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 IS = 20 μA = 20 × 10-6 A
VD = 0.2 V
η = 2
VT = 26 mV
The forward current
\({I_D} = {I_S}\left( {{e^{\frac{{{V_D}}}{{η {V_T}}}}} - 1} \right)={20\times 10^{-6}}\left( {{e^{\frac{{{0.2}}}{{2 \times 26 \times 10^{-3} }}}} - 1} \right)\)
\(I_D={20\times 10^{-6}}\left( {{e^{3.85}} - 1} \right)={20\times 10^{-6}}\left( {47 - 1} \right)=920\times 10^{-6}\; A\)
ID = 920 μA