= 125 + 273 = 398 °K

and k = 8.62×10^‒5 = Boltzmann's constant

and

V_T = 398×8.62×10^‒5

 = 0.034307 V = 34.3076 mV

The dynamic resistance is given by,

 r = ηV_T  / I₀e^V/ηVT

Assume germanium diode, η = 1

For forward bias, V = +0.2 V

 r_f = ( 1×0.0343076 ) / ( 30×10^‒6 × e^{0.2/0.0343076} )

 = 3.3612 Ω

For reverse bias, V = ‒ 0.2 V

 r_r = ( 1×0.0343076 ) / ( 30×10^‒6 ×e^{‒0.2/0.0343076} )

 = 389.078 ΚΩ

 

Effect of Temperature on Diode Behaviour

Ex. 9: The reverse saturation current of a silicon diode is 5 mA at room temperature. Find the diode current at i) 40 °C and a forward voltage of 0.3 V. ii) 60 °C and a forward voltage of 0.5 V.

Solution:

 I₀₁ = 5 mA at room temperature of T₁ = 27 °C = 300 °K

i) At T₂ = 40 °C = 40+ 273 = 313 °K and V = 0.3 V

ΔT = T₂‒T₁ = 313‒300 = 13

(I₀₂) = (2^ΔT/10) I₀₁

= (2^13/10) × 5 × 10^-3

= 12.3114 mA

 η = 2 for silicon

V_T = T₂ / 11600 = 313 / 11600

= 0.02698 A

I = I₀₂ [e^V/ηVT ‒1]

= 12.3114 [e^{0.3/2×0.02698} ‒1]

= 3185.428 mA = 3.185 A

i) At T₃ = 60 °C =  333 °K and V = 0.5 V

ΔT = T₃‒T₁ = 333‒300 = 13

(I₀₃) = (2^ΔT/10) I₀₁

= (2^33/10) × 5 × 10^-3

= 49.2457 mA

V_T = T₃ / 11600 = 333 / 11600

= 0.0287 A

I = I₀₃ [e^V/ηVT ‒1]

= 49.2457 [e^{0.5/2×0.0287} ‒1]

= 298778.75 mA

= 298.7787 A

 

Ex. 10: The reverse saturation current of a germanium diode is 100 μA at room temperature of 27 °C. Calculate the current in forward biased condition, if forward bias voltage is 0.2 V at room temperature. If temperature is increased by 20 °C, calculate the reverse saturation current and the forward current, for same current and the forward voltage, at new temperature.

Solution:

At T₁ = 27 °C = 300 °K, V_T =  26 mV

 (I₀)1 = 100 μA,

V = 0.2 V

I = I₀ [e^V/ηVT ‒1]

= 100×10^‒6 [e^{0.2/1×26×10−3} −1]

= 219.04 mA

Using the approximate result,

 (I₀)₂ = 2^{T2‒T1 / 10]} ×(I₀)₁

= 2^(20/10) ×(I₀)₁

           …… T₂‒T₁ =20°C

= 4×100 = 400 μА

New reverse current at 27 + 20 = 47°C is 400 μA. So for V = 0.2 V, the new forward current can be calculated using current equation of diode.

  V_T at T₂ = 47 + 273 = 320 °K = KT

= 8.62×10^‒5×320 = 27.584 mV

I = I₀ [e^V/ηVT ‒1]

= 400 × 10^‒6 [e^{0.2/1×27.584 × 10−3} ‒ 1 ]

= 563.158 mA

 

Transition Capacitance (C_T)

 

Ex. 11: A germanium diode has a contact potential of 0.2 V and concentration of acceptor impurity is 3×10²⁰/m³. Area of cross‒section of junction is 1 mm² and ε_r = 16. Calculate width of depletion region and transition capacitance for reverse bias of a) 0.1 V and b) 10 V (Assume charge on each electron as 1.6 × 10^‒19C)

Solution:

 V₁ = 0.2 V, N_A = 3 × 10²⁰/m³, A = 1 mm², ε_r = 16

a) V = ‒0.1 V

       ……… Negative as reverse bias

 V_B = V_J‒ V = 0.2 − (− 0.1) = 0.3 V

 V_B = qN_A/ε . W²/2

 i.e.

 0.3 = (1.6 × 10^‒19 × 3 × 10²⁰) / (8.854 × 10^‒12 × 16) . W²/2

 W = 1.33 μm,

 С_т = εA/W = (ε×1×10^‒19) / (1.33×10^‒6) = 106.45 pF

b) V = ‒10V

hence

V_B = V_J ‒ V = 0.2 ‒ (‒10) = 10.2 V

 10.2 = [ 1.6 × 10^‒19 × 3 × 10²⁰ ] / [8.854 × 10^‒12 × 16] × [W²/2]

i.e.

W = 7.75 μm

Ст = εA

= ( 8854 × 10^‒12 × 16 × 1 × 10^‒6 ) / (7.75 × 10^‒6)

= 18.25 pF

Thus as reverse bias voltage increases, width of depletion region increases while transition capacitance decreases.