PN Junction Diode Current Equation and Nature of Characteristics

Diode Current Equation and Nature of Characteristics

• The relationship between applied voltage V and the diode current I is exponential and is mathematically (given by the equation called diode current equation. It is expressed as,

 I = I₀ [e^{V/ηVT ‒1}] A

where

 I₀ = Reverse saturation current in amperes,

V = Applied voltage

 η = 1 for germanium diode

 and = 2 for silicon diode

  V_T = Voltage equivalent of temperature in volts

• The voltage equivalent of temperature indicates dependence of diode current on temperature.

• The voltage equivalent of temperature V_T for a given diode at temperature T is calculated as,

 V_T = KT volts

where k = Boltzmann's constant = 8.62×10^-5eV/°K

and T = Temperature in °K.

• Thus at room temperature of 27 °C i.e. T = 27+ 273 = 300°K the value of V_T is,

 V_T = 8.62 × 10^‒5 × 300 = 0.02586 V ≈ 26 mV

• The diode current equation is applicable for all the conditions of diode i.e. unbiased, forward biased and reverse biased.

For forward biased, V must be taken positive and we get current I positive which is forward current. For reverse biased, V must be taken negative and we get negative current I which indicates that it is reverse current.

1. Nature of V‒I Characteristics from Equation of Diode

• Consider a current equation of diode as,

 I = I₀(e^V/ηVT −1)

• Now for a forward biased condition, the bias voltage V is considered positive and hence exponential index has positive sign. Due to this, 1 <<< e^V/ηVT hence neglecting 1 we get the equation for a forward current as,

 I_f = I₀e^V/ηVT

• This indicates that once bias voltage exceeds cut‒in voltage, the forward current increases exponentially. In reverse biased condition, the bias voltage V is treated negative and due to this exponential index has negative sign. So e^‒V/ηVT << 1, hence neglecting exponential term we get,

 αI_R ≡ I₀(−1) ≡ – I₀

• The above equation indicates that under reverse biased condition, the current is reverse saturation current which is negative indicating that it flows in opposite direction to that of forward current and almost constant.

• Such nature of diode characteristics is already been discussed and it is as shown in Fig. 2.9.1. The dashed portion represents breakdown region.

Ex: 2.9.1: 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. 2.9.2: 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. 2.9.3: 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.2.9.4: 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

Review Questions

1. Write the diode current equation of a p‒n junction and explain the V‒I characteristics from it.

2. The current of germanium diode is 100 μA at a voltage of ‒1 V, at room temperature. Determine the magnitude of the current for the voltages of ± 0.2 V at room temperature.

[Ans.: 291 mA, 99.95 μA ]

3. A silicon diode has a reverse saturation current of 60 nA. 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. [Ans. : 0.6252 V]

4. Explain the theory of PN junction diode and derive its diode current equation.