Correct Answer - Option 3 : 920 μA

An ideal diode equation or Shockley equation is given by

\({I_D} = {I_S}\left( {{e^{\frac{{q{V_D}}}{{η kT}}}} - 1} \right)\;\)

Where IS is the reverse saturation current

q is the charge on the electron

VD is applied forward-bias voltage across the diode

η is an ideality factor = 1 for indirect semiconductors

= 2 for direct semiconductors

k is the Boltzmann’s constant

T is the temperature in Kelvin

kT/q is also known as thermal voltage (VT).

At 300 K (room temperature), kT/q = 25.9 mV ≈ 26 mV

Now, the Shockley equation is given by

\({I_D} = {I_S}\left( {{e^{\frac{{{V_D}}}{{η {V_T}}}}} - 1} \right)\;\;\)

**Application:**

Given I_{S} = 20 μA = 20 × 10^{-6} A

V_{D} = 0.2 V

η = 2

V_{T} = 26 mV

The forward current

\({I_D} = {I_S}\left( {{e^{\frac{{{V_D}}}{{η {V_T}}}}} - 1} \right)={20\times 10^{-6}}\left( {{e^{\frac{{{0.2}}}{{2 \times 26 \times 10^{-3} }}}} - 1} \right)\)

\(I_D={20\times 10^{-6}}\left( {{e^{3.85}} - 1} \right)={20\times 10^{-6}}\left( {47 - 1} \right)=920\times 10^{-6}\; A\)

**I**_{D} = 920 μA