Question

In a p-n junction diode, the current I can be expressed asI=I0exp{(eV/2kBT)-1}is called the reverse saturation current, V is the voltageacross the diode and is positive for forward bias and negative forreverse bias, and I is the current through the diode, kBis theBoltzmann constant (8.6×10–5 eV/K) and T is the absolutetemperature. 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 thediode 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 Vto 2 V?

Answer 1

Given: For a given diode, reverse saturation current is 5× 10 −12  A and the absolute temperature is 300 K.

a)

The expression for the current is given as,

I= I 0 exp( eV k B T −1 )

Where, I 0 is the reverse saturation current, V is the voltage across the diode, k B os the Boltzmann constant and T is the absolute temperature.

By substituting the given values in the above equation, we get

I=5× 10 −12 exp( 1.6× 10 −19 ×0.6 8.6× 10 −5 ×1.6× 10 −19 ×300 −1 ) =5× 10 −12 ×1.259× 10 10 =6.295× 10 −2  A =62.95 mA

Thus, the forward current is 62.95 mA.

b)

The forward current when the voltage across the diode is 0.7 V is,

I ′ =5× 10 −12 exp( 1.6× 10 −19 ×0.7 8.6× 10 −5 ×1.6× 10 −19 ×300 −1 ) =5× 10 −12 ×6.07× 10 11 =3.035 A

The increase in the current is,

I ′ −I=3.035−0.0629 =2.972 A

Thus, the increase in the current is 2.972 A.

c)

The dynamic resistance is given as,

R= ΔV ΔI

By substituting the values in the above equation, we get

R= 0.7−0.6 2.972 =33.65 mΩ

Thus, the dynamic resistance is 33.65 mΩ.

d)

The current in the diode be almost equal when voltage changes, which shows almost infinite dynamic resistance in reverse bias.

Hence, the current in the diode for both voltages will be,

I≈ I 0 =( −5× 10 −12 )Α