Question

In a p-n junction diode, the current I can be expressed as

I = `"I"_0 exp ("eV"/(2"k"_"BT") - 1)`

where I₀ is called the reverse saturation current, V is the voltage across the diode and is positive for forward bias and negative for reverse bias, and I is the current through the diode, k_Bis the Boltzmann constant (8.6×10⁻⁵ eV/K) and T is the absolute temperature. If for a given diode I₀ = 5 × 10⁻¹² A and T = 300 K, then

(a) What will be the forward current at a forward voltage of 0.6 V?

(b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V?

(c) What is the dynamic resistance?

(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?

Answer 1

a) In a p-n junction diode, the expression for current is given as:

I = `"I"_0 exp("eV"/(2"k"_"B" "T") - 1)`

Where,

I₀ = Reverse saturation current = 5 × 10⁻¹² A

T = Absolute temperature = 300 K

k_B = Boltzmann constant = 8.6 × 10⁻⁵ eV/K = 1.376 × 10⁻²³ J K⁻¹

V = Voltage across the diode

(a) Forward voltage, V = 0.6 V

= `5 xx 10^(-12)[exp ((1.6xx 10^(-19) xx 0.6)/(1.376 xx 10^(-23) xx 300))-1]`

∴ Current, I 

`= 5 xx 10^(-12) xx   exp [22.36] = 0.0256 A`

Therefore, the forward current is about 0.0256 A.

(b) For forward voltage, V^’ = 0.7 V, we can write:

I' =`5 xx 10^(-12) [exp ((1.6 xx 10^(-19) xx 0.7)/(1.376 xx 10^(-23) xx 300)) - 1]`

`= 5 xx 10^(-12) xx exp [26.25] = 1.257` A

Hence, the increase in current, ΔI = I^' − I

= 1.257 − 0.0256 = 1.23 A

(c) Dynamic resistance  = `"Change in voltage"/"Change in Current"`

`= (0.7 - 0.6)/1.23 = 0.1/1.23 = 0.081 "Ω"`

(d) If the reverse bias voltage changes from 1 V to 2 V, then the current (I) will almost remain equal to I₀ in both cases. Therefore, the dynamic resistance in the reverse bias will be infinite.