Effect of Temperature on PN Junction Diode Behaviour

Effect of Temperature on Diode Behaviour

• The temperature has following effects on the diode parameters,

1. The cut‒in voltage decreases as the temperature increases. The diode conducts at smaller voltages at large temperature.

2. The reverse saturation current increases as temperature increases.

This increase in reverse current I₀ is such that it doubles at every 10 °C rise in temperature. Mathematically,

 I₀₂ = 2^(ΔT/10)I₀₁

 ……… ΔT = (T₂‒ T₁)

where I₀₂ = Reverse current at T₂ °C

and

I₀₁ = Reverse current at T₁ °C

3. The voltage equivalent of temperature V_T also increases as temperature increases.

4. The reverse breakdown voltage increases as temperature increases.

5. The maximum safe value of power dissipation is mentioned in the datasheet of the diode as (P_D)_max which is specified at normal room temperature of 25 °C. At higher temperatures, as the device junction temperature is higher, it can dissipate less power. Thus maximum power dissipation of the device decreases at higher temperatures.

• The effect of characteristics is shown in Fig. 2.12.1.

Ex. 2.12.1: 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. 2.12.2: 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

Review Questions

1. What is the effect of temperature on cut‒in voltage and reverse saturation current ?

2. Explain the effect of temperature on the diode characteristics.

3. A p‒n junction silicon diode has a reverse room saturation current of 50 nA at the temperature 27 °C. If the new reverse saturation current is observed to be 160 nA, calculate the new temperature. [Ans.: 44.19°C]

4. The reverse of saturation current silicon diode is found to be 50 nA at 27 °C. In a certain circuit the junction temperature is observed to rise up to 105 °C. Calculate the value of new reverse current. [Ans.: 9.79 μA]

5. A silicon diode operates at a forward voltage of 0.4 V. Calculate the factor by which the current will be multiplied when the temperature is increased from 25 °C to 150 °C. [Ans. : 476.29]