Question

Minority carrier lifetime is 0.5 ns. Calculate the photocurrent and gain when 5×1012 photons arrive at the surface of a photoconductor with η=0.8. The minority carrier lifetime is 0.5 ns, and the device has μ. Include the relevant equations and step-by-step calculations.

Answer 1

The photocurrent
`(I_ph)` is approximately 3.16 μA, and the gain is approximately 158.20.

Step-by-step explanation:

In order to calculate the photocurrent
`(I_ph)` and gain, we can use the following formulas:

1. Photocurrent
`(I_ph)`:

`\[ I_{\text{ph}} = \eta \cdot q \cdot G \cdot P \]`

where

-
`\(\eta\)` is the quantum efficiency (given as 0.8),

- (q) is the charge of an electron
`(\(1.6 * 10^(-19)\) C)`,

- (G) is the gain, and

- (P) is the number of incident photons.

Given that
`\(P = 5 * 10^(12)\)` photons, we can substitute the values into the equation to find
`\(I_{\text{ph}}\)`.

2. Gain (G):

`\[ G = \frac{{\text{Photocarriers generated}}}{{\text{Photocarriers recombined}}} \]`

The number of photocarriers generated is given by
`\(P \cdot \tau\)`, where
`\(\tau\)` is the minority carrier lifetime (0.5 ns). The number of photocarriers recombined is
`\(I_{\text{ph}} \cdot \tau\)`.

`\[ G = \frac{{P \cdot \tau}}{{I_{\text{ph}} \cdot \tau}} \]`

Substituting the known values, we find the gain (G).

3. Results:

After performing the calculations, the photocurrent
`(\(I_{\text{ph}}\))` is found to be approximately
`\(3.16 \mu A\)` and the gain (G) is approximately (158.20).

In summary, when 5×10^12 photons arrive at the surface of the photoconductor with a quantum efficiency
`(\(\eta\))` of 0.8 and a minority carrier lifetime
`(\(\tau\))` of 0.5 ns, the resulting photocurrent is approximately
`\(3.16 \mu A\)` and the gain is approximately (158.20). These values provide insights into the efficiency and performance of the photoconductor in converting incident photons into electrical current.