Answer:
a) i) Vd = 358 mV , Vd = 417.5 mV, Vd = 477.2 mV
ii) Vd = -17.9 mV
b) Vd = 477.2 mV, Vd = 536.8 mV, Vd = 596.5 mV
Vd = 161 mV
Explanation:
We will use the ideal diode equation:
[tex]I_{D} = I_{s} (e^{\frac{qv_{d} }{nkT} }-1)[/tex]
Where Id = Diode current
Is = Reverse Saturation Current
Vd = Diode Voltage
q = charge of electron
n = ideality factor
k = Boltzmann constant
T = Temperature in Kelvin
(a) We are given:
Is = 10⁻¹¹ A
i) Id = 10 μA
Ideally, kT/q = 25.9mV and n=1. So,
q/nkT = 38.6
Id = Is e^(38.6Vd) -1
Id/Is + 1 = e^(38.6Vd)
ln(Id/Is + 1) = 38.6Vd
Vd = ln(Id/Is + 1)/38.6
= ln [(10*10⁻⁶)/10⁻¹¹ + 1] / 38.6
= 13.815/38.6
Vd = 0.3579
Vd = 358 mV
Id = 100 µA
Vd = ln(Id/Is + 1)/38.6
= ln [(100*10⁻⁶)/10⁻¹¹ + 1] / 38.6
= 16.118/38.6
Vd = 0.4175
Vd = 417.5 mV
Id = 1 mA
Vd = ln(Id/Is + 1)/38.6
= ln [(1*10⁻³)/10⁻¹¹ + 1] / 38.6
= 18.420/38.6
Vd = 0.4772
Vd = 477.2 mV
ii) Id = -5 x 10⁻¹² A
Vd = ln(Id/Is + 1)/38.6
= ln [(-5 x 10⁻¹²)/10⁻¹¹ + 1] / 38.6
= -0.693/38.6
Vd = -0.0179 V
Vd = -17.9 mV
(b) Is = 10⁻¹³ A
i) Id = 10 µA
Vd = ln(Id/Is + 1)/38.6
= ln [(10 x 10⁻⁶)/10⁻¹³ + 1] / 38.6
= 18.42/38.6
Vd = 0.4772 V
Vd = 477.2 mV
Id = 100 µA
Vd = ln(Id/Is + 1)/38.6
= ln [(100 x 10⁻⁶)/10⁻¹³ + 1] / 38.6
= 20.723/38.6
= 0.5368 V
Vd = 536.8 mV
Id = 1 mA
Vd = ln(Id/Is + 1)/38.6
= ln [(1 x 10⁻³)/10⁻¹³ + 1] / 38.6
= 23.025/38.6
= 0.5965
Vd = 596.5 mV
ii) Id = -5 x 10⁻¹² A
Vd = ln(Id/Is + 1)/38.6
= ln [(-5 x 10⁻¹²)/10⁻¹³ + 1] / 38.6
= ln (-49) / 38.6
Not possible.
Is = -10⁻¹⁴
Id = -5 x 10⁻¹² A
Vd = ln(Id/Is + 1)/38.6
= ln [(-5 x 10⁻¹²)/-10⁻¹⁴ + 1] / 38.6
= 6.216/38.6
Vd = 0.161 V
Vd = 161 mV
