Answer:
[tex]V_{D} = 2 volts[/tex]
Explanation:
First we need to find the value of diode current:
[tex]Diode\ Current = I_{D} = (25\ A/cm^2)(4\ cm^2)\\I_{D} = 100\ A[/tex]
Now, we use ideal diode equation:
[tex]I_{D} = I_{S}(e^{\frac{qV_{D}}{nkT}}-1)\\[/tex]
where,
Is = saturation current = 2 x 10⁺¹⁵ A
q = charge on electron = 1.6 x 10⁻¹⁹ C
V_D = Diode Voltage = ?
n = ideality factor = 1 (for ideal diodes)
k = Boltzman constant = 1.38 x 10⁻²³ J/k
T = Temperature = 300 k
Therefore,
[tex]100\ A = (2\ x\ 10^{-15}\ A)(e^{[\frac{1.6\ x\ 10^{-19} C\ V_{D}}{(1)(1.38\ x\ 10^{-23})(300\ k)}]}-1)\\\\\frac{100\ A}{2\ x\ 10^{-15}\ A} = e^{[(38.64 V^{-1})(V_{D})]}-1[/tex]
taking natural log on both sides:
[tex]ln(5\ x\ 10^{16}) = ln[e^{(38.6\ V^{-1}\ V_{D})} - 1]\\38.45 = (38.6\ V^{-1})(V_{D}) - 1\\\\V_{D} = \frac{38.45}{38.6\ V^{-1}} + 1\\\\[/tex]
[tex]V_{D} = 2\ volts[/tex]
