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

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

where I_{0} 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_{B}is the Boltzmann constant (8.6×10^{−5} eV/K) and T is the absolute temperature. If for a given diode I_{0} = 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?

#### Solution

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,

I_{0} = 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 I_{0} in both cases. Therefore, the dynamic resistance in the reverse bias will be infinite.