4 Unwanted and wanted photodiode effectsĪ photodiode is a PIN structure or p–n junction.Photodiodes can be used to form an optocoupler, allowing transmission of signals between circuits without a direct metallic connection between them, allowing isolation from high voltage differences. A photodiode can be used as the receiver of data encoded on an infrared beam, as in household remote controls. Photodiodes are used in scientific and industrial instruments to measure light intensity, either for its own sake or as a measure of some other property (density of smoke, for example). A solar cell used to generate electric solar power is a large area photodiode. A photodiode is designed to operate in reverse bias. Photodiodes usually have a slower response time as their surface area increases. Devices designed for use specially as a photodiode use a PIN junction rather than a p–n junction, to increase the speed of response. The package may include lenses or optical filters. The package of a photodiode allows light (or infrared or ultraviolet radiation, or X-rays) to reach the sensitive part of the device. It produces current when it absorbs photons. This will make your dark current insignificant.One Ge (top) and three Si (bottom) photodiodesĪ photodiode is a light-sensitive semiconductor diode. In your case, if your source of UVA comes from a single, compact source (like an LED) and that source is stationary, you can point a 2 cm lens at it with a 100 micron photodiode and get the same photocurrent as using a 2 cm detector. Try it on a cloudy day (when the light comes from clouds at all angles) and the lens does nothing. Try to focus sunlight on a sunny day works well. A lens trades (via the Fourier transform) angle for position, so focusing only works if the source has a narrow range of angles (which you can Fourier transform into a narrow range of positions). The reason larger photodiodes exist is due to radiance. The smaller diode would also be cheaper, faster and introduce less electronic noise into your opamp. Since dark current scales with area, this would allow you to divide your dark current by a factor of hundreds or thousands at no loss of irradiance. If you really care about irradiance, then you would get a lens and use it to focus down onto a tiny photodiode. The difference is that radiance cannot be increased by optics (or anything), but irradiance can be using a simple lens. The maximum radiance of a photodiode goes up with the active area, but the number you are looking at (10uW/cm^2) has units irradiance, which does not have the same relationship. This is a great optical engineering question. If so, what's the point of choosing a larger sensor (for applications where you're not blasting a laser at it)? For very low light they expect you to use photovoltaic mode.īecause there is a version of this photodiode with much smaller area and the dark current is lower but so too is photocurrent and they just pretty much scale with each other. Note that the photodiode is advertised for use in photovoltaic mode, but the dark current is speced with a reverse bias (so in photoconductive mode where dark current is a bigger problem). Yet, the minimum optical power source range that this photodiode is recommended to be used at is 1nW/cm^2 which is way, way below the 10uW/cm^2 that was being used in the estimate above. You will additionally need to calibrate your particular detector as dark current varies between detectors. Be sure to size your integration capacitor to account for the dark current. Increase integration time to obtain whatever SNR you need. Adjusting the integration time allows you to increase the signal strength proportional to integration time, but the noise only increases as the square root of integration time. If you wanted to measure photocurrents below the dark current, you can do so using a capacitive TIA, which is just a regular TIA with a capacitor in feedback, also known as an op-amp integrator. Sum all of these components, multiply by your bandwidth, and take the square root to get a measure of your expected noise in A. The third is something you may have to measure, unless you have good noise models in your simulator. Where \$S\$ is the input referred noise power spectral density in A 2/Hz. The first two can be easily calculated as: The three big noise components you are likely to encounter are dark current shot noise, photocurrent shot noise, and noise from your readout circuit. The concern when your dark current approaches your photocurrent is SNR.
0 kommentar(er)
