An ideal photo-diode of unit quantum efficiency, at room temperature, is illuminated with 8 mW of radiation at 0.65 µm wavelength. Calculate

Question

An ideal photo-diode of unit quantum efficiency, at room temperature, is illuminated with 8 mW of radiation at 0.65 µm wavelength. Calculate the current and voltage output when the detector is used in the photo-conductive and photovoltaic modes respectively. The reverse saturation current (Is) is 9 nA.

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Thiên Di 5 months 2021-09-01T03:12:33+00:00 1 Answers 0 views 0

Answers ( )

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    2021-09-01T03:13:51+00:00

    Answer:

    I = 4.189 mA    V = 0.338 V

    Explanation:

    In order to do this, we need to apply the following expression:

    I = Is[exp^(qV/kT) – 1]   (1)

    However, as the junction of the diode is illuminated, the above expression changes to:

    I = Iopt + Is[exp^(qV/kT) – 1]   (2)

    Now, as the shunt resistance becomes infinite while the current becomes zero, we can say that the leakage current is small, and so:

    I ≅ Iopt

    Therefore:

    I ≅ I₀Aλq / hc  (3)

    Where:

    I₀A = Area of diode (radiation)

    λ: wavelength

    q: electron charge (1.6×10⁻¹⁹ C)

    h: Planck constant (6.62×10⁻³⁴ m² kg/s)

    c: speed of light (3×10⁸ m/s)

    Replacing all these values, we can get the current:

    I = (8×10⁻³) * (0.65×10⁻⁶) * (1.6×10⁻¹⁹) / (6.62×10⁻³⁴) * (3×10⁸)

    I = 4.189×10⁻³ A or 4.189 mA

    Now that we have the current, we just need to replace this value into the expression (2) and solve for the voltage:

    I = Is[exp^(qV/kT) – 1]

    k: boltzman constant (1.38×10⁻²³ J/K)

    4.189×10⁻³ = 9×10⁻⁹ [exp(1.6×10⁻¹⁹ V / 1.38×10⁻²³ * 300) – 1]

    4.189×10⁻³ / 9×10⁻⁹ = [exp(38.65V) – 1]

    465,444.44 + 1  = exp(38.65V)

    ln(465,445.44) = 38.65V

    13.0508 = 38.65V

    V = 0.338 V

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