Question

Use the ideal diode equation to calculate the voltage required to produce a current density of 25 A/cm2 for a junction of cross-sectional area 10-4 cm2, a reverse saturation current of 2x10-15 A and at 300 K

Answer 1

Answer:

$V_{D} = 2 volts$

Explanation:

First we need to find the value of diode current:

$Diode\ Current = I_{D} = (25\ A/cm^2)(4\ cm^2)\\I_{D} = 100\ A$

Now, we use ideal diode equation:

$I_{D} = I_{S}(e^{\frac{qV_{D}}{nkT}}-1)$

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,

$100\ A = (2\ x\ 10^{-15}\ A)(e^{[\frac{1.6\ x\ 10^{-19} C\ V_{D}}{(1)(1.38\ x\ 10^{-23})(300\ k)}]}-1)\\\\\frac{100\ A}{2\ x\ 10^{-15}\ A} = e^{[(38.64 V^{-1})(V_{D})]}-1$

taking natural log on both sides:

$ln(5\ x\ 10^{16}) = ln[e^{(38.6\ V^{-1}\ V_{D})} - 1]\\38.45 = (38.6\ V^{-1})(V_{D}) - 1\\\\V_{D} = \frac{38.45}{38.6\ V^{-1}} + 1$

$V_{D} = 2\ volts$