The quantum efficiency (η) and responsivity (R) at a wavelength λ (in μm) in a p-i-n photodetector are related by

Understanding Photodetector Relationship: Responsivity and Quantum Efficiency

This solution explains the relationship between a photodetector's quantum efficiency ($\eta$) and its responsivity ($R$) at a specific wavelength ($\lambda$). We will use the fundamental definitions and constants to derive the correct formula.

Key Concepts in Photodetection

- **Quantum Efficiency ($\eta$): This measures how efficiently incident photons are converted into charge carriers that are successfully collected by the detector. It is calculated as the ratio of collected charge carriers (like electrons) to the total number of incident photons. While often expressed as a percentage, it's used as a dimensionless ratio (between 0 and 1) in calculations.
- Responsivity ($R$)**: This quantifies the detector's performance by measuring the output electrical signal (photocurrent, $I_{ph}$) generated per unit of incident optical power ($P_{in}$). It is typically measured in units of Amperes per Watt (A/W).
- **Photon Energy ($E_{photon}$)**: Each photon carries a specific amount of energy determined by its wavelength ($\lambda$). Shorter wavelengths correspond to higher photon energy.

Derivation of the Relationship

To find the link between responsivity and quantum efficiency, let's start with their definitions:

The incident optical power ($P_{in}$) represents the total energy delivered to the detector per unit time. If $N_{ph}$ is the number of photons arriving per second, and each photon has energy $E_{photon}$, then:

$$P_{in}=N_{ph}\times E_{photon}$$

The photocurrent ($I_{ph}$) is the flow of charge generated by these photons. If $\eta$ represents the quantum efficiency (as a ratio, 0 to 1) and $q$ is the elementary charge (the magnitude of charge on a single electron, approximately $1.602\times 10^{-19}$ Coulombs), then the number of charge carriers collected per second is $\eta \times N_{ph}$. The photocurrent is:

$$I_{ph}=(\eta \times N_{ph})\times q$$

Responsivity ($R$) is defined as the ratio of photocurrent to incident power:

$$R=\frac{I_{ph}}{P_{in}}$$

Now, we substitute the expressions for $I_{ph}$ and $P_{in}$ into the responsivity formula:

$$R=\frac{(\eta \times N_{ph})\times q}{N_{ph}\times E_{photon}}$$

Notice that the number of photons per second ($N_{ph}$) cancels out:

$$R=\frac{\eta \times q}{E_{photon}}$$

The options provided involve a constant factor related to wavelength in micrometers ($\mu$ m). The energy of a photon ($E_{photon}$) is related to wavelength ($\lambda$) by the equation $E_{photon}=\frac{hc}{\lambda }$, where $h$ is Planck's constant and $c$ is the speed of light. When wavelength is measured in micrometers ($\mu$ m) and energy is expressed in electron-volts (eV), the constant $\frac{hc}{q}$ is approximately 1.24 eV·$\mu$ m.

Therefore, the photon energy can be written as:

$$E_{photon}(\text{in eV})\approx \frac{1.24}{\lambda (\text{in }\mu m)}$$

To use this in our responsivity formula $R=\frac{\eta \times q}{E_{photon}}$, we need $E_{photon}$ in Joules. We convert eV to Joules using $E_{photon}(\text{Joules})=E_{photon}(\text{eV})\times q$:

$$R=\frac{\eta \times q}{\left(E_{photon}(\text{eV})\times q\right)}$$

The charge $q$ cancels out:

$$R=\frac{\eta }{E_{photon}(\text{eV})}$$

Finally, substitute the expression for $E_{photon}$ in terms of wavelength:

$$R=\frac{\eta }{\left(\frac{1.24}{\lambda }\right)}$$

Simplifying this expression yields the relationship:

$$R=\frac{\eta \times \lambda }{1.24}$$

In this formula, $R$ is the responsivity in A/W, $\eta$ is the quantum efficiency ratio (0 to 1), and $\lambda$ is the wavelength in micrometers ($\mu$ m). The constant 1.24 effectively incorporates the values of Planck's constant, the speed of light, and the charge of an electron, along with unit conversions.

Conclusion

The derived relationship shows that responsivity ($R$) is directly proportional to both the quantum efficiency ($\eta$) and the wavelength ($\lambda$), with the proportionality constant being approximately $1/1.24$ when $\lambda$ is in $\mu$ m. This matches the first option.